Cambridge Maths Extension 1 Year 11 Textbook

11 MATHEMATICS EXTENSION 1 CambridgeMATHS NSW STAGE 6 SECOND EDITION Bill Pender David Sadler, Derek Ward Brian Dorofaeff, William McArthur CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Shaftesbury Road, Cambridge CB2 8EA, United Kingdom One Liberty Plaza, 20th Floor, New York, NY 10006, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia 314–321, 3rd Floor, Plot 3, Splendor Forum, Jasola District Centre, New Delhi – 110025, India 103 Penang Road, #05–06/07, Visioncrest Commercial, Singapore 238467 Cambridge University Press & Assessment is a department of the University of Cambridge. We share the University’s mission to contribute to society through the pursuit of education, learning and research at the highest international levels of excellence. www.cambridge.org © Bill Pender, David Sadler, Derek Ward, Brian Dorofaeff and Julia Shea 2019 © Bill Pender, David Sadler, Derek Ward, Brian Dorofaeff and William McArthur 2025 This publication is in copyright. Subject to statutory exception and to the provisions of relevant collective licensing agreements, no reproduction of any part may take place without the written permission of Cambridge University Press & Assessment. First published 2019 Second Edition 2025 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 1 Cover designed by Sardine Design Text designed by Vtex Typeset by Lumina Datamatics Printed in China by C & C Offset Printing Co. Ltd. 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CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Contents Introduction and overview ix Acknowledgements xi About the authors xii 1 Algebra review 1 1A Expanding brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1B Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1C Algebraic fractions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 1D Solving quadratic equations . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1E Solving simultaneous equations . . . . . . . . . . . . . . . . . . . . . . . . 21 Review of Chapter 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 2 Numbers and surds 26 . . . . . . . . . . . . . . . . . . . . . . . . . . 27 2A Real numbers and intervals 2B Surds and their arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . 33 2C Further simplification of surds . . . . . . . . . . . . . . . . . . . . . . . . . 36 2D Rationalising the denominator . . . . . . . . . . . . . . . . . . . . . . . . . 39 Review of Chapter 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43 3 Functions and graphs 45 3A Functions and function notation . . . . . . . . . . . . . . . . . . . . . . . . 46 3B Functions, relations, and graphs . . . . . . . . . . . . . . . . . . . . . . . . 51 3C Review of linear graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 3D Quadratic functions — factoring and the graph . . . . . . . . . . . . . . . . 64 3E Completing the square and the graph . . . . . . . . . . . . . . . . . . . . . 71 3F The quadratic formulae and the graph 3G Powers, cubics, and circles . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 3H Two graphs that have asymptotes . . . . . . . . . . . . . . . . . . . . . . . 90 3I Direct and inverse variation . . . . . . . . . . . . . . . . . . . . . . . . . . 95 . . . . . . . . . . . . . . . . . . . . 79 Review of Chapter 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. iv Contents 4 Equations and inequations 105 4A Linear equations and inequations . . . . . . . . . . . . . . . . . . . . . . . 106 4B Quadratic equations and inequations . . . . . . . . . . . . . . . . . . . . . 110 4C The discriminant 4D Quadratic identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 Review of Chapter 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125 5 Transformations and symmetry 127 5A Translations of known graphs . . . . . . . . . . . . . . . . . . . . . . . . . 128 5B Reflection in the y-axis and x-axis . . . . . . . . . . . . . . . . . . . . . . . 137 5C Even and odd symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 5D Horizontal and vertical dilations . . . . . . . . . . . . . . . . . . . . . . . . 149 5E The absolute value function 5F Composite functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 5G Combining transformations 5H Continuity and piecewise-defined functions . . . . . . . . . . . . . . . . . . 178 . . . . . . . . . . . . . . . . . . . . . . . . . . 158 . . . . . . . . . . . . . . . . . . . . . . . . . . 171 Review of Chapter 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 6 Further graphs Extension 1 189 6A Solving two particular inequations . . . . . . . . . . . . . . . . . . . . . . . 190 6B The sign of a function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 196 6C Sketching reciprocal functions . . . . . . . . . . . . . . . . . . . . . . . . . 202 6D Sketching sums and differences . . . . . . . . . . . . . . . . . . . . . . . . 209 6E Modifying a function using absolute value . . . . . . . . . . . . . . . . . . . 216 6F Inverse relations and functions . . . . . . . . . . . . . . . . . . . . . . . . 221 6G Inverse function notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226 6H Defining functions and relations parametrically . . . . . . . . . . . . . . . 231 Review of Chapter 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Contents 7 Trigonometry 240 7A Trigonometry with right-angled triangles . . . . . . . . . . . . . . . . . . . 241 7B Problems involving right-angled triangles . . . . . . . . . . . . . . . . . . . 249 7C Trigonometric functions of a general angle . . . . . . . . . . . . . . . . . . 254 7D Quadrant, sign, and related acute angle . . . . . . . . . . . . . . . . . . . . 262 7E Given one trigonometric function, find another . . . . . . . . . . . . . . . . 268 7F Trigonometric identities 7G Trigonometric equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277 7H The sine rule and the area formula 7I The cosine rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291 7J Problems involving general triangles . . . . . . . . . . . . . . . . . . . . . 297 . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 . . . . . . . . . . . . . . . . . . . . . . 282 Review of Chapter 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 8 Lines in the coordinate plane 307 8A Lengths and midpoints of line segments . . . . . . . . . . . . . . . . . . . . 308 8B Gradients of line segments and lines . . . . . . . . . . . . . . . . . . . . . . 313 8C Equations of lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322 8D Further equations of lines . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 8E Using pronumerals in place of numbers . . . . . . . . . . . . . . . . . . . . 332 Review of Chapter 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 9 Exponential and logarithmic functions 339 9A Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 9B Fractional indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 9C Logarithms 9D The laws for logarithms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357 9E Equations involving logarithms and indices . . . . . . . . . . . . . . . . . . 362 9F Exponential and logarithmic graphs . . . . . . . . . . . . . . . . . . . . . . 368 9G Applications of these functions . . . . . . . . . . . . . . . . . . . . . . . . . 377 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352 Review of Chapter 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 381 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. v vi Contents 10 Differentiation 383 10A Tangents and the derivative . . . . . . . . . . . . . . . . . . . . . . . . . . 384 10B The derivative as a limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 10C A rule for differentiating powers of x 10D The notation 10E The chain rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 10F Differentiating powers with negative indices . . . . . . . . . . . . . . . . . 415 10G Differentiating powers with fractional indices 10H The product rule 10I The quotient rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 10J Rates of change . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432 10K Average velocity and average speed . . . . . . . . . . . . . . . . . . . . . . 440 10L Instantaneous velocity and speed . . . . . . . . . . . . . . . . . . . . . . . 447 . . . . . . . . . . . . . . . . . . . . . 397 dy for the derivative . . . . . . . . . . . . . . . . . . . . . . . 405 dx . . . . . . . . . . . . . . . . 419 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424 Review of Chapter 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 453 11 Polynomials Extension 1 456 11A The language of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . 457 11B Graphs of polynomial functions 11C Division of polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 467 11D The remainder and factor theorems . . . . . . . . . . . . . . . . . . . . . . 471 11E Consequences of the factor theorem . . . . . . . . . . . . . . . . . . . . . 476 11F Sums and products of zeroes 11G Geometry using polynomial techniques . . . . . . . . . . . . . . . . . . . . 491 . . . . . . . . . . . . . . . . . . . . . . . . 462 . . . . . . . . . . . . . . . . . . . . . . . . . 483 Review of Chapter 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 12 Euler’s number 497 12A The exponential function base e . . . . . . . . . . . . . . . . . . . . . . . . 498 12B Transformations of exponential functions . . . . . . . . . . . . . . . . . . . 505 12C The logarithmic function base e . . . . . . . . . . . . . . . . . . . . . . . . 509 Review of Chapter 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 516 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Contents 13 Radian measure of angles vii 517 . . . . . . . . . . . . . . . . . . . . . . . . . 518 13A Radian measure of angle size 13B Solving trigonometric equations . . . . . . . . . . . . . . . . . . . . . . . . 524 13C Arcs and sectors of circles . . . . . . . . . . . . . . . . . . . . . . . . . . . 528 13D Trigonometric graphs in radians . . . . . . . . . . . . . . . . . . . . . . . . 536 Review of Chapter 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 541 14 Probability 543 14A Sets and Venn diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 14B Probability and sample spaces . . . . . . . . . . . . . . . . . . . . . . . . . 552 14C Sample space graphs and tree diagrams . . . . . . . . . . . . . . . . . . . 561 14D Venn diagrams and the addition theorem . . . . . . . . . . . . . . . . . . . 567 14E Multi-stage experiments and the product rule . . . . . . . . . . . . . . . . 573 14F Probability tree diagrams 14G Conditional probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586 . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 Review of Chapter 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 599 15 Data and probability 602 15A Random variables and frequency tables . . . . . . . . . . . . . . . . . . . . 603 15B Cumulative frequency 15C Grouped data . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 621 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 613 Review of Chapter 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 629 16 Further trigonometry Extension 1 632 16A Three-dimensional trigonometry . . . . . . . . . . . . . . . . . . . . . . . . 633 16B Trigonometric functions of compound angles . . . . . . . . . . . . . . . . . 641 16C The double-angle formulae . . . . . . . . . . . . . . . . . . . . . . . . . . . 646 16D Trigonometric equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 650 16E The sum of sine and cosine functions . . . . . . . . . . . . . . . . . . . . . 656 Review of Chapter 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 664 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. viii Contents 17 Combinatorics Extension 1 667 17A Factorial notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668 17B Ordered selections with and without repetition . . . . . . . . . . . . . . . . 673 17C Ordered selections — three more principles . . . . . . . . . . . . . . . . . 680 17D Ordered selections with identical elements . . . . . . . . . . . . . . . . . . 685 17E Counting unordered selections 17F Using counting in probability . . . . . . . . . . . . . . . . . . . . . . . . . . 699 17G Arrangements in a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 706 . . . . . . . . . . . . . . . . . . . . . . . . 690 Review of Chapter 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 711 18 The binomial theorem and Pascal’s triangle Extension 1 713 . . . . . . . . . . . . . . . . . . 714 18A Binomial expansions and Pascal’s triangle 18B Binomial expansions with several variables . . . . . . . . . . . . . . . . . . 721 18C The binomial theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725 18D Using the general term . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 731 18E Identities in Pascal’s triangle . . . . . . . . . . . . . . . . . . . . . . . . . . 735 18F Further identities in Pascal’s triangle . . . . . . . . . . . . . . . . . . . . . 741 Review of Chapter 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 747 Answers 749 Index 877 Online appendices See the Interactive Textbook for guides to spreadsheets, the Desmos graphing calculator, and links to scientific calculator guides. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Introduction and overview CambridgeMATHS for NSW Mathematics Extension 1 Year 11 Second Edition provides complete and aligned coverage of the NESA syllabus to be implemented 2026, and forms part of a continuum of learning from Year 7 to Year 12. Its four components – the print book, downloadable PDF textbook, online Interactive Textbook (ITB) and Online Teaching Resource (OTS) – contain a huge range of resources, including worked solutions and additional worksheets, available to schools in a single package (the OTS is included with class adoptions, conditions apply). New features in the second edition: • Learning intentions at the start of each section complemented by a downloadable Skills Checklist for each chapter that allows students to check understanding and tick off your achievements. • Auto-marked quiz questions in the interactive textbook for every section to let students test understanding in a self directed way. • Foundation Worksheets available in the interactive textbook give additional practice at core skills within the chapter. Linked closely to examples and learning intentions to help guide students towards mastery. The Second Edition also features significantly revised and updated material from the first edition, including: Review of Years 9 and 10: The first two chapters in particular can be considered as a review of material covered in Years 9 and 10. The topics covered in these chapters are essential knowledge before proceeding with the rest of the book. Grading of exercises: Exercises are graded based on difficulty into three groups, Foundation, Development and Enrichment to give students more paths for progressing through the content. Foundation questions provide a gentle start to each exercise and cover the core skills required for the syllabus. Development is typically the longest section and ramp up from straightforward to more complex problems like the ones you may see towards the end of the HSC exam. Enrichment questions are designed to be more difficult and often require a deeper understanding of the concepts as well as a mix of other topics. These questions are meant to enrich students, typically beyond the level that is required by the syllabus. They don’t need to always be attempted and should serve as a challenge for content that is already well understood. Worked examples and Summary boxes: Plenty of worked examples throughout that cover every key skill from the syllabus in a clear and concise manner. They are coloured with titles to make them easy to find and refer back to when studying. To accompany the worked examples, spread throughout each section there are also numbered and coloured boxes that highlight and summarise the key concepts within a section. To help assist note taking and make later revision straightforward. Assessment practice and review: Every question in the book has been carefully crafted and refined to give students the widest coverage and number of HSC style questions possible, in order to prepare for the exam. chapter review exercises at the end of every chapter also serve to assess learning and harder practice of key syllabus concepts needed for HSC success. Additional practice questions, including Exam style multiple-choice questions, are also available in the Interactice Textbook. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. x Introduction and overview Interactive Textbook (ITB) The Interactive Textbook (ITB) is an online HTML version of the print textbook powered by the HOTmaths platform, included with the print book or available as a separate purchase. Updated and revised for the new syllabus, the Interactive Textbook includes: • Video demonstrations of all worked examples to encourage independent learning. • Quick quizzes containing auto-marked multiple-choice questions have been added to every section, enabling • • • • • students to quickly check their understanding of a section. Chapter quizzes provide a more comprehensive review of the chapter with auto-marked HSC style multiplechoice questions. A success criteria checklist at the end of each chapter with linked Worksheet questions and examples available for download, allows students to check their understanding of the syllabus as they progress through the course in a self-directed way. Comprehensive worked solutions for all questions are provided in the Interactive Textbook as an option that teacher can choose to enable for their students. Downloadable Worksheets containing additional foundation level questions, can be used for homework or in class to focus on the learning intention skills. Questions are linked to sections and modelled on the learning intentions and worked examples to help with follow-up. Interactive Widgets based on Desmos technology, allow students to explore and visualise key concepts in a dynamic way. Desmos tools are also embedded in the Interactive Textbook and can be used to complete calculations, create custom graphs, shapes and other visual representations. The Online Teaching Suite (OTS) The Online Teaching Suite is automatically enabled with a teacher account and is integrated with the teacher’s copy of the Interactive Textbook. All the teacher resources are in one place for easy access. The features include: • A teacher’s view of a student’s working and self-assessment which enables them to modify the student’s self-assessed marks, and respond where students flag that they had difficulty. • The task manager allowing to direct students on a custom activity sequence based on their scores in • • • • measurable activities Quickly create customised tests from a bank of multiple-choice questions using the test generator. Tests are auto-marked in the Interactive Textbook or can be printed and used for homework or assessment practice. An expanded and revised suite of chapter tests and assessment Editable curriculum grids and teaching programs. A brand-new Exam Generator, allowing the creation of customised printable and online trial exams (see below for more). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Introduction and overview xi More about the Exam Generator The Online Teaching Suite, available from 2025, will include a comprehensive bank of NESA exam questions, augmented by exam-style questions written by experts, to allow teachers to create custom trial exams. Custom exams can model end-of-year exams, or target specific topics or types of questions that students may be having difficulty with. Features include: • Filtering by question-type, topic and difficulty • Searchable by key words • Answers provided to teachers • Worked solutions for all questions • Multiple-choice exams can be auto-marked if completed online, with filterable reports • All custom exams can be printed and completed under exam-like conditions or used as revision. Acknowledgements The author and publisher wish to thank the following sources for permission to reproduce material: Cover: © Getty Images / Meaghan Skinner Photography. Images: © Getty Images / Tuomas A. Lehtinen, Chapter 1 Opener / lvcandy, Chapter 2 Opener / Xuanyu Han, Chapter 3 Opener / da-kuk, Chapter 4 Opener / yuanyuan yan, Chapter 5 Opener / enjoynz, Chapter 6 Opener / Westend61, Chapter 7 Opener / chaiwatgesee, Chapter 8 Opener / zhengshun tang, Chapter 9 Opener / zhengshun tang, Chapter 10 Opener / calvindexter, Chapter 11 Opener / Govindanmarudhai, Chapter 12 Opener / Sylverarts, Chapter 13 Opener / Flavio Coelho, Chapter 14 Opener / twomeows, Chapter 15 Opener / traffic_analyzer, Chapter 16 Opener / Tuomas A. Lehtinen, Chapter 17 Opener / nilo75, Chapter 18 Opener. Every effort has been made to trace and acknowledge copyright. The publisher apologises for any accidental infringement and welcomes information that would redress this situation. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. About the authors Dr Bill Pender is retired Subject Master in Mathematics at Sydney Grammar School, where he taught from 1975 to 2009. He has an MSc and PhD in Pure Mathematics from Sydney University, with theses in geometry and group theory, and a BA(Hons) in Early English from Macquarie University. He spent a year at Bonn University in Germany, and he has lectured and tutored at Sydney University, and at the University of NSW where he was a Visiting Fellow in 1989. He was a member of the NSW Syllabus Committee in Mathematics for two years and subsequently of the Review Committee for the Years 9–10 Advanced Syllabus, and has contributed extensively to syllabus discussions over the last 30 years. He was a member of the AMSI Mathematics Education Committee for several years, and an author of the ICE-EM Years 7–10 textbooks for the National Curriculum. David Sadler is a retired Mathematics teacher. He taught for 36 years at Sydney Grammar School and was Head of Mathematics for 7 years. He also taught at UNSW for one year. He was an HSC marker for many years and has been a presenter at various conferences and professional development courses. He has a strong passion for excellence in mathematics education and has previously co-authored several senior texts for Cambridge University press. Derek Ward taught Mathematics at Sydney Grammar School for 31 years, and was Master in Charge of Examination Statistics for 20 years. He has an MSc in Applied Mathematics and a BScDipEd, both from the University of NSW, where he was subsequently Senior Tutor for three years. He has a keen interest in mathematics education and has co-authored several other texts for Cambridge University press, as well as internal publications for Sydney Grammar School. Derek has an AMusA in Flute, and has an extensive background in singing. He sings in various choirs and conducts a choir in New Zealand, where he now lives. Dr Brian Dorofaeff is currently Subject Master in Mathematics at Sydney Grammar School. He is an experienced classroom teacher, having taught for over 20 years in New South Wales. He has previous experience in HSC marking and has been on the HSC Mathematics Examination Committee. Brian holds a Ph.D. in Mathematics from the University of New South Wales. William McArthur has been a Mathematics teacher for 15 years, and has taught at Sydney Grammar School since 2016. He has a BSc in Mathematics, a BEd, and an MEd(Assessment and Evaluation), all from the University of NSW. He has presented at HSC Mathematics revision seminars, and worked with pre-service Mathematics teachers at the University of Notre Dame. William has experience in HSC marking, and has been involved with the NSW Stage 6 Mathematics curriculum review. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1 Algebra review Chapter introduction Fluency in algebra, particularly in factoring, is absolutely vital for everything in mathematics. This chapter is intended as a review of earlier algebraic techniques, and readers should do as much or as little of it as necessary. Some readers will work steadily through it. Others will remember that this review is here, and work through parts of it as necessary. The course properly begins with Chapter 2. The word ‘algebra’ has an interesting history. Al-Khwarizmi was a famous and influential Persian mathematician who worked in Baghdad during the early ninth century when the Baghdad Caliphate excelled in science and mathematics. The Arabic word ‘algebra’ comes from al-jabr, a word in the title of his most important work, and means ‘the restoration of broken parts’ — a reference to the balancing of terms on both sides of an equation. Al-Khwarizmi’s own name came into modern European languages as ‘algorithm’. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2 1A Chapter 1 Algebra review 1A Expanding brackets Learning intentions • Work with like and unlike terms in algebraic expressions. • Expand brackets in algebraic expressions. • Develop the three special expansions, and work with them. This chapter reviews the manipulation of algebraic expressions involving pronumerals, where a pronumeral is a letter or symbol that stands ‘for a number’. The pronumeral may stand for a known or unknown constant, or for an unknown number in an equation, or it may be a variable and stand for any one of a whole set of possible numbers. Pronumerals are numbers, so they can take part in all the normal operations of arithmetic, including addition, subtraction, multiplication, division (except by zero), and forming powers. Algebraic expressions — like and unlike terms An algebraic expression consists of pronumerals, numbers, and the operations of arithmetic. Here an algebraic expression is simplified by collecting like terms: x2 + 2xy + 3x2 − 4xy − 3 = 4x2 − 2xy − 3 This particular algebraic expression can be simplified by combining like terms. • The two like terms x2 and 3x2 can be combined to give 4x2 . • Another pair of like terms 2xy and −4xy can be combined to give −2xy. • This yields three unlike terms, 4x2 , −2xy and −3, which cannot be combined. Assuming that these very basic foundations are well known, we now begin reviewing further methods that have been developed to handle algebraic expressions. Expanding brackets in arithmetic and algebra Expanding brackets is routine in arithmetic. For example, calculate 7 × 61 as 7 × (60 + 1) = 7 × 60 + 7 × 1, which quickly gives the result 7 × 61 = 420 + 7 = 427. The algebraic version of this procedure can be written as: 1 Expanding brackets in algebraic expressions a(x + y) = ax + ay and (x + y)a = xa + ya In longer expressions, there may be like terms to collect after the expansion. Example 1 Expanding brackets and simplifying Expand and simplify each expression. a 3x(4x − 7) b 5a(3 − b) − 3b(1 − 5a) Solution a 3x(4x − 7) = 12x2 − 21x b 5a(3 − b) − 3b(1 − 5a) = 15a − 5ab − 3b + 15ab = 15a + 10ab − 3b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1A Expanding brackets Expanding the product of two bracketed terms Expand one pair of brackets, then expand the other pair of brackets. Then collect any like terms. Example 2 Expanding more than one set of brackets Expand and simplify each expression. b (3 + x)(9 + 3x + x2 ) a (x + 3)(x − 5) Solution a (x + 3)(x − 5) b (3 + x)(9 + 3x + x2 ) = x(x − 5) + 3(x − 5) = 3(9 + 3x + x2 ) + x(9 + 3x + x2 ) = x2 − 5x + 3x − 15 = 27 + 9x + 3x2 + 9x + 3x2 + x3 = x2 − 2x − 15 = 27 + 18x + 6x2 + x3 Special expansions These three identities are important and must be memorised. Examples of these expansions occur constantly, and knowing the formulae greatly simplifies the working. They are proven in the following exercise. 2 Special expansions • Square of a sum: • Square of a difference: • Difference of squares: Example 3 (A + B)2 = A2 + 2AB + B2 (A − B)2 = A2 − 2AB + B2 (A + B)(A − B) = A2 − B2 Using the three special expansions Use the three special expansions above to simplify: b (s − 3t)2 a (x + 4)2 c (x + 3y)(x − 3y) Solution a (x + 4)2 = x2 + 8x + 16 b (s − 3t) = s − 6st + 9t 2 2 2 c (x + 3y)(x − 3y) = x2 − 9y2 (the square of a sum) (the square of a difference) (the difference of squares) Exercise 1A 1 2 FOUNDATION Expand: a 3(x − 2) b 2(x − 3) c −3(x − 2) d −2(x − 3) e −3(x + 2) f −2(x + 3) g −(x − 2) h −(2 − x) i −(x + 3) a 3(x + y) b −2(p − q) c 4(a + 2b) d x(x − 7) e −x(x − 3) f −a(a + 4) g 5(a + 3b − 2c) h −3(2x − 3y + 5z) i xy(2x − 3y) Expand: CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3 4 1A Chapter 1 Algebra review 3 4 Expand and simplify: a 2(x + 1) − x b 3a + 5 + 4(a − 2) c 2 + 2(x − 3) d −3(a + 2) + 10 e 3 − (x + 1) f b + c − (b − c) g (2x − 3y) − (3x − 2y) h 3(x − 2) − 2(x − 5) i 4(2a − 3b) − 3(a + 2b) j 4(s − t) − 6(s + t) k 2x(x + 6y) − x(x − 5y) l 5(2a − 5b) − 6(−a − 4b) Expand and simplify: a (x + 2)(x + 3) b (y + 4)(y + 7) c (t + 6)(t − 3) d (x − 4)(x + 2) e (t − 1)(t − 3) f (2a + 3)(a + 5) g (u − 4)(3u + 2) h (4p + 5)(2p − 3) i (2b − 7)(b − 3) j (5a − 2)(3a + 1) k (6 − c)(c − 3) l (2d − 3)(4 + d) DEVELOPMENT 5 a By expanding (A + B)(A + B), prove the special expansion (A + B)2 = A2 + 2AB + B2 . b Similarly, prove the special expansions: i (A − B)2 = A2 − 2AB + B2 ii (A − B)(A + B) = A2 − B2 6 7 8 Use the special expansions to expand: a (x + y)2 b (x − y)2 c (x − y)(x + y) d (a + 3) e (b − 4) f (c + 5)2 2 2 g (d − 6)(d + 6) h (7 + e)(7 − e) i (8 + f )2 j (9 − g)2 k (h + 10)(h − 10) l (i + 11)2 m (2a + 1)2 n (2b − 3)2 o (3c + 2)2 p (2d + 3e)2 q (2 f + 3g)(2 f − 3g) r (3h − 2i)(3h + 2i) s (5 j + 4) 2 t (4k − 5) u (4 + 5m)(4 − 5m) v (5 − 3n) w (7p + 4q) 2 Expand and simplify: 1 2 a t+ t b t− 1 2 t 2 2 x (8 − 3r)2 c t+ 1 1 t− t t By writing 102 as (100 + 2), and adopting a similar approach for the other two parts, use the special expansions to find (without using a calculator) the value of: a 1022 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b 9992 c 203 × 197 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1A Expanding brackets 9 10 Expand and simplify: a (a − b)(a2 + ab + b2 ) b (x + 2)2 − (x + 1)2 c (a − 3)2 − (a − 3)(a + 3) d (2x + 3)(x − 1) − (x − 2)(x + 1) e (x − 2) f (p + q + r)2 − 2(pq + qr + rp) 3 Expand and simplify: a (x + 4)3 b (x + y + z)2 − 2(xy + yz + zx) c (x + y − z)(x − y + z) d (a + b + c)(a2 + b2 + c2 − ab − bc − ca) 11 a Subtract a(b + c − a) from the sum of b(c + a − b) and c(a + b − c). b Subtract the sum of 2x2 − 3(x − 1) and 2x + 3(x2 − 2) from the sum of 5x2 − (x − 2) and x2 − 2(x + 1). c If X = x − a and Y = 2x + a, find the product of Y − X and X + 3Y in terms of x and a. ENRICHMENT 1 1 = 3. Find the value of x2 + 2 without solving for x. x x 12 Suppose that x + 13 Prove these identities: a (a + b + c)(ab + bc + ca) − abc = (a + b)(b + c)(c + a) b (ax + by)2 + (ay − bx)2 + c2 (x2 + y2 ) = (x2 + y2 )(a2 + b2 + c2 ) 14 If (a + b)2 + (b + c)2 + (c + d)2 = 4(ab + bc + cd), prove that a = b = c = d. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5 6 1B Chapter 1 Algebra review 1B Factoring Learning intentions • Factor using HCF, difference of squares, quadratic factoring, and grouping. • Recognise quadratic expressions that are perfect squares. Factoring is the reverse process of expanding brackets. It is needed routinely throughout the course, and very often is the key step in solving a problem. Four standard methods of factoring There are four standard methods, but in every situation, common factors should always be taken out first. 3 The four basic methods of factoring • Highest common factor: Always try this first. • Difference of squares: This involves two square terms separated by minus. • Quadratics: This involves three terms. Quadratics that are perfect squares: Recognise any crucial special cases. • Grouping: This involves four or more terms. Factoring should continue until each factor is irreducible, meaning that it cannot be factored further. Factoring by taking out the highest common factor Always look first for any common factors of all the terms. Then take out the HCF (highest common factor), or take out common factors in successive steps. Example 4 Factoring by taking out the highest common factor Factor each expression by taking out the highest common factor. a 4x3 + 3x2 b 9a2 b3 − 15b3 Solution a The HCF of 4x3 and 3x2 is x2 , b The HCF of 9a2 b3 and 15b3 is 3b3 , so 4x + 3x = x (4x + 3). so 9a2 b3 − 15b3 = 3b3 (3a2 − 5). 3 2 2 Factoring using the difference of squares If the expression has two terms, both squares, with a minus between them, reverse the special expansion from Section 1A labelled ‘Difference of squares’: A2 − B2 = (A + B)(A − B) Sometimes a common factor must be taken out first. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1B Factoring Example 5 Factoring using the difference of squares Use the difference of squares to factor each expression. a a2 − 36 b 80x2 − 5y2 Solution a a2 − 36 = (a + 6)(a − 6) b 80x2 − 5y2 = 5(16x2 − y2 ) = 5(4x − y)(4x + y) (take out the highest common factor) (use the difference of squares) Factoring monic quadratics A quadratic is called monic if the coefficient of x2 is 1. Suppose that we want to factor the monic quadratic expression x2 − 13x + 36. Look for two numbers: • whose sum is −13 (the coefficient of x), and • whose product is +36 (the constant term). Example 6 Factoring monic quadratics Factor these monic quadratics. a x2 − 13x + 36 b a2 + 12a − 28 Solution a The numbers with sum −13 and product +36 are −9 and −4, so x2 − 13x + 36 = (x − 9)(x − 4). b The numbers with sum +12 and product −28 are +14 and −2, so a2 + 12a − 28 = (a + 14)(a − 2). Factoring non-monic quadratics In a non-monic quadratic such as 2x2 + 11x + 12, where the coefficient of x2 is not 1, look for two numbers: • whose sum is 11 (the coefficient of x), and • whose product is 12 × 2 = 24 (the constant times the coefficient of x2 ). Then split the middle term into two terms. Example 7 Factoring non-monic quadratics Factor these non-monic quadratics. a 2x2 + 11x + 12 b 6s2 − 11s − 10 Solution a The numbers with sum 11 and product 12 × 2 = 24 are 8 and 3, so 2x2 + 11x + 12 = (2x2 + 8x) + (3x + 12) (split 11x into 8x + 3x) = 2x(x + 4) + 3(x + 4) (take out the HCF of each group) = (2x + 3)(x + 4). (x + 4 is a common factor) Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7 8 1B Chapter 1 Algebra review b The numbers with sum −11 and product −10 × 6 = −60 are −15 and 4, so 6s2 − 11s − 10 = (6s2 − 15s) + (4s − 10) (split −11s into −15s + 4s) = 3s(2s − 5) + 2(2s − 5) (take out the HCF of each group) = (3s + 2)(2s − 5). (2s − 5 is a common factor) Recognising quadratics that are perfect squares A quadratic is called a perfect square if it is the square of a binomial expression. The special expansions in Section 1A gave the perfect square formula in two forms: Square of a sum: A2 + 2AB + B2 = (A + B)2 Square of a difference: A2 − 2AB + B2 = (A − B)2 where we have reversed the sides to make them into identities for factoring: x2 + 6x + 9 = (x + 3)2 x2 − 18xy + 81y2 = (x − 9y)2 and There is a test to check if x2 + 6x + 9 is a perfect square: Halve the coefficient 6 of x to get 3, then square it to get 9. Because 9 is the constant term, this means it is a perfect square. The previous method of factoring a quadratic will still work, but recognising perfect squares is quicker, and is an important skill in itself. We will look again at perfect squares in Section 1D. Factoring by grouping When there are four or more terms, it is sometimes possible to factor the expression by grouping. • Split the expression into groups. • Then factor each group in turn. • Then factor the whole expression by taking out a common factor or by some other method. Example 8 Factoring by grouping Factor each expression by grouping. b s2 − t 2 + s − t a 12xy − 9x − 16y + 12 Solution a 12xy − 9x − 16y + 12 = 3x(4y − 3) − 4(4y − 3) (take out the HCF of each pair) = (3x − 4)(4y − 3) (4y − 3 is a common factor) b s − t + s − t = (s + t)(s − t) + (s − t) 2 (factor s2 − t2 using difference of squares) 2 = (s − t)(s + t + 1) (s − t is a common factor) Exercise 1B 1 FOUNDATION Factor by taking out any common factors. a 2x + 8 b 6a − 15 c ax − ay d 20ab − 15ac e x + 3x f p2 + 2pq g 3a2 − 6ab h 12x2 + 18x i 20cd − 32c j a b+b a k 6a + 2a l 7x3 y − 14x2 y2 2 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 2 2 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1B Factoring 2 Factor by grouping in pairs. a mp + mq + np + nq b ax − ay + bx − by c ax + 3a + 2x + 6 d a + ab + ac + bc e z −z +z−1 f ac + bc − ad − bd g pu − qu − pv + qv h x − 3x − xy + 3y i 5p − 5q − px + qx j 2ax − bx − 2ay + by k ab + ac − b − c l x3 + 4x2 − 3x − 12 m a3 − 3a2 − 2a + 6 n 2t3 + 5t2 − 10t − 25 o 2x3 − 6x2 − ax + 3a 2 3 3 2 2 Factor using the difference of squares. a a2 − 1 b b2 − 4 c c2 − 9 d d2 − 100 e 25 − y2 f 1 − n2 g 49 − x2 h 144 − p2 i 4c2 − 9 j 9u2 − 1 k 25x2 − 16 l 1 − 49k2 m x − 4y n 9a2 − b2 2 2 o 25m2 − 36n2 4 Factor each quadratic expression. They are all monic quadratics. a a2 + 3a + 2 b k2 + 5k + 6 c m2 + 7m + 6 d x2 + 8x + 15 e y2 + 9y + 20 f t2 + 12t + 20 g x − 4x + 3 h c − 7c + 10 i a2 − 7a + 12 j b2 − 8b + 12 k t2 + t − 2 l u2 − u − 2 m w2 − 2w − 8 n a2 + 2a − 8 o p2 − 2p − 15 p y2 + 3y − 28 q c2 − 12c + 27 r u2 − 13u + 42 s x2 − x − 90 t x2 + 3x − 40 u t2 − 4t − 32 v p2 + 9p − 36 w u2 − 16u − 80 x t2 + 23t − 50 2 5 6 7 p 81a2 b2 − 64 2 Factor these perfect squares. a x2 + 4x + 4 b y2 + 2y + 1 c p2 + 14p + 49 d m2 − 12m + 36 e t2 − 16t + 64 f x2 + 20x + 100 g u2 − 40u + 400 h a2 − 24a + 144 Copy and complete: a x2 + 6x + · · · = (x + · · · )2 b y2 + 8y + · · · = (y + · · · )2 c a2 − 20a + · · · = (a − · · · )2 d b2 − 100b + · · · = (b − · · · )2 e u2 + u + · · · = (u + · · · )2 f t2 − 7t + · · · = (t − · · · )2 g m2 + 50m + · · · = (m + · · · )2 h c2 − 13c + · · · = (c − · · · )2 Factor each quadratic expression. They are all non-monic quadratics. a 3x2 + 4x + 1 b 2x2 + 5x + 2 c 3x2 + 16x + 5 d 3x2 + 8x + 4 e 2x2 − 3x + 1 f 5x2 − 13x + 6 g 5x2 − 11x + 6 h 6x2 − 11x + 3 i 2x2 − x − 3 j 2x2 + 3x − 5 k 3x2 + 2x − 5 l 3x2 + 14x − 5 m 2x2 − 7x − 15 n 2x2 + x − 15 o 6x2 + 17x − 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9 10 1B Chapter 1 Algebra review p 6x2 − 7x − 3 q 6x2 + 5x − 6 r 5x2 + 23x + 12 s 5x2 + 4x − 12 t 5x2 − 19x + 12 u 5x2 − 11x − 12 v 5x + 28x − 12 w 9x − 6x − 8 x 3x2 + 13x − 30 2 2 DEVELOPMENT 8 Use the techniques of the previous questions to factor each expression. a a2 − 25 b b2 − 25b c c2 − 25c + 100 d 2d2 + 25d + 50 e e3 + 5e2 + 5e + 25 f 16 − f 2 g 16g2 − g3 h h2 + 16h + 64 i i − 16i − 36 j 5 j2 + 16 j − 16 k 4k2 − 16k − 9 l 2k3 − 16k2 − 3k + 24 m 2a2 + ab − 4a − 2b n 6m3 n4 + 9m2 n5 o 49p2 − 121q2 p t2 − 14t + 40 q 3t2 + 2t − 40 r 5t2 + 54t + 40 s 5t2 + 33t + 40 t 5t3 + 10t2 + 15t u u2 + 15u − 54 v 3x3 − 2x2 y − 15x + 10y w (p + q)2 − r2 x 4a2 − 12a + 9 2 9 10 Factor each expression as fully as possible. (Take out any common factors first.) a 3a2 − 12 b x 4 − y4 c x3 − x d 5x2 − 5x − 30 e 25y − y3 f 16 − a4 g 4x2 + 14x − 30 h a4 + a 3 + a 2 + a i c3 + 9c2 − c − 9 j x3 − 8x2 + 7x k x4 − 3x2 − 4 l ax2 − a − 2x2 + 2 Factor as fully as possible: a 4p2 − (q + r)2 b a2 − b2 − a + b c a3 − 10a2 b + 24ab2 d 6x4 − x3 − 2x2 e 4x4 − 37x2 + 9 f 40 − 18x − 40x2 g 4x3 − 12x2 − x + 3 h x2 + 2ax + a2 − b2 i x4 − x2 − 2x − 1 ENRICHMENT 11 12 Factor fully: a a2 + b(b + 1)a + b3 b (x2 + xy)2 − (xy + y2 )2 c (a2 − b2 )2 − (a − b)4 d 4x4 − 2x3 y − 3xy3 − 9y4 e (a2 − b2 − c2 )2 − 4b2 c2 f (ax + by)2 + (ay − bx)2 + c2 (x2 + y2 ) g a4 + a2 b2 + b4 h a4 + 4b4 Given that 3a2 + 4b2 + 18c2 − 4ab − 12ac = 0, prove, by completing squares, that a = 2b = 3c. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1C Algebraic fractions 11 1C Algebraic fractions Learning intentions • Simplify algebraic fractions using the HCF of numerator and denominator. • Add, subtract, multiply, and divide algebraic fractions. • Simplify compound algebraic fractions. An algebraic fraction is a fraction that contains pronumerals. Algebraic fractions are manipulated in the same way as arithmetic fractions, and factoring may play a major role. Cancelling algebraic fractions The key step here is to factor the numerator and denominator completely before cancelling factors. 4 Cancelling algebraic fractions • First factor the numerator and denominator. • Then cancel all common factors. Always cancel each algebraic fraction in the expression before combining them. Example 9 Factoring and cancelling algebraic fractions Simplify each algebraic fraction. 6x + 8 a 6 b x2 − x x2 − 1 Solution a 6x + 8 2(3x + 4) = 6 6 3x + 4 (or x + 43 ) = 3 b x2 − x x(x − 1) = 2 x − 1 (x + 1)(x − 1) x = x+1 Adding and subtracting algebraic fractions When adding or subtracting algebraic fractions, they need to have a common denominator. Finding the lowest common denominator may involve factoring each denominator. 5 Adding and subtracting algebraic fractions • First factor each denominator. • Then work with the lowest common denominator. If necessary, cancel the resulting algebraic fraction. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12 1C Chapter 1 Algebra review Example 10 Adding and subtracting algebraic fractions Use a common denominator to add and subtract these algebraic fractions. x x 5x 11x 2 3 2 − + − + a b c 2 3 6 4 3x 5x 15x d 1 1 − x−4 x Solution a c x x 3x 2x − = − 2 3 6 6 x = 6 2 3 2 10 9 2 − + = − + 3x 5x 15x 15x 15x 15x 3 = 15x 1 = 5x Example 11 5x 11x 10x 33x + = + 6 4 12 12 43x = 12 1 1 x − (x − 4) d − = x−4 x x(x − 4) 4 = x(x − 4) b Factoring before adding and subtracting Factor the denominators of 2+x 5 , then simplify the expression. − 2 x −x x−1 Solution 2+x 2+x 5 5 = − − 2 x − x x − 1 x(x − 1) x − 1 2 + x − 5x = x(x − 1) 2 − 4x = x(x − 1) Multiplying and dividing algebraic fractions These processes are done exactly as for arithmetic fractions. 6 Multiplying and dividing with algebraic fractions Multiplying algebraic fractions: • First factor all numerators and denominators completely. • Then cancel common factors. Dividing by an algebraic fraction: • To divide by an algebraic fraction, multiply by its reciprocal. For example: 3 4 3 y ÷ = × . x y x 4 4 y • The reciprocal of the fraction is . y 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1C Algebraic fractions Example 12 13 Multiplying and dividing algebraic fractions Simplify these products and quotients of algebraic fractions. 2a 12x 6x a−3 ÷ 2 a 2 b × 5a x + 1 x + 2x + 1 a −9 Solution 2a a−3 a−3 = × (factor a2 − 9) 5a (a − 3)(a + 3) 5a 2 = (cancel a − 3 and a) 5(a + 3) 6x x2 + 2x + 1 12x 12x ÷ 2 × (multiply by the reciprocal) b = x + 1 x + 2x + 1 x + 1 6x (x + 1)2 12x × (factor x2 + 2x + 1) = x+1 6x = 2(x + 1) (cancel x + 1 and 6x) a 2a a2 − 9 × Simplifying compound fractions A compound fraction is a fraction in which either the numerator or the denominator is itself a fraction. 7 Simplifying compound fractions • Find the lowest common multiple of the denominators on the top and bottom. • Multiply top and bottom by this lowest common multiple. This will clear all the fractions from the top and bottom together. Example 13 Simplifying compound arithmetic and algebraic fractions Simplify each compound fraction. 1 1 2 − 3 a 1 1 4 + 6 b 1 t +1 1 t −1 b 1 t +1 = 1 t −1 Solution a 1 1 2 − 3 1 1 4 + 6 1 = 21 − 13 × 1 4 + 6 = 12 12 6−4 2 = 3+2 5 = 1 t t +1 × 1 t t −1 1+t 1−t Exercise 1C 1 2 Simplify: x a x Simplify: x 3 a × 3 x 3x 2 × 2 e 4 x FOUNDATION b 2x x c a a ÷ 4 2 5 ÷ 10 f a b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x 2x d a a2 c x2 × g e 3 x 2ab 6 × 2 3 ab 3x2 9xy f 12ab 4a2 b 1 × b2 2b 8ab 4ab ÷ h 5 15 d © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 14 1C Chapter 1 Algebra review 3 4 5 6 Write as a single fraction: x a x+ b 2 x y e f − 8 12 y y + 4 2 2a 3a + 3 2 Write as a single fraction: 1 1 + a a a 1 1 d − 2x 3x Simplify: x+1 x+2 a + 2 3 x+2 x+3 d − 2 3 m m − 3 9 7b 19b g − 10 30 n n + 2 5 xy xy h − 30 12 c 1 2 − x x 3 4 e + 4a 3a d 1 1 + a 2a 5 1 f − 6x 3x b c 2x − 1 2x + 3 + 5 4 2x + 1 2x − 3 e − 4 5 b Factor where possible and then simplify: 2p + 2q 3t − 12 a b p+q 2t − 8 x+3 x−3 + 6 12 2x − 1 2x + 1 f − 3 6 c c x2 + 3x 3x + 9 d a ax + ay e 3a2 − 6ab 2a2 b − 4ab2 f x2 + 2x x2 − 4 g a2 − 9 a2 + a − 12 h x2 + 2x + 1 x2 − 1 i x2 + 10x + 25 x2 + 9x + 20 DEVELOPMENT 7 8 9 Simplify: 1 1 + a x x+1 2 3 + d x−3 x−2 1 1 − x x+1 3 2 − e x+1 x−1 1 1 + x+1 x−1 2 2 − f x−2 x+3 b Simplify: 3x + 3 x2 a × 2 2x x −1 c b a2 + a − 2 a2 − 3a × 2 a+2 a − 4a + 3 c c2 + 5c + 6 c + 3 ÷ c−4 c2 − 16 d x2 − x − 20 x2 − x − 2 x+1 × ÷ 2 2 2 x − 25 x + 2x − 8 x + 5x e ax + bx − 2a − 2b 9x2 − 1 × 3x2 − 5x − 2 a2 + 2ab + b2 f 2x2 + x − 15 x2 + 6x + 9 6x2 − 15x ÷ ÷ 2 x2 + 3x − 28 x2 − 4x x − 49 Simplify: b−a a a−b 1 1 − d a−b b−a CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 v2 − u 2 u−v m n + e m−n n−m b x2 − 5x + 6 2−x x−y f 2 y + xy − 2x2 c © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1C Algebraic fractions 10 11 Simplify: 1 1 a 2 + x + x x2 − x c 2x − y 1 + 2 x − y x − y2 d e x x − a2 − b2 a2 + ab f 1 1 + x2 − 4 x2 − 4x + 4 3 x2 + 2x − 8 f b 1 + 12 t − 1t g t + 1t c 5 − 23 1 1 1 b + a h 1 1 + 10 y x y + x y x y − x − 2 x2 + x − 6 1 1 1 + − x2 − 4x + 3 x2 − 5x + 6 x2 − 3x + 2 Study the exercise on compound fractions and then simplify: 1 1 2 + 13 1 − 12 2 − 5 a 12 b 15 d i 17 3 20 − 4 4 3 5 − 10 1 1 − x+1 1 1 x + x+1 e j 1 x 1 + 2x 3 2 x+2 − x+1 5 4 x+2 − x+1 1 1 y and y = and z = , show that z = t. t 1−x y−1 If x = ENRICHMENT 13 Simplify: (a − b)2 − c2 c ac − bc + c2 a × ÷ ab − b2 − bc a2 + ab − ac a2 − (b − c)2 b c d 14 8x2 + 14x + 3 12x2 − 6x 18x2 − 6x × ÷ 8x2 − 10x + 3 4x2 + 5x + 1 4x2 + x − 3 4y x2 + 2xy − 3x − 2y 3x + 2 xy xy + 2y 1 2 3x − 2 1 + − − x − 1 x + 1 x2 − 1 x2 + 2x + 1 Simplify as fully as possible: 1 1 1 + + a (a − b)(a − c) (b − c)(b − a) (c − a)(c − b) 45 26 65 8 − 3− + b 1+ x−8 x−6 x+7 x−2 ⎞ ⎛ ⎟⎟⎟ ⎜⎜⎜⎜ 2 2 ⎟⎟⎟ ⎜⎜⎜ 1 3n 9n − 2m 1 ⎟⎟⎟ + 2 c 2− ÷ ⎜⎜⎜ − ⎜⎜⎝ m m m + 2mn 4n2 ⎟⎟⎟⎠ m − 2n − m+n 1 d x+ 1 x+2 × 1 x+ 1 x−2 ÷ x− 4 x x2 − 2 + CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 x2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 16 1D Chapter 1 Algebra review 1D Solving quadratic equations Learning intentions • Solve a quadratic equation by factoring. • Solve a quadratic equation by completing the square. • Solve a quadratic equation by the formula. This section will review the three approaches to solving a quadratic equation: • Factoring — this only works in special cases. • Completing the square — this works for every quadratic equation. • The quadratic formula — developed from completing the square. Solving linear equations It is assumed that readers can solve linear equations, by following these two principles: • The same term can be added to or subtracted from both sides of an equation. • Both sides can be multiplied or divided by the same expression, with the exception that this expression cannot be zero. These procedures will be used throughout the book, but are not reveiwed here. Solving a quadratic equation by factoring This method is the simplest, but it only works in special cases. 8 Solving a quadratic equation by factoring • Get all the terms on the left, then factor the left-hand side. • Then use the principle that if A × B = 0, then A = 0 or B = 0. Example 14 Solving a quadratic equation by factoring Solve the quadratic equation 5x2 + 34x − 7 = 0 by factoring. Solution 5x2 + 34x − 7 = 0 5x2 + 35x − x − 7 = 0 (35 and −1 have sum 34 and product −7 × 5 = −35.) 5x(x + 7) − (x + 7) = 0 (5x − 1)(x + 7) = 0 5x − 1 = 0 or x + 7 = 0 x = 15 or x = −7 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (the LHS is now factored) (one of the factors must be zero) (there are two solutions) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1D Solving quadratic equations 17 Completing the square in an expression x2 + bx + · · · In Section 1B, we expanded the quadratic (x + 3)2 , (x + 3)2 = x2 + 6x + 9, and noticed that the coefficient of x is two times 3, and the constant is 3 squared. Reversing the process, the constant term in a perfect square can be found by taking half the coefficient of x and squaring the result. 9 Completing the square in an expression x2 + bx + · · · Halve the coefficient b of x, and then square the result — this is the constant term. Completing the square in an expression x2 + bc + · · · Example 15 Complete the square in each expression: a x2 + 16x + · · · b x2 − 3x + · · · Solution a The coefficient of x is 16, half of 16 is 8, and 82 = 64, so x2 + 16x + 64 = (x + 8)2 . b The coefficient of x is −3, half of −3 is −1 12 , and (−1 12 )2 = 2 14 , so x2 − 3x + 2 14 = (x − 1 12 )2 . Solving a quadratic equation by completing the square Solving a quadratic equation by factoring is only possible in special cases. However, completing the square gives a way to solve every quadratic equation. 10 Solving a quadratic equation by completing the square • Divide both sides by the coefficient of x2 . • Then complete the square in the quadratic by adding the same to both sides. Example 16 Solving a quadratic equation by completing the square Solve each quadratic equation by completing the square. a 3t2 + 24t = 60 b x2 − x − 1 = 0 c x2 − 14x + 49 = 0 d x2 + x + 1 = 0 Solution 3t2 + 24t = 60 a x2 − x − 1 = 0 b ÷3 t2 + 8t = 20 +1 x2 − x = 1 +16 t2 + 8t + 16 = 36 + 14 x2 − x + 14 = 1 14 (t + 4)2 = 36 t + 4 = 6 or t + 4 = −6 t = 2 or −10 (x − 12 )2 = 54 √ √ x − 12 = 12 5 or x − 12 = − 12 5 √ √ x = 12 + 12 5 or 12 − 12 5 Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 18 1D Chapter 1 Algebra review c x2 − 14x + 49 = 0 d x2 + x + 1 = 0 This is already a perfect square. (x − 7)2 = 0 x2 + x + 14 = − 34 (x + 12 )2 = − 34 , x=7 The equation has just one solution. However, a square can’t be negative, so the equation has no solutions. Solving a quadratic equation by the formula This method works whether the solutions are rational numbers or involve surds. It will be proven in the last Development question of Exercise 3E. 11 The quadratic formula • The solutions of ax2 + bx + c = 0 are: √ √ −b − b2 − 4ac −b + b2 − 4ac or x= . x= 2a 2a • Always calculate b2 − 4ac first. (Later, this quantity will be called the discriminant and given the symbol Δ.) Example 17 Solving a quadratic equation by the formula Solve each quadratic equation using the quadratic formula. a 5x2 + 2x − 7 = 0 b 3x2 + 4x − 1 = 0 c x2 − 8x + 16 = 0 d x2 − 8x + 10 = 0 Solution 5x2 + 2x − 7 = 0, a = 5, b = 2 and c = −7. Hence b2 − 4ac = 22 + 140 a For 3x2 + 4x − 1 = 0, a = 3, b = 4 and c = −1. Hence b2 − 4ac = 42 + 12 b For = 144 = 12 , −2 − 12 −2 + 12 or x= 10 10 = 1 or −1 25 . 2 so x2 − 8x + 16 = 0, a = 1, b = −8 and c = 16. Hence b2 − 4ac = (−8)2 − 64 c For = 0, so x = 82 = 4 is the only solution. The LHS is a perfect square (x − 4)2 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 = 28 = 4 × 7, √ −4 + 2 7 −4 − 2 7 so x = or 6 6 √ √ −2 + 7 −2 − 7 = or . 3 3 d For x2 − 8x + 10 = 0, a = 1, b = −8 and c = 10. Hence b2 − 4ac = (−8)2 − 40 √ = −24. There are no square roots of −24, so the equation has no solutions. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1D Solving quadratic equations Exercise 1D 1 2 3 FOUNDATION Solve: a x2 = 9 b y2 = 25 c a2 − 4 = 0 d c2 − 36 = 0 e 1 − t2 = 0 f x2 = 94 g 4x2 − 1 = 0 h 9a2 − 64 = 0 i 25y2 = 16 Solve by factoring: a x2 − 5x = 0 b y2 + y = 0 c c2 + 2c = 0 d k2 − 7k = 0 e t2 = t f 3a = a2 g 2b2 − b = 0 h 3u2 + u = 0 i 4x2 + 3x = 0 j 2a2 = 5a k 3y2 = 2y l 3n + 5n2 = 0 Solve by factoring: a x2 + 4x + 3 = 0 b x2 − 3x + 2 = 0 c x2 + 6x + 8 = 0 d a2 − 7a + 10 = 0 e t2 − 4t − 12 = 0 f c2 − 10c + 25 = 0 g n2 − 9n + 8 = 0 h p2 + 2p − 15 = 0 i a2 − 10a − 24 = 0 j y2 + 4y = 5 k p2 = p + 6 l a2 = a + 132 m c2 + 18 = 9c n 8t + 20 = t2 o u2 + u = 56 p k2 = 24 + 2k q 50 + 27h + h = 0 r a2 + 20a = 44 2 4 Solve by factoring: a 2x2 + 3x + 1 = 0 b 3a2 − 7a + 2 = 0 c 4y2 − 5y + 1 = 0 d 2x2 + 11x + 5 = 0 e 2x2 + x − 3 = 0 f 3n2 − 2n − 5 = 0 g 3b2 − 4b − 4 = 0 h 2a2 + 7a − 15 = 0 i 2y − y − 15 = 0 j 3y2 + 10y = 8 k 5x2 − 26x + 5 = 0 l 4t2 + 9 = 15t m 13t + 6 = 5t2 n 10u2 + 3u − 4 = 0 o 25x2 + 1 = 10x p 6x2 + 13x + 6 = 0 q 12b2 + 3 + 20b = 0 r 6k2 + 13k = 8 2 5 19 Solve each equation using the quadratic formula. Give exact answers, followed by approximations correct to four significant figures where appropriate. a x2 − x − 1 = 0 b y2 + y = 3 c a2 + 12 = 7a d u2 + 2u − 2 = 0 e c2 − 6c + 2 = 0 f 4x2 + 4x + 1 = 0 g 2a2 + 1 = 4a h 5x2 + 13x − 6 = 0 i 2b2 + 3b = 1 j 3c2 = 4c + 3 k 4t2 = 2t + 1 l x2 + x + 1 = 0 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 20 1D Chapter 1 Algebra review 6 Solve each quadratic equation by completing the square. a x2 − 2x = 3 b x2 − 6x = 0 c a2 + 6a + 8 = 0 d x2 + 4x + 1 = 0 e x2 − 10x + 20 = 0 f y2 + 3y = 10 g b2 − 5b − 14 = 0 h y2 − y + 2 = 0 i a2 + 7a + 7 = 0 DEVELOPMENT 7 Solve by factoring: x+2 a x= x c y+ 8 Find the exact solutions of: 1 a x= +2 x c a= 9 10 2 9 = y 2 a+4 a−1 d (5b − 3)(3b + 1) = 1 b 4x − 1 =x x d 5m 1 =2+ 2 m a 2x2 − 4x − 1 = 0 b 2x2 + 8x + 3 = 0 c 3x2 + 6x + 5 = 0 d 4x2 + 4x − 3 = 0 e 4x2 − 2x − 1 = 0 f 2x2 − 10x + 7 = 0 Find a in terms of b if: b 3a2 + 5ab − 2b2 = 0 Find y in terms of x if: a 4x2 − y2 = 0 12 10 =7 a Solve, by dividing both sides by the coefficient of x2 and then completing the square: a a2 − 5ab + 6b2 = 0 11 b a+ Solve each equation. 5k + 7 = 3k + 2 a k−1 c y+1 3−y = y+2 y−4 2 a + 3 10 + = a+3 2 3 √ 3t g 2 = 3 t −6 e b x2 − 9xy − 22y2 = 0 b u+3 2u − 1 = 2u − 7 u−3 d 2(k − 1) = 4 − 5k k+1 f k + 10 10 11 − = k−5 k 6 h 3m + 1 3m − 1 − =2 3m − 1 3m + 1 ENRICHMENT 13 3 7 2 + = . 3x − 2c 2x − 3c 2c a2 b b a2 b Find x in terms of a and b if 2 + 1 + a = 2b + . x x x a Find x in terms of c, given that CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1E Solving simultaneous equations 21 1E Solving simultaneous equations Learning intentions • Solve simultaneous equations by substitution. • Solve simultaneous equations by elimination. There are two algebraic approaches to solving simultaneous equations — substitution and elimination. They can be applied to both linear and non-linear simultaneous equations. Solving by substitution This method can be applied whenever one of the equations can be solved (or rearranged) for one of the variables. 12 Solving simultaneous equations by substitution • Solve one of the equations for one of the variables. • Then substitute it into the other equation. Example 18 Solving simultaneous equations by substitution Solve each pair of simultaneous equations by substitution. a 3x − 2y = 29 (1) 4x + y = 24 (2) b y = x2 (1) y= x+2 (2) Solution y = 24 − 4x. a Solving (2) for y, (2A) 3x − 2(24 − 4x) = 29 Substituting (2A) into (1), x = 7. Substituting x = 7 into (1), 21 − 2y = 29 y = −4. Hence x = 7 and y = −4. (This should be checked in the original equations.) x2 = x + 2 b Substituting (1) into (2), x2 − x − 2 = 0 (x − 2)(x + 1) = 0 x = 2 or −1. From (1), when x = 2, y = 4, and when x = −1, y = 1. Hence x = 2 and y = 4, or x = −1 and y = 1. (Check in the original equations.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 22 1E Chapter 1 Algebra review Solving by elimination This method, when it can be used, usually involves less algebraic manipulation. 13 Solving simultaneous equations by elimination Take suitable multiples of the equations so that one variable is eliminated when the equations are added or subtracted. Example 19 Solving simultaneous equations by elimination Solve each pair of simultaneous equations by elimination. a 3x − 2y = 29 4x + 5y = 8 (1) b x2 + y2 = 53 (1) (2) x − y = 45 (2) 2 2 Solution a Taking 4 × (1) and 3 × (2), b Adding (1) and (2), 12x − 8y = 116 (1A) 12x + 15y = 24. 2x2 = 98 (2A) x2 = 49. Subtracting (1A) from (2A), 23y = −92 y = −4. ÷ 23 Substituting into (1), Subtracting (2) from (1), 2y2 = 8 y2 = 4. Hence x = 7 and y = 2, 3x + 8 = 29 or x = 7 and y = −2, x = 7. or x = −7 and y = 2, Hence x = 7 and y = −4. or x = −7 and y = −2. Exercise 1E 1 2 3 FOUNDATION Solve by substituting the first equation into the second. a y = x and 2x + y = 9 b y = 2x and 3x − y = 2 c y = x − 1 and 2x + y = 5 d a = 2b + 1 and a − 3b = 3 e p = 2 − q and p − q = 4 f v = 1 − 3u and 2u + v = 0 Solve by either adding or subtracting the two equations. a x + y = 5 and x − y = 1 b 3x − 2y = 7 and x + 2y = −3 c 2x + y = 9 and x + y = 5 d a + 3b = 8 and a + 2b = 5 e 4c − d = 6 and 2c − d = 2 f p − 2q = 4 and 3p − 2q = 0 Solve by substitution: a y = 2x and 3x + 2y = 14 b y = −3x and 2x + 5y = 13 c y = 4 − x and x + 3y = 8 d x = 5y + 4 and 3x − y = 26 e 2x + y = 10 and 7x + 8y = 53 f 2x − y = 9 and 3x − 7y = 19 g 4x − 5y = 2 and x + 10y = 41 h 2x + 3y = 47 and 4x − y = 45 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 1E Solving simultaneous equations 4 23 Solve by elimination: a 2x + y = 1 and x − y = −4 b 2x + 3y = 16 and 2x + 7y = 24 c 3x + 2y = −6 and x − 2y = −10 d 5x − 3y = 28 and 2x − 3y = 22 e 3x + 2y = 7 and 5x + y = 7 f 3x + 2y = 0 and 2x − y = 56 g 15x + 2y = 27 and 3x + 7y = 45 h 7x − 3y = 41 and 3x − y = 17 i 2x + 3y = 28 and 3x + 2y = 27 j 3x − 2y = 11 and 4x + 3y = 43 DEVELOPMENT 5 6 7 Solve by substitution: a y = 2 − x and y = x2 b y = 2x − 3 and y = x2 − 4x + 5 c y = 3x2 and y = 4x − x2 d x − y = 5 and y = x2 − 11 e x − y = 2 and xy = 15 f 3x + y = 9 and xy = 6 Solve simultaneously: x y y x − = 1 and + = 10 a 4 3 2 5 b 4x + x−3 y−2 = 12 and 3y − =6 3 5 Solve simultaneously: a x + y = 15 and x2 + y2 = 125 b x − y = 3 and x2 + y2 = 185 c 2x + y = 5 and 4x2 + y2 = 17 d x + y = 9 and x2 + xy + y2 = 61 e x + 2y = 5 and 2xy − x2 = 3 f 3x + 2y = 16 and xy = 10 ENRICHMENT 8 9 Solve simultaneously: 7 5 2 25 − = 3 and + = 12 a x y x 2y b 9x2 + y2 = 52 and xy = 8 Consider the equations 12x2 − 4xy + 11y2 = 64 and 16x2 − 9xy + 11y2 = 78. a By letting y = mx, show that 7m2 + 12m − 4 = 0. b Hence, or otherwise, solve the two equations simultaneously. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 24 Chapter 1 Algebra review Chapter 1 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 1 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Review Chapter Review Exercise 1 Expand and simplify: a 4(x + 3) + 5(2x − 3) b 8(a − 2b) − 6(2a − 3b) c −(a − b) − (a + b) d −4x2 (x + 3) − 2x2 (x − 1) e (n + 7)(2n − 3) f (r + 3)2 g (y − 5)(y + 5) h (3x − 5)(2x − 3) i (t − 8) j (2c + 7)(2c − 7) 2 k (4p + 1) 2 3 4 l (3u − 2)2 2 Factor: a 18a + 36 b 20b − 36 c 9c2 + 36c d d2 − 36 e e2 + 13e + 36 f f 2 − 12 f + 36 g 36 − 25g2 h h2 − 9h − 36 i i2 + 5i − 36 j 2 j2 + 11 j + 12 k 3k2 − 7k − 6 l 52 − 14 + 8 m 4m2 + 4m − 15 n mn + m + pn + p o p3 + 9p2 + 4p + 36 p qt − rt − 5q + 5r q u2 w + vw − u2 x − vx r x2 − y2 + 2x − 2y Simplify: x x + a 2 4 3a 2a + e 2b 3b x y + i y x x x − 2 4 3a 2a − f 2b 3b x y − j y x b Simplify: x+4 x−5 a + 5 3 x+1 x−4 c − 2 5 x x+3 − e 2 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x x × 2 4 3a 2a × g 2b 3b x y × k y x c x x ÷ 2 4 3a 2a ÷ h 2b 3b x y ÷ l y x d 5 3 + x+4 x−5 2 5 d − x+1 x−4 4 2 − f x x+3 b © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 1 review 5 x2 + 2x − 3 x2 − 5x + 4 a+b e 2 a + 2ab + b2 2x2 + 3x + 1 2x3 + x2 + 2x + 1 3x2 − 19x − 14 f 9x2 − 4 c 6 d Solve each quadratic equation by factoring the left-hand side. a a2 − 49 = 0 b b2 + 7b = 0 c c + 7c + 6 = 0 d d2 + 6d − 7 = 0 e e2 − 5e + 6 = 0 f 2f2 − f − 6 = 0 g 2g2 − 13g + 6 = 0 h 3h2 + 2h − 8 = 0 2 7 8 Review Factor each expression where possible, then simplify it. 6a + 3b 2x − 2y a b 2 10a + 5b x − y2 25 Solve using the quadratic formula. Write the solutions in simplest exact form. a x2 − 4x + 1 = 0 b y2 + 3y − 3 = 0 c t2 + 6t + 4 = 0 d 3x2 − 2x − 2 = 0 e 2a2 + 5a − 5 = 0 f 4k2 − 6k − 1 = 0 Solve each quadratic equation by completing the square on the left-hand side. a x2 + 4x = 6 b y2 − 6y + 3 = 0 c x2 − 2x = 12 d y2 + 10y + 7 = 0 The following questions are more difficult 9 If x + y = 5 and xy = 7, find the value of x2 + y2 without attempting to find x or y. 10 Factor x4 + x2 + 1 by adding and subtracting x2 . 11 Simplify 12 Simplify 13 Solve the equation 14 Solve y = x2 − 4 and x2 + y2 = 4 simultaneously. x3 + 2x2 + 4x + 8 x3 − 4x2 + 4x × . 2x3 − 8x x4 − 16 5x x2 + 3x + 2 + 3x x2 + 4x + 3 − 8x x2 + 5x + 6 . 4 1 7 − = . 3x − 1 x + 1 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2 Numbers and surds Chapter introduction Arithmetic is the study of the properties of numbers and of operations on them. This short chapter reviews the four number systems that we will be dealing with: whole numbers, integers, rational numbers, real numbers. Real numbers are central to this course, and almost all the functions we study act on real numbers. The arithmetic of surds also needs attention, because they arise all the time in quadratics and trigonometry. Most of this material will be familiar from earlier years, but the alternative notation for intervals on the number line might be new. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2A Real numbers and intervals 27 2A Real numbers and intervals Learning intentions • Relate whole numbers, integers, and rational numbers to counting. • Show that there are irrational numbers, and define real numbers geometrically. • Distinguish various types of intervals on the real number line. • Write intervals as inequalities and with bracket interval notation. Our ideas about numbers arise from the two quite distinct sources: • The whole numbers, integers, and rational numbers arise from counting and rhythms. • The real numbers are developed from geometry and the number line. Real numbers are the subject of this section. But first, we briefly review the whole numbers, the integers, and the rational numbers. The whole numbers Counting is the first operation in arithmetic. Counting things such as people in a room requires zero (if the room is empty) and then the successive numbers 1, 2, 3, . . . , generating all the whole numbers: 0, 1, 2, 3, 4, 5, 6, . . . There is no last number, because every number is followed by another number, distinct from all previous numbers. The list is therefore called in-finite, which means that it never ‘finishes’. The symbol N is used for the set of whole numbers. A non-zero whole number can be factored, in one and only one way, into the product of prime numbers, which are the whole numbers greater than 1 whose only divisors are itself and 1. The primes form a sequence whose distinctive pattern has confused every mathematician since Greek times: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, . . . The whole numbers greater than 1 and not prime are called composite numbers. The whole numbers 0 and 1 are special cases, being neither prime nor composite. The number 0 is divisible by every whole number except zero. 1 The set N of whole numbers • The whole numbers N are 0, 1, 2, 3, 4, 5, 6, . . . • Every whole number except 0 and 1 is either prime or composite, and every composite number can be factored into primes in one and only one way. • When whole numbers are added or multiplied, the result is a whole number. The integers Any two whole numbers can be added or multiplied, and the result is another whole number. Subtraction, however, requires the negative integers as well: . . . , −6, −5, −4, −3, −2, −1 so that calculations such as 5 − 7 = −2 can be completed. The symbol Z (from German ‘Zahlen’ meaning ‘numbers’) is used for the set of integers. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 28 2A Chapter 2 Numbers and surds 2 The set Z of integers • The integers Z are . . . , −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, . . . • When integers are added, subtracted, or multiplied, the result is an integer. The rational numbers A problem such as, ‘Divide 7 cakes into 3 equal parts’, leads naturally to fractions, where the whole is ‘fractured’ or ‘broken’ into pieces. Thus we have the system of rational numbers, which are numbers that can be written as the ‘ratio’ of two integers. Here are some examples of rational numbers written as single fractions: 1 −1 5 372 4 7 2 13 = − = 30 ÷ 24 = 3.72 = 4= 3 3 3 4 100 1 The symbol Q for ‘quotient’ is used for the set of rational numbers. Operations on the rational numbers Addition, multiplication, subtraction and division (except by 0) can all be carried out within the rational numbers, as is well known from earlier years. • Rational numbers are simplified by dividing top and bottom by their HCF (highest common factor). 21 21 ÷ 7 3 = = 35 35 ÷ 7 5 • Rational numbers are added and subtracted using the LCM (lowest common multiple) of their denominators. For example, 6 and 8 have LCM 24, so 1 5 1 × 4 5 × 3 19 1 5 1×4 5×3 11 + = + = and − = − =− 6 8 24 24 24 6 8 24 24 24 • Fractions are multiplied by multiplying the numerators and multiplying the denominators, after first cancelling out any common factors. To divide by a fraction, multiply by its reciprocal. 10 9 2 3 6 8 3 8 4 32 × = × = and ÷ = × = 21 25 7 5 35 21 4 21 3 63 3 The set Q of rational numbers a • The rational numbers Q are the numbers that can be written as fractions , where a and b are integers b and b 0. a • Every integer a can be written as a fraction , so every integer is a rational number. 1 • When rational numbers are added, subtracted, multiplied, and divided (but not by zero), the result is a rational number. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2A Real numbers and intervals 29 There are numbers that are not rational At first glance, it would seem reasonable to believe that all the numbers on the number line are rational, because the rational numbers are clearly spread ‘as finely as we like’ along the whole number line. Between 0 and 1 there are 9 rational numbers with denominator 10: 0 1 10 2 10 3 10 4 10 5 10 6 10 7 10 8 10 9 10 1 Between 0 and 1 there are 99 rational numbers with denominator 100: 0 50 100 1 Most points on the number line, however, represent numbers that cannot be written as fractions, and are called √ irrational numbers. Some of the most important numbers in this course are irrational, such as 2 and π, and Euler’s number e that will be introduced in Chapter 12. The square root of 2 is irrational √ √ The number 2 is particularly important, because by Pythagoras’ theorem, 2 is the diagonal of a unit square. √ Here is a proof by contradiction that 2 is an irrational number. √ Suppose that 2 were a rational number. √ a Then 2 could be written as a fraction in lowest terms. b √ a 1 That is, 2 = , where a and b have no common factor, and we know that b > 1 2 b √ because 2 is not a whole number. a2 1 Squaring, 2 = 2 , where b2 > 1 because b > 1. b a a2 Because is in lowest terms, 2 is also in lowest terms, which is impossible, b b a2 2 because 2 = 2, but b > 1. b √ This is a contradiction, so 2 cannot be a rational number. The Greek mathematicians were greatly troubled by the existence of irrational numbers. Their concerns can still be seen in modern English, where the word ‘irrational’ means both ‘not a fraction’ and ‘not reasonable’. The real numbers and the number line The whole numbers, the integers, and the rational numbers are based on counting. The existence of irrational numbers, however, means that this approach to arithmetic is inadequate, and a more general idea of number is needed. We have to turn away from counting now, and make an appeal to geometry. 4 Definition of the set R of real numbers • The real numbers R are defined to be all the points on the number line. √ • All rational numbers are real, but real numbers such as 2 and π are irrational. At this point, geometry replaces counting as the basis of our arithmetic. An irrational real number cannot be written as a fraction, or as a terminating or recurring decimal. In this course, √ an irrational number is usually specified in exact form, such as x = 2 or x = π, or as a decimal approximation correct to a certain number of significant figures, such as x 1.4142 or x 3.1416. Very occasionally a fractional approximation is useful or traditional, such as π 3 17 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 30 2A Chapter 2 Numbers and surds The real numbers are often referred to as the continuum, because the rationals, despite being dense, are scattered √ along the number line like specks of dust, but do not ‘join up’. For example, the rational multiples of 2, which are all irrational, are just as dense on the number line as the rational numbers. It is only the real line itself that is completely joined up, to be the continuous line of geometry rather than falling apart into an infinitude of discrete points. The real line looks exactly the same under any magnification. But zoom in on physical matter, and eventually you get to discrete atoms, then nuclei, then quarks — zoom way beyond that, and space itself breaks up into a ‘quantum foam’, where distance has no meaning. Open and closed intervals Any connected part of the real number line is called an interval. • An interval such as 13 ≤ x ≤ 3 is called a closed interval because it contains all its 1 3 endpoints. • An interval such as −1 < x < 5 is called an open interval because it does not contain any of its endpoints. • An interval such as −2 ≤ x < 3 is closed on the left, but open on the right (the word half-closed is sometimes used). 3 x 5 x −1 −2 3 x In diagrams, an endpoint is represented by a closed circle (•) if it is contained in the interval, and by an open circle (◦) if it is not contained in the interval. Bounded and unbounded intervals The three intervals above are bounded because they have two endpoints, which bound the interval. An unbounded interval in contrast is either open or closed, and the direction that continues towards ∞ or −∞ is represented by an arrow. • The unbounded interval x ≥ −5 is a closed interval because it contains all its endpoints (it only has one). • The unbounded interval x < 2 is an open interval because it does not contain its single endpoint. x −5 2 x • The real line itself is an unbounded interval without any endpoints. 5 Intervals • An interval is a connected part of the number line. • A closed interval such as 13 ≤ x ≤ 3 contains all its endpoints. • An open interval such as −1 < x < 5 does not contain any of its endpoints. • An interval such as −2 ≤ x < 3 is neither open nor closed. • A bounded interval has two endpoints, which bound the interval. • An unbounded interval such as x ≥ −5 continues towards ∞ or −∞ (or both). Bracket interval notation The three bounded intervals above can be written in an alternative notation: 1 3 ≤ x ≤3 can also be written as [ 13 , 3]. −1 < x < 5 can also be written as (−1, 5). −2 ≤ x < 3 can also be written as [−2, 3). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2A Real numbers and intervals 31 This notation places the endpoints of the interval inside brackets, using a square bracket when the endpoint is included, and a round bracket otherwise. The two unbounded intervals on the previous page are written using the symbols ∞ and −∞: x ≥ −5 can also be written as [−5, ∞). x<2 can also be written as (−∞, 2). Always use a round bracket for −∞ or ∞, because the symbols ∞ and −∞ are just notation, and are not numbers at all. The whole real line is given by (−∞, ∞). Degenerate intervals — for clarification only For readers who ask too many questions: A single number, [7, 7], and the empty set, say (7, 7), are often regarded as degenerate intervals. Both the whole real line, (−∞, ∞), and the empty set, are both open and closed — each contains all its endpoints, and each contains none of its endpoints, because they have none! These matters are not concerns of our course, however. Exercise 2A 1 Classify these real numbers as rational or irrational. Express those that are rational in the form ab in lowest terms, where a and b are integers. √ √3 √ a −3 b 1 12 c 3 d 4 e 27 √4 4 f 8 g h 0.45 i 12% j 0.333 9 k 0.3̇ 2 FOUNDATION mπ l 3 17 n 3.14 o 0 Write these intervals using inequalities. a The real numbers greater than 4. b The real numbers less than or equal to 3. c The real numbers greater than −2 and less than 2. d The real numbers greater than or equal to −3 and less than or equal to zero. 3 Write these intervals using bracket interval notation. a The real numbers greater than 0 and less than 7. b The real numbers greater than or equal to 2 and less than or equal to 5. c The real numbers greater than or equal to 4. d The real numbers less than −4. 4 Graph these intervals on the real number line. a x ≤ −1 5 c −3 < x ≤ 4 Graph these intervals on the real number line. a [3, ∞) 6 b x > −2 b (−∞, 1) c [−5, 5) Write these intervals using bracket interval notation. a −3 ≤ x ≤ 2 b −3 < x < 2 c −3 < x ≤ 2 d −3 ≤ x < 2 e x≤2 f x > −3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 32 2A Chapter 2 Numbers and surds 7 Write these intervals using inequalities. a (−6, 6) b [−6, 6] c [−6, 6) d (−6, 6] e (6, ∞) f (−∞, −6] DEVELOPMENT 8 Between which two consecutive integers does each of these irrational numbers lie? (Do NOT use a calculator.) √ √ √3 a 15 b 150 c 15 d π 9 a Classify each interval as open or closed or neither (that is, half-closed). i 0≤x≤7 ii x > 5 iii x ≤ 7 iv 5 < x ≤ 15 v x < −1 vi −4 < x < 10 vii x ≥ 6 viii −4 ≤ x < −3 b Classify each interval in part a as bounded or unbounded. 10 a Continue the list of primes at the start of the section to all primes less than 100. b Factor into primes: i 30 11 ii 98 iii 144 iv 365 v 2025 vi 1 000 000 000 a Write down all the integer factors of: ii −28 i 6 b Write down three distinct integers whose product is: i −6 12 ii 3 a Write down all rational numbers in the interval [0, π] with denominators less than 4. 1 1 1 b What is the lowest common denominator required to calculate 12 + 30 + 75 ? ENRICHMENT 13 Write each of these sets of real numbers in simplest form using bracket interval notation. a x > −1 or x < 1 14 15 b x ≥ −1 and x ≤ 1 √ √ Prove that 3 is irrational. (Adapt the given proof that 2 is irrational.) Suppose that a and b are positive irrational numbers, where a < b. Choose any positive integer n such that p 1 < b − a, and let p be the greatest integer such that < a. n n p+1 a Prove that the rational number lies between a and b. n 1 1 and b = √ , find the least possible value of n and the corresponding value of p. b If a = √ 1001 1000 1 1 and √ . c Hence use part a to find a rational number between √ 1001 1000 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2B Surds and their arithmetic 33 2B Surds and their arithmetic Learning intentions √ • Understand the meaning of the term surd, and of the notation x . • Simplify surds by taking out square divisors and collecting like terms. √ √ Numbers such as 2 and 3 occur constantly in this course because they occur when solving quadratic equations and when using Pythagoras’ theorem. These last three sections of the chapter review various methods of dealing with them. Square roots and positive square roots The square of any real number is positive, except that 02 = 0. Hence a negative number cannot have a square root, and the only square root of 0 is 0 itself. A positive number, however, has two square roots, which are the opposites of each other. For example, the square roots of 9 are 3 and −3. √ Warning: The well-known symbol x does not mean ‘the square root of x’. It is defined to mean the positive square root of x (or zero, if x = 0). 6 Definition of the symbol √ x √ • For x > 0, x means the positive square root of x. √ • For x = 0, 0 = 0. √ • For x < 0, x is not defined. √ For example, 25 = 5, even though 25 has two square roots, −5 and 5. The symbol for the negative square root √ of 25 is − 25. Cube roots √ Cube roots are less complicated. Every number has exactly one cube root, so the symbol 3 x simply means ‘the cube root of x’. For example, √3 √3 √3 8=2 and −8 = −2 and 0 = 0. What is a surd? The word surd is very commonly used to refer to any expression involving a square or higher root. More √3 precisely, however, surds do not include expressions such as 49 and 8 that can be simplified to rational numbers. 7 Surds √ An expression n x, where x is a real number and n ≥ 2 is a whole number, is called a surd if it is not itself a rational number. The word ‘surd’ is related to ‘absurd’ — surds are irrational. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 34 2B Chapter 2 Numbers and surds √ It was proven in the last section that 2 is irrational, and in the same way, ‘most’ roots of rational numbers are irrational. Here is the precise result for square roots, which won’t be proven formally, and applies similarly to higher roots: √ ‘If b 0 and a are whole numbers with no common factor, then a/b is rational if and only if both a and b are squares of whole numbers.’ Simplifying expressions involving surds Here are some laws from earlier years for simplifying expressions involving square roots. The first pair restate the definition of the square root, and the second pair are easily proven by squaring. 8 Laws concerning surds Let a ≥ 0 and b ≥ 0 be real numbers. Then: √ √ √ √ a × b = ab 2 a =a √ and √ 2 a a a =a , √ = b b provided that b 0. Taking out square divisors √ A surd such as 500 is not regarded as being simplified, because 500 is divisible by the square number 100, so √ √ 500 can be written as 10 5: √ √ √ √ √ 500 = 100 × 5 = 100 × 5 = 10 5. 9 Simplifying a surd • Check the number inside the square root for divisibility by one of the squares 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, . . . • Continue until the number inside the square root sign has no more square divisors (apart from 1). Example 1 Simplifying surds by taking out squares Simplify these expressions involving surds. √ √ a 108 b 5 27 Solution a √ √ 36 × 3 √ √ = 36 × 3 √ =6 3 108 = Example 2 √ c √ b 5 27 = 5 9 × 3 √ √ =5× 9× 3 c √ √ 216 √ √ 4 × 54 √ √ √ = 4× 9× 6 √ =6 6 216 = = 15 Simplifying surds by collecting like terms Simplify the surds in these expressions, then collect like terms. √ √ √ √ √ a 44 + 99 b 72 − 50 + 12 Solution a √ √ √ √ 44 + 99 = 2 11 + 3 11 √ = 5 11 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b √ √ √ 72 − 50 + 12 √ √ √ =6 2−5 2+2 3 √ √ = 2+2 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2B Surds and their arithmetic Exercise 2B 1 2 3 4 5 FOUNDATION Write down the value of: √ √ a 16 b 36 √ √ e 144 f 400 √ 81 √ g 2500 Simplify: √ a 12 √ e 28 √ i 54 √ m 80 √ 20 √ g 32 √ k 60 √ o 800 √ 121 √ h 10 000 c √ 18 √ f 40 √ j 200 √ n 98 b Simplify: √ √ a 3+ 3 √ √ d −3 2 + 2 √ √ √ √ g 7 6+5 3−4 6−7 3 Simplify: √ a 3 8 √ e 3 45 35 d √ 27 √ h 99 √ l 75 √ p 1000 c √ √ √ √ √ e 4 3+3 2−2 3 √ √ √ √ h −6 2 − 4 5 + 3 2 − 2 5 b 5 7−3 7 √ √ d √ √ 5 √ √ √ f −5 5 − 2 7 + 6 5 √ √ √ √ i 3 10 − 8 5 − 7 10 + 10 5 c 2 5− √ √ √ √ b 5 12 c 2 24 d 4 44 f 6 52 g 2 300 h 2 96 √ √ √ √ Write each expression as a single square root. For example, 3 2 = 9 × 2 = 18. √ √ √ √ a 2 5 b 5 2 c 8 2 d 6 3 √ √ √ √ e 5 5 f 4 7 g 2 17 h 7 10 DEVELOPMENT 6 7 8 Simplify: √ √ a 8+ 2 √ √ d 54 + 24 √ √ √ g 27 + 75 − 48 √3 j 125 Simplify fully: √ √ √ a 600 + 300 − 216 √ √ 12 − 3 √ √ e 45 − 20 √ √ √ h 45 + 80 − 125 √4 k 81 b √ √ √ b 4 18 + 3 12 − 2 50 Find the value of x if: √ √ √ a 63 − 28 = x √ √ √ c 2 150 − 3 24 = x √ √ 50 − 18 √ √ √ f 90 − 40 + 10 √ √ √ i 2 + 32 + 72 √5 l 32 c √ √ √ c 2 175 − 5 140 − 3 28 √ √ √ 80 − 20 = x √ √ √ √ d 150 + 54 − 216 = x b ENRICHMENT √ 7 < 3. √3 √4 √ b Without using a calculator, show that 7 + 7 + 7 < 7. 9 a Without using a calculator, show that 10 Explain why, if we only used rational numbers when drawing graphs in the coordinate plane, the graph of y = x2 − 2 would lie above and below the x-axis, but never intersect it. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 36 2C Chapter 2 Numbers and surds 2C Further simplification of surds Learning intentions • Expand and simplify expressions with surds using algebraic techniques. This section deals with the simplification of more complicated expressions with surds. Simplifying products of surds The product of two surds is found using the identity √ √ √ a × b = ab . It is important to check whether the answer needs further simplification. Example 3 Simplifying products of surds Simplify each product. √ √ a 15 × 5 Solution a √ √ √ √ √ b 5 6 × 7 10 √ √ 15 × 5 = 75 √ = 25 × 3 √ =5 3 √ b 5 6 × 7 10 = 35 60 √ = 35 4 × 15 √ = 35 × 2 15 √ = 70 15 Simplifying surds using binomial expansions Expanding binomial products can be applied to expressions with surds. Example 4 Simplifying surds using binomial expansions Expand these products and then simplify them. √ √ a 15 + 2 3−3 Solution √ b √ √ 15 − 6 2 √ √ √ 3 − 3 = 15 3 − 3 + 2 3 − 3 √ √ √ = 45 − 3 15 + 2 3 − 6 √ √ √ = 3 5 − 3 15 + 2 3 − 6 √ √ √ b 15 − 6 2 = 15 − 2 90 + 6 (using the identity (A − B)2 = A2 − 2AB + B2 ) √ = 21 − 2 × 3 10 √ = 21 − 6 10 a 15 + 2 √ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2C Further simplification of surds Exercise 2C 1 2 3 4 5 37 FOUNDATION Simplify: √ a ( 3 )2 √ √ c 7× 7 √ e 2×3 2 √ √ g 2 3×3 5 √ 2 i (2 3 ) √ √ k 5 2×3 2 √ √ 2× 3 √ √ d 6× 5 √ f 2 5×5 √ √ h 6 2×5 7 √ 2 j (3 7 ) √ √ l 6 10 × 4 10 b Simplify: √ √ a 15 ÷ 3 √ c 3 5÷3 √ √ e 3 10 ÷ 5 √ √ g 10 14 ÷ 5 2 √ √ 42 ÷ 6 √ √ d 2 7÷ 7 √ √ f 6 33 ÷ 6 11 √ √ h 15 35 ÷ 3 7 b Expand: √ √ a 5( 5 + 1) √ √ √ d 2 2( 5 − 2 ) b √ √ 2( 3 − 1) √ √ e 7(7 − 2 7 ) c Simplify fully: √ √ a 6× 2 √ √ d 2 × 2 22 b √ √ 5 × 10 √ √ e 4 12 × 3 c Expand and simplify: √ √ √ a 2( 10 − 2 ) √ √ d 6( 8 − 2) b √ √ 6(3 + 3 ) √ √ e 3 3(9 − 21 ) c √ √ 3(2 − 3 ) √ √ √ f 6(3 6 − 2 5 ) √ √ 3 × 15 √ √ f 3 8×2 5 √ √ 5( 15 + 4) √ √ √ f 3 7( 14 − 2 7 ) DEVELOPMENT 6 Expand and simplify: √ √ a ( 3 + 1)( 2 − 1) √ √ d ( 6 − 1)( 6 − 2) √ √ √ √ e ( 7 − 2)(2 7 + 5) b ( 5 − 2)( 7 + 3) √ √ √ √ 2 )( 3 + 2 ) √ √ √ f (3 2 − 1)( 6 − 3 ) c ( 5+ 7 Use the special expansion (a + b)(a − b) = a2 − b2 to expand and simplify: √ √ √ √ a ( 5 + 1)( 5 − 1) b (3 − 7 )(3 + 7 ) √ √ √ √ √ √ √ √ c ( 3 + 2 )( 3 − 2 ) d (3 2 − 11 )(3 2 + 11 ) √ √ √ √ e (2 6 + 3)(2 6 − 3) f (7 − 2 5 )(7 + 2 5 ) 8 Expand and simplify each expression using the special expansions (a + b)2 = a2 + 2ab + b2 √ √ √ d ( 7 − 5 )2 √ √ g (2 7 + 5 )2 a ( 3 + 1)2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 and (a − b)2 = a2 − 2ab + b2 . √ √ √ b ( 5 − 1)2 c ( 3 + 2 )2 √ √ e (2 3 − 1)2 f (2 5 + 3)2 √ √ √ √ h (3 2 − 2 3 )2 i (3 5 + 10 )2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 38 2C Chapter 2 Numbers and surds 9 Simplify fully: √ 40 a √ √10 √ 5 7× 3 d √ 28 √ b √ √ √ 2 6× 5 c √ √ 10 √ 6 3×8 2 f √ √ 32 × 27 18 √50 √ 15 × 20 e √ 12 10 Use Pythagoras’ theorem to find the hypotenuse of the right-angled triangle in which the lengths of the other two sides are: √ √ a 2 and 7 √ √ b 5 and 2 5 √ √ c 7 + 1 and 7 − 1 √ √ √ √ d 2 3 − 6 and 2 3 + 6 11 Simplify by forming the lowest common denominator: 1 1 +√ a √ 3+1 3−1 3 3 b √ √ − √ √ 2 5− 7 2 5+ 7 12 Given that x and y are positive, simplify: a x 2 y3 b x x 2 y6 √ d x3 + 2x2 + x e x2 y4 (x2 + 2x + 1) √ a Find a pair of values of a and b for which a2 + b2 a + b. √ b For what values of a and b is it true that a2 + b2 = a + b? 13 c f √ √ x2 + 6x + 9 x4 + 2x3 + x2 ENRICHMENT 14 Determine, without using a calculator, which is the greater number in each pair. √ √ √ √ a 2 3 or 11 b 7 2 or 3 11 √ √ √ √ c 3 + 2 2 or 15 − 7 2 d 2 6 − 3 or 7 − 2 6 15 a Write down the expansion of (a + b)2 . b Use the expansion in part a to square c Hence simplify 16 17 √ 6 + 11 − √ 6 + 11 − √ 6 − 11. √ 6 − 11. √ Given that x − y = 8 2 and xy = 137, where x and y are both positive, find the value of x + y without finding x or y. √ Suppose that a = 1 + 2. a Show that a2 − 2a − 1 = 0. b Hence show that a = 2 + √ 1 . a c Show that a 2 = a + 1, and hence that d Deduce that √ 2=1+ √ 2=1+ 1 2+ 1 . a 1 2+ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 2+ 1 2 + ··· © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2D Rationalising the denominator 39 2D Rationalising the denominator Learning intentions • Simplify expressions with surds by rationalising the denominator. • Use conjugates where necessary when rationalising the denominator. When dealing with surdic expressions, it is usual to remove any surds from the denominator, a process called rationalising the denominator. There are two cases. The denominator has a single term In the first case, the denominator is a surd or a multiple of a surd. 10 Rationalising a single-term denominator √ √ 7 In an expression such as √ , multiply top and bottom by 3 . 2 3 Example 5 Simplifying surds by rationalising the denominator √ a 7 √ 2 3 55 b √ Solution 11 √ 55 11 55 b √ = √ ×√ 11 11 11 √ 55 11 = √11 = 5 11 √ √ √ 7 7 3 a √ = √ ×√ 2 3 2 3 3 √ 21 = 2×3 √ 21 = 6 The denominator has two terms The second case involves a denominator with two terms, one or both of which contain a surd. The method uses the difference of squares identity (A + B)(A − B) = A2 − B2 to square the unwanted surds and convert them to integers. 11 Rationalising a binomial denominator √ 3 √ , multiply top and bottom by 5 − 3 . 5+ 3 • Then use the difference of squares. • In an expression such as CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 40 2D Chapter 2 Numbers and surds Example 6 a Rationalising a binomial surd denominator 3 √ 5+ 3 b 1 √ 2 3−3 2 √ Solution √ 3 5− 3 3 a Using the difference of squares, √ = √ × √ √ √ √ 5+ 3 5+ 3 5− 3 √ 5 + 3 5 − 3 = 52 − 3 2 15 − 3 3 = = 25 − 3. 25 − 3 √ = 22. 15 − 3 3 = 22 √ √ 1 1 2 3+3 2 b √ √ = √ √ × √ √ 2 3−3 2 2 3−3 2 2 3+3 2 √ √ 2 3+3 2 Using the difference of squares, = √ √ √ √ √ √ 4×3−9×2 √ √ 2 3−3 2 2 3+3 2 = 2 3 2− 3 2 2 2 3+3 2 =− = 4 × 3 − 9 × 2. 6 = −6. Conjugates of expressions with surds A new word is often used when describing rationalising the denominator. √ √ The expressions 5 + 3 and 5 − 3 are called conjugates of each other, and √ √ (5 + 3)(5 − 3) = 52 − 3, by the difference of squares, = 22. 12 Conjugates of surdic expressions √ √ √ √ • The expressions a b + c d and a b − c d are conjugates of each other. • The product of a expression with surds and its conjugate has no surds: √ √ √ √ (a b + c d ) (a b − c d ) = a2 b − c2 d, by the difference of squares. ‘Conjugate’ is a Latin word meaning ‘joined together’. Exercise 2D 1 2 FOUNDATION Write down the conjugate of: √ √ a 3−1 b 2 5+3 c √ √ 7− 2 √ √ d 3 11 + 4 6 Write down the result when each expression in the previous question is multiplied by its conjugate. √ √ √ √ √ √ a 3−1 b 2 5+3 c 7− 2 d 3 11 + 4 6 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 2D Rationalising the denominator 3 4 5 6 Rewrite each fraction with a rational denominator. 1 1 a √ b √ 3 7 √ √ 2 5 e √ f √ 3 7 Rewrite each fraction with a rational denominator. 1 1 a √ b √ 3−1 7+2 1 1 e √ f √ √ √ 5− 2 10 + 6 3 5 c √ d √ 5 √ g 2 11 √ 5 1 √ 3+ 5 1 g √ 2 3+1 c Simplify each expression by rationalising the denominator. 2 5 6 a √ b √ c √ 3 2 5 3 5 8 e √ f √ g √ 6 15 6 Rewrite each fraction with a rational denominator. 1 1 a √ b √ 3 7 2 5 e 10 √ 3 2 f 9 √ 4 3 c 41 2 √ h 3 7 √ 2 1 √ 4− 7 1 h √ 5−3 2 d 21 d √ 7 14 h √ 10 3 √ d 5 2 √ 3 g √ 2 10 2 √ 7 3 √ 2 11 h √ 5 7 DEVELOPMENT 7 Rewrite each fraction with a rational denominator. 4 3 a √ b √ √ 5+1 2 2− 3 √ √ 2 7 5 e √ f √ √ 2 7−5 10 − 5 √ √ √ 3− 7 3 2+ 5 i j √ √ √ 3+ 7 3 2− 5 √ c 7 √ 5− 7 √ 3−1 g √ 3+1 √ √ 10 − 6 k √ √ 10 + 6 8 Simplify each expression by rationalising the denominator. √ √ √ 3−1 2 5− 2 a b √ √ √ 2− 3 5+ 2 9 Show that each expression is rational by first rationalising the denominators. 1 3 3 2 a √ + b √ √ +√ 2 2+ 2 3+ 6 6 1 4 8 6 c d √ + √ √ − √ 3− 7 2 7−5 2+ 2 3−2 2 √ 5+1 1 If x = , show that 1 + = x. 2 x √ √ √ 6+1 The expression √ √ can be written in the form a 3 + b 2. Find a and b. 3+ 2 10 11 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 √ 3 3 d √ √ 5+ 3 √ √ 5+ 2 h √ √ 5− 2 √ 7 + 2 11 l √ 7 − 2 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 42 2D Chapter 2 Numbers and surds 12 13 1 Evaluate a + for these values of a: a √ a 1+ 2 √ b 2− 3 √ 3− 3 c √ 3 + 3√ √ x+ 2−x d √ √ x− 2−x Rationalise the denominator of √ 1 x+h+ √ , where x ≥ 0 and x + h ≥ 0. x 2 1 a Expand x + . x √ √ b Suppose that x = 7 + 6. √ 1 i Show that x + = 2 7. x 14 ii Use the result in part a to find the value of x2 + 1 . x2 ENRICHMENT 15 √ The value of 17 is 4.12, correct to two decimal places. a Substitute this value to determine an approximation for √ 1 √ 1 17 − 4 . 17 + 4 , and that this last result gives a more accurate value for the approximation 17 − 4 than that found in part a. b Show that √ 16 Express √ = 1 √ with a rational denominator. √ 2+ 3+ 5 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 2 review 43 Chapter 2 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 2 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise 2 a Classify each of these real numbers as rational or irrational. Express those that are rational in the form , b where a and b are integers. √ √ a 7 b −2 14 c 9 d 10 √3 √4 e 15 f 16 g −0.16 h π Graph these intervals on the real number line. a x≤2 3 5 6 7 b x > −1 c −2 < x ≤ 5 Graph these intervals on the real number line. a [−2, ∞) 4 Review 1 b (−∞, 0) c [−3, 3) Write these intervals using bracket interval notation. a −4 ≤ x ≤ 6 b −4 < x < 6 c −4 < x ≤ 6 d −4 ≤ x < 6 e x≤6 f x > −4 Write these intervals using inequalities. a (−5, 3) b [−5, 3] c [−5, 3) d (−5, 3] e (3, ∞) f (−∞, −5] Simplify: √ a 24 √ d 500 b √ 45 √ e 3 18 c Simplify: √ √ a 5+ 5 √ √ √ d 2 5+ 7−3 5 √ √ g 8× 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 √ √ 5× 5 √ √ e 35 ÷ 5 √ √ h 10 × 2 b √ 50 √ f 2 40 √ √ √ f 6 55 ÷ 2 11 √ √ i 2 6 × 4 15 c (2 7 )2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 44 Chapter 2 Numbers and surds Review 8 9 10 11 12 Simplify: √ √ a 27 − 12 √ √ √ c 3 2 + 3 8 − 50 b Expand: √ √ a 7(3 − 7 ) √ √ c 15( 3 − 5) b √ √ 18 + 32 √ √ √ √ d 54 − 20 + 150 − 80 √ √ √ 5(2 6 + 3 2 ) √ √ √ d 3( 6 + 2 3 ) Expand and simplify: √ √ a ( 5 + 2)(3 − 5 ) √ √ d ( 10 − 3)( 10 + 3) √ √ g ( 5 + 2 )2 √ √ √ c ( 7 − 3)(2 5 + 4) e (2 6 + f ( 7 − 2)2 √ √ 11 )(2 6 − 11 ) √ h (4 − 3 2 )2 d 1 √ Write with a rational denominator: 1 1 a √ b √ √ 3− 7 5√+ 2 3 3 d √ e √ √ 3+1 11 + 5 15 Simplify √ 3 5−2 +√ 2 5+2 e 5 3 14 √ 5 √ f 2 3 10 2 7 √ √ 1 √ 2 6√− 3 3 7 f √ √ 2 5− 7 c √ 3 3+5 b √ 3 3−5 by forming the lowest common denominator. √ 16 Find the values of p and q such that √ 17 Show that 18 √ Write with a rational denominator: √ 3 3 1 a √ b √ c √ 5 2 11 Rationalise √ the denominator of each fraction. √ 7− 2 a √ √ 7+ 2 √ √ √ Find the value of x if 18 + 8 = x. 13 √ √ √ b (2 3 − 1)(3 3 + 5) 5 5−2 √ = p + q 5. 1 is rational by first rationalising each denominator. √ − √ 6−3 3 2 3+3 √ Suppose that x = 3 + 10. 1 a Find the value of x + in simplest form. x 1 b Hence find the value of x2 + 2 . x 2 19 By considering the reciprocals of the left-hand and right-hand sides, prove without a calculator that √ √ √ 6 − 5 < 5 − 2. 20 Without a calculator, find √ 21 a Show that 1 1 1 1 √ +√ √ + ··· + √ √ . √ +√ 1+ 2 2+ 3 3+ 4 15 + 16 √ √ 4 + 2 3 = 1 + 3. √ 3 1+ 1+x . , use part a to simplify √ b If x = 2 1+x √ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3 Functions and graphs Chapter introduction The principal purpose of our course is the study of functions. Now that the real numbers have been reviewed, this chapter develops the idea of functions and relations involving real numbers, and their graphs. A variety of known graphs are then discussed, with particular emphasis on quadratics and their parabolic graphs, on circles, powers, and factored cubics, and on two graphs that have asymptotes. The final section deals with direct and inverse variation, which involves practical applications of some of the graphs in the chapter. Curve-sketching software is useful in emphasising the basic idea that a function has a graph, and is very useful in quickly sketching a large number of graphs and recognising their relationships — everyone with access to the technology should be doing this. But the ability to sketch the graph of a given function without technology is a central concern of this course, and will require a large variety of methods, which will be developed gradually as the course progresses. Most readers will already know many things in this chapter. Few readers will have mastered the details. Be very careful. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 46 3A Chapter 3 Functions and graphs 3A Functions and function notation Learning intentions • Understand a function as a rule, displayed using a table of values. • Identify independent and dependent variables of a function written as y = . . . • Identify any natural restrictions on the variables of a function. • Work fluently with function notation f (x) = . . . • Substitute numbers and expressions into a function. Most of the graphs studied in previous years are examples of functions. We need to make the idea of functions more precise, and introduce some new notation. Functions Here is a situation that naturally leads to a function. An electrician charges $100 to visit a home, and then charges $40 for each power point that he installs. Let x be the number of power points that he installs. Let y be the total cost, in dollars. Then y = 100 + 40x. This is an example of a function. We say that y is a function of x because the value of y is determined by the value of x, and we call x and y variables because they take many different values. The variable x is called the independent variable of the function, and the variable y is called the dependent variable because its value depends on x. Thus a function is a rule. We input a value of x, and the rule produces an output y. In this example, x must be a whole number 0, 1, 2, . . . , and we can add this restriction to the rule, describing the function as: ‘The function y = 100 + 40x, where x is a whole number.’ A table of values is a useful tool — a few values of the function are selected and arranged in a table. Here is a representative table of values showing the total cost y dollars of installing x power points: 1 x 0 1 2 3 4 5 6 y 100 140 180 220 260 300 340 A function is a rule • A variable y is a function of a variable x when y is completely determined by x as a result of some rule. • The variable x is called the independent variable of the function, and the variable y is called the dependent variable because it depends on x. • The function rule is almost always an equation, possibly with restrictions. For example: ‘The function y = 100 + 40x, where x is a whole number.’ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3A Functions and function notation The function machine Input x The function and its rule can be regarded as a ‘machine’ with inputs and outputs. For example, the numbers in the right-hand column are the outputs from the function y = 100 + 40x, when the numbers 0, 1, 2, 3 and 4 are the inputs. 0 −→ −→ 100 1 −→ −→ 140 2 −→ −→ 180 −→ 220 This model of a function has become far more intuitive in the last few decades because computers and calculators routinely produce output from a given input. 4 −→ −→ 260 ... ... 3 −→ x −→ 47 Output y f −→ 100 + 40x A question in the following exercise presents a variation of this diagram. Function notation In the diagram above, we gave the name f to our function. We can now write the results of the input–output routines as follows: f (0) = 100, f (1) = 140, f (2) = 180, f (3) = 220, ... This is read aloud as, ‘ f of zero is equal to one hundred’, and so on. When x is the input, the output is 100 + 40x, so using the well-known notation introduced by Euler in 1735, we can write the function rule as f (x) = 100 + 40x, where x is a whole number. This f (x) notation will be used throughout the course alongside y = . . . notation. The previous table of values can therefore also be written as x 0 1 2 3 4 5 6 f (x) 100 140 180 220 260 300 340 Example 1 Substituting into a function A function is defined by the rule f (x) = x2 + 5x. Find f (3), f (0), and f (−3). Solution f (3) = 32 + 5 × 3 f (0) = 02 + 0 × 3 = 9 + 15 =0+0 = 9 − 15 = 24 =0 = −6 Example 2 f (−3) = (−3)2 + 5 × (−3) Finding the formula for a function described verbally A function g(x) is defined by the rule, ‘Cube the number and subtract 7’. Write down its function rule as an equation, then draw up a table of values for −2, −1, 0, 1 and 2. Solution The function rule is g(x) = x3 − 7. −2 −1 0 1 2 g(x) −15 −8 −7 −6 1 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 48 3A Chapter 3 Functions and graphs Example 3 Substituting algebraic expressions into a function If f (x) = x2 + 5, find and simplify f (a), f (a + 1), and f (a + h). Solution f (a) = a2 + 5 f (a + 1) = (a + 1)2 + 5 f (a + h) = (a + h)2 + 5 = a2 + 2a + 1 + 5 = a2 + 2ah + h2 + 5 = a2 + 2a + 6 Restrictions on the variables of a function With many functions that describe practical situations, the variables have restrictions arising from the situation. Any such restriction should be made part of the function by adding the restriction after the formula, as in the following example. Example 4 Restricting the variables of a function Sadie the snail is crawling at a steady 10 cm per minute vertically up a wall 3 metres high, starting at the bottom. Write down the height y metres as a function of the time x minutes of climbing, adding the restriction on x. Solution 1 The snail’s height y after x minutes is 10 x metres, and the snail will take 30 minutes to get to the top, so the function is 1 x, y = 10 where 0 ≤ x ≤ 30. Exercise 3A 1 FOUNDATION Let p(x) = x2 − 2x − 3. Determine the following. b p(4) a p(0) 2 b 5 5 c −2 d −1 c f (x) = x3 + 8 d f (x) = 2 x Find f (2), f (0) and f (−2) for each function. a f (x) = 3x − 1 4 d p(−2) Find the value of the function y = 5 + 2x − x2 at each value of x. a 0 3 c p(3) b f (x) = 4 − x2 Find h(−3), h(1) and h(5) for each function. a h(x) = 2x + 2 b h(x) = 1x c h(x) = 3x − x2 d h(x) = √ x+4 Write down g(t) for each of the following functions. a g(x) = 3x + 1 b g(x) = x2 − 2x + 8 c g(x) = x − √ x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3A Functions and function notation 6 49 Copy and complete the table of values for each function. a y = x2 − 2x x −1 0 1 2 3 −2 −1 y b f (x) = x3 − 4x x −3 0 1 2 3 f (x) DEVELOPMENT 7 For the function L(x) = 3x + 1, determine: a L(1) − 2 b 3L(−1) c L(1) + L(2) d L(9) ÷ L(2) 8 I think of a number, then subtract the square of that number from 6. Let x be the number and let f (x) be the final answer. a Write down the function rule for this arithmetic. b Copy and complete the table of values for your function. x −2 −1 0 1 2 f (x) 9 10 Given that P(x) = x2 − 2x − 4, find the value of: √ a P(1 + 5 ), √ b P( 3 − 1). a Write the equation 3x + 4y + 5 = 0 as a function with independent variable x. b Write the equation 3x + 4y + 5 = 0 as a function with independent variable y. c Write the equation 4 + xy = 0 as a function with dependent variable y. d The volume of a cube with side length s is V = s3 , and its surface area is A = 6s2 . Write each formula as a function with dependent variable s. e If a rectangle has area 100 m2 and sides and b, then b = 100. Write this formula as a function with: i as the dependent variable, ii as the independent variable. 11 A restaurant offers a special deal to groups by charging a cover fee of $50, then $20 per person. Write down C, the total cost of the meal in dollars, as a function of x, the number of people in the group. 12 In each case explain why the function value cannot be found. √ a F(0), where F(x) = x − 4. √ b H(3), where H(x) = 1 − x2 . 1 . c g(−2), where g(x) = 2+x 1 d f (0), where f (x) = . x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 50 3A Chapter 3 Functions and graphs 13 Find g(a), g(−a) and g(a + 1) for each function. a g(x) = 2x − 4 b g(x) = 2 − x c g(x) = x2 d g(x) = 14 1 x−1 Find F(t) − 2 and F(t − 2) for each function. a F(x) = 5x + 2 b F(x) = √ x c F(x) = x + 2x d F(x) = 2 − x2 2 15 If f (x) = x2 + 5x, find in simplest form: f (1 + h) − f (1) a h f (p) − f (q) b p−q f (x + h) − f (x) c h 16 a If f (x) = x4 + 2x2 + 3, show that f (−x) = f (x) for all values of x. 4 x b If g(x) = x3 + , show that g(−x) = −g(x) whenever x 0. c If h(x) = x x2 + 1 , show that h( 1x ) = h(x) whenever x 0. ENRICHMENT x 1 17 Evaluate e(x) = 1 + on your calculator for x = 1, 10, 100, 1000 and 10 000, giving your answer to x two decimal places. What do you notice happens as x gets large? 18 Let c(x) = 3 x + 3−x 3 x − 3−x and s(x) = . Show that c(x) 2 − s(x) 2 = 1. 2 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3B Functions, relations, and graphs 51 3B Functions, relations, and graphs Learning intentions • Draw the graph of a function from a table of values. • Understand a function as a set of ordered pairs satisfying a condition. • Understand a relation as any set of ordered pairs, and draw its graph. • Find the domain and range of a function or relation. A graph is the most important and helpful way to represent a function or relation. Most things that we will discover about a function or relation can be seen on its graph, so the sketching of graphs is important. A function and its graph A ball is thrown vertically upwards. While the ball is in the air, its height y metres, x seconds after it is thrown, is y = 5x(6 − x). y A table of values of the function is shown. x (time) 0 1 2 3 4 5 6 y (height) 0 25 40 45 40 25 0 Each x value and its corresponding y value can be put into an ordered pair ready to plot on a graph of the function. The seven ordered pairs calculated here are: 45 40 25 (0, 0), (1, 25), (2, 40), (3, 45), (4, 40), (5, 25), (6, 0) and the graph is sketched opposite. The seven representative points have been plotted, but there are infinitely many such ordered pairs, and they all join up to make the smooth curve shown to the right. 3 6 x Like all graphs of functions, this graph has a crucial property — no two points have the same x-coordinate. This is because at any one time, the ball can only be in one position. In function-machine language, no input can have two outputs. 2 The graph of a function • The graph of a function consists of all the ordered pairs (x, y) plotted on a pair of axes, where x and y are the values of the input and output variables. • No two points on the graph ever have the same x-coordinate. • In most graphs in this course, the points join up to make a smooth curve. A function as a set of ordered pairs These ideas of the graph and its ordered pairs allow a more formal definition of a function — a function can be defined simply as a set of ordered pairs satisfying the crucial property mentioned above. 3 A function is a set of ordered pairs satisfying a condition A function is a set of ordered pairs (x, y) in which: • no two ordered pairs have the same x-coordinate. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 52 3B Chapter 3 Functions and graphs Domain and range There are two restrictions on the time x in our previous example: • The time x cannot be negative, because the ball had not been thrown then. • The time cannot be greater than 6, because the ball hits the ground after 6 seconds. The domain of the function is the set of possible x-values, so the domain is the closed interval 0 ≤ x ≤ 6. Thus the function is more correctly written as y = 5x(6 − x), where 0 ≤ x ≤ 6. The endpoints of the graph are marked with closed (filled-in) circles • to indicate that these endpoints are included in the graph. If they were not included, they would be marked with open circles ◦. These are the same conventions that were used with intervals in Section 2A. From the graph, we can see that the height of the ball ranges from 0 on the ground to 45 metres. The range is the set of possible y-values, so the range is the interval 0 ≤ y ≤ 45, which includes the two endpoints y = 0 and y = 45. Domain and range here are intervals, so we can also use bracket interval notation: domain = [0, 6] and range = [0, 45]. But be careful when using this notation! The domain here is an interval on the x-axis, whereas the range is an interval on the y-axis. 4 The domain and range of a function • The domain of a function is the set of all possible x-coordinates. • The range of a function is the set of all possible y-coordinates. It is usually easier to find the range after the graph has been drawn. Reading the domain and range from the graph Sometimes we have the graph of a function, but not its equation or rule. We can usually read the domain and range off such a graph. 5 Reading the domain and range from the graph • Domain: All values on the x-axis that have graph points above or below. • Range: All values on the y-axis that have graph points to the left or right. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3B Functions, relations, and graphs Example 5 53 Reading the domain and range off the graph Write down the domain and the range of the functions whose graphs are sketched below. y a y 8 b 4 -2 2 x x -8 Solution a Domain: all real x, range: y ≤ 4, OR domain = (−∞, ∞), range = (−∞, 4]. b Domain: −2 ≤ x ≤ 2, range: −8 ≤ y ≤ 8, OR domain = [−2, 2], range = [−8, 8]. The natural domain When a function is given as an equation with no restrictions, we assume as a convention that the domain is all the x-values that can validly be substituted into the equation. This is called the natural domain. 6 The natural domain If no restrictions are given, the domain is all x-values that can validly be substituted into the equation. This is called the natural domain. There are many reasons why a number cannot be substituted into an equation. So far, the two most common reasons are: • We cannot divide by zero. • We cannot take square roots of negative numbers. Example 6 Finding the domain of a function, given its formula Find the natural domain of each function: 1 a y= x−2 b y= √ x−2 Solution a We cannot divide by zero. Hence the domain is that is, x−20 x 2. b We cannot take square roots of negative numbers. Hence the domain is that is, x−2≥0 x ≥ 2. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 54 3B Chapter 3 Functions and graphs Relations y We shall often be dealing with graphs such as the circle sketched to the right. This graph is a set of ordered pairs. But it is not a function, because, for example: 10 The points (5, 0) and (5, 10) have the same x-coordinate, so that the input x = 5 has the two outputs y = 0 and y = 10. The graph thus fails the crucial property that no input can have more than one output. 5 The more general word ‘relation’ is used for any graph in the plane, whether a function or not. In terms of ordered pairs: 7 5 10 x Relations A relation is any set of ordered pairs. • Like a function, a relation has a graph. • Like a function, a relation has a domain and a range. • Unlike a function, a relation may have two or more points with the same x-coordinate. A function is thus a special type of relation, just as a square is a special type of rectangle. Example 7 Identifying why a particular relation is not a function In each part, show that the relation is not a function by writing down two ordered pairs on the graph with the same x-coordinate. Illustrate this by connecting the two points by a vertical line. Then write down the domain and range. y a b 5 y 1 5 x -5 1 x -1 -5 x2 + y2 = 25 y2 = x Solution a The points (0, 5) and (0, −5) on the graph have the same x-coordinate x = 0. Thus when x = 0 is the input, there are two outputs, y = 5 and y = −5. The vertical line x = 0 meets the graph at (0, 5) and at (0, −5). Domain: −5 ≤ x ≤ 5, range: −5 ≤ y ≤ 5 OR domain = [−5, 5], range = [−5, 5]. y y 5 1 5 x -5 1 x -1 -5 b The points (1, 1) and (1, −1) on the graph have the same x-coordinate x = 1. Thus when x = 1 is the input, there are two outputs y = −1 and y = 1. The vertical line x = 1 meets the graph at (1, −1) and (1, 1). Domain: x ≥ 0, range: all real y OR domain = [0, ∞), range = (−∞, ∞). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3B Functions, relations, and graphs 55 The vertical line test The previous example shows that we can easily use vertical lines on any graph to see whether or not it is a function. To show that the graph is not a function, we need to identify just two points with the same x-coordinate. That means, we need to draw just one vertical line that crosses the graph twice. 8 The vertical line test If at least one vertical line meets the graph of a relation more than once, then the relation is not a function. The word ‘map’ Functions (but not relations in general) are also called maps or mappings. The word map may also be used as a verb. Thus in the function f (x) = 2 x , if we substitute x = 3, we may say that ‘3 is mapped to 8’. In a map of NSW, every point on the ground in NSW corresponds to a unique point on the map. A relation really can be any finite or infinite set of ordered pairs Most relations that arise in this course are like the circles and sideways parabolas in the examples above. The following worked example looks at some different types of relations. Example 8 Examining unfamiliar examples of relations a State whether or not each set of ordered pairs, or set of points in a graph, is a relation. If it is a relation, state whether or not it is a function, giving reasons if it is not a function. i (1, 3), (2, 7), (3, 5), (5, 3), (3, 1), (0, 0) 1 1 ii (−1, 4), (10, −6), (−6, 10), (2 , 3 ) 2 2 2 iii The region inside the circle x + y2 = 25. iv The region between the lines x = 1 and x = 4. v The region inside a cube. b State the domain and range of those that are relations. Solution a i A relation, but not a function — it contains (3, 5) and (3, 1). ii A relation, and a function. iii A relation, but not a function — it contains (0, 0) and (0, 4). iv A relation, but not a function — it contains (2, 0) and (2, 5). v Not a relation, because points in 3D space are not ordered pairs. b i Domain: 0, 1, 2, 3, 5. Range: 0, 1, 3, 5, 7. 1 1 2 2 iii Domain: −5 < x < 5. Range: −5 < y < 5. iv Domain: 1 < x < 4. Range: all real y. ii Domain: −6, −1, 2 , 10. Range: −6, 3 , 4, 10. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 56 3B Chapter 3 Functions and graphs Exercise 3B 1 FOUNDATION In each case, copy the graph, then draw a vertical line to show that the graph does not represent a function. y a y b 3 2 3x -3 x x 2 -2 -2 y y e 2 y f 3 1 -1 4 x -2 2 2 -2 -3 d y c x x -1 Use the vertical line test to find which of the following graphs represent functions. y a y b 1 y d 1 8 1 2 x 10 x -10 -1 2 -1 x -1 y f y g y h 2 2 -1 -2 2 x -1 -10 y 3 e y c -2 x 1 x -2 2 x 0 1 2 x -2 -3 3 What are the domain and range of each relation in Question 2? 4 For each of the following functions: i Copy and complete the table of values. ii Plot the points in the table and hence sketch the function. iii Then write down the domain and range. a y = x2 + 2x + 1 x −3 −2 b y = 2x −1 0 1 y 5 x −2 −1 2 y Use the fact that division by zero is undefined to find the natural domain of each function. 1 1 1 a f (x) = b f (x) = c f (x) = x x−3 2+x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3B Functions, relations, and graphs 6 Use the fact that a negative number does not have a square root to find the natural domain of each function. √ √ √ a f (x) = x b f (x) = x − 2 c f (x) = 5 + x 7 Which of these relations are also functions? Explain your answers. a x f -2 2 -1 1 0 b y x 1 4 1 2 2 1 x g -2 2 -1 1 0 4 c y p 4 4 2 1 1 0 d y x y q -3 -2 2 -1 1 0 57 3 -4 4 -5 5 0 DEVELOPMENT 8 The following relations are not functions. Write down the coordinates of two points on each graph that have the same x-coordinate. y a y b 3 c 1 y 5 y d 2 3 -3 3 x -1 -3 9 -2 1 2 4 3 2x − 1 2 e f (x) = √ 1−x b f (x) = d f (x) = √ 4 − 2x Let R(x) = √ c f (x) = 2 x -2 x Find the natural domain of each function. a f (x) = 7 − 3x 10 x -1 √ x+4 f f (x) = √ 1 2x − 3 x . a What is the natural domain of R(x)? b Copy and complete the table of values. Use a calculator to approximate values correct to one decimal place where necessary. x 0 12 1 2 3 4 5 R(x) c Plot these points and join them with a smooth curve starting at the origin. This curve may look similar to a curve you know. Describe it. 11 2 . x a What is the natural domain of h(x)? b Copy and complete the table of values. Why is there a star for the value where x = 0? x −4 −2 −1 − 12 0 12 1 2 4 Let h(x) = h(x) ∗ c Plot these points and join them with a smooth curve in two parts. This curve is called a rectangular hyperbola. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 58 3B Chapter 3 Functions and graphs 12 Jordan is playing with a 20 cm piece of copper wire, which he bends into the shape of a rectangle, as shown. Let x be the length of a side of the rectangle. a Find the area A of the rectangle as a function of x. x b Use the fact that lengths must be positive to find the domain of A. c Sketch the graph of A for the domain you found in part b. 13 14 Solve each equation for y and hence show that it represents a function. a 2x − y + 3 = 0 b xy = 4 c xy − 2y = 3 d y+2= e f (x) = f f (x) = √ √ x2 − 4 State the natural domain of each function. x a f (x) = √ x+2 1 c f (x) = 2 x +x √ e f (x) = x2 − 4 √ 9 − x2 1 1 − x2 2 x2 − 4 1 d f (x) = 2 x − 5x + 6 1 f f (x) = √ 1 − x2 b f (x) = ENRICHMENT 15 a i What are the domain and range of the parabolic function y = 4 − x2 ? ii Use part i and the properties of square roots to find the domain and range of y = iii Hence determine the domain and range for y = √ 16 √ 4 − x2 . 1 . 4 − x2 b Likewise find the domain and range of each of these functions. 1 1 i √ ii √ 2 2 3 − 2x − x x + 2x + 3 1+x . Consider the function ath(x) = log2 1−x 2x = 2 ath(x). a What is the domain of ath(x)? b Show that ath 1 + x2 Note: A function name does not have to be a single letter. In this case the function has been given the name ‘ath’ because it is related to the arc hyperbolic tangent, which is studied in some university courses. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3C Review of linear graphs 59 3C Review of linear graphs Learning intentions • Work with linear functions and relations and their graphs. • Solve simultaneous linear equations graphically and algebraically. The next few sections will review some functions and relations that have been introduced in previous years, as well as the sketching of their graphs. Linear graphs are briefly reviewed in this section, and are the main subject of Chapter 8. Linear functions A function is called linear if its graph is a straight line. It can be written in the form y = 2x − 3, or in function form, f (x) = 2x − 3, with a term in x and a constant term. We often write a linear function with all its terms on the left — the equation of the function above then becomes 2x − y − 3 = 0, where the coefficient of y cannot be zero, because we must be able to solve for y. 9 Linear functions • A linear function has a graph that is a straight line. • The equation of a linear function can be written in gradient–intercept form, y = mx + b, or in function form, f (x) = mx + b. • Alternatively, the equation of a linear function can be written in general form, ax + by + c = 0, where the coefficient of y is non-zero. Sketching linear functions When all three terms of the equation are non-zero, the easiest way to sketch a linear function is to find the two intercepts with the axes. 10 Sketching a linear function whose equation has three non-zero terms • Find the x-intercept by putting y = 0. • Find the y-intercept by putting x = 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 60 3C Chapter 3 Functions and graphs Example 9 Solving simultaneous linear equations graphically a Sketch these linear functions on one set of axes by finding the x-intercept and the y-intercept. ii y = x − 3 i x + 2y − 6 = 0 b Estimate from the graph where the two lines intersect. Solution a i x + 2y − 6 = 0. Consider When y = 0, y 3 x − 6= 0 1 x = 6. When x = 0, 3 4 y = 3. ii 6 -3 y = x − 3. Consider When y = 0, x 2y − 6 = 0 x − 3= 0 x = 3. When x = 0, y = −3. b From the diagram, the lines appear to meet at (4, 1). Using simultaneous equations The methods of solving simultaneous equations were reviewed in Section 1E. Example 10 Solving simultaneous linear equations algebraically Solve the simultaneous equations in the previous example, and check that the solution agrees with the estimate from the graph. Solution The equations are Subtracting (2) from (1), x + 2y − 6 = 0 (1) x − y − 3 = 0. (2) 3y − 3 = 0 y = 1. Substituting into (1), x + 2 − 6= 0 x = 4. Check this by substitution into (2). Thus the lines meet at (4, 1), as seen in the graph in the previous example. Linear relations A linear relation is a relation whose graph is a straight line. By the vertical line test, every linear relation is a function except for vertical lines, which fail the vertical line test. The equation of such a relation has no term in y, such as the relation: x = 3. Its graph consists of all points whose x-coordinate is 3, giving a vertical line, which is sketched in the next example. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3C Review of linear graphs 61 Three special cases The two-intercept method won’t work for sketching the graph of ax + by + c = 0 if any of the constants a, b or c is zero. 11 Sketching special cases of linear graphs ax + by + c = 0 Horizontal lines: If a = 0, the equation can be put into the form y = k. • Its graph is a horizontal line with y-intercept k. Vertical lines: If b = 0, the equation can be put into the form x = . • Its graph is a vertical line with x-intercept . • This is the only type of linear graph that is a relation, but not a function. Lines through the origin: If c = 0, and the line is neither horizontal nor vertical, then the equation can be put into the form y = mx, where m 0. • Both intercepts are zero, so the graph passes through the origin. • Find one more point on the line by substituting a value such as x = 1. Example 11 Sketching three special cases of linear relations Sketch these three lines. Which of the three is not a function? b x−3=0 a y+2=0 c x + 2y = 0 Solution y a y b 2 c y 2x + y = 0 2 1 -2 2 -2 x y = -2 -2 2 x -2 - 12 x x=3 a The line y + 2 = 0, or y = −2, is horizontal with y-intercept −2. b The line x − 3 = 0, or x = 3, is vertical with x-intercept 3. c The line x + 2y = 0 passes through the origin, and when x = 1, y = − 12 . The vertical line in part b is a relation, but not a function — it fails the vertical line test. Domain and range of linear functions and relations The following remarks will be clear from the diagram above, and from the other sketches in the section. • A horizontal line f (x) = c has domain all real x and range y = c. • A vertical line x = k, which is a relation but not a function, has domain x = k and range all real y. • An oblique line f (x) = mx + b, where m 0, has domain all real x and range all real y. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 62 3C Chapter 3 Functions and graphs Exercise 3C 1 FOUNDATION Follow these steps for the linear function y = 2x − 2. a Find the y-intercept by putting x = 0. b Find the x-intercept by putting y = 0. c Plot these intercepts and hence sketch the line. 2 3 Repeat the steps in the previous question for each line. a y= x+1 b y = 4 − 2x c x+y−1=0 d x − 2y − 4 = 0 e 2x − 3y − 12 = 0 f x + 4y + 6 = 0 A linear function has the equation y = −2x. a Show that the x-intercept and the y-intercept are both zero. b Substitute x = 1 to find a second point on the line, then sketch the line. 4 Repeat the steps in Question 3 for each line. a y = 3x 5 b x+y=0 c x − 2y = 0 Graph y = f (x) where: a f (x) = −3x − 6 b f (x) = 12 x − 2 c f (x) = − 12 x 6 7 Sketch the following vertical and horizontal lines. a x=1 b y=2 c x = −2 d y=0 e 2y = −3 f 3x = 5 a State which lines in Question 6 are not functions. b For each line that is not a function, write down the coordinates of two points on the line with the same x-coordinate. 8 For each of parts c to f in Question 2, solve the given equation for y and hence write the equation of the line using function notation. 9 Determine, by substitution, whether or not the given point lies on the given line. y = 20 − 2x d (−5, 3) 2x + 3y + 1 = 0 f (−6, −4) 4x − 5y − 4 = 0 y= x−2 c (1, −2) y = −3x + 1 e (−1, −4) 3x − 2y − 5 = 0 b (7, 4) a (3, 1) DEVELOPMENT 10 Consider the lines x + y = 5 and x − y = 1. a Graph the lines on a number plane, using a scale of 1 cm to 1 unit on each axis. b Read off the point of intersection of the two lines. c Confirm your answer to part b by solving the two equations simultaneously. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3C Review of linear graphs 11 Repeat Question 10 for the following pairs of lines. a x+y=2 x−y=3 c x + 2y = −4 2x + y = 0 2x − y = −3 b x − y = −4 12 63 Looksmart Shirts charges $60 for one shirt and $50 for each shirt after this. a Find, as a function of n, the cost C in dollars of n shirts. b Delivery costs $10 for one shirt and $2 for each subsequent shirt. i Find, as a function of n, the cost D in dollars of delivering n shirts. ii Find, as a function of n, the total cost T in dollars of buying n shirts and having them delivered. 13 Consider the linear equation y = 12 x + c. a Sketch on one number plane the four lines corresponding to the following values of c: i c = −2 ii c = −1 iii c = 1 iv c = 2 b What do you notice about all these lines? 14 Consider the linear equation y − 2 = m(x − 1). a Sketch on one number plane the four lines corresponding to the following values of m: i m=1 ii m = 2 iii m = − 12 iv m = 0 b Which point in the number plane do all these lines pass through? c Prove that the line y − 2 = m(x − 1) passes through the point found in the previous part, regardless of the value of m. ENRICHMENT 15 [Two-intercept form] x y + = 1. What do you notice? a b b Use the result of part a to sketch these lines quickly. y x y x y i ii −x + = 1 iii − =1 + =1 6 3 2 2 5 a Find the x- and y-intercepts of the equation 16 a Find the coordinates of the point M where x + 2y − 6 = 0 and 3x − 2y − 6 = 0 meet. b Show that the new line (x + 2y − 6) + k(3x − 2y − 6) = 0 always passes through M, regardless of the value of k. c Hence find the equation of the line PM where P = (2, −1). 17 The line Ax + By + C = 0 passes through the two fixed points P(x1 , y1 ) and Q(x2 , y2 ). Let R(x, y) be a variable point on the line. a Substitute the coordinates of P, Q and R into the equation of the line to get three equations. b Solve these equations simultaneously by eliminating A, B and C in order to show that the equation of the line can also be written as (y1 − y)(x2 − x) = (y2 − y)(x1 − x). c Use this formula to write down the equation of the line through P(1, 2) and Q(3, −4). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 64 3D Chapter 3 Functions and graphs 3D Quadratic functions — factoring and the graph Learning intentions • Work with quadratic functions and their terminology. • Use factoring to find the zeroes, vertex, and axis of symmetry, and sketch the graph. • Use correctly the language of zeroes and roots. • Use correctly the language of double and simple zeroes and roots, and perfect squares. • Use correctly the language of equal and distinct zeroes and roots. The material in Sections 3D–3F will be partly review and partly new work, with the emphasis on graphing. Readers confident with quadratics from earlier years will not need to work so thoroughly through the earlier questions in the exercises. Quadratic functions A quadratic function is a function that can be written in the form: f (x) = ax2 + bx + c, where a, b and c are constants, and a 0. A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, where a, b and c are constants, and a 0. The requirement that a 0 means that the term in x2 cannot vanish. Thus linear functions and equations are not regarded as special cases of quadratics. The word ‘quadratic’ comes from the Latin word quadratus, meaning ‘square’, and reminds us that quadratics tend to arise as the areas of regions in the plane. Monic quadratics A quadratic function f (x) = ax2 + bx + c is called monic if a = 1. Calculations are usually easier in monic quadratics than in non-monic quadratics. 12 Monic quadratics A quadratic is called monic if the coefficient of x2 is 1. For example, y = x2 − 8x + 15 is monic and y = −x2 + 8x − 15 is non-monic. Zeroes and roots The solutions of a quadratic equation are called the roots of the equation, and the x-intercepts of a quadratic function are called the zeroes of the function. This distinction is often not strictly observed, however, because questions about quadratic equations are so closely related to questions about quadratic functions and their graphs. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3D Quadratic functions — factoring and the graph 65 Five questions about the graph of a quadratic The graph of any quadratic function y = ax2 + bx + c is a parabola, as seen in earlier years. Before sketching the parabola, five questions need to be answered: 13 Five questions about the parabola y = ax2 + bx + c Answer: Look at the sign of a. 2 What is the y-intercept? Answer: Put x = 0, and then y = c. 3 What are the x-intercepts, or zeroes, if there are any? 4 What is the axis of symmetry? 5 What is the vertex? Method: Substitute the axis back into the quadratic. 1 Which way up is the parabola? The first two questions are easy to answer: • If a is positive, the curve is concave up. If a is negative, it is concave down. • To find the y-intercept, put x = 0, then y = c. • And once the axis of symmetry is found, the y-coordinate of the vertex can be found by substituting back into the quadratic. However, finding the x-intercepts, and finding the axis of symmetry, need careful work — the three standard approaches are discussed in this and the next two sections: • Section 3D: Factoring • Section 3E: Completing the square • Section 3F: Formulae Factoring and the zeroes Factoring of monic and non-monic quadratics were reviewed in Section 1B. Most quadratics cannot easily be factored, or can’t be factored at all. When straightforward factoring is possible, the quickest approach to find the zeroes is to make y = 0 and use the principle: If A × B = 0 A = 0 or B = 0. then For example, y = x2 − 2x − 3 is a quadratic function. 1 Its graph is concave up because a = 1 is positive. (Next heading for the sketch.) 2 Substituting x = 0 gives so the y-intercept is 3 Factoring, Substituting y = 0 gives y=0+0−3 y = −3. y = (x + 1)(x − 3), 0 = (x + 1)(x − 3) x + 1 = 0 or x − 3 = 0 so the x-intercepts are x = −1 and x = 3. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 66 3D Chapter 3 Functions and graphs Finding the axis of symmetry and the vertex from the zeroes The axis of symmetry is always the vertical line midway between the x-intercepts. Thus its x-intercept is the average of the zeroes. Continuing with our example of y = x2 − 2x − 3, which factorises as y = (x + 1)(x − 3), and so has zeroes −1 and 3: 4 Taking the average of the zeroes, the axis of symmetry is y -1 x = 12 (−1 + 3) 3 x -3 x = 1. (1, -4) 5 Substituting x = 1 into the factored quadratic, y = (1 + 1)(1 − 3) = −4, so the vertex is at (1, −4). 14 The zeroes and intercepts of a factored quadratic Suppose that we have managed to factor a quadratic as y = a(x − α)(x − β). • Its x-intercepts (zeroes) are x = α and x = β. • Its axis of symmetry is the line x = 12 (α + β). Take the average of the zeroes. • Substitute the axis into the factored form of the quadratic to find the y-coordinate of the vertex. Example 12 Sketching a non-monic quadratic Sketch the curve y = −x2 − 2x + 3. Solution 1 Because a < 0, the curve is concave down. 2 When x = 0, (-1, 4) y = 3. y = −(x2 + 2x − 3) 3 Factoring, 3 -3 = −(x + 3)(x − 1). When y = 0, 1 x + 3 = 0 or x − 1 = 0, so the zeroes are y x x = −3 and x = 1. 4 Taking the average of the zeroes, the axis of symmetry is x = 12 (−3 + 1) x = −1. 5 When x = −1, y = −(−1 + 3) × (−1 − 1) (substitute into the factored form) = 4, so the vertex is at (−1, 4). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3D Quadratic functions — factoring and the graph 67 Double zeroes, double roots, and perfect squares The quadratic y = −2x2 + 4x − 2 factors as y 2 x 1 y = −2(x − 1) , 2 which is −2 times the perfect square (x − 1)2 . -2 To find the zeroes, put y = 0 −2(x − 1)2 = 0 x = 1. so Should we write the last line as ‘x = 1 or 1’? That could be confusing. Instead, we say that x = 1 is a double zero of the function, and that x = 1 is a double root of the quadratic equation −2(x − 1)2 = 0. The graph of the quadratic is tangent to the x-axis at x = 1. 15 Double zeroes, double roots, and perfect squares • The quadratic function y = −2(x − 1)2 is said to have a double zero at x = 1. • The quadratic equation −2(x − 1)2 = 0 is said to have a double root at x = 1. • Square expressions such as (x − 1)2 and (x + 6)2 are called perfect squares. Some other terminology used only with quadratics: ‘Equal roots’ and ‘distinct roots’ Warning: Use this terminology only with quadratics. Never use it with cubics, quartics, and higher degrees. • The quadratic equation (x − 4)2 = 0 is often said to have equal roots, (and the quadratic function f (x) = (x − 4)2 is often said to have equal zeroes). • The quadratic equation (x − 3)(x − 5) = 0 is often said to have distinct roots, (and the function f (x) = (x − 3) (x − 5) is often said to have distinct zeroes). Note: These plural forms ‘equal zeroes’ and ‘equal roots’ can be misleading, because there is still only one zero or root. Quadratics with given zeroes If a quadratic f (x) has zeroes at x = α and x = β, then its equation has the form y f (x) = a(x − α)(x − β), where a is the coefficient of x2 . By taking different values of the coefficient a, this equation forms a family of quadratics, all with the same x-intercepts. The sketch to the right shows four of the curves in the family, two concave up with positive values of a, and two concave down with negative values of a. a b x 16 The family of quadratics with given zeroes The quadratics with zeroes at x = α and x = β form a family of parabolas with the equation y = a(x − α)(x − β), for some non-zero value of a. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 68 3D Chapter 3 Functions and graphs Example 13 Finding a family of quadratics with given zeroes a Write down the family of quadratics with zeroes x = −2 and x = 4. b Find the equation of such a quadratic if: i the y-intercept is 16, ii the curve passes through (5, 1). Solution a The family of quadratics with zeroes −2 and 4 is y = a(x + 2)(x − 4). b 16 = a × 2 × (−4) i Substituting the point (0, 16) gives so the quadratic is y = −2(x + 2)(x − 4). a = −2, 1=a×7×1 ii Substituting the point (5, 1) gives so the quadratic is y = 17 (x + 2)(x − 4). a = 17 , Domains and ranges of quadratic functions As with linear functions, for a quadratic function f (x) with vertex (h, k): • If f (x) is concave up, the domain is all real x, and the range is y ≥ k. • If f (x) is concave down, the domain is still all real x, but the range is y ≤ k. Exercise 3D 1 FOUNDATION a The parabola with equation y = (x − 1)(x − 3) is concave up. i Explain why the parabola is concave up. ii Write down its y-intercept. iii Put y = 0 to find the x-intercepts. iv Hence determine the equation of the axis of symmetry. v Use the axis of symmetry to find the coordinates of the vertex. vi Sketch the parabola, showing these features. vii Does the quadratic function have equal zeroes or are the zeroes distinct? b Follow the steps of part a to sketch these concave-up parabolas. ii y = (x − 1)(x + 1) i y = (x − 1)(x + 3) 2 a The parabola y = −x(x − 2) is concave down. i Explain why the parabola is concave down. ii Write down its y-intercept. iii Put y = 0 to find the x-intercepts. iv Hence determine the equation of the axis of symmetry. v Use the axis of symmetry to find the coordinates of the vertex. vi Sketch the parabola, showing these features. vii Does the quadratic function have equal zeroes or are the zeroes distinct? b Follow the steps of part a to sketch these concave-down parabolas. i y = (2 + x)(2 − x) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii y = (x + 2)(4 − x) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3D Quadratic functions — factoring and the graph 3 69 a The parabola y = (x − 1)2 is a perfect square. i Explain whether the parabola is concave up or concave down. ii Write down its y-intercept. iii Put y = 0 to find its single x-intercept at the vertex. iv Use symmetry and the y-intercept to find another point on the parabola. v Sketch the parabola showing these points. vi Does the quadratic function have equal zeroes or are the zeroes distinct? b Follow similar steps to part a to sketch these parabolas involving perfect squares. ii y = −(x − 2)2 i y = (x + 1)2 4 a The graph of y = 12 x2 passes through the origin and has no other x-intercept. i Plot the point corresponding to x = 2 on this parabola. ii What other point must lie on the parabola, and why? iii Hence graph the parabola y = 12 x2 . b Use a similar approach to graph these parabolas. ii y = 13 x2 i y = −x2 5 [Technology] a Use computer graphing software to plot on the one number plane the parabola y = a(x − 1)(x − 3) for the following values of a. ii a = 1 i a=2 iii a = −1 iv a = −2 b Which two points do all these parabolas pass through? 6 Write down, in factored form, the equation of the monic quadratic function with zeroes: 7 c −3 and 5 b 0 and 3 a 4 and 6 d −6 and −1 Write down, in factored form, the equation of each quadratic function y = f (x) sketched below. In each case, the coefficient of x2 is either 1 or −1. a y y b y c 3 3 x -2 1 y d -5 x -2 -1 x 3 -2 x -10 DEVELOPMENT 8 9 Use factoring to find the zeroes of each quadratic function. Hence sketch the graph of y = f (x), showing all intercepts and the coordinates of the vertex. a f (x) = x2 − 9 b f (x) = x2 + 4x − 5 c f (x) = x2 + 4x − 12 d f (x) = 4x − x2 e f (x) = −x2 + 2x + 3 f f (x) = 8 − 2x − x2 Sketch these parabolas involving perfect squares. Hint: Begin by factoring the quadratic. In each case, symmetry will be needed to find a third point on the parabola. a y = x2 − 6x + 9 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b y = −x2 + 2x − 1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 70 3D Chapter 3 Functions and graphs 10 11 Use factoring to sketch the graphs of the following non-monic quadratic functions, clearly indicating the vertex and the intercepts with the axes. a y = 2x2 + 7x + 5 b y = 2x2 + 5x − 3 c y = 3x2 + 2x − 8 d y = 2x2 − 18 e y = 3x2 + x − 4 f y = 7x − 3 − 4x2 Find the equations of the quadratic functions sketched below. y a y b y c 6 4 x -2 -1 2 -3 x 2 x 13 14 -2 2 x -24 -2 12 y 2 d Find, in factored form, the equations of the parabolas with the given intercepts. a x = 1, 3 b x = −2, 1 c x = −1, 5 d x = −2, −4 y=6 y=4 y = 15 y=2 The general form of a quadratic with zeroes x = 2 and x = 8 is y = a(x − 2)(x − 8). Find the equation of such a quadratic for which: a the coefficient of x2 is 3 b the y-intercept is −16 c the vertex is (5, −12) d the curve passes through (1, −20). The graph of y = ax(x − α) passes through the origin. (Why?) Find the values of a and α in this quadratic given that: a it is monic and passes through (−3, 0) b there is one x-intercept and it passes through (2, 6) c the vertex is (1, 4) d the axis of symmetry is x = −3 and the coefficient of x is −12. 15 The general form of a quadratic with zeroes α and β is y = a(x − α)(x − β). Find a in terms of α and β if: a the y-intercept is c, 16 b the coefficient of x is b, c the curve passes through (1, 2). Consider the quadratic function f (x) = x2 − 2x − 8. a Factor the quadratic and hence find the equation of the axis of symmetry. b i Expand and simplify f (1 + h) and f (1 − h). What do you notice? ii What geometric feature of the parabola does this result demonstrate? ENRICHMENT 17 In each case, factor the quadratic to find the x-intercepts. Hence find the equation of the axis of symmetry. a y = x2 + 2x + 1 − p2 b y = x2 − 2px − 1 + p2 c y = x2 − 2x − 2p − p2 18 Let f (x) = a(x − α)(x − β). In each case, prove the given identity and explain how it demonstrates that x = 12 (α + β) is the axis of symmetry. a f 12 (α + β) + h = f 12 (α + β) − h b f (α + β − x) = f (x) 19 Show that the quadratic equation ax2 + bx + c = 0 cannot have more than two distinct roots when a 0. (Hint: Suppose that the equation has three distinct roots α, β and γ. Substitute α, β and γ into the equation and conclude that a = b = c = 0.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3E Completing the square and the graph 71 3E Completing the square and the graph Learning intentions • Complete the square in monic and non-monic quadratics. • Use completing the square to find the axis of symmetry, the vertex, and the zeroes. Completing the square is the most general method of dealing with quadratics. It works in every case, whereas factoring only works in some cases. We first review the algebra of completing the square, and then extend it to non-monic quadratics. Once this has been done, sketching the graph follows easily. Completing the square in a monic quadratic The quadratic f (x) = x2 + 6x + 5 is not a perfect square, but it can be made the sum of a perfect square and a constant. Adapting the procedure developed in Section 1D to a function rather than an equation: • Look just at the two terms in x, that is, x2 + 6x. • Halve the coefficient 6 of x to get 3, then square to get 9. • Add and subtract 9 on the RHS to produce a perfect square plus a constant. f (x) = x2 + 6x + 5 = (x2 + 6x + 9) − 9 + 5 (add and subtract 9) = (x + 3)2 − 4 17 Completing the square in a monic quadratic • Take the coefficient of x, halve it, then square the result. • Add and subtract this number to produce a perfect square plus a constant. Example 14 Completing the square in a monic quadratic Complete the square in each quadratic. a f (x) = x2 − 4x − 5 b f (x) = x2 + x + 1 Solution a Here f (x) = x2 − 4x − 5. The coefficient of x is −4. Halve it to get −2, then square to get (−2)2 = 4. (add and subtract 4) Hence f (x) = (x2 − 4x + 4) − 4 − 5 = (x − 2)2 − 9. b Here f (x) = x2 + x + 1. The coefficient of x is 1. Halve it to get 12 , then square to get ( 12 )2 = 14 . (add and subtract 14 ) Hence f (x) = (x2 + x + 14 ) − 14 + 1 = (x + 12 )2 + 34 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 72 3E Chapter 3 Functions and graphs Completing the square in a non-monic quadratic For a non-monic quadratic such as y = 2x2 − 12x + 16, where the coefficient of x2 is not 1, you must divide through by the coefficient of x2 before completing the square. This slightly more difficult procedure was not covered in Chapter 1. 18 Completing the square in a non-monic quadratic • Divide through by the coefficient of x2 so that the coefficient of x2 is 1. • Complete the square in the resulting monic quadratic. This method runs more cleanly using y = . . . rather than f (x) = . . . Example 15 Completing the square in a non-monic quadratic Complete the square in each quadratic. a y = 2x2 − 12x + 16 b y = −x2 + 8x − 15 Solution a y = 2x2 − 12x + 16 y = x2 − 6x + 8 2 y = (x2 − 6x + 9) − 9 + 8 2 y = (x − 3)2 − 1 2 y = 2(x − 3)2 − 2 b (divide through by the coefficient 2 of x2 ) (complete the square on the RHS) (multiply by 2 to make y the subject again) y = −x + 8x − 15 2 −y = x2 − 8x + 15 (divide through by the coefficient −1 of x2 ) −y = (x2 − 8x + 16) − 16 + 15 (complete the square on the RHS) −y = (x − 4)2 − 1 y = −(x − 4)2 + 1 (multiply by −1 to make y the subject again) Finding the vertex from the completed square The completed square allows the vertex to be found using one fundamental fact about squares: 19 A square can never be negative • x2 = 0, when x = 0, • x2 > 0, when x 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3E Completing the square and the graph 73 y Thus, in part a of Example 15, where y = 2(x − 3)2 − 2, the term 2(x − 3)2 can never be negative. Substituting x = 3 gives y=0−2 2 = −2, x 4 (3, -2) but if x 3, then y > −2. Hence the graph passes through the point V(3, −2), but never goes below it. This means that V(3, −2) is the vertex of the parabola, and x = 3 is its axis of symmetry. y But in part b of Example 15, where y = −(x − 4) + 1, (4, 1) 2 3 the term −(x − 4)2 can never be positive. Substituting x = 4 gives 5 x y=0+1 = 1, but if x 4, then y < 1 Hence the graph passes through the point V(4, 1), but never goes above it. This means that V(4, 1) is the vertex of the parabola, and x = 4 is its axis of symmetry. There is no need to repeat this argument every time. The result is simple: 20 Finding the axis and vertex from the completed square For the quadratic y = a(x − h)2 + k, the axis is x = h and the vertex is V(h, k). In Chapter 5, we will interpret this result as the effect of translations and dilations on y = x2 . Some readers may have done this already in earlier years. Finding the zeroes from the completed square The completed square also allows the zeroes to be found in the usual way: 21 Finding the zeroes from the completed square To find the x-intercepts from the completed square, put y = 0. • There may be two zeroes, in which case they may or may not involve surds. • There may be no zeroes. • There may be exactly one zero, in which case the quadratic is a perfect square. The vertex sits on the axis, and the zero is called a double zero of the quadratic. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 74 3E Chapter 3 Functions and graphs These methods are illustrated in the worked examples below. Example 16 Completing the square with and without surds Complete the square to sketch the graphs of these quadratic functions. a f (x) = x2 − 4x − 5 b f (x) = x2 − 4x − 1 Solution a The curve y = x2 − 4x − 5 is concave up, with y-intercept −5. y The square was completed earlier in this section, y = (x − 2)2 − 9, 5 x -1 so the axis of symmetry is x = 2 and the vertex is at (2, −9). Put y = 0, then (x − 2)2 = 9 -5 (2, -9) x − 2 = 3 or x − 2 = −3 x = 5 or x = −1. b The curve y = x2 − 4x − 1 is concave up, with y-intercept 1. y Completing the square, y = (x − 4x + 4) − 4 − 1, 2 2 + Ö5 -1 2 - Ö5 = (x − 2) − 5, so the axis of symmetry is x = 2 and the vertex is at (2, −5). Put y = 0, then (x − 2)2 = 5 √ √ x − 2 = 5 or x − 2 = − 5 √ √ x = 2 + 5 or x = 2 − 5 . 2 x (2, -5) Note: With this method, the axis of symmetry and the vertex are read directly from the completed-square form — the zeroes are then calculated afterwards. Compare this with factoring, where the zeroes are found first and the axis of symmetry and the vertex can then be calculated from them. Example 17 Completing the square with no zeroes, and with a double zero Complete the square to sketch the graphs of these quadratics. Then write down the domain and range. a y = x2 + x + 1 b y = x2 + 6x + 9 Solution a The curve y = x2 + x + 1 is concave up, with y-intercept 1. y The square was completed earlier in this section, y = (x + 12 )2 + 34 , so the axis is x = − 12 , and the vertex is at (− 12 , 34 ) . Put y = 0, then (x + 12 )2 = − 34 . Because negative numbers do not have square roots, this equation has no solutions, so there are no x-intercepts. Domain: All real x OR (−∞, ∞). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 Range: y ≥ 34 OR (- 12 , 34 ) -1 1 x [ 34 , ∞). © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3E Completing the square and the graph b y = x2 + 6x + 9 is concave up, with y-intercept 9. 75 y This quadratic is already a perfect square, (-6, 9) 9 y = (x + 3) + 0, 2 so the axis is x = −3 and the vertex is at (−3, 0). Put y = 0, then (x + 3)2 = 0 x+3=0 x -3 x = −3 (a double zero). Domain: All real x OR (−∞, ∞). Range: y ≥ 0 OR [0, ∞). The symmetric point (−6, 9) has been plotted so that the parabolic graph has at least three points on it. Note: The symmetric point (−6, 9) has been plotted so that the parabolic graph has at least three points on it. Example 18 Sketching the graph from the completed square a Use the previous completed square to sketch y = −x2 + 8x − 15. b Write down its domain and range, in both notations Solution a y = −x2 + 8x − 15 is concave down, with y-intercept −15. y Completing the square, y = −(x − 4) + 1, so the axis is x = 4 and the vertex is at (4, 1). Put y = 0, then (x − 4)2 = 1 2 x − 4 = 1 or x − 4 = −1 (4, 1) x 3 5 -15 x = 5 or x = 3. b Domain: All real x OR (−∞, ∞). Range: y ≤ 1 OR (−∞, 1]. The family of quadratics with a common vertex y A quadratic with vertex (h, k) has an equation of the form y = a(x − h)2 + k where a is the coefficient of x2 . This equation gives a family of parabolas all with vertex (h, k), as different values of a are taken. The sketch to the right shows six curves in the family, three with a positive, and three with a negative. k h x 22 The family of quadratics with a common vertex The quadratics with vertex (h, k) form a family of parabolas, all with the equation y = a(x − h)2 + k, for some non-zero value of a. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 76 3E Chapter 3 Functions and graphs Example 19 Finding the family of quadratics with common vertex Write down the family of quadratics with a vertex of (−3, 2). Then find the equation of such a quadratic: a if x = 5 is one of its zeroes, b if the coefficient of x is equal to 1. Solution y The family of quadratics with vertex (−3, 2) is y = a(x + 3)2 + 2. (∗) 3 12 a Substituting (5, 0) into (∗) gives 0 = a × 64 + 2, (-3, 2) 1 1 , and the quadratic is y = − 32 (x + 3)2 + 2. so a = − 32 b Expanding the equation (∗), y = ax2 + 6ax + (9a + 2), 134 -11 5 so 6a = 1. Hence a = 16 , and the quadratic is y = 16 (x + 3)2 + 2. Exercise 3E 1 x FOUNDATION The equation of a parabola in completed square form is y = (x − 2)2 − 1. a What is the concavity of this parabola? b Substitute x = 0 to find the y-intercept. c Use the results of Box 20 to write down the equation of the axis of symmetry and the coordinates of the vertex. d Use your answers to parts a and c to explain why there must be x-intercepts, then find them by putting y = 0. e Hence sketch this parabola, showing all key features. 2 3 4 5 Repeat the steps of Question 1 in order to sketch the graphs of these quadratic functions. a y = (x + 1)2 − 4 b y = (x − 1)2 − 9 c y = −(x + 2)2 + 4 d y = −(x − 2)2 + 9 Complete the square in each monic quadratic function. a f (x) = x2 − 4x + 5 b f (x) = x2 + 6x + 11 c f (x) = x2 − 2x + 8 d f (x) = x2 − 10x + 1 e f (x) = x2 + 2x − 5 f f (x) = x2 + 4x − 1 Follow the steps in Question 1 to sketch these quadratics. The x-intercepts involve surds. a y = (x + 1)2 − 3 b y = (x − 4)2 − 7 c y = 2 − (x − 3)2 d y = 5 − (x + 1)2 Find the zeroes of each quadratic functions by completing the square. Then show that the same answer is obtained by factoring. a f (x) = x2 − 4x + 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b f (x) = x2 + 2x − 3 c f (x) = x2 − x − 2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3E Completing the square and the graph 6 77 Use the formula in Box 22 to write down the equation of each of the quadratic functions sketched below. In each case, the coefficient of x2 is either 1 or −1. y a y b 1 3 x (1, 2) (-2, -3) x c y y d (3, 4) (2, -1) x x -5 -5 7 Write down the equation of the monic quadratic with vertex: a (2, 5) 8 b (0, −3) c (−1, 7) d (3, −11) [Technology] a Use computer graphing software to plot on the one number plane the parabola y = a(x − 1)2 − 2 for the following values of a. i a=2 ii a = 1 iii a = −1 iv a = −2 b Which point do all these parabolas pass through? c For which values of a does the parabola have x-intercepts? d Explain your answer to part c geometrically. DEVELOPMENT 9 Complete the square of each quadratic function in order to find its zeroes. Each zero involves a surd. a f (x) = x2 − 4x + 1 10 11 12 b f (x) = 2x2 − 4x − 2 c f (x) = −x2 − 6x + 1 Complete the square in each quadratic. Then sketch the graph of each function, showing the vertex and the intercepts with the axes. a y = x2 − 2x b y = x2 − 4x + 3 c y = x2 − 2x − 5 d y = x2 + 2x − 1 e y = x2 + 2x + 2 f y = x2 − 3x + 4 Explain why y = a(x + 4)2 + 2 is the general form of a quadratic with vertex (−4, 2). Then find the equation of such a quadratic for which: a the quadratic is monic, b the coefficient of x2 is 3, c the y-intercept is 16, d the curve passes through the origin. Write down the coordinates of the vertex and state the concavity for each parabola. Hence determine the number of x-intercepts. a y = 2(x − 3)2 − 5 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b y = −3(x + 2)2 − 1 c y = 4(x + 1)2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 78 3E Chapter 3 Functions and graphs 13 14 Complete the square for these non-monic quadratics. (In each case, notice that the coefficient of x2 is not 1.) Then sketch each curve, showing the vertex and any intercepts. a y = −x2 + 4x + 1 b y = 2x2 − 4x + 3 c y = 2x2 + 6x + 2 d y = −2x2 − 8x − 11 e y = −3x2 + 6x + 3 f y = 5x2 − 20x + 23 Complete the square for each quadratic function. Hence write each quadratic in factored form. Your answers will involve surds. a f (x) = x2 + 2x − 1 15 16 b f (x) = x2 − 4x + 1 c f (x) = −x2 − 2x + 4 Write down the general form of a monic quadratic whose axis of symmetry is x = −2. Hence find the equation of such a quadratic which also: a passes through (0, 0), b passes through (5, 1), c has a zero at x = 1, d has y-intercept −6, e touches the line y = −2, f has range y ≥ 7. Find the equations of the quadratic functions sketched below. a y y b y c (3 , 2) d (-1, 4) y x x 3 -2 -7 (1, 1) x (-2, -4) x 17 1 a Find the zeroes of the monic quadratic y = (x + d)2 − e, where e > 0. b Find an expression for the difference between the two zeroes. c Hence find the condition for the difference between the two zeroes to be 2, and describe geometrically the family of quadratics with this property. 18 Consider the general quadratic function y = ax2 + bx + c. a Complete the square in x in order to show that b y=a x+ 2a 2 − b2 − 4ac . 4a b Hence write down the coordinates of the vertex, and the equation of the axis. c Put y = 0 in order to find the x-intercepts. d Write the quadratic in factored form. ENRICHMENT 19 Complete the square to find the vertex and x-intercepts of the function y = x2 + px + q. Then sketch a possible graph of the function if: a p > 0 and p2 > 4q, 20 b p > 0 and p2 < 4q, c p > 0 and p2 = 4q. Consider the quadratic f (x) = a(x − h)2 + k with vertex (h, k). Prove the following identities and hence establish that x = h is the axis of symmetry. a f (h + t) = f (h − t) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b f (2h − x) = f (x). © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3F The quadratic formulae and the graph 79 3F The quadratic formulae and the graph Learning intentions • Use the discriminant to work with the formula for the zeroes of a quadratic. • Use the formula for the axis of symmetry, and hence find the vertex. • Use the discriminant and leading coefficient to answer questions about the quadratic. • Use the discriminant to ‘discriminate’ (determine) the number of zeroes. Completing the square for the general quadratic gives formulae for its axis of symmetry and for its zeroes. These formulae are extremely useful, and like factoring and completing the square, allow the graph to be sketched. The algebra of completing the square for a general quadratic is shown in Question 18 at the end of Development in the previous Exercise 3E. The formula for the axis of symmetry Question 18 in the previous Exercise 3E gives the axis of symmetry: 23 The axis of symmetry of y = ax2 + bx + c b . 2a • Substitute this x-value back into the quadratic to find the y-coordinate of the vertex. • The axis of symmetry is the line x = − Readers should remember only the formula for the axis of symmetry, and find the y-coordinate of the vertex by substituting back into the quadratic. The formula for the zeroes Further working in Question 18 of Exercise 3E shows that putting y = 0 into y = ax2 + bx + c gives √ √ −b − Δ −b + Δ or x = , where Δ = b2 − 4ac x= 2a 2a The quantity Δ = b2 − 4ac is very important in quadratics. It is called the discriminant because it discriminates, and has the symbol Δ (Greek uppercase letter delta, corresponding to the Latin letter ‘D’). Using Δ in this formula makes it much easier to deal with and to remember. 24 The zeroes of the quadratic function y = ax2 + bx + c √ √ −b − Δ −b + Δ and x = , x= 2a 2a where Δ = b2 − 4ac. • Always calculate the discriminant first when finding the zeroes of a quadratic. • If Δ > 0, there are two zeroes, because positives have two square roots. • If Δ = 0, there is one double zero, because 0 is the only square root of 0. In this situation, the quadratic is a perfect square, and the vertex sits on the x-axis which is tangent to the graph. • If Δ < 0, there are no zeroes, because negatives don’t have square roots. • If Δ > 0, the quadratic has distinct zeroes (meaning ‘two distinct zeroes’). • If Δ = 0, the quadratic has two equal zeroes (noting that there is only zero). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 80 3F Chapter 3 Functions and graphs Example 20 Using the quadratic formula to sketch a quadratic Use the quadratic formulae to sketch each quadratic. Give any irrational zeroes first in simplified surd form, then approximated correct to three decimal places. a y = −x2 + 6x + 1 b y = 3x2 − 6x + 4 Solution a The curve y = −x2 + 6x + 1 is concave down, with y-intercept 1. The formulae are now applied with a = −1, b = 6, and c = 1. b First, the axis of symmetry is x = − 2a 6 x = − −2 x = 3. When x = 3, y = −9 + 18 + 1 y (3, 10) = 10, so the vertex is at (3, 10). Secondly, Δ = b2 − 4ac 1 = 40 so y = 0 when = 4 × 10 (take out square factors), √ √ −b + Δ −b − Δ x= or 2a √ 2a √ −6 − 2 10 −6 + 2 10 or = −2 √ √ −2 = 3 − 10 or 3 + 10 3 - Ö10 3 + Ö10 x −0.162 or 6.162. b The curve y = 3x2 − 6x + 4 is concave up, with y-intercept 4. Apply the formulae with a = 3, b = −6 and c = 4. b First, the axis of symmetry is x = − 2a x = − −6 6 x = 1, and substituting x = 1, the vertex is at (1, 1). Secondly, Δ = 36 − 48 y 4 (2, 4) (1, 1) x = −12 which is negative, so there are no zeroes, because negative numbers do not have square roots. Notice again that the symmetric point (2, 4) has been plotted so that the parabolic graph has at least three points on it. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3F The quadratic formulae and the graph Example 21 81 Using the discriminant to find the number of zeroes a Use the discriminant to find the number of zeroes of f (x) = 5x2 − 20x + 20. b What might be a better approach to this question? Solution a f (x) = 5x2 − 20x + 20 b Better, take out the common factor: f (x) = 5x2 − 20x + 20 Δ = 20 − 4 × 5 × 20 2 = 5(x2 − 4x + 4) = 0, so there is exactly one zero. Example 22 = 5(x − 2)2 , giving 5× a perfect square. Using the discriminant to test for a double zero Find m if the quadratic f (x) = 2x2 − 4mx + 10m has a double zero (or ‘two equal zeroes’), and write down the resulting quadratic function(s) in factored form. Solution Here a = 2, b = −4m, and c = 10m, so Δ = 16m2 − 80m = 16m(m − 5). The condition for a double root is Δ = 0, that is, m = 0 or 5. Hence there are two quadratics; f (x) = 2x2 and f (x) = 2(x − 5)2 . Exercise 3F 1 FOUNDATION Consider the parabola y = x2 − 2x − 1. a i Use the value of a to determine the concavity of the parabola. ii Write down the value of the y-intercept. iii Use the formula x = −b 2a to find the axis of symmetry. iv Use the axis of symmetry to find the y-coordinate of the vertex. v Calculate the discriminant Δ = b2 − 4ac. vi Explain why this parabola must have x-intercepts. √ √ −b − Δ −b + Δ or . vii Find the x-intercepts using the quadratic formula x = 2a 2a b Sketch the parabola, showing these features. 2 Find the discriminant Δ = b2 − 4ac of each quadratic function, then find the zeroes. Give the zeroes first in surd form, then correct to two decimal places. a f (x) = x2 + 2x − 2 b f (x) = x2 − 4x + 1 c f (x) = −x + 3x − 2 d f (x) = −x2 − 2x + 4 e f (x) = 3x2 − 2x − 2 f f (x) = 2x2 + 4x − 1 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 82 3F Chapter 3 Functions and graphs 3 4 Sketch a graph of each parabola by following similar steps to those outlined in Question 1. a y = x2 + 6x + 4 b y = x2 − 4x + 5 c y = −x2 + 2x + 2 d y = −2x2 + 4x − 3 e y = 3x2 + 6x − 1 f y = 2x2 + 2x − 1 g y = −x2 − 2x − 4 h y = −x2 + 2x + 5 i y = x2 + 2x + 3 In each case, find the zeroes of the quadratic function first by factoring, then by completing the square, and finally by using the quadratic formula. Observe that the answers to all three methods are the same for each function. a f (x) = x2 − 3x − 4 b f (x) = x2 − 5x + 6 c f (x) = −x2 + 4x + 12 5 a Consider the parabola y = x2 + 2. i Calculate Δ and explain why this parabola has no x-intercept. ii What do you notice about the y-intercept and the vertex? iii Sketch the parabola showing the y-intercept and vertex. Then add to your sketch the point where x = 1. iv Complete the sketch with another point found by symmetry. b Follow similar working to sketch these parabolas. i y = −x2 − 1 ii y = 12 x2 + 1 DEVELOPMENT 6 Use the quadratic formula to find the roots α and β of each quadratic equation. Hence show in each case that both α + β = − ba and αβ = ac . a x2 − 6x + 1 = 0 7 b x2 − 2x − 4 = 0 c −3x2 + 10x − 5 = 0 Find the discriminant Δ = b2 − 4ac for each quadratic. Use this and the concavity to state how many zeroes the function has, without drawing its graph. a f (x) = x2 + 3x − 2 b f (x) = 9x2 − 6x + 1 c f (x) = −2x2 + 5x − 4 8 Consider the parabola with equation y = −x2 + 2x + 3. a Use algebra to find the x-coordinates of any intersection points of this parabola with each of the following lines. i y=2 ii y = 4 iii y = 6 b Graph the situation. c For what values of k does the parabola intersect the line y = k twice? 9 Use the quadratic formula to find the zeroes α and β of each quadratic function. Hence write the function in factored form, f (x) = (x − α)(x − β). a f (x) = x2 − 6x + 4 b f (x) = x2 + 2x − 1 c f (x) = x2 − 3x + 1 d f (x) = 3x2 + 6x + 2 e f (x) = −x2 + 3x + 1 f f (x) = −2x2 − x + 1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3F The quadratic formulae and the graph 10 b . The parabola y = ax2 + bx + c has an axis of symmetry at x = − 2a b Δ a Use this value to show that vertex of the parabola is V = − , − . 2a 4a b For each of the following quadratics, use the formula in part a to find the axis of symmetry. Then find the coordinates of the vertex, first by substitution into the quadratic, then using the formula in part a. i y = x2 + 4x + 1 11 83 ii y = x2 − 6x + 10 iii y = −3x2 − 12x + 1 Let f (x) = ax2 + bx + c, where a 0. a Show that f (− ba ) = f (0). b Explain this in terms of the symmetry of the parabola y = ax2 + bx + c. 12 The interval PQ has length p, and the point A lies between the points P and Q. Find PA when PQ × QA = PA2 . 13 a Expand f (x) = (x − h)2 + k and show that Δ = −4k. Then use the quadratic formulae to show that the axis is x = h (as expected) and to find an expression for the zeroes. b Expand f (x) = (x − α)(x − β) and show that Δ = (α − β)2 . Hence show that the zeroes are x = α and x = β ⎛ 2 ⎞ ⎜⎜⎜ α + β ⎟⎟⎟ α − β ⎟⎟⎠. ,− (as expected) and that the vertex is ⎜⎜⎝ 2 2 14 a Find the axis of symmetry and the vertex of y = x2 + bx + c. b Find the zeroes, and then find the difference between them. c What condition on the constants b and c must be satisfied for the difference to be exactly 1? d Hence show that the family of such quadratics is the family of parabolas with vertices on the line y = − 14 . ENRICHMENT 15 [The golden mean] Sketch y = x2 − x − 1, showing the vertex and all intercepts. √ a Let α = 12 ( 5 + 1). Show that: D 1 =α−1 i α2 = α + 1 ii iii α6 = 8α + 5 α b ABCD is a rectangle with length and breadth in the ratio α : 1. It is divided into a square APQD and a second rectangle PBCQ, as shown. Show that the length and breadth of rectangle PBCQ are also in the A ratio α : 1. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 Q C P B © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 84 3G Chapter 3 Functions and graphs 3G Powers, cubics, and circles Learning intentions √ • Sketch graphs involving y = x , and understand their properties. • Sketch cubic graphs, and understand some of their properties. • Sketch circles and semicircles. This section deals with the graphs of cubes and some higher powers of x, and of the square root of x, then sketches cubic functions that have been factored into linear factors. Finally, it reviews circles and semicircles. The function y = √ √ x The graph of y = x is the upper half of a parabola on its side, as can be seen by √ squaring both sides to give y2 = x. Remember that the symbol x means the positive or zero square root of x, so the lower half of the parabola y2 = x is excluded: y= √ x 0 x y 0 1 4 1 2 1 2 √ 2 1 y 2 4 2 2 6 x 4 The domain is x ≥ 0, and the range is y ≥ 0. The cube of x Graphed to the right is the cubic function f (x) = x3 . It has a zero at x = 0, it is positive when x is positive, and is negative when x is negative. y=x 3 x −2 −1 − 12 0 −8 −1 − 18 0 y 1 2 1 8 1 2 1 8 y = x3 y 2 -2 2 x -2 The curve becomes flat at the origin. You can see this by substituting a small number such as x = 0.1. The corresponding value of y is y = 0.001, which is far smaller. The origin is called a horizontal inflection of the curve — ‘inflection’ means that the curve ‘flexes’ smoothly from concave down on the left to concave up on the right, and ‘horizontal’ means that the curve is momentarily horizontal. Inflections in general will be studied in Year 12. The graphs of odd powers f (x) = x3 , f (x) = x5 , f (x) = x7 , . . . , look similar. The origin is always a horizontal inflection. As the index increases, the curves become flatter near the origin, and steeper further away. The fourth power of x The graph to the right shows f (x) = x4 . It has a zero at x = 0, and is positive elsewhere. y = x4 x −2 y 16 −1 − 12 0 1 1 16 0 1 2 1 16 1 2 1 16 y 6 y = x4 4 All even powers y = x6 , y = x8 , . . . look similar. Like odd powers, as the index increases they become flatter near the origin and steeper further away. 2 -2 2 x The origin is called a turning point of the curve because at the origin, the curve turns smoothly from decreasing on the left to increasing on the right. Turning points in general are an important topic in Year 12. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3G Powers, cubics, and circles 85 Polynomials Polynomials are expressions such as 3x4 + 4x2 − 2x x3 − x2 and and 7 − 2x3 + 34 x6 that can be written as the sum of multiples of x, x2 , x3 , . . . and a constant. A polynomial is usually written with the powers in descending or ascending order, as above. The index of the power with highest index (and non-zero coefficient) is called the degree of the polynomial, so the polynomials above have degrees 4, 3, and 6 respectively. Thus familiar quadratics such as 3x2 + 4x + 5 are polynomials of degree 2. This section introduces cubics such as x3 − x2 , but the discussion is limited here only to cubics that have been factored completely, or to cubics that are easily factored such as x3 − x2 = x2 (x − 1). Chapter 11 is a systematic treatment of polynomials. Sketching a cubic factored into linear factors At this stage, not much graphing can be done of polynomial functions of degree 3 and higher. If, however, the polynomial has already been factored into linear factors, its basic shape can be established. This is done by drawing up a table of values to test its sign. For example, consider the cubic y = (x + 1)(x − 2)(x − 4) . The function has zeroes at x = −1, x = 2 and x = 4. Its domain is all real x, and cubic curves don’t have any breaks, so the three zeroes are the only places where it can change sign. y We can draw up a table of values dodging these zeroes, x −2 −1 0 2 3 4 5 y −24 0 8 0 −4 0 18 sign − 0 + 0 − 0 + 8 A -1 2 B 4 x We also know that, as with any cubic, y becomes very large positive or negative for large positive or negative values of x. Combining this with the table above allows us to draw a sketch. • Be careful! We cannot yet find the two turning points marked A and B on the curve, because that requires calculus. Notice that if we draw tangents there, the tangents are horizontal. It’s important to note that the points A and B are NOT midway between the zeroes — this curve is not a parabola. Mark the zeroes and the y-intercept, and nothing else. • Even more importantly, there is a point of inflection on the curve midway between A and B. We will cover this in Year 12. Cubics with square factors The cubic in this next example has a square factor, so that there are only two zeroes, one of which is a double zero. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 86 3G Chapter 3 Functions and graphs Example 23 Sketching cubics with square factors Sketch y = −(x − 1)2 (x − 4). What is happening at x = 1? Solution The cubic has zeroes at x = 1 and x = 4. y Here is a table of test values dodging around these zeroes, 4 x 0 1 2 4 5 y 4 0 2 0 −16 sign + 0 + 0 − 1 4 x y r P(x,y) At x = 1, the cubic has a double zero, but does not cross the x-axis. Instead, the x-axis is a tangent to the curve at (1, 0). Circles The equation of a circle, centre O(0, 0) and radius r, is found by Pythagoras’ theorem, in the form of the distance formula. A point P(x, y) in the plane lies on this circle -r OP = r if and only if OP2 = r2 r x -r (x − 0)2 + (y − 0)2 = r2 (distance formula) x 2 + y2 = r 2 . Four important semicircles The circle graph fails the vertical line test, so it is not a function. This can also be seen algebraically — solving for y gives two values of y for −r < x < r: √ √ y = r2 − x2 or y = − r 2 − x2 . These two graphs are the first two graphs below — the upper and lower semicircles — and in contrast to the circle, they are clearly both functions. Their domains are both −r ≤ x ≤ r, and ranges are 0 ≤ y ≤ r and −r ≤ y ≤ 0. y y r y y r r -r r x -r r x -r x -r -r r x -r Solving instead for x, and again separating the positive and negative square roots, x= r 2 − y2 or x = − r 2 − y2 , These two graphs are the last two graphs above — the left and right semicircles. They are relations, but not functions, because they clearly fail the vertical line test.Their domains are −r ≤ x ≤ 0 and 0 ≤ x ≤ r, and ranges are both −r ≤ y ≤ r. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3G Powers, cubics, and circles Exercise 3G 1 FOUNDATION Write down the coordinates of the centre and the radius of each circle. a x2 + y2 = 16 2 b x2 + y2 = 49 c x2 + y2 = 19 d x2 + y2 = 1.44 Sketch graphs of these circles, marking all intercepts with the axes, then write down the domain and range of each. a x 2 + y2 = 1 3 87 b x 2 + y2 = 9 c x2 + y2 = 14 d x2 + y2 = 94 Consider the curve y = x3 . a Copy and complete the following table of values: x −1.5 −1 −0.5 0 0.5 1 1.5 y b Plot the points in the table, using a scale of 2 cm to 1 unit on each axis, and then join the points with a smooth curve. 4 Repeat the previous question for the curve y = x4 . 5 [Technology] a Use computer graphing software to plot on the one number plane y = x, y = x3 and y = x5 . b Which three points do all these graphs pass through? c Which curve is nearest the x-axis for: ii x > 1? i 0 < x < 1, d Which curve is nearest the x-axis for: ii x < −1? i −1 < x < 0, e Rotate each curve by 180◦ about the origin. What do you notice? f Try finding other powers of x which have the same feature found in part e. What do you notice about the index of these functions? 6 [Technology] a Use computer graphing software to plot on the one number plane y = x2 , y = x4 and y = x6 . b Which three points do all these graphs pass through? c Which curve is nearest the x-axis for: ii x > 1? i 0 < x < 1, d Which curve is nearest the x-axis for: ii x < −1? i −1 < x < 0, e Reflect each curve in the y-axis. What do you notice? f Try finding other powers of x which have the same feature found in part e. What do you notice about the index of these functions? 7 Write down the zeroes of each cubic, use a table of values to test its sign, then sketch it, showing the y-intercept. a y = (x − 1)(x − 3)(x − 5) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b y = −3(x + 4)x(x − 2) c y = 2x2 (3 − x) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 88 3G Chapter 3 Functions and graphs 8 Write down the equation of each circle. a b y 2 -2 x 2 c y Ö5 y (3 , 4) Ö5 x -Ö5 -2 d y x x -Ö5 (1 , -3) DEVELOPMENT 9 Consider the circle x2 + y2 = 25. a Copy and complete the following table of values, correct to one decimal place where necessary. (Remember that a positive number has two square roots.) x 0 1 2 3 4 5 y≥0 y≤0 b Plot the points in the table, using a scale of 1 cm to 1 unit on each axis. c Reflect the points plotted in part b in the y-axis, and so sketch the entire circle. 10 Consider the curve y = √ x. a Copy and complete the following table of values: x 0 0.25 1 2.25 4 6.25 y b Plot the points in the table, using a scale of 2 cm to 1 unit on each axis, and then join the points with a smooth curve. √ c Repeat the process for the function y = − x 11 a Use the results of Question 10 to sketch the graphs of y = √ √ x and y = − x on the same number plane. b What shape has been formed? c Explain why this has happened. 12 13 14 Sketch each semicircle, and state the domain and range: √ √ √ a y = 4 − x2 b y = − 4 − x2 c y = − 1 − x2 d y= 25 2 4 −x Sketch these semicircles which are not functions. State the domain and range of each. a x = 1 − y2 b x = − 4 − y2 c x = − 9 − y2 d x= 9 2 4 −y Write down the equation of each semicircle. y a b 3 -3 y c Ö3 1 3 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 -1 -1 y Ö3 x x -Ö3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3G Powers, cubics, and circles y d y e 89 y f (-3, 4) x x (-1, -1) x (2, -1) 15 [Technology] a Graph these cubic polynomials using computer graphing software. i y = 14 x3 + 2 ii y = 12 (x3 − 6x2 + 9x) iii y = 21 (x3 − 2x2 − 5x + 6) b In each case, check that the y-intercept is equal to the constant term. c Read the x-intercepts from the screen and then check these values by substituting them into the corresponding polynomial. d Write each cubic in parts ii and iii in the form y = ax3 + bx2 + cx + d. Then take the product of all the zeroes found in part a and compare the result with da in each case. What do you notice? 16 Each graph shows a cubic of the form y = a(x − α)(x − β)(x − γ) for some constant a and where α ≤ β ≤ γ. Use the graph to determine these values and hence write down the equation of the cubic. a y y b 2 -2 -1 17 c y 2 1 x -1 2 x x 2 -2 Write down the radius of each circle or semicircle. Also state any points on each curve whose coordinates are both integers. √ a x 2 + y2 = 5 b y = − 2 − x2 c x = 10 − y2 d x2 + y2 = 17 ENRICHMENT 18 The diagram shows a ladder of length 2λ leaning against a wall so that the foot of the ladder is distant 2α from the wall. y B a Find the coordinates of B. b Show that the midpoint P of the ladder lies on a circle with centre at the origin. What is the radius of this circle? c Knowing that the point P lies on a circle, how could the radius have been found A more easily? 19 x [Technology] As discovered in the previous sections, the graphs of quadratic functions fall into one of two categories, concave up and concave down. In contrast, cubic functions have six basic types. a Use computer software or a table of values to plot accurately the cubic y = c(x) for: i c(x) = x3 + x ii c(x) = x3 iii c(x) = x3 − x b These three curves have some similarities. i For large positive and negative values of x, which two quadrants does the graph of y = c(x) lie in? ii Rotate each curve by 180◦ about the origin. What do you notice? iii By considering your investigation in Question 5, explain why that might be. c Look carefully at each curve as it passes through the origin. What distinguishes each graph there? d Conclude this investigation by graphing y = −c(x) for each cubic function in part a. Then look for any similarities and differences. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 90 3H Chapter 3 Functions and graphs 3H Two graphs that have asymptotes Learning intentions • Sketch exponential functions, and identify their asymptotes. • Sketch rectangular hyperbolas and identify their asymptotes. • Use limit notation for exponential graphs and rectangular hyperbolas. This section reviews simple exponential graphs and rectangular hyperbolas. They are grouped together because both types of graphs have asymptotes, which need further discussion. Exponential functions Functions of the form f (x) = a x , where the base a is positive and a 1, are called exponential functions, because the variable x is in the exponent or index. Here is a sketch of the function f (x) = 2 x . y = 2x y x −4 −3 −2 −1 0 1 2 3 4 1 16 1 8 1 4 1 2 1 2 4 8 16 y 2 1 Three key features should be shown when sketching this graph: • The y-intercept is y = 1, because 20 = 1. • When x = 1, y = 2, which is the base 2, because 2 = 2. 1 1 2 -1 1 x • The x-axis is a horizontal asymptote, as discussed below. The domain of an exponential function is all real x, and the range is y > 0. Limits and asymptotes of exponential functions On the far left of the graph, as x is becoming a very large negative number, y = 2 x becomes very small. Indeed, we can make y ‘as small as we like’ by choosing sufficiently large negative values of x. We say that ‘as x approaches negative infinity, y approaches the limit zero’, and write As x → −∞, y → 0 or lim y = 0. x→ −∞ The x-axis is called an asymptote of the curve (from the Greek word asymptotos, meaning ‘apt to fall together’), because the curve gets ‘as close as we like’ to the x-axis for sufficiently large negative values of x. In the right-hand direction, there is no limit, because ‘As x → ∞, y → ∞’. Thus y = 2 x has domain all real x, and range y > 0, or in bracket interval notation: Domain: (−∞, ∞). Range: (0, ∞). Rectangular hyperbolas 1 The reciprocal function y = can also be written as xy = 1, and has a graph that is called a rectangular x hyperbola. This graph has two parts, called branches, which in this case are disconnected. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3H Two graphs that have asymptotes 91 y x 0 y 1 y= x ∗ 1 5 1 10 1 2 5 10 1 2 1 5 1 10 2 5 2 1 x −10 −5 −2 −1 − 12 − 15 1 − 10 1 − 10 − 15 − 12 −1 −2 −5 −10 y 10 1 2 -2 (-1, -1) (1, 1) 2 x -2 The star (∗) at x = 0 means that the function is not defined there. Limits and asymptotes of rectangular hyperbolas The x-axis is an asymptote to this curve on both sides of the graph. We can make y ‘as small as we like’ by choosing sufficiently large positive or negative values of x. We say that ‘as x approaches ∞, y approaches zero’, and ‘as x approaches −∞, y approaches zero’, and we write As x → ∞, y → 0 or As x → −∞, y → 0 and lim y = 0 lim y = 0 or x→∞ x→−∞ The y-axis is a second asymptote to the graph. On the right-hand side of the origin, when x is a very small positive number, y becomes very large. We can make y ‘as large as we like’ by taking sufficiently small but still positive values of x. We say that ‘as x approaches zero from the right, y approaches ∞’, and write As x → 0+ , y → ∞. On the left-hand side of the origin, y is negative and can be made ‘as large negative as we like’ by taking sufficiently small negative values of x. We say that ‘as x approaches zero from the left, y approaches −∞’, and write As x → 0− , y → −∞. Domain and range 1 has domain x 0, and range y 0. x Neither domain nor range is an interval, so bracket interval notation cannot be used here as an alternative. From the sketch, it is clear that y = Exercise 3H FOUNDATION Note: If computer graphing software is not available, the two technology questions can be completed using tables of values. 1 a Copy and complete the following table of values of the hyperbola y = x −4 −2 −1 y − 12 0 1 2 1 2 2 . x 4 ∗ b Plot the points on a number plane, using a scale of 1 cm to 1 unit on each axis, then sketch the hyperbola. c Which two quadrants do the branches of the curve lie in? d Write down the equations of the two asymptotes of the hyperbola. e Write down the domain and range of the function. 2 Construct a table of values for each hyperbola, then sketch it. Write down the domain and range of each hyperbola and the equations of the two asymptotes. Also state which quadrants the branches lie in. 3 4 a y= b y= x x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 92 3H Chapter 3 Functions and graphs 3 [Technology] Use computer graphing software (if unavailable, tables of values will also do) to plot the following functions on the one number plane: 4 9 1 and y= and y= . y= x x x a Which two quadrants do the branches of each hyperbola lie in? b Write down the equations of the two asymptotes of each hyperbola. c Write down the domain and range of each hyperbola. d Write down the coordinates of the points on each hyperbola closest to the origin. What do you notice? e Add the line y = x to the graph and observe that the graphs are symmetric in this line. What other line are the graphs symmetric in? 4 Consider the exponential curve y = 3 x . a Copy and complete the following table of values of the exponential function y = 3 x . Give answers correct to one decimal place where necessary. x −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y b Plot these points on a number plane, using scales of 1 cm to 1 unit on both axes, then sketch the curve. c What is the y-intercept of the function? d What is the y-coordinate when x = 1? e Write down the equation of the asymptote. f Write down the domain and range of the function. 5 Construct a table of values for each exponential function, then sketch its graph. Write down the domain and range in each part and the equation of the asymptote. Also state the y-intercept and the y-coordinate at x = 1. b y = 1.5 x a y = 4x 6 [Technology] Use computer graphing software to plot on the one number plane y = 2 x , y = 3 x and y = 4 x . a Which point is common to all three graphs? b Write down the equation of the asymptote of each exponential curve. c Write down the domain and range of each exponential function. d Confirm by observation that at x = 1, the y-coordinate equals the base. e Which curve increases more rapidly to the right of the y-axis, and why? f Which curve approaches the asymptote more quickly to the left of the y-axis? Why? 7 a Construct a table of values for each hyperbola then sketch it. 4 3 2 ii y = − iii y = − x x x b Compare these equations and graphs with those in Questions 1 and 2. i y=− i Which quadrants do the graphs of part a lie in? ii What has changed in the equation to cause this difference? c The domain and range of each graph in part a are the same. i Write down the domain and range. ii Are they the same as in Questions 1 and 2? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3H Two graphs that have asymptotes 8 93 a Construct a table of values for each exponential function then use the table to sketch the function. i y = 3−x ii y = 4−x iii y = 1.5−x b Compare these equations and graphs with those in Questions 4 and 5. i Has the y-intercept changed? ii Has the asymptote changed? iii For the graphs in part a, at what value of x is the y-coordinate equal to the base? iv Heading to the right along each of these curves, describe how the y-coordinate changes. v What has changed in the equation to cause these differences? c The domain and range of each graph in part a are the same. i Write down the domain and range. ii Are they the same as in Questions 4 and 5? DEVELOPMENT 9 Make y the subject of each equation, then sketch its graph. b xy = −6 a xy = 12 10 For each hyperbola in the previous question: i write down the coordinates of the points closest to the origin, ii list any points with integer coordinates. 11 Sketch these exponential graphs without resorting to a table of values. Ensure the key features are shown. b y = 2−x a y = 5x 12 This question requires the language of limits from the first two pages of this section. a In Question 4, the line y = 0 is an asymptote to y = 3 x . Write a statement using limits to justify this. b In Question 6a ii, the line y = 0 is an asymptote to the exponential curve y = 4−x . Write a statement using limits to justify this. c In Question 1, the lines y = 0 and x = 0 are asymptotes to the hyperbola y = using limits to justify this. 13 a Use the index laws to explain why y = ( 12 ) x has the same graph as y = 2−x . b Hence sketch the graph of the function y = 14 2 . Write four statements x x 1 2 a Where does the hyperbola xy = c2 intersect the line y = x? b Confirm your answer by plotting the situation when c = 2. 15 Does the equation xy = 0 represent a hyperbola? Explain your answer. 16 a Show that (x + y)2 − (x − y)2 = 4 is the equation of a hyperbola. Sketch it. b Show that (x + y)2 + (x − y)2 = 4 is the equation of a circle. Sketch it. c Solve these two equations simultaneously. Begin by subtracting the equation of the hyperbola from the equation of the circle. d Sketch both curves on the same number plane, showing the points of intersection. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 94 3H Chapter 3 Functions and graphs ENRICHMENT 17 Some curves can cross their asymptote. a Complete the following table of values for y = x −8 −4 −2 −1 −0.5 0 2x x2 + 1 . 0.5 1 2 4 8 y b Plot the points, using a scale of 1 cm to 1 unit on each axis, and join them with a smooth curve. c What is the horizontal asymptote of this curve? d Where does the curve cross its asymptote? 18 a The line y = − 14 b2 x + b has intercepts at A and B. Find the coordinates of P, the midpoint of AB. 1 . x c Show that the area of OAB, where O is the origin, is independent of the value of b. b Show that P lies on the hyperbola y = 19 The curve y = 2 x is approximated by the parabola y = ax2 + bx + c for −1 ≤ x ≤ 1. The values of the constants a, b and c are chosen so that the two curves intersect at x = −1, 0, 1. a Find the values of the constant coefficients. √ 1 2 and √ . 2 c Compare the values found in part b with the values obtained by a calculator. Show that the percentage errors are approximately 1.6% and 2.8% respectively. b Use this parabola to estimate the values of CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3I Direct and inverse variation 95 3I Direct and inverse variation Learning intentions • Understand ‘varies with’ as a synonym for ‘is proportional to’. • Understand variation with higher powers of x, and with square roots. • Work with the functions arising from variation in scientific and practical examples. Direct variation and inverse variation were introduced in previous years, where we usually used the language of ‘is proportional to’ rather than ‘varies with’ Direct and inverse variation 25 Direct and inverse variation Direct variation: A variable y varies directly with a variable x if y = kx, for some non-zero constant k of proportionality. • The graph of y as a function of x is thus a line through the origin. Inverse variation: A variable y varies inversely with a variable x if k y = , for some non-zero constant k of proportionality. x • The graph of y as a function of x is thus a rectangular hyperbola whose asymptotes are the x-axis and the y-axis. The term ‘varies with’ is often used as a synomym for ‘is proportional to’. Inverse variation is more commonly called ‘indirect variation’, and common usage often says ‘is indirectly proportional to’ rather than ‘varies inversely with’. Most applications begin by finding the constant of proportionality, as in the next two examples. Example 24 Using direct variation in a practical situation a Garden mulch is sold in bulk, with the cost C proportional to the volume V in cubic metres. Write this algebraically. b The shop quotes $270 for 7.5 m3 . Find the constant of proportionality, and graph the function. c How much does 12 m3 cost? d How much can you buy for $600? Solution a C = kV, for some constant k. b Substituting, C 270 = k × 7.5 k = 36. Adding the units, 270 k = $36/m3 . c C = 36 × 12 = $432 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 d 600 = 36V V = 16 23 m3 . 7×5 V © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 96 3I Chapter 3 Functions and graphs Note: The functions y = kx associated with direct variation with x are not just linear functions, but linear functions through the origin. Think y = mx + b, where b = 0, so that we only need to find the single constant m (the gradient), conventionally written k in variation. The constant of proportionality and units In the previous example, we can solve the equation C = kV for k to give k = CV . Thus the constant k has units of Cost (in dollars) over Volume (in cubic metres). Hence the units of k are ‘dollars per cubic metre’, so we wrote that k = $36/m3 . But the constant may have no units — in a circle, the circumference C varies directly with the radius r, and the radius r varies directly with the circumference: C = 2πr and 1 r = 2π C. 1 Both C and r are lengths, so the two constants 2π and 2π are dimensionless. Example 25 Using inverse variation in a practical situation a The wavelength λ in metres of a musical tone is inversely proportional to its frequency f in vibrations per second. Write this algebraically. b The frequency of middle C is about 260 s−1 (‘260 vibrations per second’), and its wavelength is about 1.319 m. Find the constant of proportionality. c Find the wavelength of a sound wave with frequency 440 s−1 . d Find the frequency of a sound wave with wave length 1 metre. e What is the approximate speed of sound in air, and why? Solution l a λ = kf , for some constant k. b Substituting, Adding the units, k 260 k 343 1.319 = k 343 m/s. k 440 0.779 m c λ= 1×319 d 1= k f 260 f f 343 s−1 e About 343 m/s, because 343 waves, each 1 metre long, go past in 1 second. (Very roughly, this is 3 seconds per kilometre, which is extremely useful for estimating how far away a loud lightning strike is.) Direct and inverse variation are closely related Direct and inverse variation are closely related, despite their contrasting graphs. For example, if a rectangle has area A and adjacent sides x and y, then A A = xy and y= . x A y x • If the side y is constant, the area A is directly proportional to the other side x. • If the area A is constant, the side y is inversely proportional to the side x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3I Direct and inverse variation 97 Direct and inverse variation with other powers A quantity can vary directly and indirectly with powers of x, typically x2 and x3 . √ √ It can also vary with the square root x and the cube root 3 x of x. In other situations that we will not consider until later chapters, a quantity Q varies with other functions of time t. • With rates, Q typically varies with an exponential function such as 2t . • With periodic phenomena such as waves and air temperature and tides, a quantity Q typically varies with say sin t or cos t. Example 26 Varying with the cube root of a quantity Spherical balls are being made solid, from just one material. a Explain why the mass of the ball is proportional to the cube of the radius. b Write part a as a formula, and hence show the radius is proportional to the cube root of the mass. c Write part b as a formula, and hence find (nearest millimetre) the radius of a ball with mass 19 kg if a ball of mass 0.9 kg has a radius of 5 cm. Solution Let the ball have mass m kg, radius r cm, and volume V cm3 . a We know that V = 43 πr3 , so volume is proportional to the cube of the radius. Also mass is proportional to volume, because only one material is used. Hence the mass is also proportional to the cube of the radius. b Thus M = kr3 , for some constant k > 0, √3 1 so r = M , where = √3 is also a positive constant. k √3 c Substituting, 5 = 0.9 5 = √3 0.9 = 5.178720 . . . (store in the calculator). √3 When M = 19, r = 19 13.8 cm. Gravitational force involves four different proportions In our universe (and on Earth), bodies are attracted to each other by gravity. The force F of attraction between two bodies is known to be: • directly proportional to each mass m1 and m2 of the two bodies, and • inversely proportional to the square of the distance r between them. This can be written as a single function: Gm1 m2 F= , r2 where G is a constant called the universal gravitational constant (ignore the units). We can solve this equation for each of the other variables: Fr2 m1 = Gm2 and Fr2 m2 = Gm1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 and r= Gm1 m2 F © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 98 3I Chapter 3 Functions and graphs Hence m1 varies directly with the force F and with the square of the distance r, and varies inversely with the other mass m2 . Similarly m2 . Also r varies directly with the square root of each mass m1 and m2 , and varies indirectly with the square root of the force F. This equation is one of the most famous in all of physics, and goes back to Sir Isaac Newton (1635-1727). At the time, this law of universal gravity was more often thought of in terms of proportionality than as an equation. Exercise 3I 1 FOUNDATION It is known that y varies directly with x. a Write this algebraically using k as the constant of proportionality. b Find the value of k given that: i y = 10 when x = 2 ii y = −6 when x = 3 iii y = 6 when x = 4 2 It is known that y is inversely proportional to x. a Write this algebraically using k as the constant of proportionality. b Find the value of k given that: i y = 2 when x = 3 ii y = 12 when x = −3 iii y = −2 when x = 4 3 a The value of y is in proportion with x. What happens to the value of y when the value of x is: ii halved? i doubled? b The value of y varies inversely with x. What happens to the value of y when the value of x is: ii halved? i doubled? 4 The quantity y varies inversely with the square of x. a i Write this algebraically using k as the constant of proportionality. ii It is found that y = 2 when x = 3. Determine the constant k. iii Evaluate y when x = 6. b What happens to the value of y when the value of x is: ii halved? i doubled? 5 The area A cm2 of a triangle varies directly with its height h cm. a Write this algebraically using k as the constant of proportionality. b Determine k if A = 12 cm2 when h = 3 cm. c What are the units of k? d Explain the units of k using the formula for the area of a triangle. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3I Direct and inverse variation 6 99 In a scale model of an Spitfire (aeroplane), each length m on the model is in varies directly with the corresponding length s on the Spitfire. a Write this algebraically. b The length of the Spitfire is 9.12 m and the model has length 0.38 m. What is the constant of proportionality? c The height of the Spitfire on the runway is 3.48 m. What is the height of the model? d The wing span of the model is 468 mm. What is the wingspan of the Spitfire in metres? 7 a The amount of paint P, in litres, used to paint a building is directly proportional to the area A, in square metres, to be covered. Write this algebraically. b A certain building requires 48 L to cover an area of 576 m2 . Find the constant of proportionality, and graph the function. c A larger building has an area of 668 m2 to be painted. How many litres of paint will this require? d A paint supplier sells 40 L buckets and 4 L tins of paint. How many buckets and tins must be bought in order to paint the building completely? 8 The owners of the Fizgig Manufacturing Company use a uniformly elastic demand curve to model their sales. Thus if the price of a Fizgig is p and the quantity sold per year is q then the turnover from sales is constant. That is, pq = T , for some constant T . a Last year, the price of a Fizgig was p = $6 and the quantity sold was q = 400 000. Find T . b The company’s board of directors want to raise the price to p = $8 next year. How many Fizgigs can the company expect to sell if this happens? c Under this model, what will happen to the sales if the price is doubled instead? d Sketch the graph of the demand curve with q on the horizontal axis and p on the vertical axis. DEVELOPMENT 9 If a gas behaves in an ideal way then the relationship between pressure, volume and temperature can be approximated with simple equations. a When the gas is kept at a constant pressure, the volume V it occupies varies directly with its temperature T when that is expressed in Kelvin instead of Celsius. i Write down an equation relating V and T . ii A certain amount of gas occupies 2 m3 when the temperature is T = 273 Kelvin (= 0◦ C). What volume will it occupy when the temperature is T = 298 Kelvin (= 25◦ C)? Give your answer correct to the nearest cm3 . b When the gas is kept at a constant temperature, the pressure P varies inversely with the volume. i Write down an equation relating P and V. ii A certain amount of gas is at twice atmospheric pressure when it occupies 11.2 L. What is the pressure when the volume is 30 L? Express your answer as a multiple of the atmospheric pressure correct to two decimal places. 10 An object is initially stationary and accelerates. The acceleration is constant. a It is found that the velocity v m/s varies directly with the time t seconds. i After 4 seconds the speed is 5 m/s. Find the constant of proportionality. ii What is the speed of the object at t = 12 s? iii When will the object be travelling at 12.5 m/s? b The distance x metres that the object travels is proportional to the square of the time. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 100 3I Chapter 3 Functions and graphs c Write this algebraically using k as the constant of proportionality. i After 4 seconds the distance is x = 10 m. Show that the constant of proportionality is half that in part a. ii How far has the object travelled at t = 12 s? iii How much time has elapsed when x = 160 m? iv What happens to x when the value of t is doubled? Check your answer by comparing x when t = 8 s with part i. 11 The acceleration a due to gravity varies inversely with the square of the distance r to the centre of the earth. a Write this algebraically using k as the constant of proportionality. b At the surface of the earth the acceleration due to gravity is approximately 9.81 m s−2 . The radius of the earth at sea level is approximately 6 378 137 m. Find the value of k correct to 3 significant figures. c Galileo dropped a cannonball from the tower of Pisa Cathedral, approximately 55 m above sea level. Show that the acceleration due to gravity at this height is still 9.81 m s−2 correct to two decimal places. d A satellite is in geosynchronous orbit above the equator at an altitude of 35 786 km above sea level. What is the acceleration due to gravity at this height? Give your answer correct to 2 decimal places. e What happens to a when the distance from the centre of the earth is doubled? 12 The formula for the simple interest $I paid when a principal $P is invested at an interest rate R for n years is I = nPR. a Suppose that R and n are fixed. What is the relationship between I and P? b Suppose that I and P are fixed. What is the relationship between n and R? c Suppose that PR = k is constant so that I = kn. What are the units of k? 13 In simple cases, the time t it takes for a tank to empty is found to be proportional to a power of the initial depth h of the water. a The time taken to empty a cylindrical tank is proportional to the square-root of the initial depth. i Write this algebraically using k as the constant of proportionality. ii When the initial depth was 9 m the time taken was 8448 s. How long will it take to empty this tank if it is filled to a depth of 4 m? b The time taken to empty a conical tank is proportional to the square-root of the fifth power of the initial depth. i Write this algebraically using k as the constant of proportionality. ii When the initial depth was 3 m the time taken was 956 s. How long will it take to empty this tank if it is filled to a depth of 4 m? Give your answer correct to the nearest second. 14 The lateral force F required to keep a car on the road when driving around a bend with constant radius is proportional to the square of the speed V. a Write this algebraically using k as the constant of proportionality. b A formula one car drives at 302.4 km/h around the corner named 130R at Suzuka in Japan. The force keeping the car on the road is 15, 120 newtons. Find k correct to 6 decimal places. c On the next lap the car takes the corner at 288 km/h. What is the force for this lap, correct to the nearest newton? d The driver goes slowly around the corner on the third lap. On the fourth lap the car travels around 130R at twice the speed of lap 3. By what factor has the force increased from lap 3 to lap 4? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 3I Direct and inverse variation 15 101 In a special type of pendulum, called a conical pendulum, the time taken to complete one oscillation, which is called the period P, varies directly with the square-root of the length L of the pendulum. a Write this algebraically using k as the constant of proportionality. b When the length of the pendulum is L = 0.29 m the period is P = 1.005 s. What will be the period of a conical pendulum with length L = 2 m? Give your answer in seconds rounded to two decimal places. 16 [Inverse Variation] An architect is designing a building. The client has insisted that one of the rooms in the building must have an area of 48 m2 . For ease of design, the length and the breadth b must each be a whole number of metres. Because of the furniture that must go into the room, no wall may be less than 4 m. What are the possible dimensions of the room? ENRICHMENT 17 The orbit of a planet can be closely approximated with an ellipse, which is a stretched circle. The semi-major axis is half the maximum distance across the ellipse. Kepler’s third law of orbits states that the square of the period T (the time to complete one orbit) is proportional to the cube of the semi-major axis a. a Write this algebraically. b The Earth takes one year to orbit the sun and has semi-major axis a = 149.6 × 106 km. Find the value of the constant of proportionality correct to four significant figures. c The semi-major axis of Mars is 227.9 × 106 km. Calculate the period of Mars in years correct to two decimal places. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 102 Chapter 3 Functions and graphs Chapter 3 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 3 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Review Chapter Review Exercise 1 Determine which of the following relations are functions. y a y b y c d y 2 2 2 x -2 -2 2 x x -2 2 Write down the domain and range of each relation in Question 1. 3 Find f (3) and f (−2) for each function. 5 a f (x) = x2 + 4x b f (x) = x3 − 3x2 + 5 Find the natural domain of each function. √ 1 a f (x) = b f (x) = x − 1 x−2 c f (x) = 1 2−x Find the x-intercept and the y-intercept of each line, then sketch use the intercepts to sketch each line. b x − 3y + 6 = 0 Sketch each of these lines through the origin. a y = 2x 8 d f (x) = √ b F(x) = x2 − 3x − 7 a y = 2x + 2 7 √ 3x + 2 Find F(a) − 1 and F(a − 1) for each function. a F(x) = 2x + 3 6 x -2 -2 4 2 b x + 2y = 0 a Sketch these two lines. i y = −1 ii x − 3 = 0 b Where do the lines intersect? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 3 review 9 11 12 a f (x) = 16 − x2 b f (x) = x(x + 2) c f (x) = (x − 2)(x − 6) d f (x) = −(x + 5)(x − 1) e f (x) = x + x − 6 f f (x) = −x2 + 2x + 8 2 Complete the square for each quadratic. Then sketch the graph of each function, showing the vertex and the intercepts with the axes. a y = x2 + 2x − 5 b y = −x2 + 6x − 6 c y = −x2 + 2x − 3 d y = x2 + 6x + 10 Sketch the graph of each parabola. Use the discriminant to determine whether or not there are any x-intercepts. Show the vertex and the intercepts with the axes. a y = −x2 − 2x + 1 b y = x2 − 4x + 2 c y = x2 − 4x + 8 d y = −x2 + 6x − 15 Use a table of values to test the sign of each cubic at different points, and then sketch each cubic polynomial. b y = x2 (x − 3) a y = (x − 1)(x − 3)(x − 6) 13 Sketch each circle. a x 2 + y2 = 9 14 b x2 + y2 = 25 Sketch each semicircle, and write down the domain and range in each case. √ a y = 16 − x2 b x = 9 − y2 c x = − 4 − y2 √ d y = − 25 − x2 15 Construct a table of values for each hyperbola, then use the table to sketch the function. Write down the domain and range in each case. 4 8 a y= b y=− x x 16 Construct a table of values for each exponential function, then use the table to sketch the function. Write down the domain and range in each case. b y = 3−x a y = 2x 17 Review 10 Use factoring where necessary to find the zeroes of each quadratic function. Hence sketch the graph of y = f (x), showing all intercepts and the coordinates of the vertex. Also state the domain and range. 103 [A revision medley of curve sketches] Sketch each set of graphs on a single pair of axes, showing all significant points. A table of values may help in certain cases. a y = 2x, b y = − 12 x, 2 c y=x , d x + y = 0, e y = x2 , f x − y = 0, g x2 + y2 = 4, h y = 3x, i y=2 , x j y = −x, k y = x − x, 2 l y = x2 − 1, y = 2x + 3, y = − 12 x + 1, y = (x + 2)2 , x + y = 2, y = 2x2 , x − y = 1, x 2 = 1 − y2 , x = 3y, y = 3x , y = 4 − x, y = x2 − 4x, y = 1 − x2 , CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y = 2x − 1 y = − 12 x − 2 y = (x − 1)2 x + y = −3 y = 12 x2 x − y = −2 y2 = 25 − x2 y = 3x + 1, y = 4x y = x − 4, y = x2 + 3x y = 4 − x2 , x = 3y + 1 x = −4 − y y = −1 − x2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 104 Chapter 3 Functions and graphs y = (x + 2)2 − 4, n y = x2 − 1, y = x2 − 4x + 3, √ √ o y = 9 − x2 , y = − 4 − x2 , 2 1 y= , p y= , x x √ √ q y = x, y = 2 − x, r y = x3 , y = x3 + 1, s y = x4 , y = (x − 1)4 , x t y = 2−x , y = 12 , Review m y = (x + 2)2 , 18 y = (x + 2)2 + 1 y = x2 − 8x + 15 √ y = 1 − x2 3 y=− x √ y= 1−x y = (x + 1)3 y = x4 − 1 1 y= x 2 In simple spring theory, the change in length x of a spring varies directly with the mass m hung from it. a Write this algebraically using k as the constant of proportionality. b For a certain spring, x = 2 cm when m = 3 kg. Evaluate k, giving its units in your answer. c What will be the change in length of the spring when m = 4.5 kg? d What happens x when the mass is double? 19 A common formula learnt in science is D = S × T , that is, distance equals the product of speed and time. a How does distance vary with time when S is constant? b i Give an example of inverse variation using this equation. ii In your example, how will one variable change when the other is doubled? 20 When an object is thrown up a hill, its maximum range R is the greatest distance it can reach. The maximum range of an object varies directly with the square of the speed V it is thrown at. a Write this algebraically using k as the constant of proportionality. b It is found that R = 5.505 m, correct to four significant figures, when V = 9 m s−1 . Evaluate k correct to four significant figures. c What is R when V = 15 m s−1 ? Give your answer correct to the nearest centimetre. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4 Equations and inequations Chapter introduction This short chapter introduces some further approaches to linear inequations, and to quadratic equations and inequations. The discriminant of a quadratic is developed further, and quadratic identities are discussed with a clear statement of the relevant theorem leading the discussion. Practical applications particularly are developed in the exercises. The algebraic ideas are in the foreground, but the parabolic graphs of quadratics are always in the background of every discussion, and should be drawn whenever there are any questions about what it all means. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 106 4A Chapter 4 Equations and inequations 4A Linear equations and inequations Learning intentions • Solve linear inequations, reversing the inequality sign when necessary. Most of this chapter involves solving inequations. The necessary algebraic operations on an inequation are the same as for equations, with a single extra rule — reverse the inequality sign when multiplying or dividing by a negative. In this section, the linear inequations can be solved by these operations alone. Algebraic manipulation of equations and inequations The rules for operations on equations are well known, and have been used without comment in earlier chapters. One extra rule is needed for inequations. 1 Algebraic operations on equations and inequations • Equations: Add or subtract the same number on both sides. Multiply or divide both sides by the same nonzero number. • The extra rule for inequations: Reverse the inequality when multiplying or dividing by a negative number. The extra rule for inequations is easily justified by two simple examples: We know that 6 ≥ −4. We know that 8 < 12. × (−1) −6 ≤ 4. ÷ (−4) −2 > −3. Writing the operation on the left draws attention to the needed reversal. Example 1 Solving a linear inequation and graphing the solution Solve these linear inequations, then graph their solutions on a number line. 4 − 7x ≥6 a x − 12 < 5 + 3x b 13 ≥ 3 Solution − 3x −2x − 12 < 5 ×3 4 − 7x ≥6 3 39 ≥ 4 − 7x ≥ 18 + 12 −2x < 17 x − 12 < 5 + 3x a 13 ≥ b −4 35 ≥ −7x ≥ 14 ÷ (−2) x > −8 12 ÷ (−7) −5 ≤ x ≤ −2 -8 12 0 x -5 -2 0 x In each case, the inequalities were reversed when we divided by a negative. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4A Linear equations and inequations Example 2 107 Solving a practical problem using a linear inequation A school wants to buy some computers from a technology fund of $30 000. a If the n computers cost $820 each, how much money will remain in the fund? b If at least $14 500 must remain, how many computers can they buy? Solution Total cost = 820n, a so amount remaining = 30 000 − 820n. b Put 30 000 − 820n ≥ 14 500 − 30 000 −820n ≥ −15 500 ÷ (−820) n ≤ 18.9024 . . . So they can buy 18 computers. Exercise 4A 1 2 3 FOUNDATION Graph each set of real numbers on a number line. a x>1 b x≤2 c x ≥ −3 d 0<x<3 e −2 < x ≤ 1 f −5 ≤ x < 10 Solve each inequation, and graph your solution on a number line. a x−2<3 b 3x ≥ −6 c 5x + 5 > 15 d 4x − 3 ≤ −7 e 2x + 3 ≥ x + 7 f 6x − 5 < 3x − 17 a −2x < 6 b −5x ≥ −50 c −x ≤ 1 d 3 − 2x > 7 e 11 − 3x < 2 f −4 − x ≥ 1 g 2 − x > 2x − 4 h 3 − 3x ≤ 19 + x i 12 − 7x > −2x − 18 Solve each inequation. DEVELOPMENT 4 5 6 Solve each double inequation, and graph your solution on a number line. a 3< x+2<6 b −5 < x − 3 ≤ 4 c −8 ≤ 4x < 12 d −1 ≤ 2x ≤ 3 e −10 < 3x − 1 < 2 f −7 < 5x + 3 ≤ 3 Solve each double inequation. Write your answers using bracket interval notation. a −4 < −2x < 8 b −2 ≤ −x ≤ 1 c −7 ≤ 5 − 3x < 4 d −4 < 1 − 13 x ≤ 3 Solve each inequation. Write your answers using bracket interval notation. x x <3+ a b 14 x ≥ 12 x − 1 5 2 2x − 1 x + 1 c d 16 (2 − x) − 13 (2 + x) > 2 ≥ −1 3 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 108 4A Chapter 4 Equations and inequations 7 In each case, use the given graph to help solve the inequation. Then confirm your answer by solving it algebraically. b 1 + 12 x ≤ 2x − 2 a 6 − 3x > 0. 6 y y 2 1 1 2 x -2 2 x 8 a i Graph the line y = 2x − 3. ii How can this graph be used to solve 2x − 3 ≥ 0? Write down the solution. b i Graph the two lines y = 2x − 1 and y = 2 − x on the same number plane. ii How can this graph be used to solve 2x − 1 < 2 − x? Write down the solution. 9 Write down and solve a suitable inequation to find the values of x for which the line y = 5x − 4 is below the line y = 7 − 12 x. 10 a Sketch the lines y = 1 − x, y = 2 and y = −1 on a number plane and find the points of intersection. b Solve the inequation −1 < 1 − x ≤ 2 and relate the answer to the graph in part a. c Write down similar graphical explanations of Questions 4a and 5a. 11 An investment of $200 000 made on the 2nd of January is earning $823 per month in simple interest. The interest is paid into the account on the 1st of each month. a Write down an equation for the balance B of the account on the 1st of the nth month. b Find the last time that the total balance of the account is no more than $205 000 c Find the first time that the balance is greater than $215 000. 12 An electrical company is installing new lights in a new house. The company will charge the customer a call-out fee of $211 plus $55 per light. This fee covers both the purchase of the lights and their installation. The customer has a budget of $650. What is the maximum number of lights that can be fitted? 13 Jack and Jill were running for president of the Hill Climb Club. There were 21 members of the club who voted. The number x of members that voted for Jack was more than twice the number y that voted for Jill. What was the minimum number of votes that Jack received? Show how this can be determined graphically. 14 When Jack and Jill’s daughter Janine was born they made two investments for her. They put $600 in a term deposit earning 5% simple interest and another $500 in a managed fund earning 7% simple interest. In each case the interest is paid on Janine’s birthday each year. a Write down an expression for each investment giving the balance y after x years. b For how many years is the balance of the term deposit greater than that of the managed fund? Show this graphically. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4A Linear equations and inequations 15 109 [Break-even point] A certain business has fixed costs of $900 plus costs of $30 per item sold. The sale price of each item is $50. If enough items are sold then the company is just able to pay its total costs. That point is called the break-even point. Companies may use several different methods to graph this information. Here are two such methods. In each case, let n be the number of items sold. a i The gross profit per item is $50 − $30 = $20. Sketch the graph of the gross profit for n items where y = $20 × n. ii On the same graph, sketch the fixed costs y = $900. iii The point where these two lines cross is the break-even point. How many items need to be sold to break even? b i On a new graph, draw y = $50 × n, the total sales for n items. ii On the same graph, draw y = $900 + $30 × n, the total cost for n items. iii Does the break-even point for this graph agree with the result of part a? c The net profit is the income from sales minus the cost. A positive net profit indicates that the business makes a profit and a negative net profit indicates a loss. i Write down a formula for the net profit $P in terms of n. ii The company has sufficient funds to cover a loss of up to $400 and still stay in business. How many items are sold if the company does indeed remain in business? iii The company has a profit target of $750. How many items are sold given that the company exceeds the profit target? ENRICHMENT 16 If −1 ≤ t < 3, find the range of values of: a 4t d 2t − 1 17 b −t e 1 2 (t + 1) c t+7 f 1 2 (3t − 1) Cartons are stored in a large room. In order to allow access to the boxes, the available space in the room is a 7 m by 7 m square. Cartons cannot be stacked on top of each other. a i Some cartons have a base 3 m by 2 m. By comparing areas, what is the maximum number of these cartons that can be stored in this room? ii Draw a diagram to show how this can be achieved. b i Other cartons have a base 4 m by 2 m. By comparing areas, what is the maximum number of these cartons that can be stored in this room? ii It is not possible to store the number of boxes found in part i. What is the actual maximum number of these boxes that can be stored in this room? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 110 4B Chapter 4 Equations and inequations 4B Quadratic equations and inequations Learning intentions • Solve quadratic equations by the three standard methods. • Solve quadratic inequations using the graph. • Solve quadratic inequations using a table of test points. This section is mostly about the inequations. First, we quickly review from Chapter 3 the three methods of solving quadratic equations. Three ways to solve a quadratic equation For comparison of the methods, here is x2 − 9x + 14 = 0 solved in the three ways. Factoring is quick when it works, but is only available in special cases: Factoring: Completing the square: Formula: x − 9x + 14 = 0 x2 − 9x + 14 = 0 x2 − 9x + 14 = 0 (x − 7)(x − 2) = 0 81 x2 − 9x + 81 4 = −14 + 4 2 so x = 7 or 2. Δ = 81 − 56 = 52 (x − 92 )2 = 25 4 x − 4 12 = 2 12 or − 2 12 9−5 x = 9+5 2 or 2 x = 7 or 2 = 7 or 2 We have also seen in Chapter 3 how all three methods can identify a double root, otherwise called two equal roots, and how completing the square and the discriminant both show when an equation has no roots. Solving a quadratic inequation — graphical method The clearest way to solve a quadratic inequation is to read its solution off the graph. First, put the inequation in the form ax2 + bx + c > 0, then graph y = LHS, and see where the curve is above the x-axis. And similarly for the other three inequality signs. Example 3 Solving a quadratic inequation graphically Use the graphs to solve these inequations: a x2 − 2x − 3 ≥ 0 y -1 b x2 + 1 ≤ 2x y 3 c 6x − 4 > x2 y x (2, 1) 1 -3 (1, -4) y = x2 − 2x − 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 y = x2 − 2x + 1 4 (2, 4) (1, 1) x x y = 3x2 − 6x + 4 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4B Quadratic equations and inequations 111 Solution a x2 − 2x − 3 ≥ 0 (x − 3)(x + 1) ≥ 0 so x ≤ −1 or x ≥ 3. 2 x2 + 1 ≤ 2x b 3x2 − 6x + 4 < 0 Here Δ = −12 < 0. so there are no solutions. x2 − 2x + 1 ≤ 0 − 2x 6x − 4 > 3x2 c (x − 1)2 ≤ 0 so x = 1 is the only solution. Graphical solution of a quadratic inequation • Rearrange the inequation into the form ax2 + bx + c > 0 (or < , or ≥ , or ≤). • Graph y = LHS, and observe where y is positive, zero, and negative. • Hence read the solution off the graph. Solving a quadratic inequation — zeroes and test points • The graphs show that everything depends on finding the zeroes of the LHS. • Once the zeroes are found, a table of test points dodging around the zeroes can be used to establish the sign of the function for different values of x — because zeroes are the only places where a quadratic function can change sign. 3 Solution of a quadratic inequation using zeroes and test points • With the inequation in the form ax2 + bx + c > 0, find the zeroes of LHS. • Make a table of test points of the LHS dodging around the zeroes. • This is because a quadratic function can only change sign at a zero. Example 4 Solving quadratic inequations using zeroes and test points Solve the inequations found in Example 3 using their known zeroes and a table of values, a x2 − 2x − 3 ≥ 0 b x2 − 2x + 1 ≤ 0 c 3x2 − 6x + 4 ≤ 0 b The only zero is 1. c There are no zeroes. Solution a The zeroes are −1 and 3. x −2 −1 0 3 4 x 0 1 2 x 0 y 5 0 −3 0 5 y 1 0 1 y 4 sign + 0 − 0 + sign + 0 + sign + Hence x ≤ −1 or x ≥ 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 Hence x = 1. There are no solutions. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 112 4B Chapter 4 Equations and inequations Example 5 A practical application of quadratic inequations A ball is thrown vertically upwards from the ground, and its height h metres after t seconds is given by h = −5t2 + 30t. When is the ball: b above 40 metres? a on the ground again? c in the air and below 25 metres? Solution h=0 a Put −5t2 + 30t > 40 b Solve −5t2 + 30t = 0 ÷ (−5) −5t(t − 6) = 0 (t − 4)(t − 2) < 0 The zeroes of the LHS are 2 and 4. t = 0 or 6. Hence the ball hits the ground again after 6 seconds. Thus it is in the air for 0 < t < 6. t2 − 6t + 8 < 0 t 0 2 3 4 5 (t − 4)(t − 2) 8 0 −1 0 3 sign + 0 − 0 + Hence 2 < t < 4. c Solve −5t2 + 30t < 25 ÷ (−5) t2 − 6t + 5 ≤ 0 (t − 1)(t − 5) ≤ 0 The zeroes of the LHS are 1 and 5. t 0 1 3 5 6 (t − 1)(t − 5) 5 0 −4 0 5 sign + 0 − 0 + Hence 0 < t < 1 or 5 < t < 6. Exercise 4B 1 FOUNDATION Write down the values of x for which each parabola is: i below the x-axis ii above the x-axis y a b 10 y c y 2 -2 -1 -1 2 1 x 1 x 5 x 5 x 2 For each parabola, state the values of x for which: ii y > 0 i y=0 y a iii y < 0 y b c y 3 -4 1 3 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 2 -8 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4B Quadratic equations and inequations 3 113 a Use the given graph to help solve each inequation. ii (x − 3)(x + 1) ≥ 0 y i x(x − 4) < 0 y 4 x -1 3 iii x(2 − x) ≤ 0 y x 2 x -3 b Confirm your answers to part a by using tables of values. 4 a Sketch the associated parabola and hence solve: i (x + 2)(x − 4) < 0 ii (3 − x)(x + 1) > 0 iii (2 − x)(x − 5) ≤ 0 iv (x + 1)(x + 3) ≥ 0 v (x + 1)(1 − x) < 0 vi (x + 1)(x + 2) ≤ 0 b Confirm your answers to part a by using tables of values. DEVELOPMENT 5 6 7 8 Factorise the LHS, then sketch an appropriate parabola in order to solve: a x2 + 2x − 3 < 0 b x2 − 5x + 4 ≥ 0 c x2 + 6x + 8 > 0 d x2 − x − 6 ≤ 0 e 2x2 − x − 3 ≤ 0 f 4 + 3x − x2 > 0 a x2 ≤ 1 b x2 > 3x c x2 ≥ 144 d x2 > 0 e x2 + 9 ≤ 6x f 4x − 3 ≥ x2 Solve: Solve these inequations with quadratics on both sides. a 2x2 + x − 1 < x2 + 4x + 3 b x2 + 5x + 6 ≤ −x2 + x + 12 c x2 + 4x + 3 ≥ −x2 + 5x + 6 d 4x2 − 7x − 2 > x2 − 2x In each case solve the quadratic inequation by using the given graph of the LHS. a 6 + x − x2 > 4 y 6 5 4 3 2 1 3 -2 -1 1 2 4 x b 2 + x − x2 ≤ −4 y 2 -1 1 2 -2 -1 -2 -3 -4 1 3 4 x c How could the graph in part a be used to solve the problem in part b? 9 a Graph the parabola y = x2 − x and the line y = 1 − x on the same number plane. b How can this graph be used to solve x2 − x < 1 − x? Write down the solution. 10 Write down and solve a suitable inequation to find the values of x for which the line y = 2x − 4 is below the parabola y = x2 − 3x + 2. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 114 4B Chapter 4 Equations and inequations 11 Two positive numbers have a sum of 30. Let one of the numbers be x. a Find an expression for the product P of the two numbers. b Find the values of x for which P ≥ 200. c List the distinct pairs of integers that have a sum of 30 and a product of at least 200. 12 A rectangle has perimeter 180 mm. Let one side have length of x m. a Write down an expression for the area A of this rectangle in terms of x. b What values may x take if A > 20 cm2 ? Take care with the units. 13 A farmer needs to build a rectangular holding pen for some sheep. The side of a barn will be used for one side of the pen and there is 56 m of fencing available for the remaining three sides. Let x be the length of each side perpendicular to the barn wall. a Find an expression for the third side in terms of x and draw a diagram showing all this information. b The area A of the pen must be at least 392 m2 so that the sheep are not overcrowded. What are the possible dimensions of the pen? 14 The surface area of a cylinder with fixed height 3 cm is given by SAcyl = 2πr(r + 3). What values may r take if SAcyl is no greater than 8π cm2 ? ENRICHMENT 15 What range of values may x2 + 1 take if: b −1 < x ≤ 3 a 2<x<4 16 A business-owner leases a shop in a shopping centre. There is only one type of product sold by this business. a It is found that the number n of items sold each week varies directly with the difference between $84 and the price $p. That is, when p = 84 there are no items sold. When the price is $60 there are 192 items sold. Find n as a function of p. b The turnover T for the product is the number sold times the price. Find T as a function of p. c In order to stay in business, the turnover must be at least $11 800. What price may be charged for the product in order to stay in business? d Because of the business-owner’s lease agreement, if turnover is greater than $14 080 then extra rent must be paid. What prices can be charged without paying extra rent but still staying in business? 17 Square ABCD has side length L. The point E is on DA with DE = x. The point F is on AB with AF = x. a Draw a sketch of the situation and state any restrictions on x. b Find the area of CEF in terms of x. c What values my x take if the area of the triangle is at most 38 L2 . d By completing the square or otherwise, and noting the restrictions on x, what values does the area of CEF lie between? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4C The discriminant 115 4C The discriminant Learning intentions • Understand the various ways in which the discriminant discriminates. • Apply the discriminant to solve problems. Chapter 3 established that the zeroes of a quadratic y = ax2 + bx + c are √ √ −b − Δ −b + Δ and x = , where Δ = b2 − 4ac. x= 2a 2a This section returns to the discriminant Δ = b2 − 4ac, and develops further methods of analysing quadratics using the power of the discriminant to discriminate. The discriminant tells us the number of zeroes The most obvious purpose of the discriminant is to find the zeroes, but even more importantly, it tells us how many zeroes there are. This is because: • Only positive numbers have two square roots. • Zero has just one square root, namely zero. • Negative numbers have no square roots. 4 The Zeroes ( x-intercepts) of the quadratic y = ax2 + bx + c Always find the discriminant Δ = b2 − 4ac before finding the zeroes. √ √ −b + Δ −b − Δ and x = . • If Δ > 0, there are two zeroes, x = 2a 2a b . • If Δ = 0, there is just one double zero, x = − 2a In this situation, the quadratic is a perfect square, and the x-axis is tangent to the graph. • If Δ < 0, there are no zeroes. We can say ‘equal zeroes’ when Δ = 0, and ‘distinct zeroes’ when Δ > 0. The discriminant tells us whether the zeroes are rational or irrational Suppose now that all three coefficients of y = ax2 + bx + c are rational numbers. Then because we need to take the square root of Δ, the zeroes will also be rational numbers if Δ is a square. But if Δ is not a square and Δ > 0, the zeroes will involve a surd and be irrational. 5 The discriminant and rational zeroes Suppose that y = ax2 + bx + c, where a, b, and c are rational. • If Δ is a square, then the zeroes are rational. • If Δ is positive and not a square, then the zeroes are irrational. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 116 4C Chapter 4 Equations and inequations Example 6 Using the discriminant to describe the zeroes of a quadratic Use the discriminant to describe the zeroes of each quadratic (do not solve). a y = 5x2 − 2x − 3 b y = 3x2 − 12x + 12 c y = 8 + 3x − 2x2 Solution y = 5x2 − 2x − 3, a For b For Δ = 22 − 4 × 5 × (−3) y = 3x2 − 12x + 12, Δ = 122 − 4 × 3 × 12 = 64, which is a square, so there are two rational zeroes. = 0, so there is one rational double zero. y = 8 + 3x − 2x2 , c For Δ = 32 − 4 × (−2) × 8 = 73, which is positive, but not a square, so there are two irrational zeroes. Example 7 Using the discriminant to investigate tangents Use the discriminant to find the values of m for which the line y = mx is a tangent to the circle (x − 10)2 + y2 = 25. Solution Solve the line and circle simultaneously by substituting the line into the circle: (x − 10)2 + m2 x2 = 25 (1 + m2 )x2 − 20x + 75 = 0 Δ = 202 − 4 × 75(1 + m2 ) = 100 − 300m2 The line is a tangent when Δ = 0, because then line and circle meet in just one point. Hence Example 8 100 − 300m2 = 0 √ √ m = 13 3 or − 13 3 Using the condition for a quadratic to have zeroes Find the values of k for which y = x2 + kx + 9 has zeroes. Solution Δ = k2 − 4 × 9 Here the discriminant is = k2 − 36. Δ≥0 The quadratic has zeroes when k − 36 ≥ 0 2 (k − 6)(k + 6) ≥ 0. The LHS has zeroes at x = 6 and x = −6, so we construct a table of test points: Hence k ≤ −6 or k ≥ 6. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 k −7 −6 0 6 7 (k − 6)(k + 6) 13 0 −36 0 13 sign + 0 − 0 + © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4C The discriminant Example 9 117 Using the condition for a quadratic to have zeroes For what values of λ does x2 − (λ + 5)x + 9 = 0 have: a equal roots? b no roots? Solution Here the discriminant is Δ = (λ + 5)2 − 36. Δ a Δ = 0 when (λ + 5) = 36 2 −11 λ + 5 = 6 or λ + 5 = −6 −5 1 λ λ = 1 or − 11, so there are equal roots when λ = 1 and when λ = −11. b There are no roots when Δ is negative, so from the graph of Δ as a function of λ, there are no roots for −11 < λ < 1. −11 Exercise 4C FOUNDATION 1 2 3 4 −36 In each part below, the discriminant of a quadratic function with rational coefficients is given. State whether there are any zeroes. If there are zeroes then state whether they are rational or irrational, and whether they are equal or are distinct. a Δ=2 b Δ = −2 c Δ=0 d Δ=1 e Δ = −3 f Δ=4 Find the discriminant Δ for each equation. Hence state how many roots there are, and whether or not they are rational. a x2 − 4x + 3 = 0 b x2 + 2x − 7 = 0 c x2 + 3x + 4 = 0 d 4x2 + 4x + 1 = 0 e x2 − 3x + 5 = 0 f 6x2 + 11x − 10 = 0 In each case the discriminant Δ of the quadratic function involves the pronumeral k. Put Δ = 0 in order to find the values of k for which the quadratic has exactly one zero. a y = x2 + 10x + k b y = kx2 − 4x + 1 c y = 2x2 − 3x + (k + 1) d y = (k − 2)x2 + 6x + 1 In each quadratic equation the discriminant Δ involves the pronumeral p. Solve Δ ≥ 0 to find the values of p for which the roots are real numbers. a x2 + 2x + p = 0 b px2 − 8x + 2 = 0 c 3x2 − 4x + (p + 1) = 0 d (2p − 1)x2 − 5x + 2 = 0 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 118 4C Chapter 4 Equations and inequations 5 In each equation the discriminant Δ of the quadratic involves λ. Solve Δ < 0 to find the values of λ for which each quadratic equation has no zeroes. a x2 + 6x + λ = 0 b λx2 − 10x + 1 = 0 c 2x2 − 5x + (λ + 3) = 0 d (λ − 1)x2 + 6x + 3 = 0 DEVELOPMENT 6 Determine the discriminant Δ, then solve Δ = 0 to find the values of g for which each quadratic function has exactly one zero. a y = gx2 − 8x + g b y = 4x2 + 4gx + (6g + 7) c y = 9x2 − 2(g + 1)x + 1 d y = (2g − 3)x2 + (5g − 1)x + (3g + 2) 7 A parabola has equation y = gx2 − gx + 1. a Determine the discriminant. b Find the values of g which make Δ = 0. c Explain why the lesser value of g is not valid in this case. d Write down the equation of the parabola that has a double zero. 8 For each quadratic equation, find the discriminant Δ as a function of k. Then solve Δ ≥ 0 to determine the values of k that result in real roots for x. (Hint: Either a graph of Δ versus k or a table of test points may help.) a x2 + kx + 4 = 0 b x2 − 3kx + 9 = 0 c x2 − (3 + k)x + 1 = 0 d x2 + (k − 2)x + 1 = 0 9 In each case, find the discriminant Δ and solve Δ < 0 to find any values of λ for which the parabola has no x-intercepts. (Hint: Either a graph of Δ versus λ or a table of test points may help.) a y = x2 + λx + 1 b y = λx2 + 6x + λ c y = x2 + (1 − λ)x − λ d y = λx2 − 2λx − (λ − 2) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4C The discriminant 10 119 Questions 8 and 9 used the relationship between the discriminant and the zeroes of the quadratic y = ax2 + bx + c, as given in Box 4. Here you will prove those results. a Rearrange the equation of the parabola to show that b y−c x= , a a then add an appropriate quantity to both sides in order to complete the square on the left hand side. b Explain why there are no zeroes when Δ < 0. That is, explain why there is no solution if y = 0 and b2 − 4ac < 0. c Explain why there is a double zero when Δ = 0. That is, explain why there is one solution if y = 0 and b2 − 4ac = 0. d i Explain why there are two zeroes when Δ > 0. That is, explain why there are two solutions if y = 0 and b2 − 4ac > 0. ii Determine the formula for the zeroes in this case. x2 + 11 a Use the discriminant of a suitable quadratic equation to determine whether or not the parabola y = x2 + x + 1 intersects the line y = 7x − 6. If they do intersect then how many points are there and what are the x-coordinates? b Use a similar method to investigate these parabolas and straight lines. i y = x2 − 2x + 7 y = 4x − 2 ii y = x2 + 3x + 6 y = −x + 1 12 a Use the discriminant of a suitable quadratic equation to determine whether or not the parabola y = x2 + 5x + 1 intersects the parabola y = −x2 + x − 2. If they do intersect then how many points are there and what are the x-coordinates? b Use a similar method to investigate these parabolas lines. i y = x2 + 3x + 1 y = −x2 − 5x − 1 ii y = 12 x2 − x + 2 y = − 12 x2 − 3x + 1 13 a Consider the circle x2 + y2 = 4 and the line y = x + 1. i Use substitution to obtain a quadratic equation in x. ii Find the discriminant for this quadratic and hence state how many times the line intersects the circle. (Do NOT solve the quadratic equation.) b Likewise, determine how many times the line intersects the circle in each case. i x2 + y2 = 1, y = x + 2 ii x2 + y2 = 5, y = −2x + 5 14 a Use substitution to show that y = x + b and y = 2x2 − 7x + 4 intersect when 2x2 − 8x + (4 − b) = 0. b Solve Δ = 0 to find the value of b for which the line is a tangent to the curve. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 120 4C Chapter 4 Equations and inequations 15 The straight line function y = mx passes through the origin and is a secant of the circle x2 + (y − 2)2 = 1, intersecting it twice. a What values may m take? b There is one secant of the circle which passes through the origin but is not a function. Write down its equation. 16 The line y = mx + b is tangent to the parabola y = x2 . a Show that b = − 14 m2 . b Hence find the coordinates of the point of contact in terms of m. 17 There are two lines y = mx + b that pass through (1, −3) and are tangent to y = x2 . Find the equations of these lines. 18 The line y = mx + b passes through (−4, −4). What values may m take if this line does not intersect the −9 ? hyperbola y = x 19 a Prove that the roots of each equation are real and distinct for all real values of λ. (Hint: Find Δ and write it in such a way that it is obviously positive.) i x2 + λx − 1 = 0 ii x2 + (λ + 1)x + (λ − 2) = 0 b In the case of part a i, use the concavity and y-intercept of the associated parabola to give an alternative explanation. ENRICHMENT 20 [Other Methods] Each problem in Question 13 could be easily solved by graphing the line and circle. How could parts a and b i be solved using geometry? 21 Let p and q be integers with p + q 0. Show that the quadratic equation (p + q)2 x2 + 2pqx − 14 (p − q)2 = 0 always has distinct rational solutions for x. 22 A certain quadratic equation with rational coefficients has an irrational root. Explain why there must be a second root, and why it must be irrational. 23 The quadratic equation ax2 + bx + c = 0 has integer coefficients. a Suppose that the roots are rational. Explain why 4ac is the difference of two squares. b Now suppose that the roots are integers. By considering separately the sum and product of the roots, explain why both b and c are divisible by a. 24 Consider the quadratic equation λx2 − (λ + b)x + c = 0 where both b and c are real numbers with b > 0. It is known that the equation has two roots for all real values of λ 0. Show that 0 < c < b. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4D Quadratic identities 121 4D Quadratic identities Learning intentions • Distinguish between identities and equations. • Show that the graph of a quadratic determines its quadratic coefficients. • Find coefficients of quadratics, given sufficient data. This section continues the attention on quadratics, and particularly on quadratic identities and the coefficients. Identities and equations When we are doing algebra, we are constantly using identities, such as: 3 3x2 − 2 2 , for x 0 − 3 = x x x3 These are algebraic statements true for all values of x (apart from troublesome values of x that we specifically exclude, or are understood to be excluded). (x − 5)2 = x2 − 10x + 25 and Compare these with equations, which have solutions: the values of x for which they are true. For example: 3x + 4 = 7 has solution x = 1 6 and x2 = 49 has solutions x = 7 and x = −7. Identities and equations • A statement of equality between two algebraic expressions is called an identity if it is true for all values of the variables (apart from troublesome values that we specifically exclude, or are understood to be excluded). We solve equations, and we prove identities. • Any statement of equality between two algebraic expressions is called an equation, and an equation may or may not have one or more solutions. Thus we prove identities, and we solve equations. The formula for a quadratic is unique We have assumed so far that any quadratic function f (x) has exactly one expression in the form f (x) = ax2 + bx + c, and that if you change the coefficients, you change the function. This is actually a theorem that needs to be proven: Theorem. Suppose that two quadratics are equal for all values of x, that is, ax2 + bx + c = Ax2 + Bx + C, for all values of x, (∗) where a, b, c, A, B, and C are constants. Then a = A, b = B, and c = C. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 122 4D Chapter 4 Equations and inequations Proof We only need to substitute three values of x into (∗) to prove the result. Substitute x = 0, then c=C (1) Substitute x = 1, then a+b+c = A+ B+C (2) Substitute x = −1, then a − b + c = A − B + C (3) Substituting (1) into (2) and (3), and cancelling: a+b= A+B (2A) a−b= A−B (3A) Then adding and subtracting (2A) and (3A) gives a = A and b = B, as required. Application of this theorem to finding coefficients Because of this theorem, if two quadratic expressions are equal, there are now two ways of generating equations for finding unknown coefficients. 7 Two methods of generating equations for finding coefficients • Equate coefficients of like terms. • Substitute carefully chosen values of x. We actually used the second method three times in the proof of the theorem above. Now that theorem justifies the first method. Example 10 Finding coefficients of a quadratic function a Express n2 as a quadratic in (n − 3). b Find r, s, and t if 3(x − 1)2 + 2(x − r) + 6 = sx2 + tx + 7, for all x. Solution n2 = a(n − 3)2 + b(n − 3) + c a Let Equate coefficients of n2 , 1 = a, (1) substitute n = 3, then n − 3 = 0, so 9 = c, (2) substitute n = 0, 0 = 9a − 3b + c. (3) Now substituting (1) and (2) into (3), 0 = 9 − 3b + 9 b = 6, n2 = (n − 3)2 + 6(n − 3) + 9. so b We know that for all x, 3(x − 1)2 + 2(x − r) + 6 = sx2 + tx + 7. 3 = s, Equate coefficients of x2 , substitute x = 0, 3 − 2r + 6 = 7 r = 1, substitute x = 1, (1) 0 + 2(1 − r) + 6 = s + t + 7. Now substituting (1) and (2) into (3), (2) (3) 6=3+t+7 t = −4. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 4D Quadratic identities Exercise 4D 1 123 FOUNDATION It is known that a(x + 2) + b(x − 1) = 5x + 1 for all values of x. a Substitute x = 1 to find a. b Substitute x = −2 to find b. c Why were the values x = 1 and x = −2 chosen? d Equate the coefficients of like terms to get a pair of simultaneous equations in a and b, then solve them. Do you get the same answer? 2 It is given that a(x + 3) + b(x − 2) = 3x − 1 for all values of x. a Substitute x = 2 to find a. b Substitute x = −3 to find b. c Why were the values x = 2 and x = −3 chosen? d Equate the coefficients of like terms to get a pair of simultaneous equations in a and b, then solve them. Do you get the same answer? 3 Let 2x2 − 5x + 3 = a(x − 2)2 + b(x − 2) + c. a Equate the coefficients of x2 to find the value of a. b Substitute x = 2 into the identity to find the value of c. c Substitute x = 3 into the identity to find the value of b. d Why was the value x = 3 chosen in part c? 4 Suppose that x2 + x + 1 = a(x − 1)2 + b(x − 1) + c for all values of x. a Equate the coefficients of x2 to find the value of a. b Substitute x = 1 into the identity to find the value of c. c Substitute x = 2 into the identity to find the value of b. d Why was the value x = 2 chosen in part c? 5 The quadratic f (x) = a(x + 1)2 + b(x + 1) + c has the same graph as g(x) = 2x2 + 3x − 1. a Compare the coefficients of x2 to find the value of a. b Substitute x = −1 to find the value of c. c Substitute another suitable value of x to determine b. d Write down the resulting quadratic identity. DEVELOPMENT 6 Use similar methods to Questions 3 and 4 to solve these problems. a Find a, b and c if n2 − n = a(n − 4)2 + b(n − 4) + c for all values of n. b Find a, b and c if 2x2 + 4x + 5 = a(x − 3)2 + b(x − 3) + c for all values of x. 7 Express x2 in the form: a A(x + 1)2 + B(x + 1) + C 8 b A(x − 4)2 + B(x − 4) + C c A(x + 2)2 + B(x + 2) + C a Express 2x2 + 3x − 6 in the form A(x + 1)2 + B(x + 1) + C. b Express 3x2 − 2x − 1 in the form p(x − 1)2 + q(x − 1) + r. 9 Determine the equation of the parabola y = ax2 + bx + c that passes through A(−1, 0), B(0, 4) and C(1, 6). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 124 4D Chapter 4 Equations and inequations 10 Find the values of a, b and c for which x2 = a(x − 1)2 + b(x − 2)2 + c(x − 3)2 . 11 Find the unknown constants if P(x) = ax2 + b(x + 1)2 + c(x + 2)2 and Q(x) = −x2 + 3x have the same graph. 12 a Find A and B such that Am2 + B(m − 1)2 = 2m − 1. b Hence find the sum of 1 + 3 + 5 + · · · + 61. 13 Let 13x2 − 7x + 6 = a(2 − x)2 + b(2 − x) + c. a Evaluate the constants a, b and c. b Hence evaluate 13x2 − 7x + 6 when x = 1.98 without using a calculator. ENRICHMENT 14 a Find A and B such that A(m + 1)2 + B(m − 1)2 = 4m. b Hence find the sum of the first eleven multiples of 4. 15 Let p, q and r be distinct real numbers. Substitute three appropriate values of x in order to show that (p − x)(q − x) (q − x)(r − x) (r − x)(p − x) + + =1 (p − r)(q − r) (q − p)(r − p) (r − q)(p − q) for all values of x. 16 [Another Approach] 2 a Express x2 as a quadratic in (x − a). Begin by writing x2 = (x − a) + a , then expand only the outer brackets of the RHS. b Use the formula you found in part a to quickly find x2 as a quadratic in: ii (x + 3) i (x − 1) iii (x − 11) c The formula can be used to express 2x2 − 3x + 4 in terms of (x − 3) as follows: 2x2 − 3x + 4 = 2 (x − 3)2 + 6(x − 3) + 9 − 3 (x − 3) + 3 + 4 = 2(x − 3)2 + (2 × 6 − 3)(x − 3) + (2 × 9 − 3 × 3 + 4) = 2(x − 3)2 + 9(x − 3) + 13 Redo Question 8 using this method. 17 The calculator uses several clever tricks in order to evaluate the square-root of a number. One of the tricks √ employed approximates y = x with a concave down parabola over the interval [ 12 , 1]. a Let the parabola be y = ax2 + bx + c. Evaluate the constants a, b and c so that the parabola passes through √ P( 12 , √12 ), Q( 34 , 23 ) and R(1, 1). b Use this parabola to estimate c Use √ 3 5 in terms of √ √ 2 and 3. √ 2 1.41421 and 3 1.73205 to obtain a decimal approximation correct to four decimal places. Then compare this answer with the value of CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 3 5 given by your calculator. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 4 review 125 Chapter 4 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 4 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise 2 Review 1 Solve each linear inequation, and graph your solution on a number line. a 2x + 5 < 11 b 3x + 2 > 17 c 4 − 3x ≥ 16 d 2 − 3x ≤ 5 e 6 − x < 3x − 2 f 7x + 5 ≤ 3x + 1 Solve each double inequation then give your answer using bracket interval notation. b 1 ≤ 2x + 3 < 11 a 4< x+1≤9 c −1 < 5 − 3x < 14 3 For what values of x does the line y = 12 x + 1 lie between y = −1 and y = 2? 4 Find the discriminant Δ of each equation. Hence state how many zeroes there are, and whether or not they are rational. a x2 − 6x + 9 = 0 5 b −x2 − 4x + 2 = 0 c 9x2 − 1 = 0 The parabola y = x2 − 2x − 3 is sketched on the right. By considering where the curve is in relation to the x-axis, write down the values of x for which: a y=0 -1 b y>0 c y<0 6 y 3 x -3 Solve each quadratic inequation, using the graph provided. a x2 + 2x < 0 b x2 + 2x − 8 ≥ 0 y y 1 2 x -4 -2 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 -8 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 126 Chapter 4 Equations and inequations Review 7 8 Solve each quadratic inequation using a suitable sketch or table of values. a (x + 4)(x − 3) < 0 b (x − 1)(x + 5) ≥ 0 c x2 − 4x > 0 d x2 − 9 ≤ 0 e x2 − 4x − 12 ≥ 0 f 2x2 + x − 1 < 0 Find the values of k for which the quadratic equation x2 + 6x + k = 0 has: a real roots, 9 b equal roots, c unreal roots. Find the values of m for which the quadratic equation x2 − 3mx + 9 = 0 has: a real roots, b equal roots, c unreal roots. 10 Find the values of that will make the quadratic y = ( + 6)x2 − 2x + 3 a perfect square. 11 [Break-even point] A certain business has fixed costs of $10 500 and costs of $121 per item sold. Each item sells for $416. Using either a graph or suitable inequation, find the minimum number n of items that must be sold in order to make a profit. 12 A rectangle has perimeter 22 cm. Let one side have a length of x cm. a Write down an expression for the area A of this rectangle in terms of x. b What values may x take if A > 28 cm2 ? 13 A farmer needs to build a rectangular holding pen for some sheep. The side of a barn will be used for one side of the pen and there is 60 m of fencing available for the remaining three sides. Let x be the length of each side perpendicular to the barn wall. a Find an expression for the third side in terms of x and draw a diagram showing all this information. b The area A of the pen must be at least 400 m2 so that the sheep are not overcrowded. What is the smallest value of x for these requirements? 14 a Find the values of a, b and c so that 3x2 − 5x + 7 = a(x − 1)2 + b(x − 1) + c for all x. b Hence evaluate 3x2 − 5x + 7 when x = 1.12 without using a calculator. 15 Determine the values of p, q and r so that 3x2 − 2x − 5 = px2 + q(x − 1)2 + r(x − 2)2 for all values of x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5 Transformations and symmetry Chapter introduction In the previous two chapters, various graphs of functions and relations were introduced or reviewed. Sections 5A–5D of this chapter deal with transformations of these graphs under horizontal and vertical translations, reflections in the y-axis and the x-axis, rotations of 180◦ about the origin, and horizontal dilations from the y-axis and vertical dilations from the x-axis. These procedures allow a wide variety of new graphs to be obtained quickly and simply, and relationships amongst different graphs to be discovered. Many graphs are unchanged under one or more of these transformations, which means that they are symmetric in some way. Section 5C deals with line symmetry under reflection in the y-axis, and point symmetry under rotation of 180◦ about the origin. All these transformations and symmetries are described geometrically and algebraically — the theme remains, as always, the interrelationship between the algebra and the graphs. Section 5E introduces the absolute value function together with its own transformations and reflection symmetry. Section 5F generalises the transformations earlier in this chapter to composite functions, and Section 5G examines the composition of transformations rather than of functions. The final Section 5H introduces continuity at a point and its importance with piecewise-defined functions. As always, computer sketching of curves can demonstrate quickly how the geometric features of a graph are related to the algebraic properties of its equation, and can help in becoming familiar with the variety of graphs and their relationships. The reader should apply any curve-sketching software particularly to transformations. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 128 5A Chapter 5 Transformations and symmetry 5A Translations of known graphs Learning intentions • Transform an equation so that its graph shifts vertically or horizontally. • Show that the order of performing a vertical and a horizontal translation is irrelevant. • Apply vertical and horizontal translations particularly to parabolas and circles. Once a graph is drawn, it can be shifted (or more formally translated ) vertically or horizontally to produce further graphs. These procedures work generally on all functions and relations, and greatly extends the range of functions and relations whose graphs can be quickly recognised and drawn. In particular, translations are particularly helpful when dealing with parabolas and circles, where they are closely related to completing the square. Shifting right and left The graphs of y = x2 and y = (x − 2)2 are sketched from their tables of values. y x x2 (x − 2) 2 −2 −1 0 1 2 3 4 4 1 0 1 4 9 16 16 9 4 1 0 1 4 The values for (x − 2)2 in the third row are the values of x2 in the second row shifted 2 steps to the right. 1 1 2 x Hence the graph of y = (x − 2)2 is obtained by shifting the graph of y = x2 to the right by 2 units. 1 Shifting (or translating) right and left • To shift a graph a units right, replace x by x − a. • Thus for a function y = f (x), the new function rule is y = f (x − a). Shifting a units left means shifting −a units right, so replace x by x − (−a) = x + a. Thus for a function y = f (x), the new function rule is y = f (x + a). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5A Translations of known graphs Example 1 129 Shifting (or translating) a graph right and left 1 1 and y = . x x+1 b Sketch the two graphs, and state the asymptotes of each graph. 1 1 ? c What transformation maps y = to y = x x+1 a Draw up tables of values for y = Solution −3 x a −2 −1 0 1 2 y 3 3 1 1 1 1 1 − − −1 ∗ 1 x 3 2 2 3 1 1 1 1 1 −1 ∗ 1 − x+1 2 2 3 4 1 b y = has asymptotes x = 0 and y = 0. x 1 has asymptotes x = −1 and y = 0. y= x+1 c Because x is replaced by x + 1 = x − (−1), it is a shift left of 1 unit. -3 -1 1 3 1 -1 x -3 Shifting up and down The graphs of y = x2 and y = x2 + 1 are sketched from their tables of values: x −3 −2 −1 0 1 2 3 x2 9 4 1 0 1 4 9 x2 + 1 10 5 2 1 2 5 10 y 2 1 • The values for x2 + 1 in the third row are each 1 more than the corresponding values of x2 in the second row. • Hence the graph of y = x2 + 1 is produced by shifting the graph of y = x2 upwards 1 unit. −1 1 x Rewriting the transformed graph as y − 1 = x2 makes it clear that the shifting has been obtained by replacing y with y − 1, giving a rule that is completely analogous to that for horizontal shifting: 2 Shifting (or translating) up and down • To shift a graph b units upwards, replace y with y − b. • For a function y = f (x), the new function rule is y − b = f (x), or y = f (x) + b. Shifting b units down means shifting −b units up, so replace y by y − (−b) = y + b. Thus for a function y = f (x), the new function rule is y + b = f (x), or y = f (x) − b. Example 2 Shifting (or translating) a graph up and down The graph of y = 2 x is shifted down 2 units. a Write down the equation of the shifted graph. b Construct tables of values, and sketch the two graphs. c State the asymptotes of the two graphs, and their domains and ranges. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 130 5A Chapter 5 Transformations and symmetry Solution a Replace y with y − (−2) = y + 2, so the new function is y + 2 = 2x , b y 4 y = 2x − 2 . that is, x −2 −1 0 1 2 2x 1 4 1 2 1 2 4 2x − 2 −1 34 −1 12 −1 0 2 2 1 -2 -1 x 1 2 -1 -2 c y = 2 has asymptote y = 0. Domain: all real x, range: y > 0. x y = 2 x − 2 has asymptote y = −2. Domain: all real x, range: y > −2. Combining horizontal and vertical translations When a graph is shifted horizontally and vertically, the order in which the translations are applied makes no difference. This is because: If y = f (x) is shifted a units right, then b units up, y = f (x) −→ y = f (x − a) −→ y − b = f (x − a), and if y = f (x) is shifted b units up, then then a units right, y = f (x) −→ y − b = f (x) −→ y − b = f (x − a). 3 The order of two perpendicular translations does not matter • If two perpendicular translations are applied to a graph, order does not matter. Example 3 Combining two translations of a parabola a How is the graph of y = (x + 2)3 − 4 obtained from the graph of y = x3 by a horizontal translation followed by a vertical translation? b Draw up a table of values for the two functions and the intermediate function. c Sketch the two curves, together with the intermediate graph. d Perform this by a vertical translation followed by a horizontal translation. Solution a Shifting y = x3 left 2 gives y = (x + 2)3 . Shifting y = (x + 2)3 down 4 gives b which can be written as x −4 −3 y c y + 4 = (x + 2)3 , −2 −1 y = (x + 2) − 4. 0 1 2 8 3 4 -4 -2 −64 −27 −8 −1 0 1 8 −8 −1 0 1 8 27 64 -4 (x + 2) − 4 −12 −5 −4 −3 4 23 60 -8 x3 (x + 2) 3 3 2 x d Translate y = x3 down 4. The result is y + 4 = x3 , which is y = x3 − 4. Then translate left 2. The result is the same: y = (x + 2)3 − 4. (We have already shown above in Box 3 that order is not important here.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5A Translations of known graphs Translations and the vertex of a parabola 131 y When we complete the square in a quadratic, it has the form y = a(x − h)2 + k, y − k = a(x − h)2 . that is, This is a translation of the quadratic y = ax2 . The parabola has been shifted h units right and k units up. k 4 x h This gives a clear and straightforward motivation for completing the square. The completed square and the vertex of a parabola • The completed square form of a quadratic y = a(x − h)2 + k or y − k = a(x − h)2 displays its graph as the parabola y = ax2 shifted right h units and up k units. • Thus the vertex of the parabola is (h, k). Example 4 Shifting y = x2 to any parabala In each part, complete the square in the quadratic. Then identify its graph as a translation of a parabola with vertex at the origin. Sketch its graph, and state its range. a y = x2 − 4x + 5 b y = −2x2 − 4x Solution a y = x2 − 4x + 5 y y = (x2 − 4x + 4) − 4 + 5 y = (x − 2)2 + 1, that is, y − 1 = (x − 2)2 . This is y = x2 shifted right 2 and up 1. Range: y ≥ 1. 5 1 x 2 y = −2x2 − 4x b y − 2y = x2 + 2x ÷ (−2) -2 − 2y = (x2 + 2x + 1) − 1 − 2y = (x + 1)2 − 1 2 y = −2(x + 1) + 2 × (−2) 2 -1 or x y − 2 = −2(x + 1)2 This is y = −2x2 shifted left 1 and up 2. Range: y ≤ 2. Translations and the centre of a circle The circle drawn to the right has centre (3, 2) and radius 3. To find its equation, we start with the circle with centre the origin and radius 3, y 5 2 x + y = 9, 2 2 3 then translate it 3 to the right and 2 up to give −1 x 6 (x − 3)2 + (y − 2)2 = 9 . This formula can also be established directly by Pythagoras’ theorem in the form of the distance formula, but as we saw with parabolas, translations make things clearer and more straightforward. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 132 5A Chapter 5 Transformations and symmetry Completing the squares to identify the translations When the equation of a circle is presented with any binomial squares expanded, the centre and radius can be found by completing the squares in x and in y. Example 5 Shifting any circle a Complete the squares in x and in y in the relation x2 + y2 − 6x + 8y = 0. b Identify the circle with centre at the origin that can be translated to the relation from part a, and state the translations. c Sketch both circles on the same diagram, and explain why each circle passes through the centre of the other circle. d Write down the domain and range of each circle. Solution a Completing the squares in x and in y, (x2 − 6x + 9) + (y2 + 8y + 16) = 9 + 16 (x − 3)2 + (y + 4)2 = 25. b It is the circle x2 + y2 = 52 shifted right 3 and down 4, so its centre is Z(3, −4) and y its radius is 5. c Using the distance formula, -2 1 OZ 2 = 32 + 42 -4 OZ = 5, O3 6 8 x Z -8 which is the radius of each circle. d For the circle with centre the origin, domain: −5 ≤ x ≤ 5, range: −5 ≤ y ≤ 5. For the other circle, domain: −2 ≤ x ≤ 8, range: −9 ≤ y ≤ 1. -9 Exercise 5A FOUNDATION 1 a Copy and complete the table of values for y = x2 and y = (x − 1)2 . x x −2 −1 0 1 2 3 2 (x − 1)2 b Sketch the two graphs and state the vertex of each. c What transformation maps y = x2 to y = (x − 1)2 ? 2 a Copy and complete the table of values for y = 14 x3 and y = 14 x3 + 2. x −3 −2 −1 0 1 2 3 1 3 4x 1 3 4x + 2 b Sketch the two graphs and state the y-intercept of each. c What transformation maps y = 14 x3 to y = 14 x3 + 2? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5A Translations of known graphs 3 1 In each case, the given equation is the result of shifting one of the curves y = x2 , y = or y = 2 x . State the x direction of the shift and by how much. Use this information to help sketch the curve without resorting to a table of values. a y = x2 + 2 b y = (x + 1)2 1 x−2 e y = 2 x−1 d y= 1 +1 x f y = 2x − 2 c y= 4 5 Sketch each circle by shifting x2 + y2 = 1 either horizontally or vertically. Mark all intercepts with the axes. a (x − 1)2 + y2 = 1 b x2 + (y − 1)2 = 1 c x2 + (y + 1)2 = 1 d (x + 1)2 + y2 = 1 Write down the new equation for each function or relation after the given translation has been applied. Then sketch the graph of the new curve. a y = x2 : right 2 units b y = 2 x : down 2 units c y = x3 : left 1 unit d y= 1 : right 3 units x f y = x2 − 4: left 1 unit e x2 + y2 = 4: up 1 unit 6 133 In each case an unknown function has been drawn. Draw the functions specified below each graph on the same axis. a b y 1 −2 y y = f (x) y = P(x) 1 2 2x −1 −1 1 −2 −1 2 x 1 i y = f (x − 2) i y = P(x + 2) ii y = f (x + 1) ii y = P(x + 1) c y d y = h(x) 1 −1 y 1 2 x y = g(x) 1 −1 1 x −1 i y − 1 = h(x) i y − 1 = g(x) ii y = h(x) − 1 ii y = g(x − 1) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 134 5A Chapter 5 Transformations and symmetry DEVELOPMENT 7 8 In each part, complete the square in the quadratic. Then identify its graph as a translation of either y = x2 or y = −x2 . Finally, sketch its graph. a y = x2 + 2x + 3 b y = x2 − 2x − 2 c y = −x2 + 4x + 1 d y = −x2 − 4x − 5 e y = x2 − 2x − 1 f y = x2 − 2x − 4 Describe each graph below as the parabola y = x2 or the hyperbola xy = 1 transformed by shifts, and hence write down its equation: y a b 4 y 1 -1 3 10 g 12 3 h y x x (2, -1) 1 x -3 x 3 x x y -2 -1 -1 1 y -1 1 2 9 2 3 3 x -1 f y d -2 1 x 2 y e y c -4 Use shifting, and completion of squares where necessary, to determine the centre and radius of each circle. a (x + 1)2 + y2 = 4 b (x − 1)2 + (y − 2)2 = 1 c x2 − 2x + y2 − 4y − 4 = 0 d x2 + 6x + y2 − 8y = 0 e x2 − 10x + y2 + 8y + 32 = 0 f x2 + 14x + 14 + y2 − 2y = 0 Describe each graph below as the circle x2 + y2 = r2 transformed by shifts, and hence write down its equation. a y y b -2 1 3 2 x x 2 y c -1 d 4 y x (-1, 1) x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (2, -1) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5A Translations of known graphs 11 135 a Use a table of values to sketch y = 14 x3 . Then use translations to sketch each graph below. i y = 14 x3 − 2 ii y = 14 (x − 2)3 iii y = 14 (x + 3)3 + 1 b Use a table of values to sketch y = −2x3 . Then use translations to sketch each graph below. i y = 3 − 2x3 12 ii y = −2(x + 3)3 iii y = −2(x − 1)3 − 2 Consider the straight line equation x + 2y − 4 = 0. a The line is translated 2 units left. Find the equation of the new line. b The original line is translated 1 unit down. Find the equation of this third line. c Comment on your answers. Drawing the situation may help. 13 14 1 Sketch y = , then use shifting to sketch the following graphs. Find any x-intercepts and y-intercepts, and x mark them on your graphs. 1 1 a y= b y=1+ x−2 x−2 1 1 c y= d y= −2 −1 x−2 x+1 1 1 e y=3+ f y= +4 x+2 x−3 In each part, explain how the graph of each subsequent equation is a translation of the first graph (there may be more than one answer), then sketch each function. a From y = 2x: i y = 2x + 4 ii y = 2x − 4 b From y = x2 : i y = x2 + 9 ii y = x2 − 9 iii y = (x − 3)2 ii y = −(x + 1)2 iii y = −(x + 1)2 + 2 c From y = −x2 : i y = 1 − x2 d From y = √ √ x: x+4 2 e From y = : x 2 i y= +1 x i y= 15 ii y = ii y = √ x+4 iii y = 2 x+2 iii y = √ x+4−2 2 +1 x+2 Complete squares, then sketch each circle, stating the centre and radius. By substituting x = 0 and then y = 0, find any intercepts with the axes. a x2 − 4x + y2 − 10y = −20 b x2 + y2 + 6y − 1 = 0 c x2 + 4x + y2 − 8y = 0 d x2 − 2x + y2 + 4y = 1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 136 5A Chapter 5 Transformations and symmetry ENRICHMENT 16 a The circle x2 + y2 = r2 has centre the origin and radius r. This circle is shifted so that its centre is at C(h, k). Write down its equation. b The point P(x, y) lies on the circle with centre C(h, k) and radius r. That is, P lies on the shifted circle in part a. This time use the distance formula for the radius PC to obtain the equation of the circle. 17 a Explain the point-gradient formula y − y1 = m(x − x1 ) for a straight line in terms of a translation of the line y = mx. b Hence explain why parallel lines must have the same gradient. 18 Suppose that the graph of y = f (x) has been drawn. Let H be a shift to the right by a units, and let V be a shift upwards by b units. a Write down the equations of the successive equations obtained by applying H then V. b Write down the equations of the successive equations obtained by applying V then H. c What do you conclude about the order of applying horizontal and vertical shifts? 19 A certain circle has equation (x − 1)(x + 3) + (y − 2)(y − 4) = 0. a From this equation, write down four points that lie on the circle. b What shape do these four points form? Give its dimensions. c Hence find the centre and radius of the circle. d Confirm your answer to part c by expanding, then completing the squares. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5B Reflection in the y-axis and x-axis 137 5B Reflection in the y-axis and x-axis Learning intentions • Transform an equation so that its graph is reflected in the y-axis or x-axis. • Show that the order of performing reflections in the x-axis and y-axis is irrelevant. • Apply these reflections to familiar curves. Reflecting only in the y-axis and the x-axis may seem an unnecessary restriction, but in fact these two transformations are the key to understanding many significant properties of functions, particularly the symmetry of graphs. When these two reflections are combined, they produce a rotation of 180◦ about the origin, which again is the key to the symmetry of many graphs. Reflection in the y-axis The graphs of y = 2 x and y = 2−x are sketched from their tables of values: x −3 −2 −1 0 1 2 3 x 1 8 1 4 1 2 1 2 4 8 2−x 8 4 2 1 1 2 1 4 1 8 2 y 2 1 • The second and third rows are the reverse of each other. • Hence the graphs are reflections of each other in the y-axis. 5 −1 1 x Reflection in the y-axis • To reflect a graph in the y-axis, replace x with −x. • Thus for a function y = f (x), the new function rule is y = f (−x). Reflection in the y-axis is mutual — it maps each graph to the other graph. Example 6 Reflecting in the y-axis a Sketch the parabola y = (x − 2)2 , and on the same set of axes, sketch its reflection in the y-axis. b Use the rule in Box 5 above to write down the equation of the reflected graph. c Why can this equation be written as y = (x + 2)2 ? d What are the vertices of the two parabolas? e What other transformation moves the first parabola to the second, and why? Solution y a b Replacing x with −x gives the equation y = (−x − 2)2 . 4 c Taking out the factor −1 from the brackets, y = (−x − 2)2 -2 2 x y = (−1)2 × (x + 2)2 y = (x + 2)2 . Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 138 5B Chapter 5 Transformations and symmetry d The vertices are (2, 0) and (−2, 0). e The second parabola is also the first parabola shifted left 4 units. This replaces x with x + 4, so the new equation is the same as before, that is, y = (x + 2)2 . y = (x + 4) − 2 2 , The reason why there are two possible transformations is that every parabola has line symmetry in its axis of symmetry. Reflection in the x-axis The graphs of y = 2 x and y = −2 x are sketched from their tables of values. x −3 −2 −1 0 1 2 3 x 1 8 − 18 1 4 − 14 1 2 − 12 1 2 4 8 −1 −2 −4 −8 2 −2 x y 2 1 x −2 • The values in the second and third rows are the opposites of each other. • Hence the graphs are reflections of each other in the x-axis. Rewriting the transformed graph as −y = 2 x makes it clear that the reflection has been obtained by replacing y with −y, giving a rule that is completely analogous to that for reflection in the y-axis. 6 Reflection in the x-axis • To reflect a graph in the x-axis, replace y with −y. • For a function y = f (x), the new function rule is −y = f (x), or y = − f (x). Again, reflection in the x-axis is mutual — it maps each graph to the other graph. Example 7 Reflecting in the x-axis The graph of y = (x − 2)2 is reflected in the x-axis. a Write down the equation of the reflected graph. b Construct tables of values, and sketch the two graphs. c What are the vertices of the two parabolas? d What are their domains and ranges? Solution a Replace y with −y, so the new function is b −y = (x − 2)2 , that is, y = −(x − 2)2 . x (x − 2) 2 0 1 2 3 4 4 1 0 1 4 −(x − 2) −4 −1 0 −1 −4 c They both have vertex (2, 0). d For y = (x − 2)2 , domain: all real x, range: y ≥ 0. For y = −(x − 2)2 , domain: all real x, range: y ≤ 0. y 4 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 2 x -4 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5B Reflection in the y-axis and x-axis 139 Combining the two reflections — rotating 180◦ about the origin • Draw an x-axis and y-axis on a thin, semi-transparent sheet of paper. • Hold the paper out flat, and regard it as a two-dimensional object. • Reflect in the x-axis — do this by holding the sheet steady at the two ends of the x-axis and rotating it 180◦ so that you are now looking at the back of the sheet. Then reflect in the y-axis — do this by holding the sheet at the two ends of the y-axis and rotating it 180◦ so that you are looking at the front of the sheet again. What has happened? • Reflect it in the y-axis, then in the x-axis. What happens? y y 2 Reflect in the y-axis x tsrif 2 tnardauq → x 2 2- -2 2- the x-axis Reflect in Reflect in the x-axis ↓ ↓ -2 Reflect in the y-axis → quadrant first x first quadrant 2 x 2 2 -2 -2 y 2 -2 first quadrant -2 2 y This little experiment should convince you of two things: • Performing successive reflections in the x-axis and in the y-axis results in a rotation of 180◦ about the origin. • The order in which these two reflections are done does not matter. This rotation of 180◦ about the origin is sometimes called reflection in the origin, because every point in the plane is moved along a line through the origin to a point the same distance from the origin on the opposite side. 7 Rotation of 180◦ about the origin • To rotate a graph 180◦ about the origin, replace x by −x and y by −y. • Successive reflections in the x-axis and the y-axis are the same as a rotation of 180◦ about the origin. • The order in which these two successive reflections are done does not matter. • Rotation of 180◦ about O is also called reflection in O, because every point is moved through O to a point the same distance from O on the opposite side. Rotation of 180◦ about O is also mutual — it maps each graph to the other graph. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 140 5B Chapter 5 Transformations and symmetry Example 8 Applying two perpendicular reflections √ x, deduce the graph of y = − −x using reflections. b What single transformation maps each graph to the other? c What are the domains and ranges of the two functions? a From the graph of y = Solution √ √ a The equation y = − −x can be rewritten as y √ −y = −x √ so the graph is obtained from the graph of y = x by successive reflections in the x-axis and the y-axis, where the reflections may be done in either order. b This is the same as rotation of 180◦ about the origin. √ c For y = x, domain: x ≥ 0, range: y ≥ 0. √ For y = − −x, domain: x ≤ 0, range: y ≤ 0. Exercise 5B 1 -1 1 1 -1 x FOUNDATION Consider the parabola y = x2 − 2x. a Show that when y is replaced with −y, the equation becomes y = 2x − x2 . b Copy and complete the table of values for y = x2 − 2x and y = 2x − x2 . −2 −1 0 1 2 3 4 x x − 2x 2 2x − x2 c Sketch the two parabolas and state the vertex of each. d What transformation maps y = x2 − 2x to y = 2x − x2 ? 2 Consider the hyperbola y = 2 . x−2 a Show that when x is replaced with −x the equation becomes y = − b Copy and complete the table of values for y = x 2 . x+2 2 2 and y = − . x−2 x+2 −4 −3 −2 −1 0 1 2 3 4 2 x−2 2 − x+2 ∗ ∗ c Sketch the two hyperbolas and state the vertical asymptote of each. d What transformation maps y = 3 2 2 to y = − ? x−2 x+2 a Sketch the graph of the quadratic function y = x2 − 2x − 3, showing the intercepts and vertex. b In each case, determine the equation of the result when this parabola is reflected as indicated. Then draw a sketch of the new function. i in the y-axis CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii in the x-axis iii in both axes © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5B Reflection in the y-axis and x-axis 4 141 a Sketch the graph of the exponential function y = 2−x , showing the y-intercept and the coordinates at x = −1, and clearly indicating the asymptote. b In each case, find the equation of the curve when this exponential is reflected as indicated. Then draw a sketch of the new function. ii in the y-axis i in the x-axis 5 iii in both axes a The graph of an unknown function y = f (x) has been drawn on the right. y i Which reflection produces y = f (−x)? Draw the graph of this function. ii Which reflection produces y = − f (x)? Draw the graph of this function. iii Compare the zeroes of f (x), f (−x) and − f (x). Which pair have the same y = f (x) 2 1 x 1 2 3 zeroes, and why? b The graph of an unknown function y = P(x) has been drawn on the right. y y = P(x) i Which transformation produces y = −P(x)? Draw the graph of this function. ii Which transformation produces y = −P(−x)? Draw the graph of this function. iii In this case, P(x) and −P(−x) both have the same zeroes. Why might that be? c The graph of an unknown function y = A(x) has been drawn on the right. 1 -1 y 2 i Which transformation produces y = A(−x)? Draw the graph of this function. ii Which transformation produces y = −A(−x)? Draw the graph of this function. iii Explain why the zeroes of all three functions are the same in this instance. iv What do you notice about the graphs of y = A(x) and y = −A(−x)? 6 2 x 1 -1 y = A(x) 1 1 -1 x -2 Consider the cubic function y = C(x) where C(x) = x3 . a Show that C(−x) = −C(x). Describe this relationship in terms of reflections. b Show that −C(−x) = C(x). Describe this relationship in terms of a rotation. 7 Write down the new equation for each function or relation after the given transformation has been applied. Then sketch the graph of the new curve. a y = x2 : reflect in the x-axis b y = x3 : reflect in the y-axis c y = 2 x : rotate by 180◦ d y = 2x − x2 : rotate by 180◦ e x2 + y2 = 9: reflect in the y-axis f y= 1 : reflect in the x-axis x DEVELOPMENT 8 Consider the hyperbola y = a Sketch this hyperbola. 1 − 1. x+2 b In each case, determine the reflection or rotation required to achieve the specified result. Then write down the equation of the new hyperbola and sketch it. i The vertical asymptote is unchanged, but the horizontal asymptote changes sign. ii The intercepts with the axes are positive. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 142 5B Chapter 5 Transformations and symmetry 9 a Sketch the circles (x − 3)2 + y2 = 4 and (x + 3)2 + y2 = 4. b What transformation maps each circle onto the other? c Confirm your answer by making an appropriate substitution into the first equation. d What translation maps the first circle onto the second? e Confirm your answer by making an appropriate substitution into the first equation. 10 Consider x2 + y2 = r2 , the circle with centre the origin and radius r. a Show that this equation is unchanged when reflected in either the x-axis or the y-axis. b Explain this result geometrically. 11 In each part, explain how the graph of each subsequent equation is a reflection of the first graph or a rotation of 180◦ , then sketch each one. a From y = 12 x + 1: i y = − 12 x + 1 ii y = − 12 x − 1 iii y = 12 x − 1 ii y = x + 4 iii y = −x − 4 ii y = (x + 1)2 iii y = −(x − 1)2 b From y = 4 − x: i y= x−4 c From y = (x − 1) : 2 i y = −(x + 1)2 d From y = √ x: √ i y = − −x √ √ −x ii y = − x iii y = ii y = −3−x iii y = 3−x e From y = 3 x : i y = −3 x 1 : x−1 1 i y=1− x+1 f From y = 1 + 12 ii y = −1 + 1 x+1 iii y = −1 + 1 1−x Consider the two parabolas y = x2 − 4x + 3 and y = x2 + 4x + 3. a Sketch both quadratic functions on the same set of axes. b What reflection maps each parabola onto the other? c How can the second parabola be obtained by shifting the first? d Confirm your answer to part c algebraically. e Investigate which parts of Question 11 could also have been achieved by shifting instead. 13 a Let c(x) = 2 x + 2−x . Show that c(−x) = c(x), and explain this geometrically. 2 2 x − 2−x . Show that −t(−x) = t(x), and explain this geometrically. 2 x + 2−x c [Technology] Confirm your observations in parts a and b by plotting each function using graphing software. b Let t(x) = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5B Reflection in the y-axis and x-axis 14 In each case, explain how the graph of the second function can be obtained from the first by a combination of reflections and translations (there may be more than one correct answer), then sketch each pair. a From y = x to y = 2 − x. b From y = x2 to y = 4 − x2 . 1 1 to y = x 2−x e From y = 2 x to y = 21−x − 2 d From y = x2 to y = −x2 − 2x c From y = 15 143 f From y = √ √ x to y = − 4 − x Consider the parabola y = (x − 1)2 . Sketches or plots done on graphing software may help answer the following questions. a i The parabola is shifted right 1 unit. What is the new equation? ii This new parabola is then reflected in the y-axis. Write down the equation of the new function. b i The original parabola is reflected in the y-axis. What is the new equation? ii This fourth parabola is then shifted right 1 unit. What is the final equation? c Parts a and b both used a reflection in the y-axis and a shift right 1 unit. Did the order of these affect the answer? d Investigate other combinations of shifts and reflections. In particular, what do you notice if the shift is parallel with the axis of reflection? ENRICHMENT 16 Suppose that y = f (x) is a function whose graph has been drawn. a Let U be shifting upwards a units and H be reflection in y = 0. Write down the equations of the successive graphs obtained by applying U, then H, then U, then H, and prove that the final graph is the same as the first. Confirm the equations by applying these operations successively to a square book or piece of paper. b Let R be shifting right by a units. Write down the equations of the successive graphs obtained by applying R, then H, then R, then H, and describe the final graph. Confirm using the square book. c Let V be reflection in x = 0 and I be reflection in y = x. Write down the equations of the successive graphs obtained by applying I, then V, then I, then H, and show that the final graph is the same as the first. Confirm using the square book. d Write down the equations of the successive graphs obtained by applying the combination I-followed-byV once, twice, . . . , until the original graph returns. Confirm using the square book. 17 Use combinations of shifts and reflections in the axes to demonstrate that the graph of y = f (2a − x) is the result of reflecting the graph of y = f (x) in the line x = a. (Hint: How could a reflection in x = a be turned into a reflection in the y-axis?) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 144 5C Chapter 5 Transformations and symmetry 5C Even and odd symmetry Learning intentions • Understand even and odd functions and relations geometrically and algebraically. • Test the equations of functions and relations for evenness and oddness. It is often said that all mathematics is the study of symmetry. Two simple types of symmetry occur so often in the functions of this course that every function should be tested routinely for them. y Even functions and line symmetry in the y-axis 2 A relation or function is called even if its graph has line symmetry in the y-axis. This means the graph is unchanged by reflection in the y-axis, as with the graph to the right. As explained in Section 5B, to reflect the graph of a function y = f (x) in the y-axis, the new curve has equation y = f (−x). Hence for a function to be even, the graph of y = f (x) and the graph of y = f (−x) must coincide, that is, f (−x) = f (x), 1 3 x −3 for all x in the domain. More generally, to reflect a relation in the y-axis, replace x by −x everywhere in its equation. Thus the relation is even when its equation is unchanged after replacing x by −x. 8 Even functions and line symmetry in the y-axis • A relation is called even if its graph has line symmetry in the y-axis. • Algebraically, this means that a relation is even if its equation is unchanged when x is replaced by −x. • In particular, a function f (x) is even if f (−x) = f (x), for all x in the domain. Odd functions and point symmetry in the origin y A relation or function is called odd if its graph has point symmetry in the origin. ‘Point symmetry in the origin’ means that the graph is unchanged by a rotation of 180◦ about the origin. Because a rotation of 180◦ about the origin is the same as successive reflections in the y-axis and x-axis (in any order), oddness means also that the graph is unchanged by successive reflections in the y-axis and the x-axis. 2 3 x -3 -2 Hence a relation is odd when its equation is unchanged after replacing x by −x and y by −y. In particular, reflecting a function y = f (x) in both axes gives −y = f (−x). Replacing y by f (x), and swapping sides, this condition for f (x) to be odd is f (−x) = − f (x), for all x in the domain. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5C Even and odd symmetry 9 145 Odd functions and point symmetry in the origin • Point symmetry in the origin means that the graph is mapped onto itself by a rotation of 180◦ about the origin. Equivalently, it means that the graph is mapped onto itself by successive reflections in the x-axis and the y-axis (where order does not matter). • A relation is called odd if its graph has point symmetry in the origin. • Algebraically, this means that a relation is odd if its equation is unchanged when x is replaced by −x and y is replaced by −y. • In particular, a function f (x) is odd if f (−x) = − f (x), for all x in the domain. Testing functions algebraically for evenness and oddness A single test will pick up both these types of symmetry in functions: 10 Testing a function for evenness and oddness (or neither) • Simplify f (−x) and note whether it is f (x), − f (x), or neither. Most functions are neither even nor odd. Example 9 Testing a function for evenness and oddness Test each function for evenness or oddness, then sketch it. a f (x) = x4 − 3 b f (x) = x3 c f (x) = x2 − 2x Solution a Here Substituting −x for x, f (x) = x4 − 3. y f (−x) = (−x)4 − 3 = x4 − 3 4 = f (x). Hence f (x) is an even function. 4 - Ö3 Ö3 x 1 x 2 x -3 b Here Substituting −x for x, f (x) = x3 . y f (−x) = (−x)3 1 = −x 3 = − f (x). Hence f (x) is an odd function. c Here Substituting −x for x, f (x) = x2 − 2x. -1 -1 y f (−x) = (−x) − 2(−x) 2 = x2 + 2x. Because f (−x) is equal neither to f (x) nor to − f (x), the function is neither even nor odd. (The parabola does, however, have line symmetry, not in the y-axis, but in its axis of symmetry x = 1.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 −1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 146 5C Chapter 5 Transformations and symmetry Example 10 The symmetries of a circle a Test the circle x2 + y2 = r2 for evenness and oddness. b What other rotation and reflection symmetries does the circle have? Solution a Replacing x by −x, (−x)2 + y2 = r2 y r x 2 + y2 = r 2 , so the circle is even. Replacing x by −x and y by −y, (−x)2 + (−y)2 = r2 -r x 2 + y2 = r 2 , P(x,y) r x -r so the circle is also odd. b From the diagram, the circle has reflection symmetry in every diameter, and has rotation symmetry when rotated about the centre through any angle. Exercise 5C 1 FOUNDATION Classify each function y = f (x) as even, odd or neither. a y b y x x c y d y x e y x f y x x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5C Even and odd symmetry 2 In each diagram below, complete the graph of y = f (x) so that f (x) is: ii odd i even a y y b 2 x -1 x Consider the function g(x) = x3 − 3x. b Hence show that g(x) is an odd function. Consider the function h(x) = x3 + 3x2 − 2. b Hence show that h(x) is neither even nor odd. a Simplify h(−x). 7 2 -1 b Hence show that f (x) is an even function. a Simplify g(−x). 6 x Consider the function f (x) = x4 − 2x2 + 1. a Simplify f (−x). 5 y 1 1 4 c 1 1 3 147 Simplify f (−x) for each function, and hence determine whether it is even, odd or neither. a f (x) = x2 − 9 b f (x) = x2 − 6x + 5 c f (x) = x3 − 25x d f (x) = x4 − 4x2 e f (x) = x3 + 5x2 f f (x) = x5 − 16x On the basis of the previous questions, copy and complete these sentences: a ‘A polynomial function is odd if . . . ’. b ‘A polynomial function is even if . . . ’. DEVELOPMENT 8 Factor each polynomial in Question 6 above and write down its zeroes (that is, its x-intercepts). Then use a table of values to sketch its graph. Graphing software might also be used. Confirm that the graph exhibits the symmetry established above. 9 [Algebra and Technology] In Questions 3 to 8, the odd and even functions were all polynomials. Other functions can also be classified as odd or even. In each case following, simplify f (−x) and compare it with f (x) and − f (x) to determine whether it is odd or even. Then confirm your answer by plotting the function on appropriate graphing software. 2 x − 2−x 2 x + 2−x a f (x) = b f (x) = 2 √ 22 √3 c f (x) = x d f (x) = 3 x x 2 e f (x) = 2 f f (x) = 2 x −4 x√− 4 √ g f (x) = 9 − x2 h f (x) = x 9 − x2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 148 5C Chapter 5 Transformations and symmetry 10 Determine whether each function is even, odd or neither. a f (x) = 2 x b f (x) = 2−x c f (x) = d f (x) = e f (x) = √ 3 − x2 4x 1 x2 + 1 f f (x) = 3 x + 3−x x2 + 4 g f (x) = 3 x − 3−x h f (x) = 3 x + x3 ENRICHMENT 1 1 − x is an odd function. 2 2 +1 b Confirm the result by plotting the graph. 11 a Show that f (x) = 12 a Given that h(x) = f (x) × g(x), determine what symmetry h(x) has if: i both f and g are even, ii both f and g are odd, iii one is even and the other odd. b Given that h(x) = f (x) + g(x), determine what symmetry h(x) has if: i both f and g are even, ii both f and g are odd, iii one is even and the other odd. 13 a Prove that an odd function defined at x = 0 passes through the origin. √ b Show that f (x) = x2 is odd, and explain why it does not pass through the origin. (Hint: What does its x graph look like?) c If f (x) is an even function defined at x = 0, does the graph of y = f (x) have to pass through the origin? Either prove that it does or give a counter-example. 14 1 1 f (x) + f (−x) and h(x) = f (x) − f (−x) . 2 2 a Show that f (x) = g(x) + h(x), that g(x) is even and that h(x) is odd. b Hence write each function as the sum of an even and an odd function. For any function f (x), define g(x) = ii f (x) = 2 x i f (x) = 1 − 2x + x2 c Why is there a problem with this process if f (x) = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 √ 1 or f (x) = x ? x−1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5D Horizontal and vertical dilations 149 5D Horizontal and vertical dilations Learning intentions • Transform an equation so that its graph is dilated horizontally or vertically. • Show that the order of performing a horizontal and a vertical dilation is irrelevant. • Apply horizontal and vertical dilations to familiar curves. • Obtain a given function formula by successive translations, reflections and rotations. Dilations stretch a curve so that, for example, a circle becomes an ellipse. These are the third and last of our standard transformations. Most of the functions and relations in the course can be reduced to very simple functions using a combination of translations, reflections, and dilations. Stretching a graph horizontally away from the y-axis x In the graphs sketched opposite, x has been replaced with : 3 x x 1 − 2 = 9 x(x − 6), y = x(x − 2) and y= 3 3 and the tables of values are: x −2 −1 0 1 2 3 4 x(x − 2) 8 3 0 −1 0 3 8 −6 −3 0 3 6 9 12 8 3 0 −1 0 3 8 x 3 x x 3 −2 y 12 3 6 x −1 The y-coordinates in each table are the same, but we needed to treble the x-coordinates to produce those same y-coordinates in the second table. Geometrically, this means that all points on the graph triple their distance from the y-axis, and the graph is stretched or dilated away from the y-axis in the horizontal direction by a dilation factor of 3. The points on the y-axis do not move. 11 Horizontal dilations — stretching a graph horizontally away from the y-axis x • To stretch a graph horizontally from the y-axis with factor k, replace x with . k x . • For a function y = f (x), the new function rule is y = f k Stretching a graph vertically away from the x-axis y = x(x − 2) or y = 3x(x − 2). 3 Each value in the table below for y = 3x(x − 2) is three times the corresponding value in the table for y = x(x − 2). This means that the graph of y = 3x(x − 2) is obtained from the graph of y = x(x − 2) by stretching away from the x-axis in the vertical direction by a factor of 3: Compare the graphs of y = x(x − 2) and y 1 x −2 −1 0 1 2 3 4 −1 x(x − 2) 8 3 0 −1 0 3 8 −3 3x(x − 2) 24 9 0 −3 0 9 24 2 x This time the points on the x-axis do not move, and all other points on the graph triple their distance from the x-axis. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 150 5D Chapter 5 Transformations and symmetry 12 Vertical dilations — stretching a graph vertically away from the x-axis y • To stretch a graph vertically from the x-axis by a factor of , replace y by . y • For a function y = f (x), the new function rule is = f (x), or y = f (x). Combining horizontal and vertical dilations When a graph is dilated horizontally and vertically, the order of applying the dilations doesn’t matter. The argument is almost the same as for translations. If y = f (x) is dilated k horizontally and then vertically, y = f (x) −→ y = f ( kx ) −→ y = f ( kx ), and if y = f (x) is dilated vertically and then k horizontally, y = f (x) −→ y = f (x) −→ y = f ( kx ). 13 The order of two perpendicular dilations does not matter • If two dilations are applied to a graph, the order does not matter. Example 11 Identifying dilations of a circle Obtain the graph of x 2 y2 + = 1 from the graph of the circle x2 + y2 = 1. 16 4 Solution The equation can be rewritten as x 2 y 2 + = 1, 4 2 which is the unit circle stretched vertically by a factor of 2, and horizontally by a factor of 4. y 2 −4 −1 4x 1 −2 x 2 y2 + = 1 is called an ellipse. It can be obtained from the unit circle x2 + y2 = 1 k2 2 by stretching horizontally by a factor of k and vertically by a factor of , so that its x-intercepts are k and −k and its y-intercepts are and −. Note: Any curve of the form A¢¢ Enlargements A dilation of a figure is usually not similar to the original. For example, the equilateral triangle ABC in the figure to the right with its base on the x-axis is stretched to the squat isosceles triangle A BC by a horizontal dilation with factor 2. And it is stretched to the skinny isosceles triangle A BC by a vertical dilation with factor 3. A A¢ B CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 C C¢ © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5D Horizontal and vertical dilations But if two dilations with the same factor 2, one horizontal and the other vertical, are applied in order to the equilateral triangle ABC — the order does not matter — the result is the similar equilateral triangle PBQ. Such a combined transformation is called an enlargement, and the factor 2 is called the enlargement factor or similarity factor. 151 P A In the coordinate plane, the centre of an enlargement is normally taken as the origin. B C Q 14 Enlargements • An enlargement of a figure is similar to the original. In particular, matching angles are equal, and the ratios of matching lengths are equal. • The composition of two dilations with the same factor, one horizontal and one vertical, is an enlargement with centre the origin. x y • To apply an enlargement with factor k, centre O, replace x by and y by . k k x x y • For a function, the new function rule is = f , or y = k f . k k k Example 12 Enlarging a circle y Apply an enlargement with centre the origin and factor 3 to the circle (x + 1)2 + (y − 1)2 = 1. Write down the new function, then sketch both curves. Solution The new function is × 32 = 9 x 3 2 +1 + y 3 2 −1 6 3 =1 (x + 3)2 + (y − 3)2 = 9. -6 x -3 The two circles are sketched to the right. y Stretching with a fractional or negative factor In the upper diagram to the right, a vertical dilation with factor 12 has been applied to the parabola y = x2 + 2 to yield the parabola y that is, y = 12 x2 + 1 . = x2 + 2, 1/2 The result is a compression, but we still call it a dilation. In the lower diagram to the right, vertical dilations with factors −1 and − 12 have been applied to the same parabola y = x2 + 2. The results are the parabolas y = −x2 − 2 and y = − 12 x2 − 1 . The first parabola is the reflection of the original in the x-axis. The second parabola is the reflection in the x-axis of the compressed image. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 3 2 1 -2 -2 2 y -1 -2 x 2 x -3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 152 5D Chapter 5 Transformations and symmetry 15 Dilations with a fractional or negative factor • If the dilation factor is between 0 and 1, the graph is compressed. • If the dilation factor is negative, the dilation is the composition of a dilation with the opposite positive factor and a reflection — the order does not matter. • In particular: A reflection is a dilation with factor −1. A rotation of 180◦ about the origin is an enlargement with factor −1, and is often called a reflection in the origin. Example 13 Applying dilations with negative and fractional factors Write down the new functions when each dilation is applied to the parabola y = (x − 3)(x − 5). Then sketch the four curves on one set of axes. a A horizontal dilation with factor −2. b A horizontal dilation with factor − 12 . c A vertical dilation with factor −1. (What other transformation is this?) d Which domains and ranges are changed by these transformations? Solution x , −2 x x y = (− 2 − 3)(− 2 − 5) y a Replacing x with y = 14 (x + 6)(x + 10). x b Replacing x with = −2x, −1/2 y = (−2x − 3)(−2x − 5) 15 - 23 -10 -6 - 25 y = 4(x + 1 12 )(x + 2 12 ). y c Replacing y with = −y, −1 −y = (x − 3)(x − 5) 3 5 x -15 y = −(x − 3)(x − 5). This is a vertical dilation with factor −1, so it is also a reflection in the x-axis. d In part c, the original range y ≥ −1 is changed to the range y ≤ 1. Warning: The word ‘dilation’ is often used to mean ‘enlargement’ rather than a stretching in one direction. Be careful when looking at other sources on the web. Combining different types of transformations — the order usually matters Be careful, because with different types of transformations, order usually matters. These examples also show how many complex-looking functions are generated by applying a sequence of different transformations to a very simple function. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5D Horizontal and vertical dilations Example 14 153 Describing a function as the result of a sequence of transformations Describe how y = x2 can be transformed to give y = 2(3x + 5)2 − 7. Write down the result of each successive step. Solution Shift left 5, then dilate horizontally with factor 13 , then dilate vertically with factor 2, then shift down 7. y = x2 −→ y = (x + 5)2 −→ y = (3x + 5)2 −→ y = 2(3x + 5)2 −→ y = 2(3x + 5)2 − 7 Example 15 The order in which transformations are applied usually matters Transform y = x3 in the two different orders below, showing steps. a Shift y = x3 right 3, then dilate horizontally with factor 5. b Dilate y = x3 horizontally with factor 5, then shift right 3. Solution x 3 1 − 3 , that is, y = (x − 15)3 5 125 x 3 x − 3 3 1 (x − 3)3 b y = x3 −→ y = −→ y = , that is, y = 5 5 125 a y = x3 −→ y = (x − 3)3 −→ y = Example 16 a Show that Identifying the asymptotes of a transformed function 1 21x − 62 = + 7. 3x − 9 3x − 9 b Hence identify, showing steps, a sequence of transformations that will move y = c Identify the domain, range, and asymptotes of the resulting function. 21x − 62 1 to . x 3x − 9 Solution 1 21x − 63 + 3x − 9 3x − 9 = LHS. a RHS = 1 x−9 1 1 Dilate horizontally factor 13 : −→ y= x−9 3x − 9 1 1 Shift up 7: y= −→ +7 3x − 9 3x − 9 c Domain: x 3. Range: y 7. Asymptotes: x = 3 and y = 7. b Shift right 9: CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y= 1 x −→ © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 154 5D Chapter 5 Transformations and symmetry Exercise 5D 1 Write down the new equation for each function or relation after the given dilation has been applied. Draw a graph of the image after the dilation: a y = x2 : horizontally by 12 b y = 2 x : vertically by 2 c y = x2 − 1: vertically by −1 d y = 1x : horizontally by 2 e x2 + y2 = 4: vertically by 13 f y= g y= 2 FOUNDATION √ 1 − x: horizontally by −1 √ h y = x : vertically by −2 √ 4 − x2 : horizontally by 12 In each case an unknown function has been drawn. Use dilations to draw the new functions indicated beneath each, then write down the y-intercept and zeroes of the dilated function.. a y = f (x) 1 -2 b y = P(x) y 1 -1 -1 2 y 2 1 x -2 -1 1 2 i y = f (2x) i y = P 2x ii y = 2 f (x) ii y = 12 P(x) c y = h(x) d y = g(x) y y 1 1 −1 x 1 2 x −1 1 x −1 i ii 3 y 2 = h(x) y = h 2x i 2y = g(x) ii y = g(2x) [Dilations and Intercepts] On the basis of the previous question, copy and complete these two sentences about the intercepts of the graph of a function with the axes. a A zero of a function is unchanged by a . . . dilation. b The value of f (0) is unchanged by a . . . dilation. 4 Sketch x + y = 1. Then explain how each graph below may be obtained by dilations of the first graph (there may be more than one answer), and sketch it. a 5 x 2 +y=1 b y x 2 + 4 =1 c 2x + y = 1 a The circle (x − 3)2 + y2 = 4 is dilated by a factor of 13 from the origin. Write down the new equation and draw both circles on the one set of axes. √ 1 b The hyperbola y = is enlarged by a factor of 3 from the origin. Write down the new equation and x draw both hyperbolae on the one set of axes. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5D Horizontal and vertical dilations 6 155 In each case graph the three given equations on one set of axes by using dilations: a y = x(4 + x), y = 2x(4 + x), and y = 2x (4 + 2x ). y b x2 + y2 = 36, ( 2x )2 + ( 3 )2 = 36, and (2x)2 + (3y)2 = 36. 7 [The Order of Dilations] a Graph the function y = x(x + 2) i The graph in part a is stretched horizontally by a factor of 2. Write down the equation of this new function and graph it on the same number plane. ii This graph is then stretched vertically by a factor of 3. Write down the equation of this new function and graph it on the same number plane. b On a new number plane, draw another copy of the parabola y = x(x + 2). i The graph in part b is stretched vertically by a factor of 3. Write down the equation of this new function and graph it on the same number plane. ii This graph is then stretched horizontally by a factor of 2. Write down the equation of this new function and graph it on the same number plane. c Compare your graphs and equations in parts a ii and b ii. Did the order of the two dilations change the final results? 8 [Dilations and Symmetry] a i Show that y = x3 − x is an odd function. ii This function is stretched horizontally by a factor of 2. Write down the equation of this new function. iii Is the new function odd? b i Show that y = 1 − x2 is an even function. ii This function is stretched vertically by a factor of 3. Write down the equation of this new function. iii Is the new function even? c What do you conclude about dilations and symmetry from these two examples? DEVELOPMENT 9 Sketch each group of three functions on the one set of axes. √ √ √ a y = 4 + x, y = 2 4 + x, y = 2 4 + 2x b y = 2 x , y = 2−x , y = 3 × 2−x 10 Answer the following questions about the cubic y = x3 − 3x2 + 4, which can be factored as y = (x − 2)2 (x + 1). a Graph this cubic, using graphing software if needed. Notice that the curve is momentarily horizontal at (0, 4) and (2, 0). b The cubic is dilated vertically by factor 2. i Write down the equation of this new cubic and graph it. ii Where is this new curve horizontal? Have the x-coordinates changed? c The original cubic is dilated horizontally by factor 3. i Write down the equation of this third cubic and graph it. ii Where is this third curve horizontal? Have the y-coordinates changed? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 156 5D Chapter 5 Transformations and symmetry 11 1 3 x . 8 i Explain the transformation as a horizontal stretch. ii Explain the transformation as a vertical stretch. iii Graph both functions on the same number plane. a The cubic y = x3 is transformed to obtain y = b The cubic y = x3 is transformed to obtain y = −x3 . i Explain the transformation as a horizontal stretch. ii Explain the transformation as a vertical stretch. iii Graph both functions on the same number plane. c Give two possible interpretations of transforming y = x3 to get the graph of y = functions on the same number plane. 1 3 x , then graph both 4 12 In each case identify how the graph of the second equation can be obtained from the graph of the first by a suitable dilation. 1 1 a y = x2 − 2x and y = 3x2 − 6x b y= and y = x−4 2x − 4 2 1 1 x x 4 and y = c y = 3 and y = 3 d y= x+1 x+1 13 Consider the hyperbola y = 14 Consider the parabola y = x2 . 1 . x a The hyperbola is stretched horizontally by a factor of 2. Write down its equation. b The original hyperbola is stretched vertically by a factor of 2. Write down its equation. c What do you notice about the answers to parts a and b? d Can the hyperbolae in parts a or b be achieved by an enlargement? e Investigate whether there are any other functions that exhibit similar behaviour. 1 . Write down its equation. 2 b The original parabola is dilated vertically by a factor of 4. Write down its equation. c What do you notice about the answers to parts a and b? d Can the parabolas in parts a or b be achieved by an enlargement? e Investigate whether there are any other functions that exhibit similar behaviour. a The parabola is dilated horizontally by a factor of 15 The mass M grams of a certain radioactive substance after t years is modeled by the formula M = 3 × 2− 53 t . 1 a Find the initial mass. b Find the time taken for the mass to halve, called the half-life. c Suppose now that the initial mass is doubled. i Explain this in terms of a dilation and hence write down the new equation for M. ii Show that the dilation does not change the value of the half-life. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5D Horizontal and vertical dilations 16 Show that the equation y = mx of a straight line through the origin is unchanged by any enlargement with centre the origin. 17 Describe each graph below as the given function transformed by dilations, and hence write down its equation. y a y b -2 3 x -6 -2 x 2 + y2 = 1 18 y c 1 2 -3 157 y = 3x x 3 x -3 -2 y = x2 − 4 Explain the gradient-intercept formula y = mx + b for a straight line in terms of a dilation of the line y = x + b. ENRICHMENT 19 a For each pair of curves, suggest two simple and distinct transformations by which the second equation may be obtained from the first: i y = 2 x , y = 2 x+1 1 k2 ,y= x x iii x2 + (y − 2)2 = 1, x2 + (y + 2)2 = 1 ii y = b Investigate other combinations of curves and transformations with similar ambiguity. 20 21 1 will a restore the original parabola. What other stretch will produce a new parabola that appears identical to the original parabola y = x2 ? The parabola y = x2 is stretched horizontally by factor a. Clearly a horizontal stretch by factor Determine how the curve y = x3 − x must be transformed in order to obtain the graph of y = x3 − 3x. (Hint: Only stretchings are involved.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 158 Chapter 5 Transformations and symmetry 5E 5E The absolute value function Learning intentions • Define absolute value geometrically, and apply it. • Understand absolute value as a piecewise-defined function, and apply it. • Understand absolute value as the positive square root of a square, and apply it. • Sketch graphs and solve simple equations using these approaches. Often it is the size or magnitude of a number that is significant, rather than whether it is positive or negative. Absolute value is the mathematical name for this concept. We begin the discussion with its geometric definition. Absolute value as distance Distance is the clearest way to define absolute value. 16 Absolute value as distance • The absolute value |x| of a number x is the distance from x to the origin on the number line. |x| 0 x For example, |−5| = 5 and |0| = 0 and |5| = 5. • Distance is always positive or zero, so |x| ≥ 0, for all real numbers x. • The numbers x and −x are equally distant from the origin, so |−x| = |x|, for all real numbers x. Thus absolute value is a measure of the size or magnitude of a number. In the examples above, the numbers −5 and +5 both have the same magnitude 5, and differ only in their signs. Distance between numbers Replacing x by x − a in the previous definition gives a measure of the distance from x to a on the number line. 17 Distance between numbers • The distance from x to a on the number line is |x − a|. |x - a| a x For example, the distance between 5 and −2 is |5 − (−2)| = 7. • By Box 16 above, |x − a| = |a − x|, for all real numbers x and a. Absolute value defined piecewise, using cases If x is a negative number, then the absolute value of x is −x, the opposite of x. This gives an alternative piecewise definition of |x| using cases: 18 Absolute value defined piecewise, using cases ⎧ ⎪ ⎪ ⎪ for x ≥ 0, ⎨ x, If x is any real number, then |x| = ⎪ ⎪ ⎪ ⎩−x, for x < 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5E The absolute value function 159 A piecewise definition naturally divides the graph into pieces, also called branches. Both branches here are linear functions, resulting in a sharp point at the origin where the two branches meet at right angles: x −2 −1 0 1 2 |x| 2 1 0 1 2 y 2 2 x −2 • The domain is all real numbers, and the range is y ≥ 0. • The function is even, because the graph has line symmetry in the y-axis. • The function has a zero at x = 0, and is positive for all other values of x. Graphing functions with absolute value Transformations can now be applied to the graph of y = |x| to sketch many functions involving absolute value. More complicated functions, however, require an approach involving cases. Example 17 Using transformations to graph a function with absolute value a Sketch y = |x − 2| using shifting. b Check the graph using a table of values. c Write down the equations of the two branches. Solution a This is y = |x| shifted 2 units to the right. b x 0 1 2 3 y 4 2 y 2 1 0 1 2 c From ⎧ the expression using cases, or from the graph: ⎪ ⎪ ⎪ for x ≥ 2, ⎨ x − 2, y=⎪ ⎪ ⎪ ⎩−x + 2, for x < 2. Example 18 2 4 x Using cases to graph a function with absolute value a Use cases to sketch y = |x| − x. b Check using a table of values, and state its domain and range. Solution a Considering⎧ separately the cases x ≥ 0 and x < 0, ⎪ ⎪ ⎪ for x ≥ 0, ⎨ x − x, y=⎪ ⎪ ⎪ ⎩−x − x, for x < 0, ⎧ ⎪ ⎪ ⎪ for x ≥ 0, ⎨0, that is, y = ⎪ ⎪ ⎪ ⎩−2x, for x < 0. b Checking using a table of values: x −2 −1 0 1 2 Domain: all real x, y 4 2 0 0 0 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y 2 −1 x range: y ≥ 0. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 160 5E Chapter 5 Transformations and symmetry Solving absolute value equations of the form |ax + b| = k Three observations should make everything clear: • An equation such as |3x + 6| = −21 has no solutions, because an absolute value can never be negative. • An equation such as |3x + 6| = 0 is true when 3x + 6 = 0. • An equation such as |3x + 6| = 21 can be solved by realising that: |3x + 6| = 21 3x + 6 = 21 or 3x + 6 = −21 . is true when 19 To solve the equation |ax + b| = k • If k < 0, the equation has no solutions. • If k = 0, the equation has one solution, found by solving ax + b = 0. • If k > 0, the equation has two solutions, found by solving ax + b = k Example 19 or ax + b = −k. Solving simple absolute value equations Solve these absolute value equations. b |3x + 6| = 0 a |3x + 6| = −21 c |3x + 6| = 6 Solution a |3x + 6| = −21 has no solutions, because an absolute value cannot be negative. |3x + 6| = 0 b |3x + 6| = 6 c 3x + 6 = 0 3x + 6 = 6 or 3x + 6 = −6 −6 3x = −6 −6 3x = 0 or 3x = −12 ÷3 x = −2 ÷3 x=0 or x = −4 Example 20 Solving simple absolute value equations Solve each absolute value equation. b |7 − 14 x| = 3 a |x − 2| = 3 Solution |x − 2| = 3 a +2 |7 − 14 x| = 3 b x−2=3 or x − 2 = −3 x=5 or x = −1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 7 − 14 x = 3 or 7 − 14 x = −3 −7 − 14 x = −4 or − 14 x = −10 × (−4) x = 16 or x = 40 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5E The absolute value function 161 Sketching y = |3x + 6| (or any function of the form y = |ax + b| ) Method 1: Sketch using cases, as in Example 18. Method 2: Find and plot the x-intercept and the y-intercept. Put y = 0, then |3x + 6| = 0 Put x = 0. 3x + 6 = 0 y = |0 + 6| Then x = −2. = 6. The graph is symmetric about the vertical line x = −2, so the point (−4, 6) also lies on the curve. Now join up the points in the characteristic V shape. Method 3: Draw up a small table of values — the pieces are linear: x −4 −3 −2 −1 0 y 3 0 3 6 6 y 6 -4 -2 x A good final check: The two branches of the graph should have gradients 3 and −3. Method 4: Use transformations of y = |x|. y = |x| −→ y = |x + 6| • Shift the graph left 6: • Then dilate horizontally with factor 13 : y = |x + 6| −→ y = |3x + 6|. Absolute value as the square root of the square Taking the absolute value of a number means stripping any negative sign from the number. We have algebraic √ that says ‘take the positive functions that can do of this job — square the number, then apply the function square root (or zero)’. 20 Absolute value as the positive square root of the square • For all real numbers x, |x|2 = x2 and For example, |−3|2 = 9 = (−3)2 and |−3| = |x| = √ 9= √ x2 . (−3)2 . Note: We have now defined the absolute value function y = |x| using distance, using cases, and using a formula. The phrase ‘piecewise-defined function’ does not characterise a function, but only the way that it is currently being defined. Identities involving absolute value Here are some standard identities involving absolute value. 21 Identities involving absolute value • |−x| = |x|, for all x. • |x − y| = |y − x|, for all x and y. • |xy| = |x||y|, for all x and y. x |x| • = , for all x, and for all y 0. y |y| Substitution of some positive and negative values for x and y should be sufficient to demonstrate these results. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 162 5E Chapter 5 Transformations and symmetry Exercise 5E 1 2 FOUNDATION Evaluate: a |3| b |−3| c |4 − 7| e |14 − 9 − 12| f |−7 + 8| g |3 − 5 | 2 d |7 − 4| h |11 − 16| − 8 2 In each case, use the given values of x and y to verify that: ii |x − y| = |y − x| i |x| = | − x| x |x| iv = . y |y| iii |xy| = |x||y| a x = 4, y = −2 b x = −4, y = 2 c x = −4, y = −2 3 4 5 Solve each absolute value equation, then graph the solution on a number line. a |x| = 1 b |x| = 3 c |4x| = 8 d |2x| = 10 e |2x| = 6 f |3x| = 12 Solve each equation and graph the solution on a number line. a |x − 4| = 1 b |x − 3| = 7 c |x − 3| = 3 d |x − 7| = 2 e |x + 5| = 2 f |x + 2| = 2 g |x + 1| = 6 h |x + 3| = 1 a Copy and complete the tables of values for the functions y = |x − 1| and y = |x| − 1. x −2 −1 0 1 2 x 3 |x − 1| −2 −1 0 1 2 3 |x| − 1 b Draw the graphs of the two functions on separate number planes, and observe the similarities and differences between them. c Explain how each graph is obtained by shifting y = |x|. d Write down the domain and range of each function. 6 7 Show that each statement is false when x = −2. a |x| = x b |−x| = x c |x + 2| = |x| + 2 d |x + 1| = x + 1 e |x − 1| < |x| − 1 f |x|3 = x3 Say whether these statements are true or false, and if false, give a counterexample (any difficulties will usually involve negative numbers): a |x| > 0 b | − x| = |x| c −|x| ≤ x ≤ |x| d |x + 2| = |x| + 2 e |5x| = 5|x| f |x|2 = x2 g |x|3 = x3 h |x − 7| = |7 − x| CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5E The absolute value function 163 DEVELOPMENT 8 9 In each case, use the rules of Box 19 to solve the equation for x. a |7x| = 35 b |2x + 1| = 3 c |2x − 1| = 11 d |7x − 3| = −11 e |3x + 2| = −8 f |5x + 2| = 0 g |5 − 3x| = 0 h |7 − 6x| = 5 i |5x + 4| = 6 a Use cases to help sketch the branches of y = |x|. b In each part, identify the shift or shifts of y = |x| and hence sketch the graph. Then write down the equations of the two branches. i y = |x − 3| ii y = |x + 2| iii y = |x| − 2 iv y = |x| + 3 v y = |x − 2| − 1 vi y = |x + 1| − 1 c Write down the domain and range for each graph in part b. 10 a Sketch each function using cases. Check the graph with a table of values. ii y = | 12 x| i y = |2x| b Now sketch each function using a suitable vertical dilation of y = |x|, and check that you get the same graph. 11 Find and plot the x-intercept and y-intercept of each function. Then complete each graph using symmetry. Finally, check each graph with a small table of values. b y = |6 − 3x| a y = |2x − 6| 12 When the graph of y = |x| is shifted horizontally by a the result is y = |x − a|. When this undergoes a horizontal stretch with factor 1h the final result is y = |hx − a|. Use this approach to sketch the following functions. Then check each graph with a suitable table of values. b y = |3x + 3| a y = |2x − 4| 13 c y = |5x + 5| c y = | 12 x − 1| [Technology] Use suitable graphing software to help solve these problems. a i Sketch y = |x − 4| and y = 1 on the same set of axes, clearly showing the points of intersection. ii Hence write down the solution of |x − 4| = 1. b Now use similar graphical methods to solve each of the following. ii |2x + 1| = 3 i |x + 3| = 1 iii |3x − 3| = −2 iv |2x − 5| = 0 14 Consider the absolute value function f (x) = |x|. √ a Use the result f (x) = x2 given in Box 20 to help prove that the absolute value function is even. b Why was this result obvious from the graph of y = |x|? 15 Use the fact that |−x| = |x| to decide whether these functions are odd, even or neither. a f (x) = |x| + 1 16 b f (x) = |x| + x c f (x) = x × |x| |x| undefined? x b Use a table of values of x from −3 to 3 to sketch the graph. d f (x) = |x3 − x| a For what values of x is y = c Hence write down the equations of the two branches of y = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 |x| . x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 164 5E Chapter 5 Transformations and symmetry 17 Sketch each graph by drawing up a table of values for −3 ≤ x ≤ 3. Then use cases to determine the equation of each branch of the function. a y = |x| + x b y = |x| − x c y = 2(x + 1) − |x + 1| d y = x2 − |2x| ENRICHMENT 18 Prove that f (|x|) is an even function, for all functions f (x). 19 Sketch the relation |y| = |x| by considering the possible cases. 20 Carefully write down the equations of the branches of each function, then sketch its graph: a y = |x + 1| − |x − 3| b y = |x − 2| + |x + 1| − 4 c y = 2|x + 1| − |x − 1| − 1 21 The diagram on the right shows the line ax + by + c = 0, which intersects the c c axes at A(− , 0) and B(0, − ). Let p be the perpendicular distance from O to the a b line AB. y - bc a Use distances |OA| and |OB| to find the area of AOB. 1 b The area of AOB is also p × |AB|. Use Pythagoras’ theorem to rewrite this 2 in terms of a, b, c and p. c Equate your answers to parts a and b, and so find p. d Now use shifting to find a formula for the distance from P(x1 , y1 ) to the line ax + by + c = 0. B p O A - ac x e Use this formula to find the distance from (3, 2) to the line 2x − 5y + 3 = 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5F Composite functions 165 5F Composite functions Learning intentions • Form composite functions, and work with them. • Discover many composites that are transformations of one of the functions. Shifting, reflecting, dilating, and taking absolute value, are all examples of a far more general procedure of creating a composite function from a chain of two given functions. The first example below shows how a translation 3 units left of y = x2 , and a translation 3 units up of y = x2 , are both are obtained using composites. In this section, however, attention is on the algebra rather than on the final graph. Composition of functions Suppose that we have the two functions f (x) = x + 3 and g(x) = x2 . We can put them together in a chain by placing their function machines so that the output of the first function is the input of the second function x f (x) g f (x) −2 −→ −1 −→ −→ 1 −→ −→ 2 −→ −→ 1 −→ 4 −→ g −→ 9 4 −→ Square −→ 16 5 −→ −→ 25 0 −→ f −→ 3 1 −→ Add 3 −→ 2 −→ −→ x −→ −→ x + 3 −→ −→ (x + 3)2 The middle column is the output of the first function ‘Add 3’. This output is then the input of the second function ‘Square’. The result is the composite function ‘Add 3, then square.’ g f (x) = (x + 3)2 If the two functions are composed the other way around, the result is different: ‘Square, then add 3.’ f g(x) = x2 + 3 Notice how in this example, both ways around are examples of translations: • The composite graph y = g f (x) is y = g(x) shifted left 3. • The composite graph y = f g(x) is y = g(x) shifted up 3. The domains and ranges of these composites are easily read off their equations: • For g f (x) = (x + 3)2 , the domain is all real x, and the range is y ≥ 0. • For f g(x) = x2 + 3, the domain is all real x, and the range is y ≥ 3. 22 Composition of functions • A composite function g f (x) is formed by first applying the function f to x, and then applying the function g to the output f (x). • The composite function f g(x) is formed by first applying the function g to x, and then applying the function f to the output g(x). Care: A function is written on the left of the number it acts on. Thus the order in which the functions are written is the reverse of the order in which they are applied. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 166 5F Chapter 5 Transformations and symmetry We often call these successive functions a chain of functions. Later, in calculus, we will develop a rule called the chain rule for handling composite functions. This chapter has been concerned with transformations. Many composites are transformations of one of the functions, particularly when one function is linear, as the examples below and the following exercises will make clear — but this is to be observed, not learnt. Example 21 Evaluating and simplifying composite functions Let f (x) = x2 and g(x) = x + 3. a Evaluate: i g f (5) ii g f (−2) iii f g(5) iv f g(−2) b i Find and simplify g f (x) , and write down its domain and range. ii Explain how y = g f (x) is a transformation of y = f (x). c Repeat part b for the other composite function f g(x) . Solution a i g f (5) = g(25) = 28 iii f g(5) = f (8) ii g f (−2) = g(4) =7 iv f g(−2) = f (1) = 64 =1 2 2 b i g f (x) = g(x ) = x + 3. The domain is all real x, and the range is y ≥ 3. ii g f (x) = x2 + 3 is the function f (x) = x2 shifted up 3. c i f g(x) = f (x + 3) = (x + 3)2 . The domain is all real x, and the range is y ≥ 0. ii f g(x) = (x + 3)2 is the function f (x) = x2 shifted left 3. Example 22 Taking composites of an exponential function and a linear function Let f (x) = 2 x and g(x) = 4x − 7. a Simplify g f (x) , write down its asymptotes and its domain and range, and characterise it as transformation of y = f (x). b Repeat for f g(x) . Solution a g f (x) = 4 × 2 x − 7. The line y = −7 is a horizontal asymptote on the left. The domain is all real x, and the range is y > −7. The composite g f (x) = 4 × 2 x − 7 is the function f (x) = 2 x dilated vertically with factor 4 to give y = 4 × 2 x , then shifted down 7 b f g(x) = 24x−7 . The line y = 0 is a horizontal asymptote on the left. The domain is all real x, and the range is y > 0. The composite f g(x) = 24x−7 is the function f (x) = 2 x shifted right 7 to give y = 2 x−7 , then dilated horizontally with factor 14 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5F Composite functions 167 Domain and range of a composite function It is elaborate to formulate rules for the domain and range of a composite function. The best approach is to write down the formula for the composite function, and determine its domain and range in the usual ways. See the examples above for this approach used in various situations. The empty function — for clarification only Let f (x) = −x2 − 1 and g(x) = √ g f (x) = −x2 − 1 . √ x. Then √ We have a problem here, because −x2 − 1 is undefined, whatever the value of x. Thus g f (x) has domain the empty set, and its range is therefore also the empty set. This is the empty function, whose domain and range are both the empty set. Exercise 5F 1 FOUNDATION Often a pattern emerges when a function is composed with itself. Let f (x) = x + 2. a Evaluate the following. ii f f (0) i f (0) iii f f f (0) b What number pattern is formed in part a? c Now evaluate these. ii f f (1) i f (1) iii f f f (1) d What number pattern is formed in part c? 2 Once again, let f (x) = x + 2. a Find the values of: i f f (3) b Find expressions for: i f f (x) ii f f (−1) iii f f (−8) ii f f f (x) c Find the value of x for which f ( f (x)) = 0. 3 Consider the function g(x) = 2 − x. a Find the values of g g(0) , g g(4) , g g(−2) and g g(−9) . b Show that g g(x) = x. c Show that g g g(x) = g(x). 4 Consider the function h(x) = 3x − 5. a Find the values of h h(0) , h h(5) , h h(−1) and h h(−5) . b Find expressions for h h(x) and for h h h(x) . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 168 5F Chapter 5 Transformations and symmetry 5 When a function f (x) is composed with a linear function g(x) the result is a transformation of f (x). Let f (x) = x2 − 4 and g(x) = x + 1. a Write down the domain and range of y = f (x). A sketch may help. b Find the values of: i f g(3) ii g f (3) c Find expressions for: i f g(x) d ii g f (x) i What transformation maps the graph of y = f (x) to the graph of y = f g(x) ? ii Hence write down the domain and range of y = f g(x) . e 6 i What transformation maps the graph of y = f (x) to the graph of y = g f (x) ? ii Hence write down the domain and range of y = g f (x) . Let p(x) = x(x − 2) and d(x) = 2x. When p(x) is composed with d(x) the result is a transformation since d(x) is linear. a Write down the domain and range of y = p(x). A sketch may help. b Find the values of: i p d(1) ii d p(1) c Find expressions for: i p d(x) d ii d p(x) i What transformation maps the graph of y = p(x) to the graph of y = p d(x) ? ii Hence write down the domain and range of y = p d(x) . e i What transformation maps the graph of y = p(x) to the graph of y = d p(x) ? ii Hence write down the domain and range of y = d p(x) . 7 In this question, the function Q(x) = (x − 1)2 is composed with R(x) = −x. a Write down the domain and range of y = Q(x). A sketch may help. b Find the values of: i Q R(2) ii R Q(2) c Find expressions for: i Q R(x) d ii R Q(x) i What transformation maps the graph of y = Q(x) to the graph of y = Q R(x) ? ii Hence write down the domain and range of y = Q R(x) . e i What transformation maps the graph of y = Q(x) to the graph of y = R Q(x) ? ii Hence write down the domain and range of y = R Q(x) . DEVELOPMENT 8 √ Suppose that F(x) = 4x and G(x) = x . a Find the values of F G(25) , G F(25) , F F(25) and G G(25) . b Find F G(x) . c Find G F(x) ). d Hence show that F G(x) = 2G F(x) . e State the domain and range of F G(x) . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5F Composite functions 9 10 11 169 1 Two functions f and h are defined by f (x) = −x and h(x) = . x a Find the values of f h(− 14 ) , h f (− 41 ) , f f (− 14 ) and h h(− 14 ) . b Show that for all x 0: i f h(x) = h f (x) ii f f (x) = h h(x) c Write down the domain and range of f h(x) . d Describe how the graph of h(x) is transformed to obtain the graph of h f (x) . √ Suppose that f (x) = −5 − |x| and g(x) = x. a Find f g(x) , state its domain and range, and sketch its graph. b Explain why g f (x) is the empty function. √ Let H(x) = x − 2, R(x) = −x and g(x) = 4 + x. a Draw the graph of y = g(x). b Use the graph in part a to help answer the following. i Find an expression for y = g R(x) and interpret the result as a transformation. ii Hence sketch y = g R(x) and state its domain and range. c Use your results in part b to help answer the following. i Find an expression for y = H g R(x) and interpret this as a transformation of y = g R(x) . ii Hence sketch y = H g R(x) and state its domain and range. 12 a Show that if f (x) and g(x) are odd functions then g f (x) is odd. b Show that if f (x) is an odd function and g(x) is even then g f (x) is even. c Show that if f (x) is an even function then g f (x) is even, regardless of g(x). 13 Find the composite functions g f (x) and f g(x) . a f (x) = 4, for all x, and g(x) = 7, for all x. 14 b f (x) = x, g(x) any function. a Let f (x) be any function, and let g(x) = x − a, where a is a constant. Describe each composite graph as a transformation of the graph of y = f (x). i y = g f (x) ii y = f g(x) b Let f (x) be any function, and let g(x) = −x. Describe each composite graph as a transformation of the graph of y = f (x). i y = g f (x) ii y = f g(x) c Let f (x) be any function, and let g(x) = kx, where k is a constant. Describe each composite graph as a transformation of the graph of y = f (x). i y = g f (x) ii y = f g(x) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 170 5F Chapter 5 Transformations and symmetry ENRICHMENT 15 Let f (x) = 2x + 3 and g(x) = 5x + b, where b is a constant. Find the value of b so that g f (x) = f g(x) , for all x. 16 Let f (x) = 2x + 3 and g(x) = ax + b, where b is a constant. a Find expressions for g f (x) and f g(x) . b Hence find the values of a and b so that g f (x) = x, for all x. c Show that if a and b have these values, then f g(x) = x, for all x. 17 Let f (x) = x2 + x − 3 and g(x) = |x|. a Find f (g(0)), g( f (0)), f (g(−2)) and g( f (−2)). b Write an expression for f (g(x)) without any use of absolute value, given that: i x≥0 ii x < 0 18 Let L(x) = x + 1 and Q(x) = x2 + 2x. a State the ranges of L(x) and Q(x). b Find L(Q(x)) and determine its range. c Find Q(L(x)) and determine its range. d Find the zeroes of Q(L(x)). e Show that Q(L( 19 1 (x + 2)(3x + 4) , provided that x −1. )) = x+1 (x + 1)2 In some cases there are connections between familiar functions via composite functions. Let p(x) = 4 − x2 √ and r(x) = x. a Graph y = p(x) and y = r(x), writing down their domains and ranges. b Simplify c(x) = r p(x) and graph the result. What familiar curve is this? What is the domain and range of c(x)? 20 Let f(x) be any function with domain D, and let z(x) be the zero function defined by z(x) = 0 for all x. What are these functions (specify the domain of each)? a f (x) + z(x) b f (x) × z(x) c z f (x) d f z(x) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5G Combining transformations 171 5G Combining transformations Learning intentions • Review the transformations, understanding reflections also as dilations with a factor of −1. • Examine where order does or doesn’t matter when combining transformations. We can apply transformations in succession to a graph. This immediately raises the question, ‘Does it matter in which order the transformations are applied?’ A summary of transformations First, here are the procedures for applying translations and dilations to a graph: 23 A summary of translations and dilations Transformation By replacement By function rule Shift horizontally right h Replace x by x − h y = f (x) −→ y = f (x − h) Shift vertically up k Replace y by y − k x Replace x by a y Replace y by b y = f (x) −→ y = f (x) + k x y = f (x) −→ y = f a y = f (x) −→ y = b f (x) Stretch horizontally factor a Stretch vertically factor b Secondly, we can regard reflections as dilations with factor −1, as we saw in Section 5D (Box 15). And rotating 180◦ about the origin is the combination of reflections in the x-axis and y-axis (in any order). This means that when we are examining whether order matters, we can consider just four transformations — horizontal and vertical translations, and horizontal and vertical dilations. 24 A summary of reflections and 180◦ rotations Transformation By replacement By function rule Reflect in the y-axis Replace x with −x y = f (x) −→ y = f (−x) A reflection in the y-axis is a horizontal dilation with factor −1. Replace y with −y Reflect in the x-axis y = f (x) −→ y = − f (x) A reflection in the x-axis is a vertical dilation with factor −1. Rotate 180◦ about O Replace x with −x, y with −y y = f (x) −→ y = − f (−x) ◦ A rotation of 180 about the origin is the combination of reflections in the y-axis and x-axis (in any order). Hence it is also the combination of horizontal and vertical dilations each with factor −1 (in any order). The word ‘commute’ The word ‘commute’ is not in the course, but it is a standard and very useful word that makes sentences shorter and easier to read. We will use it in this section. 25 Commuting transformations Two transformations are said to commute if the order in which they are applied does not matter, whatever graph they are applied to. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 172 5G Chapter 5 Transformations and symmetry Several results about commuting transformations were established in Section 5A (Box 3), Section 5B (Box 7) and Section 5D (Box 13). To summarise: 26 Commuting transformations • Any two translations commute. • Any two dilations commute (including reflections and rotating 180◦ about O). • A translation and a dilation commute if one is vertical and the other horizontal. The following examples are intended to illustrate these three results: Example 23 Illustrating that two translations always commute a On one set of axes, draw the three graphs to illustrate the parabola y = x2 shifted right 3, then shifted down 1. b On one set of axes, draw the three graphs to illustrate the parabola y = x2 shifted down 1, then shifted right 3. Solution a The three graphs are y = x2 , y = (x − 3)2 , and y = (x − 3) − 1. b The three graphs are y = x2 , y = x2 − 1, and y = (x − 3)2 − 1. 2 y y 3 2 -1 3 x 4 Example 24 2 -1 4 x Illustrating that two dilations always commute a On one set of axes, draw the three graphs to illustrate the circle x2 + y2 = 1 stretched vertically with a factor of 2, then horizontally with factor 3. b On one set of axes, draw the three graphs to illustrate the circle x2 + y2 = 1 stretched horizontally with a factor of 3, then vertically with a factor of 2. Solution a The three graphs are x2 + y2 = 1, x2 + -3 2 2 2 y x y = 1, and + = 1. 4 9 4 b The three graphs are x2 + y2 = 1, x2 x 2 y2 + y2 = 1, and + = 1. 9 9 4 y y 2 2 -1 1 3 x -2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 -3 -1 1 3 x -2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5G Combining transformations Example 25 173 A horizontal dilation and a vertical translation commute a Apply a reflection in the y-axis (that is, a horizontal dilation with a factor of −1), then a shift up 2, successively to the circle (x − 2)2 + (y − 1)2 = 1. b Apply the transformations in the reverse order. Solution a The three graphs are b The three graphs are (x − 2) + (y − 1) = 1, (x − 2)2 + (y − 1)2 = 1, (x + 2)2 + (y − 1)2 = 1, (x − 2)2 + (y − 3)2 = 1, 2 and 2 (x + 2)2 + (y − 3)2 = 1. and (x + 2)2 + (y − 3)2 = 1. In both parts, note that (−x − 2)2 = (x + 2)2 . y y 4 3 2 1 4 3 2 1 -3-2-1 1 2 3 x 1 2 3 x -3-2-1 Transformations that do not commute Here are some situations where the transformations do not commute. They demonstrate a general result that a translation and a dilation do not commute when they are both horizontal or both vertical. Example 26 A vertical dilation and a vertical translation don’t commute a Apply a vertical dilation with a factor of 12 , then a translation down 3, to the parabola y = x2 + 4. b Apply the transformations in the reverse order Solution a The three graphs are b The three graphs are y = x2 + 4, y = x2 + 4, y = 12 x2 + 2, y = x2 + 1, y = 12 x2 − 1. y = 12 x2 + 12 . y y 4 2 1 -Ö2-1 Ö2 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 2 4 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 174 5G Chapter 5 Transformations and symmetry Example 27 A horizontal dilation and horizontal translation don’t commute a Apply a reflection in the y-axis (that is, a horizontal dilation with a factor of −1), then a translation left 2, to the circle (x + 3)2 + y2 = 1. b Apply the transformations in the reverse order. Solution a The three graphs are b The three graphs are (x + 3)2 + y2 = 1, (x + 3)2 + y2 = 1, (x − 3)2 + y2 = 1, (x + 5)2 + y2 = 1, (x − 1)2 + y2 = 1. (x − 5)2 + y2 = 1. (−x + 3)2 = (x − 3)2 . Note that Note that y –4 –2 (−x + 5)2 = (x − 5)2 . y 2 4 x x 27 Transformations that do not commute • A vertical translation and a vertical dilation do not commute. • A horizontal translation and a horizontal dilation do not commute. Exercise 5G FOUNDATION Note: The word ‘commute’ is not in the course, but it is standard and very useful, because it makes sentences shorter and easier to read. Two transformations are said to commute if the order in which they are applied does not matter, whatever graph they are applied to. 1 Let y = x2 − 2x. Sketch the graph of this function showing the intercepts and vertex. a i The parabola is shifted right 1 unit. Sketch the situation and find its equation, expanding any brackets. ii The parabola in part i is then shifted up 2 units. Sketch the new graph and find its equation. b i The original parabola y = x2 − 2x is translated up 2 units. Sketch the result and find its equation. ii The parabola in part i is then translated 1 unit right. Sketch the situation and find its equation, expanding any brackets. c Parts a and b used the same translations, 1 unit right and 2 units up, but in a different order. Do these transformations commute? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5G Combining transformations 2 175 As in Question 1, start with the parabola y = x2 − 2x. a i The parabola is dilated by a factor of 2 horizontally. Sketch the situation and find its equation. ii The parabola in part i is then dilated by a factor of 3 vertically. Sketch the new graph and find its equation. b i The original parabola y = x2 − 2x is stretched by a factor of 3 vertically. Sketch the result and find its equation. ii The parabola in part i is then stretched by a factor of 2 horizontally. Sketch the situation and find its equation. c Parts a and b used the same dilations, a factor of 2 horizontally and a factor of 3 vertically, but in a different order. Do these transformations commute? 3 Once again, start with the parabola y = x2 − 2x. a i The parabola is dilated by a factor of 2 horizontally. Sketch the situation and find its equation. ii The parabola in part i is then translated 1 unit up. Sketch the new graph and find its equation. b i The original parabola y = x2 − 2x is shifted 1 unit up. Sketch the result and find its equation. ii The parabola in part i is then stretched by a factor of 2 horizontally. Sketch the situation and find its equation. c Parts a and b used the same transformations, stretch by a factor of 2 horizontally and shift 1 unit up, but in a different order. Do these transformations commute? 4 Let y = x2 − 2x. Sketch the graph of this function showing the intercepts and vertex. a i The parabola is shifted right by 1 unit. Sketch the situation and find its equation, expanding any brackets. ii The shifted parabola is then reflected in the y-axis. Sketch the new graph and find its equation. b i The original parabola y = x2 − 2x is reflected in the y-axis. Sketch the result and find its equation. ii The reflected parabola is then shifted 1 unit right. Sketch the situation and find its equation, expanding any brackets. c Parts a and b used a shift 1 unit right and reflection in the y-axis, but in a different order. Do these transformations commute? DEVELOPMENT 5 Which of the following pairs of transformations commute? a reflection in the y-axis and horizontal translation b vertical dilation and vertical translation c vertical dilation and reflection in the x-axis d horizontal translation and vertical translation e horizontal dilation and horizontal translation f reflection in the x-axis and horizontal translation CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 176 Chapter 5 Transformations and symmetry 6 5G Write down the new equation for each function or relation after the given transformations have been applied. Draw a graph of the image. a y = x2 : right 2 units, then dilate by a factor of 12 horizontally b y = 2 x : down 1 unit, then reflect in the y-axis c y = x2 − 1: down 3 units, then dilate by a factor of −1 vertically d y = 1x : right 3 units, then dilate by a factor of 2 vertically e x2 + y2 = 4: up 2 units, then dilate by a factor of 12 vertically √ f y = x : up 2 units, then dilate by a factor of −1 horizontally 7 The table shows pairs of transformations with V for vertical, H for horizontal, T for translation and D for dilation. HT HT VD HD VD C VT HD VT C N C C C a Copy the table and complete the entries below the diagonal. Use C if the transformations commute and N if they do not. b The table does not directly include reflections. How can the table be used to decide whether a reflection commutes with another transformation? c Check that the table agrees with your answers to Question 5. 8 Determine the equation of the curve after the given transformations have been applied in the order stated. a y = x2 : left 1, down 4, dilate horizontally by a factor of 2, b y = x2 : down 4, dilate horizontally by a factor of 2, left 1, c y = 2 x : down 1, right 1, dilate vertically by a factor of −2, d y= 9 1 : right 2, dilate by 2 vertically, up 1. x The parabola y = (x − 1)2 is shifted 2 left and then reflected in the y-axis. a Show that the new parabola has the same equation. b Investigate why this has happened. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5G Combining transformations 177 ENRICHMENT 10 Let f (x) be any function and let (x) = ax + b, a linear function. The graph of y = f (x) = f (ax + b) can be obtained by two transformations of y = f (x). This was investigated briefly in Questions 5, 6 and 7 of the previous exercise. Here the situation is investigated further. Although f (x) can be any function, the one graphed on the right is used for parts a to c. a y 1 -1 1 x i Let g(x) = f (x + 1). Describe the transformation of f (x) and so graph y = g(x). ii Let h(x) = g(2x). Describe the transformation of g(x) and so graph y = h(x). iii Hence describe h(x) as a transformation of f (x), then write h(x) in terms of the function f . b i Let p(x) = f (2x). Describe the transformation of f (x) and so graph y = p(x). ii Let q(x) = p(x + 12 ). Describe the transformation of p(x) and so graph y = q(x). iii Hence describe q(x) as a transformation of f (x), then write q(x) in terms of the function f . c The functions h(x) and q(x) are identical. In parts a and b, the horizontal stretches of f are the same but the shifts left differ. Why? d Describe y = f (ax + b) as transformations of y = f (x) in two different ways. e Now complete the situation by describing y = f (x) = a f (x) + b as transformations of y = f (x) in two different ways. 11 [Symmetry in the line x = a] Suppose that y = f (x) has symmetry in the line x = a. Let g(x) = f (x + a). a Explain why y = g(x) is even. b Hence explain why f (−x + a) = f (x + a). c Hence prove that f (2a − x) = f (x). d Now suppose that h(x) has the same algebraic property, that h(2a − x) = h(x). Prove that h(x) must be symmetric in x = a. e Review Question 9 above and Question 16 of Exercise 5B in light of these results. 12 a Let H be a horizontal translation by a units and V be a vertical translation by b units. Show that H and V commute. That is, show that for all functions y = f (x) the result of applying H followed by V is the same as applying V followed by H. b Let E be a horizontal dilation by factor a and U be a vertical dilation by factor b. Show that E and U commute. c Let F be a reflection in the y-axis and L be a reflection in the x-axis. Show that F and L commute. d Now suppose that any two of the above transformations are applied in succession. Which pairs of transformations commute and why? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 178 5H Chapter 5 Transformations and symmetry 5H Continuity and piecewise-defined functions Learning intentions • Understand and apply the informal definition of continuity at a point. • Work with piecewise-defined functions and their possible discontinuities. 1 , sketched below, has a break in the curve at its asymptote x = 0, and there are many x other such curves in the course, particularly when we deal with trigonometry. We need a workable definition of continuity at a point, and we also need to test the continuity of functions defined piecewise using cases. A curve such as y = Continuity at a point Continuity at a point means that there is no break in the curve at that point. This is an informal definition, but it will be quite enough for our purposes. 28 Continuity at a point — informal definition • A function f (x) is called continuous at x = a if its graph can be drawn through the point where x = a, on both sides, without lifting the pencil off the paper. • If there is a break in the curve on either side of x = a, the curve is not continuous (or is discontinuous) at x = a. We say then that there is a discontinuity at x = a. y y 4 2 1 3 1 2 −2 −1 −1 1 2 2 x −2 1 has a discontinuity at x = 0, x and is continuous for all other values of x. y= y −2 1 −2 −1 1 2 x 1 −2 2 x y = x2 is continuous for all values of x. −1 −1 1 y= 2 has discontinuities x −1 at x = 1 and at x = −1, and is continuous for all other values of x. A function is never continuous at an endpoint of its domain For a function f (x) be continuous at x = a, we must be able to draw the function through the point, on both sides. √ For example, the function f (x) = x has domain x ≥ 0, but is not continuous at its endpoint at x = 0. y 2 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 4 6 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5H Continuity and piecewise-defined functions 179 An assumption of continuity It is intuitively obvious that a function such as y = x2 is continuous for every value of x. Without further formality, we will make a general assumption of continuity, loosely stated as follows: 29 Assumption about continuity at a point The functions in this course are continuous for every value of x in their domain, except where there is an obvious problem. Piecewise-defined functions When discussing y = |x| in Section 5E, we introduced the term piecewise-defined function — often referred to as a function defined using cases — to mean a function defined differently in different parts of its domain, for example: ⎧ ⎪ 2 ⎪ ⎪ ⎨4 − x , for x ≤ 0, f (x) = ⎪ ⎪ ⎪ ⎩4 + x, for x > 0. A piecewise definition naturally divides the graph into pieces, also called branches. The two branches here are a parabola and line, and our first question is, ‘Do these two pieces join up, like y = |x|, or leave a gap?’. The most obvious approach is to construct two tables of values, one for each line of the function’s definition. Example 28 Do the branches of a piecewise-defined function join up? ⎧ ⎪ 2 ⎪ ⎪ ⎨4 − x , Let f (x) = ⎪ ⎪ ⎪ ⎩4 + x, for x ≤ 0, for x > 0. a Find f (0), then construct tables of values of 4 − x2 on the left of x = 0, and of 4 + x on the right, to establish whether the graph is continuous at x = 0. b Sketch the graph, and hence write down its domain and range. Solution a First, f (0) = 4 − 02 = 4 (using the top line of the definition). The values of 4 − x2 on the left of x = 0, and of 4 + x on the right of x = 0, are x 4 − x2 −2 −1 0 0 3 4 x 0 1 2 4+x 4 5 6 Hence the curve is continuous at x = 0 (where f (0) = 4). Reason: When x = 0, the values of 4 − x2 and 4 + x are both 4, which is the same as f (0), so the two branches join up. b The graph is made up of the parabolic piece on the left, and the linear piece on the right. The domain is all real x, and the range is all real y. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y 4 −2 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 180 5H Chapter 5 Transformations and symmetry Testing whether piecewise-defined functions join up A straightforward way of testing this should now be clear: 30 Testing whether the branches of a piecewise-defined functions join up Suppose that the definition of f (x) changes at x = a. • Find f (a). • Substitute x = a into the formula on the left. • Substitute x = a into the formula on the right. These three values must all exist, and all agree, for f (x) to be continuous at x = a. This method is quick, but the tables of values above may be useful when sketching. Example 29 ⎧ ⎪ ⎪ 2x , ⎪ ⎪ ⎪ ⎪ ⎨ Let f (x) = ⎪ 1, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x2 , Testing for continuity where the definition changes for x < 1, for x = 1, for x > 1. a Establish whether the graph is continuous at x = 1. b Sketch the graph. c Hence write down its domain and range, and any asymptotes. Solution a First, Secondly, f (x) = 1. b y 2 = 2, when x = 1. x Thirdly, x2 = 1, when x = 1. Hence the function has a discontinuity at x = 1. 2 1 1 x c The domain is all real x, and the range is y > 0, and the x-axis is an asymptote on the left. Note: When x = 1, the value of f (x) is 1, so the endpoint (1, 1) of the right-hand branch is included in the graph, as indicated by the closed circle. But the endpoint (1, 2) of the left-hand branch is not included in the graph, as indicated by the open circle there. 31 Open and closed circles, and arrows Use the same conventions for endpoints of curves as for the intervals in Section 2A. • A closed circle marks an endpoint that is included in the graph. • An open circle marks an endpoint that is not included in the graph. • An arrow marks a curve that continues forever. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5H Continuity and piecewise-defined functions 181 Note: The test given in Box 30 does not cover some rather subtle cases — for example, in worked Example 29, replace the formula in the third line of the function definition by definition involving limits, it is the best that we can do. x2 (x − 1) . But in the absence of a x−1 Testing piecewise-defined functions for evenness and oddness • It is best to draw the graph. Then it is usually clear whether the function is even or odd or neither. For example, the two graphs in worked Examples 27 and 28 clearly have neither line symmetry in the y-axis, nor point symmetry in the origin — if required, a single counter-example can settle the matter: — In worked Example 27: f (−2) = 0 and f (2) = 6. — In worked Example 28: f (−1) = 12 and f (1) = 1. • With no graph, when the intervals where the pieces are defined are not balanced around x = 0, it is usually easy to construct a counter-example. • With no graph, and with the intervals balanced around x = 0, then the usual test for oddness and evenness can be applied, as in the next worked example. Example 30 Testing piecewise-defined functions for evenness and oddness Test these piecewise-defined functions for evenness and oddness: ⎧ ⎧ ⎪ ⎪ 5 2 ⎪ ⎪ ⎪ ⎪ ⎨2 − 3x , for x ≥ 0, ⎨ x − x , for x > 0, a f (x) = ⎪ b g(x) = ⎪ ⎪ ⎪ ⎪ ⎪ ⎩3x2 − 2, for x < 0. ⎩ 5x − x, for x < 0. Solution a Suppose x > 0. Then −x is negative, Suppose x < 0. Then −x is positive, 5 5 so f (−x) = so f (−x) = −x − − (−x) −x −x 5 5 = −x = x− x x = f (x), because x > 0. = f (x), because x < 0. Hence f (−x) = f (x) for all non-zero values of x, so the function f (x) is even. b Suppose x ≥ 0. Then −x ≤ 0, so g(−x) = 3(−x) − 2 2 = 3x2 − 2 Suppose x < 0. Then −x > 0, so g(−x) = 2 − 3(−x)2 = 2 − 3x2 = −g(x), because x < 0. = −g(x), because x ≥ 0. This proves that g(−x) = −g(x) for all values of x, so g(x) is odd. Functions in which numerator and denominator have the same zero 3x + 3 The function f (x) = is tricky, because x = −1 is a zero of the numerator and of the denominator. The x+1 function is therefore not defined at x = −1, so it is not continuous there. But what does it look like elsewhere? In this case, we can factor top and bottom. The factors will cancel everywhere except at the common zero: 3(x + 1) f (x) = x+1 = 3, provided that x −1. The graph is drawn to the right. It is the horizontal line y = 3, with the single point (−1, 3) removed. Its domain is x −1, and its range is y = 3. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y 3 −1 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 182 5H Chapter 5 Transformations and symmetry Continuous functions — for clarification only A function f (x) is called continuous if it is continuous at every point in its domain. This concept, however, is rather unhelpful for our purposes. 1 For example, the function y = is continuous at every value of x except x = 0, x 1 which is outside its domain. Hence y = is a continuous function with a x discontinuity at x = 0 !!! y 2 1 −2 The emphasis in this course will thus be on ‘continuity at a point’ rather than on ‘continuous function’. Exercise 5H 1 2 −1 −1 1 2 x −2 FOUNDATION State the zeroes and discontinuities of each function. 5 a f (x) = 6−x 3x b f (x) = (x − 1)(x − 3)(x − 5) x(x + 1) c f (x) = (x + 2)(x + 3) ⎧ ⎪ ⎪ ⎪ ⎨1 − x, for x < 0, Let f (x) = ⎪ ⎪ ⎪ ⎩1 + x2 , for x ≥ 0. a Find f (0). Then evaluate the left branch (1 − x) and right branch (1 + x2 ) at x = 0. b Is f (x) continuous at x = 0? c Sketch the graph, and write down its domain and range. 3 ⎧ ⎪ ⎪ ⎪ ⎨2 − x, Let g(x) = ⎪ ⎪ ⎪ ⎩ x − 1, for x ≤ 1, for x > 1. a Find g(1). Then evaluate the left branch (2 − x) and right branch (x − 1) at x = 1. b Is the function continuous at x = 1? c Sketch the graph, and write down its domain and range 4 Follow similar steps to Questions 2 and 3 in order to decide whether or not each function is continuous at x = 0. Then use the graph to decide whether the function is odd, even or neither. ⎧ √ ⎪ ⎧ ⎪ −1 − −x, for x < 0 ⎪ ⎪ ⎪ 2 ⎪ ⎪ ⎪ ⎪ ⎨ x + 2x, for x < 0 ⎨ a f (x) = ⎪ b f (x) = ⎪ 0, for x = 0 ⎪ ⎪ 2 ⎪ ⎪ ⎩ x − 2x, for x ≥ 0 ⎪ √ ⎪ ⎪ ⎩ 1 + x, for x > 0 ⎧ ⎪ ⎧ ⎪ x2 + 2x, for x < 0 ⎪ ⎪ ⎪ 1 ⎪ ⎪ ⎪ ⎪ − x, for x < 0 ⎨2 ⎨ c f (x) = ⎪ d f (x) = ⎪ 0, for x = 0 ⎪ ⎪ ⎪ ⎪ ⎩ x − 12 , for x ≥ 0 ⎪ ⎪ ⎪2x − x2 , for x > 0 ⎩ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 5H Continuity and piecewise-defined functions 183 DEVELOPMENT 5 6 7 8 Factor numerator and denominator, and hence write down the zeroes and discontinuities. x x2 − 16 1 a f (x) = 2 b f (x) = 2 c f (x) = 2 x − 5x x − 5x + 6 x −9 |x| Draw up a table of values for y = , and explain whether the function is continuous at x = 0. Sketch the x curve, and write down its domain and range. Find whether each function is continuous at x = 2. Then sketch the curve and state the domain and range. ⎧ ⎪ ⎧ ⎪ for x < 2, 3x , ⎪ ⎪ ⎪ 3 ⎪ ⎪ ⎪ ⎪ x , for x ≤ 2, ⎨ ⎨ 2 a f (x) = ⎪ b f (x) = ⎪ 13 − x , for x > 2, ⎪ ⎪ ⎪ ⎪ ⎩10 − x, for x > 2. ⎪ ⎪ ⎪ ⎩4, for x = 2. ⎧ ⎧ 1 ⎪ ⎪ ⎪ ⎪ for 0 < x < 2, x, for x < 2, ⎪ ⎪ ⎪ ⎪ x , ⎪ ⎪ ⎪ ⎪ ⎨ ⎨ 1 c f (x) = ⎪ d f (x) = ⎪ 1 − 4 x, for x > 2, 2 − x, for x > 2, ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ 1 ⎩ , ⎩2, for x = 2. for x = 2. 2 ⎧ ⎪ ⎪ x + 2, for x < −1 ⎪ ⎪ ⎪ ⎪ ⎨ Consider the function f (x) = ⎪ −x, for − 1 ≤ x ≤ 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ x − 2, for x > 1 a Show that f (x) is continuous at x = −1 and x = 1. b Sketch y = f (x). c Simplify f (−x) and hence show that f (x) is an odd function. 9 ⎧ ⎪ ⎪ −2x − 1, ⎪ ⎪ ⎪ ⎪ ⎨ 2 Consider the function f (x) = ⎪ x , ⎪ ⎪ ⎪ ⎪ ⎪ ⎩2x − 1, for x < −1 for − 1 ≤ x ≤ 1 for x > 1 a Show that f (x) is continuous at x = −1 and x = 1. b Sketch y = f (x). c Simplify f (−x) and hence show that f (x) is an even function. 10 In Question 6 and 7 of Exercise 5C you discovered that a polynomial with only odd powers is an odd function, and a polynomial with only even powers is an even function. These conclusions may not be valid for piecewise functions. Follow similar procedures to Questions 8 and 9 above to determine whether these functions are odd, even or neither. Graph each function to confirm your answer. ⎧ ⎪ 3 ⎪ ⎪ ⎨ x − x , for x < 0 a f (x) = ⎪ ⎪ ⎪ ⎩ x3 − x, for x ≥ 0 ⎧ ⎪ 2 ⎪ ⎪ ⎨ x + 1, for x < 0 b g(x) = ⎪ ⎪ ⎪ ⎩ x2 − 1, for x ≥ 0 11 Cancel the algebraic fraction in each function, noting first the values of x for which the function is undefined. Then sketch the curve and state its domain and range: x2 + 2x + 1 x4 − x2 a y= b y= 2 x+1 x −1 x−3 3x + 3 c y= 2 d y= x+1 x − 4x + 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 184 5H Chapter 5 Transformations and symmetry 12 In each case ⎧ find the value of a that makes the function continuous. ⎪ 2 ⎪ ⎪ for x ≤ 1, ⎨ax , a f (x) = ⎪ ⎪ ⎪ ⎩6 − x, for x > 1. ⎧ 2 ⎪ a(x − 9) ⎪ ⎪ ⎪ , for x −3, ⎨ b g(x) = ⎪ x+3 ⎪ ⎪ ⎪12, ⎩ for x = −3. ENRICHMENT 13 a Show that the function h(x) below is discontinuous at x = 1. ⎧ ⎪ ⎪ for x < 1 2x , ⎪ ⎪ ⎪ ⎪ ⎨ h(x) = ⎪ 4, for x = 1 ⎪ ⎪ ⎪ ⎪ ⎪ ⎩3 − x, for x > 1 b What simple change to h(x) would make it continuous? 14 a Suppose that f (x) is discontinuous at x = a, but continuous at x = −a. Explain why f (x) can be neither even nor odd. b Use your answer to decide which of the functions in Question 11 are neither even nor odd. 15 Consider the function z(x) below, where a 0. ⎧ ⎪ 2 ⎪ ⎪ ⎨x , z(x) = ⎪ ⎪ ⎪ ⎩mx + b, for x < a for x ≥ a It is known that z(x) is continuous. It is also known that if the right hand branch were continued to the left 1 it would intersect the x-axis at x = a. Determine m and b in terms of a and hence write z(x) in terms of 2 a alone. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 5 review 185 Chapter 5 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 5 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise Evaluate: a |4| 2 5 6 c |−2 − (−5)| d |−2| − |−5| b |x − 2| = 4 c |2x − 3| = 5 d |3x − 4| = 7 Explain how to shift the graph of y = x2 to obtain each function. a y = x2 + 5 4 b |3 − 8| Solve for x: a |3x| = 18 3 Review 1 b y = x2 − 1 c y = (x − 3)2 d y = (x + 4)2 + 7 Write down the centre and radius of each circle. Shifting may help locate the centre. a x 2 + y2 = 1 b (x + 1)2 + y2 = 4 c (x − 2)2 + (y + 3)2 = 5 d x2 + (y − 4)2 = 64 In each case, find the function obtained by the given reflection or rotation. a y = x3 − 2x + 1: reflect in the y-axis b y = x2 − 3x − 4: reflect in the x-axis c y = 2 x − x: rotate 180◦ about the origin d y= √ 9 − x2 : reflect in the y-axis Classify each function y = f (x) as odd, even or neither. a y b y c y x x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 186 Chapter 5 Transformations and symmetry Review 7 In each diagram below, complete the graph so that: ii f (x) is even. i f (x) is odd, a y b y x 8 Sketch each graph by shifting y = |x|, or by using a table of values. a y = |x| − 2 9 x b y = |x − 2| d y = |x| + 2 Sketch each graph, showing the x-intercept and y-intercept. b y = −|2x − 8| a y = |3x + 9| 10 c y = |x + 2| c y = |4x + 13| Solve these absolute value equations: a |3x| = 15 b |x + 4| = −5 c |2x + 7| = 9 d |7x + 2| = 0 11 Find f (−x) for each function, and then decide whether the function is odd, even or neither. 1 x a f (x) = x + 3 b f (x) = 2x2 − 5 c f (x) = d f (x) = 2 x x +1 12 Complete the square and so use shifting to help sketch the graph of each quadratic function. Indicate the vertex and all intercepts with the axes. a y = x2 + 2x + 3 13 14 b y = x2 − 4x + 1 d y = x2 − x − 1 Complete the squares to find the centre and radius of each circle. a x2 + y2 − 2y = 3 b x2 + 6x + y2 + 8 = 0 c x2 − 4x + y2 + 6y − 3 = 0 d x2 + y2 − 8x + 14y = 35 Write down the equation for each function after the given translations have been applied. b y = 1x : left 2 units, down 3 units a y = x2 : right 2 units, up 1 unit 15 c y = 2 + 2x − x2 Consider the cubic with equation y = x3 − x. a Use an appropriate substitution to show that when the graph of this function is shifted right 1 unit the result is y = x3 − 3x2 + 2x. b [Technology] Plot both cubics using graphing software to confirm the outcome. 16 17 Given that f (x) = 5x − 2 and g(x) = x2 + 3, find: a f (g(0)) b g( f (0)) c f (g(4)) d g( f (4)) e f (g(a)) f g( f (a)) Find the domain and range of the composite functions f g(x) and g f (x) given that: √ 1 a f (x) = x − 1 and g(x) = x b f (x) = and g(x) = x2 + 1 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 5 review 18 The graph drawn to the right shows the curve y = f (x). Use this graph to sketch the following. b y = f (x) + 1 c y = f (x − 1) d y = f (x) − 1 e y = f (−x) f y = − f (x) g y = − f (−x) h y = 2 f (x) i y = f ( 12 x) y 2 1 -3 -2 19 In each case identify how the graph of the second equation can be obtained from the graph of the first by a suitable dilation. 1 1 a y = x2 − 2x and y = 14 x2 − x b y= and y = x−4 2x − 8 20 [A revision medley of curve sketches] Sketch each set of graphs on a single pair of axes, showing all significant points. Use transformations, tables of values, or any other convenient method. a y = 2x, y = 2x + 3, y = 2x − 2 b y = − 12 x, y = − 12 x + 1, y = − 12 x − 2 c y = x + 3, y = 3 − x, y = −x − 3 d y = (x − 2) − 1, y = (x + 2) − 1, y = −(x + 2)2 + 1 e y = x2 , y = (x + 2)2 , y = (x − 1)2 f (x − 1)2 + y2 = 1, (x + 1)2 + y2 = 1, x2 + (y − 1)2 = 1 g y = x2 − 1, y = 1 − x2 , y = 4 − x2 h y = (x + 2)2 , y = (x + 2)2 − 4, y = (x + 2)2 + 1 i y = −|x|, √ y = x, y = −|x| + 1, √ y = x + 1, y = −|x − 2| √ y= x+1 y = 2x , 1 y= , x y = x3 , y = 2 x − 1, 1 y= , x−2 y = x3 − 1, y = 2 x−1 1 y= x+1 y = (x − 1)3 o y = x4 , √ y = x, y = (x − 1)4 , √ y = − x, y = x4 + 1 √ y=2− x p y = 2−x , y = 2−x − 2, y = −2 x + 2 j k l m n 2 2 x Review a y = f (x + 1) 187 Note: In the next two questions, two transformations are said to commute if the order in which they are applied does not matter, whatever graph they are applied to. 21 Which of the following pairs of transformations commute? a reflection in the y-axis and reflection in the x-axis, b vertical reflection and vertical translation, c horizontal translation and horizontal dilation, d vertical translation and horizontal dilation. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 188 Chapter 5 Transformations and symmetry Review 22 Identify the various transformations of the standard functions and hence graph each. Make sure the transformations are applied in the correct order when they do not commute. a y = 4 − x2 b y = 12 (x − 2)2 − 1 23 ⎧ ⎪ 2 ⎪ ⎪ for x ≤ 0, ⎨x , Let f (x) = ⎪ ⎪ ⎪ ⎩ x2 + 1, for x > 0. a Is the curve continuous at x = 0? b Write down the domain and range of the function. 24 ⎧ ⎪ 2 ⎪ ⎪ ⎨ x − 4, for x < 2, Let f (x) = ⎪ ⎪ ⎪ ⎩ x − 2, for x ≥ 2. a Is the curve continuous at x = 2? b What are the domain and range of the function? 3x is even. 32x + 1 25 Prove that the function f (x) = 26 Every function f (x) defined for all real numbers can be written as the sum f (x) = g(x) + h(x), where g(x) is an even function and h(x) is an odd function. Prove this statement by finding formulae for g(x) and h(x) in terms of f (x). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6 Further graphs Chapter introduction This chapter contains various loosely related discussions about functions and their graphs. Sections 6A–6B deal with some trickier types of inequations, using algebraic and graphical–geometric methods. Sections 6C–6E introduce further transformations of the graphs of functions. The graph of the reciprocal of a graphed function is drawn — a procedure that requires a little more discussion of asymptotes. The sum and difference of two graphed functions is drawn. And the compositions of a function with the absolute value function are drawn using reflections in the x-axis and y-axis. Section 6F–6G introduce inverse relations and functions graphically, with their corresponding reflections in the diagonal line y = x. This is yet another type of transformation of a known graph. Section 6G develops the formal notation for inverse functions. The last Section 6H introduces parameters so that equations of functions, and relations in general, can be expressed and graphed in terms of functions x and y of a single parameter. As always, computer sketching of curves is very useful in demonstrating how the features of a graph are related to the algebraic properties of its equation, and in gaining familiarity with the variety of graphs and their transformations. Some questions in Sections 6C and 6H use the trigonometry of the general angle and Pythagorean identities, and these questions could be delayed until after trigonometry is reviewed and extended in Chapter 7. The chapter is conceptually demanding, particularly Section 6D. Readers may prefer to leave Section 6D until later in the year — an appropriate place could be before Chapter 16: Further trigonometry, when the sum a sin x + b cos x of two trigonometric graphs needs to be sketched. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 190 6A Chapter 6 Further graphs 6A Solving two particular inequations Learning intentions • Solve absolute value inequations of the form |ax + b| < k in three ways. • Solve inequations with x in the denominator by multiplying through by its square. This section is devoted to two particular types of inequations: • Absolute value inequations of the form |ax + b| < k. 5 ≥ 1. x−4 The next Section 6B will solve inequations using a table of test points. • Inequations where x is in the denominator, such as A review of inequations solved so far So far we have solved linear and quadratic inequations: • We solved linear inequations like linear equations, except that when multiplying or dividing by a negative number, the inequality is reversed. • We solved quadratic inequations using the graph, and also using test points after finding the zeroes. Remember that we solve inequations such as x + 2 < 3, and we prove inequalities such as (x − 2)2 ≥ 0, but that the word ‘inequality’ is often used for both meanings. Do not be surprised if you are asked to ‘solve an inequality’. Method 1 for solving an absolute value inequation — Sketch the graph Our task is to solve inequations |ax + b| < k, where a 0, b, and k are constants (and the inequality sign is < , or ≤ , or > , or ≥ ). Drawing a graph is probably the clearest method of solution. Taking as a standard example the inequation |2x + 1| > 3: 1 Solving an absolute value inequation such as |2x + 1| > 3 graphically • Solve the equation |2x + 1| = 3 using the methods of Section 5E, Box 19. • Hence draw the graphs of y = |2x + 1| and y = 3 on the one set of axes. • Read the solution off the graph. Example 1 Solving an absolute value inequation graphically Solve |2x + 1| > 3 graphically. Solution First solve the equation |2x + 1| = 3 y 2x + 1 = 3 or 2x + 1 = −3 x=1 or − 2. The graph of y = |2x + 1| is V upwards, with x-intercept x = − 12 . From the sketch of |2x + 1| and y = 3 together, the solution of the inequality is x < −2 or 3 1 -2 - 12 1 x x > 1. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6A Solving two particular inequations 191 Method 2 for solving an absolute value inequation — Use distance on the number line We can use distance on the number line to solve the inequality. But we must be careful in dealing with the coefficient of x, particularly when it is negative. First, we review formulae for absolute value using distance on the number line: 2 Absolute value and distance |x| 0 |x - a| x a x • |x| = distance from x to zero on the number line • |x − a| = distance from x to a on the number line The easy case, where the coefficient of x is 1 When the coefficient of x is 1, we can immediately use the second formula above. Example 2 The coefficient of x is 1 Solve these inequations on the number line. b |x + 3| > 4 a |x − 2| ≤ 5 Solution |x − 2| ≤ 5 a |x + 3| > 4 b (distance from x to 2) ≤ 5 |x − (−3)| > 4 x (distance from x to −3) > 4 so −3 ≤ x ≤ 7. x so x < −7 or x > 1. The general case, where the coefficient of x is not 1 There are two initial steps to get it into the form where the coefficient of x is 1. 3 Solving |ax + b| < k, for any non-zero value of a Step 1: Step 2: Step 3: Force a to be positive by writing say | − 2x + 3| as |2x − 3|. Divide through by the now positive number a. Solve using distance, as in the easy case above. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 192 6A Chapter 6 Further graphs Example 3 The coefficient of x is not 1 Solve these inequations using distance on the number line. b |7 − 14 x| ≥ 3 a | − 3x − 7| < 3 Solution | − 3x − 7| < 3 a |3x + 7| < 3 Rewrite with a positive, |7 − 14 x| ≥ 3 b Rewrite with a positive, | 14 x − 7| ≥ 3 ×4 |x − 28| ≥ 12 |x − (−2 13 )| < 1 ÷3 (distance from x to −2 13 ) < 1 3 13 2 13 (distance from x to 28) ≥ 12 1 13 x x so x ≤ 16 or x ≥ 40. so −3 13 < x < −1 13 . Method 3 for solving an absolute value inequation — Algebraic approach We saw in Section 5E how an absolute value equation of the form | f (x)| = k can be solved algebraically by rewriting the equation. Rewrite an equation | f (x)| = k as f (x) = k or f (x) = −k. A similar approach can be taken to solving an inequation such as | f (x)| < k or | f (x)| > k. 4 Solving an absolute value equation or inequation algebraically • Rewrite an equation | f (x)| = k as f (x) = k or f (x) = −k. • Rewrite an inequation | f (x)| < k as −k < f (x) < k. • Rewrite an inequation | f (x)| > k as f (x) < −k or f (x) > k. Example 4 Solving an absolute value inequation algebraically a Solve |9 − 2x| = 5. b Solve |9 − 2x| < 5. c Solve |9 − 2x| ≥ 5. Solution a As in Section 5E, the first step in solving the equation |9 − 2x| = 5 is 9 − 2x = 5 or 9 − 2x = −5, giving two solutions, x = 2 or x = 7. c Using the third dotpoint, b Using the second dotpoint, |9 − 2x| < 5 |9 − 2x| ≥ 5 −5 < 9 − 2x < 5 9 − 2x ≤ −5 or 9 − 2x ≥ 5 −9 −14 < −2x < −4 −9 ÷ (−2) 7>x>2 ÷ (−2) x ≥ 7 or x ≤ 2 that is, 2 < x < 7. that is, x ≤ 2 or x ≥ 7 . −2x ≤ −14 or −2x ≥ −4 Inequations that have no solutions, or are inequalities The absolute value | f (x)| cannot be negative. Thus if k is negative: • | f (x)| = k and | f (x)| < k have no solutions, and • | f (x)| ≥ k is true for all values of x in the domain of f (x). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6A Solving two particular inequations Example 5 193 Inequations that have no solutions, or are inequalities Solve these absolute value inequations algebraically: b |6 − x| ≥ −1 a |x − 7| < −5 Solution a The absolute value is never negative, so there are no solutions. b Absolute value is always at least zero, so every real number is a solution. Thus the inequation is an inequality. Solving inequations with x in the denominator — multiply through by its square When we try to solve this inequation, we immediately run into a problem: 5 ≥ 1. x−4 The denominator x − 4 is sometimes positive and sometimes negative. Thus if we were to multiply both sides by the denominator x − 4, the inequality symbol would reverse sometimes and not other times. To avoid using cases, the most straightforward approach is to multiply through instead by the square of the denominator, which is always positive or zero. 5 Solving an inequation with the variable in the denominator • Multiply through by the square of the denominator. Be careful to exclude the zeroes of the denominator from the solutions. Once the fractions have been cleared, there will usually be common factors on both sides. These should not be multiplied out, because the factoring will be easier if they are left unexpanded. An alternative approach using a table of signs is presented in the next section. Example 6 Solve Solving an inequation by multiplying through by the square 5 ≥ 1. x−4 Solution First, x 4, because the LHS is undefined when x = 4. y The key step is to multiply both sides by (x − 4)2 . × (x − 4)2 5(x − 4) ≥ (x − 4)2 , and x 4, Don’t expand the brackets here! You will only have to re-factor. (x − 4)2 − 5(x − 4) ≤ 0, and x 4, 4 9 x (x − 4)(x − 4 − 5) ≤ 0, and x 4, (x − 4)(x − 9) ≤ 0, and x 4. From the diagram, 4 < x ≤ 9. Notice that x = 4 is not a solution of the original inequation because substituting x = 4 into the LHS of the 5 original inequation gives , which is undefined. 4−4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 194 6A Chapter 6 Further graphs Exercise 6A 1 Solve for x using distance on the number line. a |x| = 7 2 3 4 FOUNDATION b |x| = 0 c |x| ≤ 2 d |x| > 5 e |x| < 14 f |x| ≥ 32 Solve: a |2x| < 6 b |5x| ≥ −3 c |3x| ≥ 6 d |4x| ≤ 2 e |12x| < −18 f |3x| > 5 Multiply both sides by x2 and hence solve: 3 1 a b >1 <1 x x 3 1 d 4+ ≥0 e > −1 x x 1 ≥2 x 4 f ≤ −2 x c Collect like terms where necessary then multiply by x2 to solve: 3 1 a − >2 b − ≤1 x x 2 1 d 1+ >3 e 1 − ≤ −2 x x 1 ≥1 x 2 f 3− <1 x c 2+ DEVELOPMENT 5 a Using distance on the number line, solve: i |x − 1| < 2 ii |x − 5| ≥ 4 iii |x + 1| > 3 iv |x + 8| ≤ 6 b Next, solve the inequations in parts a graphically. c Finally, solve the inequations in part a algebraically. 6 Multiply both sides of each inequation by the square of the denominator and hence solve it. Do not multiply out any common factor. 2 3 2 a b c ≤1 >1 ≥2 x+1 x−3 x+4 5 2 4 d e f <1 >1 ≤ −1 2x − 3 3−x 5 − 3x 7 In each case, first simplify the inequation, then use multiplication by the square of the denominator to solve for x. 1 1 2 > −3 <3 ≤2 a 1+ b 2− c 3+ x−3 x+4 x−2 8 a Solve by any suitable method, and graph the solution on a number line. i |x − 2| < 3 ii |3x − 5| ≤ 4 iii |x − 7| ≥ 2 iv |2x + 1| < 3 v |6x − 7| > 5 vi |5x + 4| ≥ 6 b Solve the inequations in parts ii, iv, and vi by the other two of the three methods that you did not use in part a. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6A Solving two particular inequations 9 First simplify each inequation, then solve for x. b |x − 2| − 1 > 3 a |x + 1| + 2 < 3 10 11 195 Solve for x: x−1 > −2 a x+2 5x ≥3 d 2x − 1 c |x − 3| − 1 ≥ 4 2−x ≥3 x−1 2x + 5 <1 e x+3 x+1 ≤2 x−1 4x + 7 > −3 f x−2 b c a Show that the double inequation 2 ≤ |x| ≤ 6 has solution 2 ≤ x ≤ 6 or −6 ≤ x ≤ −2. b Similarly solve: ii 1 ≤ |2x − 5| < 4 i 2 < |x + 4| < 6 12 13 Say whether each statement is true or false. If it is false, give a counterexample. a x2 > 0 b x2 ≥ x c 2x > 0 e 2x ≥ x f x+2> x g x ≥ −x 1 x h 2x − 3 > 2x − 7 c |5 − x| ≥ −6 d |3x − 5| ≤ 0 d x≥ Solve these equations and inequations. a |4 − 5x| = −2 b |3x + 2| < −7 ENRICHMENT 14 1 Consider the inequation x + < 2x. x a Explain why x must be positive. b Hence solve the inequation. 2 . x 15 Solve the inequation 1 + 2x − x2 ≥ 16 Consider the inequation |x − a| + |x − b| < c, where a < b. a If a ≤ x ≤ b, show, using distances on a number line, that there can only be a solution if b − a < c. a+b+c . 2 a+b−c . c If x < a, show, using distances on a number line, that x > 2 a + b c < or there is no solution to the original problem. d Hence show that either x − 2 2 e Hence find the solution of |x + 2| + |x − 6| < 10. b If b < x, show, using distances on a number line, that x < CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 196 6B Chapter 6 Further graphs 6B The sign of a function Learning intentions • Construct a table of test points dodging around zeroes and discontinuities. • Use the table of test points to solve various types of inequation. • Use the table of test points as a first step in curve-sketching. This section completes the story of a very general method of using test points to analyse the sign of any function, provided only that we can find its zeroes and discontinuities. The pattern of signs produced by these test points can be used to solve inequations, which are the immediate point of this section. But the pattern also provides the basis for sketching curves of every kind, and will be central particularly to curve sketching using calculus in Year 12. Where can a function change sign? Sketching factored cubics in Section 3G, and solving quadratic inequations in Section 4B, used a table of test values dodging around the zeroes to see where the function changed signs. The functions in this chapter, however, may also have breaks or discontinuities, and we need to dodge around them as well, as the diagram below demonstrates: 6 Where can a function change sign? The only places where a function may possibly change sign are zeroes and discontinuities. y e a b c f x d • The graph above has discontinuities at x = c, x = d, x = e, and x = f , and has zeroes at x = a and x = b. • The function changes sign at the zero x = a and at the discontinuities x = c and x = e, and nowhere else. • Notice that it does not change sign at the zero x = b or at the discontinuities x = d and x = f . The statement in Box 6 goes to the heart of what the real numbers are and what continuity means. In this course, the sketch above is sufficient justification. A table of signs As a consequence, we can examine the sign of a function using a table of test values dodging around any zeroes and discontinuities. We add a third row for the sign, so that the table becomes a table of signs. 7 Examining the sign of a function To examine the sign of a function, draw up a a table of signs using test values that dodge around all the zeroes and discontinuities. Finding the zeroes of a function has been a constant concern ever since quadratics were introduced in earlier years. To find discontinuities, we will be using our standard assumption that all the functions in the course are continuous everywhere in their domains, except where there is an obvious problem. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6B The sign of a function 197 Solving polynomial inequations using a table of signs A simpler form of this table of signs was introduced in Section 3G to sketch cubics. Cubics, and more generally polynomials, do not have any discontinuities, so the test values only need to dodge around their zeroes — if we can find them. Section 3G’s concern was to sketch the function, whereas here the concern is to solve inequations. The graphs are not necessary here, and our sketches will be incomplete, but the sketches allow us to see the whole situation. Example 7 Sketching a factored polynomial and solving an inequality a Draw up a table of signs of the function y = (x − 1)(x − 3)(x − 5). b State where the function is positive and where it is negative. c Solve the inequation (x − 1)(x − 3)(x − 5) ≤ 0. d Confirm the answers by sketching what is now known about the graph. Solution a There are zeroes at 1, 3 and 5, and no discontinuities. x 0 1 2 3 4 5 6 y −15 0 3 0 −3 0 15 sign − 0 + 0 − 0 + b Hence y is positive for 1 < x < 3 or x > 5, and negative for x < 1 or 3 < x < 5. c x ≤ 1 or 3 ≤ x ≤ 5. Example 8 d y 1 3 5 x -15 Factoring a polynomial, solving an inequality, and sketching the curve Solve x3 + 1 ≤ x2 + x using a table of signs. Then confirm with a sketch. Solution Move all terms to the left, then factor by grouping, y x −x −x+1≤0 3 2 x (x − 1) − (x − 1) ≤ 0 2 −1 (x − 1)(x − 1) ≤ 0 2 1 1 (x + 1)(x − 1)2 ≤ 0. x The LHS has zeroes at 1 and −1, and no discontinuities. x −2 −1 0 1 2 y −9 0 1 0 3 sign − 0 + 0 + Hence x ≤ −1 or x = 1. Solving inequations involving discontinuities When the function has discontinuities, the method is the same, except that the test values now need to dodge around discontinuities as well as the zeroes. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 198 6B Chapter 6 Further graphs Again, a sketch is useful for understanding. Because horizontal asymptotes are in Year 12, however, a clue is needed before the graph really tells the story. Example 9 Examining the sign in a function with discontinuities x−1 , and examine its sign. x−4 b i Write x − 1 = (x − 4) + 3, hence find the horizontal asymptote. ii Identify the vertical discontinuities of the function, and sketch it. a Find the zeroes and discontinuities of y = y Solution a There is a zero at x = 1, and a discontinuity at x = 4. x 0 1 2 4 5 y 1 4 0 − 12 ∗ 4 sign + 0 − ∗ + 1 1 x 4 Hence y is positive for x < 1 or x > 4, and negative for 1 < x < 4. b (x − 4) + 3 x−4 3 =1+ x−4 Hence y → 1 as x → ∞ and as x → −∞, so y = 1 is a horizontal asymptote in both directions. ii As x → 4− , y → −∞, and as x → 4+ , y → ∞, so x = 4 is a vertical asymptote. i y= Example 10 Examining the sign in a situation that is trivial Find the zeroes and discontinuities of y = 1 , and examine its sign. 1 + x2 Solution The denominator 1 + x2 is always at least 1. Hence the function is defined for all real x, and is always positive. A table of signs is now unnecessary, but we can draw one up. There are no discontinuities and no zeroes, so one test value f (0) = 1 confirms that the function is always positive. x 0 y 1 sign + Comparing the methods of Sections 6A and 6B We now have two ways to solve an inequation with x in the denominator. For comparison, here is an inequality with x in the denominator solved both ways. First it is solved by multiplying both sides by the square of the denominator. Then it is solved using a table of signs. Compare the two quite different approaches. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6B The sign of a function Example 11 Solve 199 Constructing a table of test points 3 ≤ x using a table of signs. x+2 Solution 3 − x ≤ 0, x+2 3 − x2 − 2x using a common denominator, ≤ 0, x+2 (3 + x)(1 − x) ≤ 0. and factoring, x+2 The LHS has zeroes at x = −3 and x = 1, and a discontinuity at x = −2. Collecting everything on the left, x −4 −3 −2 12 −2 0 1 2 LHS 2 12 0 −3 12 ∗ 1 12 0 −1 14 sign + 0 − ∗ + 0 − So the solution is x ≥ 1 or −3 ≤ x < −2. (No sketch here — rely just on the table of signs.) Example 12 Solve Multiplying through by the square of the denominator 3 ≤ x by first multiplying though by the square of the denominator. x+2 Solution Notice that the common factor (x + 2) is never multiplied out. That would cause a serious problem, because of the effort required to re-factor the expanded cubic. y 3 ≤x x+2 × (x + 2)2 3(x + 2) ≤ x(x + 2)2 , and x −2, x(x + 2) − 3(x + 2) ≥ 0, and x −2, (x + 2)(x + 2x − 3) ≥ 0, and x −2, (x + 2)(x + 3)(x − 1) ≥ 0, and x −2. 2 2 −2 −3 1 x The LHS can now be sketched using a table of signs. From the graph, x ≥ 1 or −3 ≤ x < −2. Exercise 6B FOUNDATION The purpose of this exercise is to solve inequations using a table of signs. In the case of polynomials, this table also allows the sketch to be drawn, as in Section 3G. A sketch makes the situation clearer, but the sketch is not necessary for obtaining the solution. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 200 6B Chapter 6 Further graphs 1 Use the given graph of the LHS to help solve each inequation. a y y b 1 c x 2 -2 y 2 y e 3 3 4 (x − 3)2 (x + 3)2 ≤ 0 a Explain why the zeroes of y = (x + 1)2 (1 − x) are x = 1 and x = −1. x(x − 3)2 (x + 3)2 ≥ 0 x −2 −1 0 1 2 y sign Use the three steps of the previous question to solve each inequation. a (x + 1)(x + 3) < 0 b x(x − 2)(x − 4) ≥ 0 c (x − 1)(x + 2)2 ≥ 0 d x(x − 2)(x + 2) ≤ 0 e (x − 2)x(x + 2)(x + 4) > 0 f (x − 1)2 (x − 3)2 ≤ 0 First factor each polynomial completely, then use the methods of Questions 2 and 3 to sketch its graph. (Hint: take out any common factors first.) a f (x) = x3 − 4x 5 x x Then copy and complete the table of signs. b Use the table of signs to solve (x + 1)2 (1 − x) ≥ 0. c Sketch the graph to confirm the solution in part b. 3 y -3 -3 2 x(x − 3)2 > 0 f 4 x x2 (x − 4) ≥ 0 x 3 4 x x(x + 2)(x − 2)(x − 4) < 0 x(x − 1)(x − 2) ≤ 0 d y b f (x) = x3 − 5x2 c f (x) = x3 − 4x2 + 4x From the graphs in the previous question, or from the tables of signs used to construct them, solve the following inequations. Begin by getting all terms onto the one side. a x3 > 4x b x3 < 5x2 c x3 + 4x ≤ 4x2 DEVELOPMENT 6 Here is Question 10 from Exercise 6A. This time, collect all terms on the LHS as a simplified single fraction. Find the zeroes and discontinuities to use in a table of signs, and hence solve each inequation. 2−x x+1 x−1 > −2 ≥3 ≤2 a b c x+2 x−1 x−1 5x 2x + 5 4x + 7 ≥3 <1 > −3 d e f 2x − 1 x+3 x−2 7 a Find the zeroes and discontinuities of y = b Hence solve x2 < 0. x−3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x2 and construct a table of signs. x−3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6B The sign of a function 8 If necessary, collect all terms on the LHS and factor. Find any zeroes and discontinuities, then draw up a table of signs with interlacing values in order to solve the inequation. x−4 ≤0 a (x − 1)(x − 3)(x − 5) < 0 b (x − 1)2 (x − 3)2 > 0. c x+2 x+3 x2 d x3 > 9x e f <0 <0 x+1 x−5 g x4 ≥ 5x3 9 201 h x2 − 4 ≥0 x i x−2 ≤0 x2 + 3x a Factor each equation completely, and hence find the x-intercepts of the graph. Factor parts ii and iii by grouping in pairs. i y = x3 − x ii y = x3 − 2x2 − x + 2 iii y = x3 + 2x2 − 4x − 8 b For each function in the previous question, examine the sign of the function around each zero, and hence draw a graph of the function. 10 Find all zeroes of these functions, and any values of x where the function is discontinuous. Then analyse the sign of the function by taking test points around these zeroes and discontinuities. x−4 x+3 x a f (x) = b f (x) = c f (x) = x−3 x+2 x+1 11 Multiply through by the square of the denominator, collect all terms on one side and then factor to obtain a factored cubic. Sketch this cubic by examining the intercepts and the sign. Hence solve the original inequation. 2 8 4 a b c ≥x <x ≤ 2x − 1 x+3 2x + 3 2x − 3 12 Solve the inequations in the previous question by the alternative method of collecting everything on the LHS, finding a common denominator, identifying zeroes and discontinuities, and drawing up a table of signs. ENRICHMENT 13 a Prove that f (x) = 1 + x + x2 is positive for all x. b Prove that f (x) = 1 + x + x2 + x3 + x4 is positive for all x. Consider separately the three cases x ≥ 0, −1 < x < 0 and x ≤ −1. Group the five terms into pairs in different ways with the second and third cases. c Use similar methods to prove that for all integers n ≥ 0, f (x) = 1 + x + x2 + · · · + x2n−1 + x2n is positive for all x. d Prove that x = −1 is the only zero of f (x) = 1 + x + x2 + · · · + x2n−1 , for all positive integers n. 14 1 1 a − with a > 1. Let f (x) = 2 x+a x−a a For what values of k does f (x) = k have a solution? b Solve f (x) < 2 using any appropriate method. c Now choose a suitable value for a and use shifting to solve CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 4 1 − < . x−1 x−7 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 202 6C Chapter 6 Further graphs 6C Sketching reciprocal functions Learning intentions • Sketch the reciprocal of a function given the graph of the function. • Determine its vertical and horizontal asymptotes, and its domain and range. Section 6C–6E present further ways to transform or combine graphs. In each case, the emphasis will be more on the relationship between the graphs than on the equations of the various functions. 1 of a graphed function f (x). Reading from the graph of This section sketches the reciprocal function g(x) = f (x) y = f (x), we develop properties of this reciprocal function y = g(x), and use these properties to sketch y = g(x). The reciprocal trig functions — cosec, sec, and cot — are developed in Chapter 7. The sign and horizontal asymptotes There are many small details involved in sketching reciprocals, so it is better to give several examples, then follow them with some general statements. Example 13 Analysing sign and horizontal asymptotes The sketch shows y = 2 x . Let g(x) = 1/ f (x) be the reciprocal function of f (x). Reasoning from the sketch: y a What sign do the values of y = g(x) have? 2 b What is the y-intercept of y = g(x)? 1 c What happens to y = g(x) as x → ∞? 1 2 d What happens to y = g(x) as x → −∞? -1 e Hence sketch y = g(x). 1 x f What are the domain and range of f (x) and g(x)? Solution a f (x) is always positive, so its reciprocal is always positive. y e b f (0) = 1, and the reciprocal of 1 is 1, so g(0) = 1. c As x → +∞, f (x) → ∞, so g(x) → 0+ , so the x-axis is a horizontal asymptote on the right. d As x → −∞, f (x) → 0+ , so g(x) → +∞. f They both have domain all real x, and range y > 0. 2 1 -1 1 x −x Note: g(x) = 2 is, of course, the reflection of y = f (x) in the y-axis. But the intention here is to argue using reciprocals. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6C Sketching reciprocal functions 8 203 Graphing the reciprocal — sign and horizontal asymptotes Let f (x) be a graphed function, and let g(x) = 1/ f (x) be its reciprocal function. • When f (x) is small and positive, g(x) is large and positive. When f (x) is large and positive, g(x) is small and positive. • When f (x) = 1, g(x) = 1, and vice versa. • If f (x) → +∞ as x → ∞, then g(x) → 0+ as x → ∞, and the x-axis is a horizontal asymptote on the right. • If f (x) → 0+ as x → ∞, then g(x) → +∞ as x → ∞. The last two dotpoints have equivalent behaviour on the left. All four dotpoints have equivalent behaviour when f (x) is negative. Zeroes and vertical asymptotes Again, an initial example should makes things clear. Example 14 Analysing zeroes and vertical asymptotes. Sketch f (x) = x − 2, then with g(x) = 1/ f (x), and reasoning from the sketch: a Where is g(x) undefined? b Does y = g(x) have any zeroes? c Where is y = g(x) above and below the x-axis? d Identify any vertical asymptotes. e Identify any horizontal asymptotes. f What is the y-intercept of y = g(x)? g Where does y = g(x) intersect with y = f (x)? h Hence sketch y = g(x) on the same set of axes. i What are the domain and range of f (x) and g(x)? Solution a Zero has no reciprocal, so g(x) is undefined at the zero x = 2 of f (x). b The number zero is not the reciprocal of any number, so g(x) is never zero. c g(x) is positive when f (x) is positive, and negative when f (x) is negative. d As x → 2+ , f (x) → 0+ , so g(x) → +∞. As x → 2− , f (x) → 0− , so g(x) → −∞. Hence x = 2 is a vertical asymptote. e As x → +∞, f (x) → ∞, so g(x) → 0+ . As x → −∞, f (x) → −∞, so g(x) → 0− . Hence the x-axis is a horizontal asymptote left and right. f When x = 0, f (0) = −2, so g(−2) = − 12 . g The only numbers that are their own reciprocal are 1 and −1. Hence y = f (x) and y = g(x) intersect at (1, −1) and (3, 1). y h 2 y f (x) y 1 f (x) x i Domain: For f (x), all real x. For g(x), x 2. Range: For f (x), all real y. For g(x), y 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 204 6C Chapter 6 Further graphs 9 Graphing the reciprocal — zeroes and vertical asymptotes Let f (x) be a graphed function, and let g(x) = 1/ f (x) be its reciprocal function. • Zero is the only number that has no reciprocal. Hence g(x) is undefined at every zero of f (x). • Zero is the only number that is not a reciprocal of any number. Hence g(x) is never zero. • When f (x) has a zero, y = g(x) normally has a vertical asymptote (but check, because there are some rather bizarre exceptions). Domain and range These deserve particular attention. 10 Graphing the reciprocal — domain and range Let f (x) be a graphed function, and let g(x) = 1/ f (x) be its reciprocal function. • The domain of g(x) is the domain of f (x) with all the zeroes of f (x) removed. • g(x) is never zero, so the range of g(x) does not contain zero. Two warnings about zero, infinity, and reciprocals • Infinity is not a number: Don’t ever be tempted to say that if f (x) has an asymptote at x = a, then the function g(x) = 1/ f (x) is zero at x = a. The symbols ∞ and −∞ are part of mathematics, but they are not numbers, and they certainly do not have reciprocals. • The reciprocal of the reciprocal may not be the original function: y y= 1 f (x) Referring to our previous example, start instead with reciprocal function, that is, 1 - 21 . Then f (x) is undefined at x = 2, so its reciprocal g(x) = 1/ f (x) let f (x) = 1 y = f (x) x−2 1 is also undefined at x = 2. x 23 -1 That is, g(x) = x − 2, where x 2. -2 This is not the original line y = x − 2 of the example, because its zero x = 2 has now been removed. Sketching the reciprocal from the graph, with no equation In the previous examples, the equation of the function was there for reference. But in the two examples below, only the sketch graph is given. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6C Sketching reciprocal functions Example 15 205 Sketching the reciprocal from a graph Given each sketch of y = f (x) below, sketch g(x) = 1/ f (x), then state the domain and range of g(x). y a y b 2 2 -4 3 -3 4 x x -2 -2 Solution y a The point (−3, 2) is called a local maximum because for a small region around x = −3, f (x) is greatest at x = −3. Similarly, (3, −2) is called a local minimum. Corresponding, g(x) has a local minimum at (−3, 12 ) and a local maximum at (3, − 12 ). The domain of g(x) is x −4, 0 or 4 (remove the zeroes), and the range of g(x) is y 0. 1 2 -4 -3 3 - 21 b As x → ∞, f (x) → 2, so g(x) → 12 , so y = 12 is a horizontal asymptote on the right. Similarly, as x → −∞, f (x) → −2, so g(x) → − 12 . Thus y = − 12 is a horizontal asymptote on the left. Domain of g(x): x 0. Range: y < − 12 or y > 12 . 4 x y 1 2 - 21 x Local maxima and minima, and horizontal asymptotes Part a introduced the idea of local maxima and minima, and their relationship with the reciprocal function. Part b showed how to deal with horizontal asymptotes other than the x-axis. Exercise 6C FOUNDATION Note: Question 9 and 13 concern the graphs of the three trigonometric functions sin θ, cos θ, and tan θ. The graphs are drawn in the questions, but readers unfamiliar with them may like to delay attempting these questions until after studying Chapter 7. 1 1 after carrying out the following steps. x−1 i State the natural domain, and find the y-intercept. ii Find any points where y = 1 or y = −1. iii Explain why y = 0 is a horizontal asymptote. iv Draw up a table of values and examine the sign. v Identify any vertical asymptotes, and use the table of signs to write down its behaviour near any vertical asymptotes. a Sketch y = b Repeat for y = 2 . 3−x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c Repeat for y = − 2 . x+2 d Repeat for y = 5 . 2x + 5 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 206 6C Chapter 6 Further graphs 2 Follow steps (i)–(v) of Question 1a to investigate the function y = Show any points where y = 1. 3 2 and hence sketch its graph. (x − 1)2 Investigate the domain, zeroes, sign and asymptotes of the function y = − graph. Show any points where y = −1. 4 1 and hence sketch its (x − 2)2 a Let f (x) = x + 1. i Graph y = f (x) showing the intercepts with the axes. ii Also show the points where f (x) = 1 and f (x) = −1. 1 . f (x) iii Hence on the same number plane sketch y = b Follow similar steps for the function g(x) = 2 − x. DEVELOPMENT 5 Let y = f (x) where f (x) = 13 (x + 1)(x − 3). √ √ a Show that y = 1 at x = 1 − 7 and x = 1 + 7 . Plot these points. b Complete the graph of y = f (x) showing the vertex, the intercepts with the axes and the points where f (x) = −1 . c What is the range of y = f (x)? 1 on the same number plane. d Hence sketch y = f (x) 1 e What is the range of y = ? f (x) 6 a Follow similar steps to Question 5 to sketch y = 14 (4 − x2 ) and y = b What is the range of y = 14 (4 − x2 )? c What is the range of y = 7 4 . 4 − x2 4 ? 4 − x2 a Sketch y = 12 (x2 + 1) showing the points where y = 1. b What is the minimum value of 12 (x2 + 1)? c Sketch y = 2 x2 + 1 on the same number plane. d Explain why the x-axis is an asymptote to y = e What is the maximum value of 8 2 x2 + 1 2 x2 + 1 . ? Let f (x) = −x2 + 2x − 3. a What is the maximum of f (x)? 1 on the same number plane. f (x) 1 . c Explain why the x-axis is an asymptote for y = f (x) 1 d What is the minimum of ? f (x) b Sketch y = f (x) and y = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6C Sketching reciprocal functions 9 207 The diagrams show first y = cos θ then y = sin θ for 0◦ ≤ θ ≤ 360◦ . y y 1 180° q 360° -1 a 1 -1 360° q 1 . cos θ i Copy the sketch of y = cos θ, and add to it the sketch of y = ii What are the domain and range of y = b 180° 1 in this interval ? cos θ i Copy the sketch of y = sin θ, and add to it the sketch of y = ii What are the domain and range of y = 1 . sin θ 1 in this interval ? sin θ 10 Prove that the symmetry of a function is preserved when taking reciprocals. That is, prove that the reciprocal of an even function is an even function, and prove that the reciprocal of an odd function is an odd function. 11 a Graph y = 12 Sketch the reciprocal of each function shown, showing all significant features. 2 2+x by first noting that y = 1 + . x x x . b Hence graph y = 2+x a y 1 1 2 3 (-2, 1) x -Ö3 (-1, -1) -3 y c y b d (1, 1) Ö3 x (2, -1) y 1 2 2 -1 1 3 x x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 208 6C Chapter 6 Further graphs 13 The sketch shows y = tan θ for 0◦ ≤ θ ≤ 360◦ . y a What is the domain of y = tan θ in this interval? 1 b Where is tan θ = 0 in this interval? 1 in this interval. c Hence state the domain of y = tan θ 1 ? d What is lim ◦ θ→90 tan θ e Copy the graph of y = tan θ, and add to it the graph of y = f What is the range of the reciprocal function? 14 -1 180° 360° q 1 . tan θ In each case the graph of y = f (x) is given. Sketch the graph of y = 1 , paying careful attention to the f (x) domain, any asymptotes, and any relevant limits. a y 1 y b 1 2 x -1 2 1 x ENRICHMENT 15 The arguments in the solution to Example 15a seems to rely on the following assertion about the graphs of a function y = f (x) and its reciprocal function y = ( f (x))−1 : ‘When one curve has a local maximum, the other has a local minimum.’ This statement is not strictly true. State the qualification that needs to be made in this statement, and give an example where the qualification is necessary. 16 Let y = 1 . Write down precisely the equation of the reciprocal function. x−2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6D Sketching sums and differences 209 6D Sketching sums and differences Learning intentions • Sketch the sum and difference of two graphed functions. The problem addressed in this section is to take the sketches of two functions f (x) and g(x), and working just from those sketches, sketch their sum and difference: s(x) = f (x) + g(x) d(x) = f (x) − g(x). and The pronumerals s(x) and d(x) used here are not any mathematical convention, they are just convenient notation in this section. The word ‘ordinate’ The ordinate of a point is the y-coordinate, or more generally, the coordinate on the vertical axis. The operations in this section and the next routinely act on the y-coordinates of points, so this shorter word is useful. The coordinate on the horizontal axis is sometimes referred to as the ‘abcissa’, plural ‘abcissae’, but that word is not necessary in the course. Sketching the sum of two sketched functions y y = f (x) The graphs of two functions f (x) and g(x) are sketched to the right, and we want to sketch the sum s(x) = f (x) + g(x). The equations of the functions will usually not be given, but it is convenient to state them here while developing the method. They are: -6 -4 f (x) = x2 − 36 and 4 6 9 x -9 g(x) = 5x , -36 and here is a tables of values: x −9 −6 −4 −1 0 1 4 6 9 f (x) 45 0 −20 −35 −36 −35 −20 0 45 g(x) −45 −30 −20 −5 0 5 20 30 45 s(x) 0 −30 −40 −40 −36 −30 0 30 90 Each ordinate of s(x) is the sum of the ordinates of f (x) and g(x). This is the key idea that leads to everything else. The second sketch adds y = s(x) to the other two graphs. Now let us consider how we could have sketched s(x) if we had not had the equations. y = g(x) y y = s(x) y = f (x) -6 -4 4 -9 6 9 x -36 y = g(x) 0 Add the ordinates where possible — the key idea. This has been done in the top diagram. 1 If one curve, say f (x), has a zero at x = a, then s(a) = 0 + g(a) = g(a) , so s(x) has the same ordinate as g(x) at x = a, and the curves meet there. This happens at both the zeroes x = 6 and x = −6 of f (x) = x2 − 36, where s(x) meets g(x). It also happens at the zero x = 0 of g(x) = 5x, where s(x) meets f (x). 2 If the ordinates are opposites at x = a, then s(a) = f (a) + g(a) = 0, so s(x) has a zero at x = a. This happens at x = −9 and at x = 4. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 210 6D Chapter 6 Further graphs 3 If the two curves meet at x = a, so that their ordinates are equal, then s(a) = f (a) + g(a) = 2 f (a) (or 2g(a)) and the ordinate of s(x) is twice the ordinate of f (x) (or of g(x)). This happens at x = −4 and at x = 9. We cannot find from the graphs alone the precise position of the minimum, so in the absence of the equations, this is not required in the sketch. But once we know the equations, the minimum is the vertex of the parabola s(x) = (x2 − 36) + 5x = x2 + 5x − 36 = (x + 9)(x − 4), where the average of the zeroes is x = 12 (4 − 9) = −2 12 , and s(−2 12 ) = −42 14 . An example of adding graphs with asymptotes Here is an example in which one curve has a horizontal and a vertical asymptote. A succession of steps allow us to sketch the sum s(x) = f (x) + g(x). 1 The curves intersect at (2, 2) and at (−1, −1), so s(2) = f (2) + g(2) = 2 + 2 = 4, s(−1) = f (−1) + g(−1) = −1 + (−1) = −2. 2 The curve y = f (x) has a zero at x = −2, so s(−2) = f (−2) + g(−2) = 0 + (−2) = −2. 3 Add the ordinates at x = 1, so s(1) = f (1) + g(1) = 3 + 1 = 4. 4 There are no values of x where ordinates are opposites, so s(x) has no zeroes. 5 Because there are no zeroes for s(x), it can only change sign at the asymptote x = 0 (next step). Hence s(x) is negative for x < 0, and positive for x > 0. Dealing with the horizontal and vertical asymptotes of f (x): 6 Vertically, on both sides of the y-axis: As x → 0+ , f (x) → +∞ and g(x) → 0, so s(x) → +∞. As x → 0− , f (x) → −∞ and g(x) → 0, so s(x) → −∞. Thus the y-axis is a vertical asymptote to y = s(x). 7 Horizontally, on the right and the left: As x → ∞, f (x) → 1 and g(x) → +∞, so s(x) → +∞. As x → −∞, f (x) → 1 and g(x) → −∞, so s(x) → −∞. Note: There is a final detail that seems beyond the course, but has been added to the diagram for completeness. You will see that a third asymptote, an oblique asymptote, has been drawn. Here is the argument for it. For large values of x, positive or negative, f (x) is almost 1, so y = s(x) is almost the same graph as y = 1 + g(x). Hence y = s(x) eventually looks like a line parallel to y = g(x). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6D Sketching sums and differences 211 Having now argued from the graphs alone, we will now reveal the equations of the two functions — they are 2 f (x) = + 1 and (obviously) g(x) = x. Their sum is x 2 x2 + x + 2 . s(x) = x + 1 + = x x This has no zeroes because the discriminant Δ = 1 − 8 = −7 is negative, and clearly the graph has a vertical asymptote at x = 0. Summary of sketching the sum of two sketched functions 11 Sketching the sum of two sketched functions • To sketch the sum s(x) = f (x) + g(x) of two sketched functions: 0 Add the ordinates wherever possible. This is the key idea. • Some systematic approaches: 1 If one curve has a zero, then s(x) meets the other curve at that value of x. 2 If the curves meet, then the ordinate of s(x) is double the ordinate of f (x). 3 If the ordinates of f (x) and g(x) are opposites, then s(x) has a zero. 4 The sign of s(x) everywhere will usually be clear now. • To clarify any vertical or horizontal asymptotic behaviour: 5 If f (x) → ∞ or g(x) → ∞, then so also does s(x). If f (x) → −∞ or g(x) → −∞, then so also does s(x). If, however, f (x) and g(x) go in opposite directions, we may not be able determine what happens with the sum. As discussed in Section 5A, a translation of y = f (x) up 5 is y = f (x) + 5. This is f (x) + g(x) where g(x) = 5 is a constant function, so translations up and down are special cases of the construction in this section. y Sketching the difference of two sketched functions y f (x) Now let us sketch the difference d(x) = f (x) − g(x) of the same two functions f (x) = x2 − 36 and g(x) = 5x , but this time we will work from the sketches alone and then confirm the sketch using a table of values. The difference is the sum of f (x) and −g(x), so it could be done by reflecting g(x) in the x-axis and then adding the graphs, but it is easier to sketch it in one step. 0 Subtract the ordinates where possible — the key idea. 1 If f (x) has a zero at x = a, then d(a) = 0 − g(a) = −g(a), so the ordinate of d(x) is the opposite of the ordinate of f (x). This happens at x = −6 and at x = 6. If g(x) has a zero at x = a, then d(a) = f (a) − 0 = f (a), so the ordinate of d(x) is the same as the ordinate of f (x), and the curve d(x) meets the curve f (x). This happens at x = 0. 2 If the ordinates of f (x) and g(x) are opposites, then the ordinate of d(x) is double the ordinate of f (x). This happens at x = −9 and at x = 4. 3 If the two curves meet at x = a, then they have equal ordinates there, so d(x) has a zero at x = a. This happens at x = −4 and at x = 9. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x y g (x) y -4 -9 -6 -36 y = g(x) y = f (x) 46 9 x y = d(x) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 212 6D Chapter 6 Further graphs Here is a table of values to confirm these arguments: x −9 −6 −4 −1 0 1 4 6 9 f (x) 45 0 −20 −35 −36 −35 −20 0 45 g(x) −45 −30 −20 −5 0 5 20 30 45 d(x) 30 0 −30 −36 −40 −40 −30 0 90 Again, we cannot find from the graphs alone the precise location of the minimum, so it is not needed in the sketch. It is the vertex of the parabola y = d(x). Readers should complete the square for d(x) and show that the vertex is (2 12 , −42 14 ). Subtracting graphs with asymptotes The procedures here are very similar to the previous example of adding graphs with asymptotes, so there is no need for another example — and keep in mind that subtracting graphs means adding the opposite of the second graph. The structured Question 10 in Exercise 6D below presents the steps in subtracting graphs, and the Enrichment Question 15 deals with the oblique asymptotes. 12 Sketching the difference of two sketched functions • To sketch the difference d(x) = f (x) − g(x) of two sketched functions. 0 Subtract the ordinates wherever possible. This is the key idea. • Some systematic approaches: 1 If f (x) has a zero at x = a, then d(a) = −g(a). If g(x) has a zero at x = a, then d(a) = f (a). 2 If the curves meet at x = a, then d(a) has a zero there. 3 If the ordinates are opposites at x = a, then d(a) = 2 f (a) = −2g(a). 4 The sign of d(x) everywhere will usually be clear now. • To clarify any vertical or horizontal asymptotic behaviour: 5 If f (x) → ∞ or g(x) → −∞, then d(x) → ∞. If f (x) → −∞ or g(x) → ∞, then d(x) → −∞. If, however, f (x) and g(x) go in the same direction, we may not be able determine what happens with the difference. Domain and range of the sum and difference These need attention, but there is no real difficulty involved. 13 Domain and range of the sum and difference Let s(x) and d(x) be the sum and difference of two functions f (x) and g(x). • s(x) and d(x) have domain the intersection of the domains of f (x) and g(x). • Sort out the ranges of s(x) and g(x) from the graphs, or the equations. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6D Sketching sums and differences Exercise 6D 1 FOUNDATION Each diagram below shows the graph of two functions, y = f (x) and y = g(x). Copy each diagram to your book and draw the graph of y = f (x) + g(x), by adding ordinates. Try to distinguish the original graphs from the graph of the sum — use different colours, or dot the original graphs. a y 4 3 2 1 -1-1 -2 y = g(x) 1 4 2 3 5 x y = f(x) b y 4 y = f(x) 3 2 1 3 -2 -1-1 1 2 4 x -2 y = g(x) c 4 y = g(x) 3 2 1 -2 -1 -3 -1 -2 y y = f(x) 1 2 3 x 2 Copy each diagram in Question 1 to a fresh number plane. Subtract ordinates to sketch the graphs of y = f (x) − g(x). 3 The diagram to the right shows the graphs of y = f (x), where f (x) = x4 , and of y = g(x), where g(x) = x2 . y 2 a Copy the diagram to your book. 1 b On the same set of axes and in a different colour, sketch y = f (x) − g(x) by subtracting ordinates. Pay careful attention to points where the graphs cross, and to the zeroes of f (x) and g(x). 4 213 -1 The diagram to the right shows the graphs of y = f (x), where f (x) = x2 , and of y = g(x), where g(x) = x. 1 x 1 x y 1 a Copy the diagram to your book. b On the same set of axes and in a different colour, sketch y = f (x) + g(x) by adding ordinates. Pay careful attention to points where g(x) = − f (x), because at those points f (x) + g(x) = 0. Notice also the zeroes of f (x) because f (x) + g(x) = f (x) at those points, and the zeroes of g(x), because f (x) + g(x) = g(x) at those points. -1 -1 DEVELOPMENT 5 a Plot y = x3 and y = x on the same number plane, noting any points of intersection. b Hence sketch the graph of the difference, y = x3 − x. 6 Sketch y = x4 and y = x(2 − x), then sketch the difference y = x4 − x(2 − x). 7 When sketching the sums of absolute value graphs in this question, it is helpful to remember that the sum of two linear functions is itself a linear function. Thus the following graphs will be made up of straight-line sections. a Graph f (x) = |x + 1| and g(x) = |x − 1|, then graph: ii y = f (x) − g(x) i y = f (x) + g(x) b Graph f (x) = |2x| and g(x) = |x − 1|, then graph: i y = f (x) + g(x) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii y = f (x) − g(x) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 214 6D Chapter 6 Further graphs 8 Graphs the functions y = x2 and y = |x| on the same set of axes. a i Explain why the graph of y = x2 + |x| must lie above both original graphs. ii Hence sketch the sum y = x2 + |x|. Take care with the shape at x = 0. b i Explain why the graph of y = x2 − |x| is on or below the x-axis for −1 ≤ x ≤ 1. ii Hence sketch the difference y = x2 − |x|. Take care with the shape at x = 0. √ 9 a Sketch y = x and y = x, paying careful attention to the domain and to the points where these graphs intersect. √ b Hence sketch y = x − x. 10 The graph on the right is a repeat of the one used in the theory to investigate the sum of functions that involves an asymptote. In this question you will investigate their difference y y f (x) y g(x) d(x) = f (x) − g(x) . a The given curves intersect at x = −1 and x = 2. What feature of the graph of y = d(x) occurs at these values? b Write down the zero of f (x) and hence evaluate d(x) there. c Evaluate d(1) by using the ordinates of f (x) and g(x). d Copy and complete each statement relating to the vertical asymptote of y = f (x). x i As x → 0+ , f (x) → . . . and g(x) → . . . so d(x) → . . . ii As x → 0− , f (x) → . . . and g(x) → . . . so d(x) → . . . e Copy and complete each statement relating to the horizontal asymptote of y = f (x). i As x → +∞, f (x) → . . . and g(x) → . . . so d(x) → . . . ii As x → −∞, f (x) → . . . and g(x) → . . . so d(x) → . . . f Briefly explain why d(x) approaches y = 1 − x for large values of x. g Sketch y = d(x) showing all these features. 11 For each pair of functions, first graph y = f (x) and y = g(x) on the same number plane. Then in a different colour graph the sum y = f (x) + g(x). 1 a f (x) = and g(x) = x x 1 1 b f (x) = and g(x) = x x+2 c f (x) = 2 x and g(x) = x − 1 d f (x) = 2 x and g(x) = 2−x 12 For each pair of functions in the previous question, sketch the difference y = f (x) − g(x). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6D Sketching sums and differences 13 215 It may be that if y = f (x) and y = g(x) exhibit even or odd symmetry then we know the symmetry of s(x) = f (x) + g(x) and of d(x) = f (x) − g(x). a Try to complete the following table, predicting the symmetry of the resulting functions, and test your claims on the examples in this exercise. f (x) even, g(x) even f (x) odd, g(x) odd f (x) even, g(x) odd s(x) d(x) b Prove formally that if f (x) and g(x) are both odd then s(x) is also odd. That is, prove that s(−x) = −s(x) for all x in its domain. ENRICHMENT 14 1 . The purpose of this question is to sketch Consider the two functions f (x) = 12 x − 3 and g(x) = 1 + x−1 the graph y = f (x) + g(x). a Find the x-coordinates of the points where f (x) = −g(x). b Plot the line y = f (x) and the hyperbola y = g(x) on the same number plane, and mark the points found in part a. c Show that f (x) + g(x) − ( 12 x − 2) → 0 as x → +∞ and as x → −∞. Add the line y = 12 x − 3 to your graph. This is an oblique asymptote to the curve y = f (x) + g(x). d Complete your sketch of the sum y = f (x) + g(x). 15 2 In the text, the sum s(x) = f (x) + g(x) was drawn for the functions f (x) = + 1 and g(x) = x, and the x difference d(x) = f (x) − g(x) was drawn in Question 10. The oblique asymptotes of s(x) and d(x) were also drawn. This question now finds the equations of the oblique asymptotes to the two curves by the method used in the previous question. 2 a Simplify s(x) = f (x) + g(x), and show that s(x) − (x + 1) = . Hence explain why y = x + 1 is an oblique x asymptote to the curve y = s(x). b Similarly simplify d(x) = f (x) − g(x), and so find its oblique asymptote. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 216 6E Chapter 6 Further graphs 6E Modifying a function using absolute value Learning intentions • Sketch y = | f (x)| and y = f (|x|), given the graph of y = f (x). • Recognise these functions as composites of f (x) with the absolute value function. • Associate the resulting graphs with reflections in the axes. The problem in this section is to take some graph y = f (x), and from it sketch y = | f (x)| and y = f (|x|) . These operations can be done very simply using reflections in the x-axis and y-axis. Identifying the composites Let g(x) = |x| be the absolute value function. Then | f (x)| = g f (x) is the composite formed by f (x) followed by g(x), and f (|x|) = f g(x) is the composite formed by g(x) followed by f (x). Thus these two modified functions y = | f (x)| and y = f (|x|) are composites of f (x) and the absolute value function in the two possible orders. A review of the absolute value function Our graphs will rely on the algebraic definition of |x| using cases: ⎧ ⎪ ⎪ ⎪ for x ≥ 0 (for example, |5| = 5 and |0| = 0) ⎨ x, |x| = ⎪ ⎪ ⎪ ⎩−x, for x < 0 (for example, | − 5| = 5). Expressed in words: • The absolute value of a positive number or zero is unchanged. • The absolute value of a negative number is the opposite, so is positive. Thus an absolute value can never be negative. Sketching y = | f (x)| from the sketch of y = f (x) y Each absolute value transformation will be illustrated in turn using the graph to the right, which is the parabola f (x) = (x + 1)(x − 3). Graphing y = | f (x)| requires two arguments: • When the graph is above or on the x-axis, f (x) is positive or zero. Hence | f (x)| = f (x), so the graph is unchanged. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 -1 -3 3 x (1, -4) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6E Modifying a function using absolute value • When the graph is below the x-axis, f (x) is negative. Hence | f (x)| = − f (x) is the y opposite of f (x). 217 (1, 4) 3 Thus replace every part of the graph below the x-axis by its reflection back above the x-axis. 3 x -1 A table of values helps to understand the situation: x −2 −1 0 1 2 3 4 f (x) 5 0 −3 −4 −3 0 5 | f (x)| 5 0 3 4 3 0 5 14 To sketch y = | f (x)| from the sketch of y = f (x) • Everything above and on the x-axis stays the same. • Replace everything below the x-axis by its reflection back above the x-axis. Sketching y = f (|x|) from the sketch of y = f (x) The procedure to graph y = f (|x|) again has two arguments: y • To the right of or on the y-axis, x is positive or zero. Hence |x| = x, so the graph is unchanged. • To the left of x-axis, x is is negative, so |x| = −x, and f (|x|) = f (−x). Thus replace every part of the graph left of the y-axis by its reflection back across the y-axis. 3 x -3 -3 The table of values illustrating this has a preliminary row for |x|: x −4 −3 −2 −1 0 1 2 3 4 |x| 4 3 2 1 0 1 2 3 4 f (|x|) 5 0 −3 −4 −3 −4 −3 0 5 (-1, -4) (1, -4) 15 To sketch y = f (|x|) from the sketch of y = f (x) • Everything to the right of the y-axis stays the same. • Replace everything left of the y-axis by its reflection back across the y-axis. The resulting function is even, that is, it has line symmetry in the y-axis. To illustrate this, the table of values is clearly symmetric about x = 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 218 6E Chapter 6 Further graphs Combining these transformations to sketch y = f (|x|) These two absolute value transformations can be combined, as in the two examples below. Example 16 Combining the two transformations Earlier in this section, we sketched f (x) = (x + 1)(x − 3) and its transformations y = | f (x)| and y = f (|x|). Use these graphs to sketch y = f (|x|): Solution Start with the earlier sketch of either | f (x)| or f (|x|). (-1, 4) y (1, 4) 3 3 x -3 Example 17 Comparing all three methods Using the graph of y = f (x) sketched to the right, sketch: a y = | f (x)| y 2 c y = f (|x|) b y = f (|x|) x -2 2 -2 Solution y a y b 2 -2 y c 2 2 2 x -2 2 -2 -2 y = | f (x)| y = f (|x|) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x -2 2 x -2 y = f (|x|), same as part a © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6E Modifying a function using absolute value Exercise 6E 1 219 FOUNDATION The graph of y = f (x) is sketched to the right. To draw each transformation, copy the graph and draw the transformed graph in a different colour on the same axes. y 3 a Replace any part of the graph below the x-axis by its reflection above the x-axis. This will give you the graph of y = | f (x)|. 2 b Graph only the parts of y = f (x) that are to the right of the y-axis. Then add the 1 3 4 reflection of these parts in the y-axis. This will give you the graph of y = f (|x|). -1 Notice the symmetry in the y-axis. c Using the graph of y = f (|x|), replace any part of the graph below the x-axis by its refection above the x-axis. This gives the graph of y = f (|x|) x d Using the graph of y = | f (x)|, graph only the parts that are to the right of the y-axis. Then add the reflection of these parts in the y-axis. What do you notice about the result of this and the answer to part c? 2 In each case, follow the steps of Question 1 and use the given graph of y = f (x) to sketch: i y = | f (x)| ii y = f (|x|) iii y = f (|x|) a y 2 2 3 4 y b 4 x 1 - Ö3 y c 2 1 -1-1 Ö3 1 -3 x -2 -1 Sketch each given function and use it to then sketch: 1 x i y = | f (x)| ii y = f (|x|) iii y = f (|x|) a y= x−1 b y = 2−x − 1 c y= 1 1−x a Explain why the graphs of y = 2 x and y = |2 x | are the same. b Write y = 2|x| using cases, then sketch its graph. DEVELOPMENT 5 a Sketch y = f (x) where f (x) = (x + 1)(x − 2), showing all x-intercepts and the vertex. b Hence sketch: i y = | f (x)| ii y = f (|x|) 6 Repeat the steps of the previous question for the function f (x) = x − 1 − 1. 7 a Graph y = f (x), where f (x) = asymptotes. b Hence sketch: i y = | f (x)| CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 iii y = f (|x|) 1 + 1. Be careful to identify any intercepts with the axes and any x−1 ii y = f (|x|) iii y = f (|x|) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 220 6E Chapter 6 Further graphs 8 a Let f (x) = x(x − 2). ii Use repeated transformations to sketch y = | f (|x|)|. i Sketch y = f (x). b Repeat part a for the function f (x) = (x + 1)(3 − x). 9 a Show that y = |2 − x| is neither even nor odd and graph it. c Hence graph y = 1 − 2 − |x|. d Finally graph y = 1 − 2 − |x|. b Show that y = 2 − |x| is even and use part a to graph this new function. 10 a Let f (x) be any function. Explain why g(x) = f (|x|) is even. b Let f (x) be an odd function. Show that h(x) = | f (x)| is even. 11 a When will the graphs of y = f (x) and y = | f (x)| be the same? b When will the graphs of y = f (x) and y = | f (x)| be symmetric in the x-axis? ENRICHMENT 12 a Read carefully the instruction in Question 1b for graphing y = f (|x|). Write a similar instruction for graphing |y| = f (x). b Test your instruction by graphing |y| = f (x) for each of the functions in Question 2. 13 Sketch |y| = |x|. 14 a Describe the graph of y = f (−|x|) in terms of the graph of y = f (x). b What type of symmetry must y = f (−|x|) possess? 15 Let f (x) be any function, and let g(x) = f (|x|) and h(x) = 12 ( f (x) + f (−x)). a Prove that both g(x) and h(x) are even functions. b Are g(x) and h(x) always the same function? If so then prove it, otherwise give a counter-example. 16 a Investigate the graphs of the sequence of functions y = |x|, y = 1 − |x|, y = 1 − 1 − |x|, ... b Show that the 2nd, 4th, 8th, . . . functions in this sequence can be simplified to y = 1 − |x|, y = 1 − 2 − |x|, CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y = 1 − 2 − 4 − |x|, ... © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6F Inverse relations and functions 221 6F Inverse relations and functions Learning intentions • Construct the formula of the inverse of a relation or function. • Use the horizontal line test to see whether the inverse relation is a function. Mathematics is full of inverse processes: • The inverse of multiplying by 7 is dividing by 7 — and the inverse of dividing by 7 is multiplying by 7. • The inverse of shifting up 3 is shifting down 3 — and the inverse of shifting down 3 is shifting up 3. • The inverse of reflecting in the y-axis is reflecting in the y-axis — this process is its own inverse. This section uses geometric and graphical methods to obtain the inverse of a relation, and to find a simple geometric condition for the inverse to be a function. Inverse relations Cube the number 5 and get 125. The inverse process is taking the cube root, which sends 125 back to 5. We can do this with any number, positive, negative or zero, so the cubing function y = x3 has a well-defined inverse √ function y = 3 x that sends any output back to its original input. When these two function machines are put together in a chain, in either order, the composition of the two functions is the identity function that maps every number to itself: x y x y 2 −→ −→ 8 8 −→ −→ 2 1 −→ −→ 1 1 −→ −→ 1 −→ −→ 1 8 1 8 −→ 1 2 −→ 0 0 −→ −→ 0 −→ −→ − 18 − 18 −1 −→ −→ −1 −1 −→ 1 2 Cube y = x3 0 −→ − 12 −→ Cube root √ y= 3x −→ −→ − 12 −→ −1 The exchanging of input and output can also be seen in the two tables of values, where the two rows are interchanged: x x 3 2 8 1 1 1 2 1 8 0 − 12 0 − 18 −1 −1 −2 −8 x 8 √3 x 2 1 1 1 8 1 2 0 − 18 −1 −8 0 − 12 −1 −2 This exchanging of input and output means that the coordinates of each ordered pair are exchanged. Remembering that a relation was defined simply as a set of ordered pairs, we are led to a definition of inverse that can be applied to any relation, whether it is a function or not: 16 Inverse relations • The inverse relation of any relation is obtained by reversing each ordered pair. • The inverse relation of the inverse relation is the original relation. The second statement follows from the first because each ordered pair returns to its original state when reversed a second time. For example, the pair (2, 8) in the original becomes the pair (8, 2) in the inverse, and reversed goes back to (2, 8). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 222 6F Chapter 6 Further graphs y Graphing the inverse relation Reversing an ordered pair means that the original first coordinate is read off the vertical axis, and the original second coordinate is read off the horizontal axis. Geometrically, this exchanging of the two coordinates can be done by reflecting the point in the diagonal line y = x. This can be seen by comparing the graphs of y = x3 √ and y = 3 x, which are drawn here on the same pair of axes. 1 −1 1 x −1 17 The graph of the inverse relation The graph of the inverse relation is obtained by reflecting the original graph in the diagonal line y = x. Domain and range of the inverse relation Exchanging x- and y-coordinates means the domain and range are exchanged: 18 Domain and range of the inverse relation • The domain of the inverse is the range of the relation. • The range of the inverse is the domain of the relation. Finding the equation and restrictions of the inverse relation When the coordinates are exchanged, the x-variable becomes the y-variable and the y-variable becomes the x-variable. Hence the method for finding the equation and restrictions of the inverse is: 19 The equation of the inverse relation • To find the equation and restrictions of the inverse relation, write x for y and y for x every time each variable occurs. • This process can be applied to any relation whose equation and restrictions are known, whether or not it is a function. For example, the inverse of the function y = x3 is x = y3 . This particular equation can then be solved for y to give √ y = 3 x, confirming that in this particular case, the inverse relation is again a function. Example 18 Graphing function and inverse together a Write down the inverse relation of the function y = x2 . b Graph both relations on the same number plane, showing the reflection line. c Write down the domain and range of both relations. d Is the inverse relation a function? Solution a Writing x for y and y for x, the inverse is x = y2 . c For the original, domain: all real x, range: y ≥ 0. For the inverse, domain: x ≥ 0, range: all real y. Notice how the domain and the range have been swapped. d The inverse is not a function — it fails the vertical line test. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b y 4 2 1 -2 -1-1 -2 12 4 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6F Inverse relations and functions Example 19 223 Graphing function and inverse together Repeat the previous questions for the function y = x3 + 2. Solution a Writing x for y and y for x, the inverse is y b x = y3 + 2, √3 which is y = x − 2 . c For both, domain and range are all real numbers. d The inverse is again a function, by the vertical line test. 2 2 x Forming the inverse when there are restrictions When there are any restrictions, then x and y must be swapped in these as well, as in the next example. Example 20 Forming the equation of the inverse with restrictions Consider the function y = 2x − 3, where x > 1. a Write down the equation and restriction of the inverse relation. b Rewrite the inverse as a function with y as the subject, changing the restriction to a restriction on x. Solution a The function is so the inverse relation is y = 2x − 3, where x > 1, x = 2y − 3, where y > 1. y = 12 (x + 3), b Solving for y, The restriction y > 1 is where y > 1. 1 2 (x + 3) > 1 x+3>2 x > −1, y = 12 (x + 3), so the inverse is the function where x > −1. Testing graphically whether the inverse relation is a function In general, the inverse of a relation is not a function. For example, the sketches below show the graphs of another function, with a restriction, and its inverse: y y 4 y=3 2 4 -2 y = 4 − x2 , 2 x where − 2 ≤ x ≤ 2 x -2 x=3 x = 4 − y2 , where − 2 ≤ y ≤ 2 The first graph is a function, passing the vertical line test. But when we read it backwards, the value y = 3 gives two answers, x = 1 and x = −1. Accordingly, when we sketch the inverse relation, we see that it fails the vertical line test, with x = 3 crossing the graph twice. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 224 6F Chapter 6 Further graphs But reflection in y = x exchanges vertical and horizontal lines! Now we can see immediately from the first graph that its inverse is not a function, because it fails the horizontal line test — the line y = 3 crosses it more than once. We didn’t need to draw the second graph to know that the inverse is not a function. All we need to know is that the first graph fails the horizontal line test. 20 Horizontal line test for whether the inverse is a function • Geometrically, the inverse relation of a given relation is a function if and only if no horizontal line crosses the original graph more than once. We could also have done it the slow way — solve the equation x = 4 − y2 of the inverse relation to give √ √ y = 4 − x or y = − 4 − x , showing again that for many values of x, there is more than one value of y, so the inverse is not a function. Exercise 6F 1 FOUNDATION Draw the inverse relation of each relation by reflecting in the line y = x. y a y b 3 2 −2 y c 2 x 1 3 −3 y d x x (-1, -1) (1, -1) x −1 y e f y y g h 2 1 −2 −1 2 x 2 x -2 y 2 1 2 x 1 x -2 2 Use the vertical and horizontal line tests to determine which relations and which inverse relations drawn in Question 1 are also functions. 3 Determine each inverse algebraically by swapping x and y and then making y the subject. a y = 3x − 2 b y = 12 x + 1 c y = 3 − 21 x d x−y+1=0 e 2x + 5y − 10 = 0 f y=2 4 For each function in the previous question, draw a graph of the function and its inverse on the same number plane to verify the reflection property. Draw a separate number plane for each part. 5 a Find each inverse algebraically by swapping x and y and then making y the subject. 1 x+2 3x 1 ii y = iii y = iv y = +1 x x+1 x−2 x+2 b For parts i and iv above, find the domain and range of the function, and the domain and range of the inverse function. i y= CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6F Inverse relations and functions 6 225 Swap x and y and solve for y to find the inverse of each function. What do you notice, and what is the geometric significance of this? 1 2x − 2 −3x − 5 a y= b y= c y= d y = −x x x−2 x+3 DEVELOPMENT 7 a Graph each relation and its inverse relation, then find the equation of the inverse relation. Which of the four original relations are functions, and which inverse relations are functions? i (x − 3)2 + y2 = 4 ii (x + 1)2 + (y + 1)2 = 9 iii y = x2 − 4 iv y = x2 + 1 b For parts i and iv above, find the domain and range of the relation, and the domain and range of the inverse relation. 8 Write down the inverse of each function, solving for y if it is a function. Sketch the function and the inverse on the same graph and observe the symmetry in the line y = x. √ √ a y = x2 b y = 2x − x2 c y=− x d y = − 4 − x2 9 Each function below has a restriction. Write down its inverse relation. Then attempt to solve it for y. If the inverse is a function, rewrite the restriction as a restriction on x. If the inverse is not a function, give a value of x that corresponds to two or more values of y. a y = 3x − 10, where x < 2 b y = 13 − 6x, where x ≥ 3 c y = x + 2, where x < 3 d y = x2 − 3, where x ≥ −2 3 10 a Factorise f (x) = x2 − 2x − 3 in order to show that the graph of y = f (x) fails the horizontal line test. b Let g(x) = x2 − 2x − 3 for x ≥ 1. Explain why this function has an inverse and find its equation. 11 a Show that the inverse function of y = b Hence show that y = ax + b b − cx is y = . x+c x−a ax + b is its own inverse if and only if a + c = 0. x+c ENRICHMENT 2 x + 2−x is not a function. 2 2 x − 2−x is a function. b Show that the inverse of y = 2 12 a Show that the inverse of y = 13 a Show that if the domain of an even function contains a non-zero number, then its inverse is not a function. b Is the inverse of an odd function always a function? If not, give a counter-example. 14 a The polynomial in Question 10a has two linear factors and fails the horizontal line test, so its inverse is not a function. Explain why all polynomials with at least two linear factors must automatically fail the horizontal line test. b Do any polynomials have an inverse which is a function? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 226 6G Chapter 6 Further graphs 6G Inverse function notation Learning intentions • Use the alternative definition of inverse function in terms of composition. • Use the standard notation for the inverse function. • Restrict a function or relation appropriately so that its inverse is a function. This section ignores relations, and deals only with functions and inverse functions. One-to-one functions A function f (x) must by definition pass the vertical line test, meaning that every value of x in the domain corresponds to exactly one y-value. For the inverse of f (x) to be a function also, f (x) must pass the horizontal line test, meaning that every value of y in the range corresponds to exactly one x-value. Thus the condition for the inverse of f (x) to be a function is that f (x) is a one-to-one correspondence between the elements of the domain and the elements of the range. Such a function is called simply one-to-one. 21 One-to-one functions, and the inverse of a function • A function f (x) is called one-to-one, or a one-to-one correspondence, if for every value of y in the range, there is exactly one element x in the domain so that f (x) = y. • The inverse relation of a function f (x) is also a function: if and only if f (x) passes the horizontal line test, and if and only if f (x) is one-to-one, and if and only if the equation of the inverse can be solved uniquely for y. The composite of a function and its inverse function is the identity Suppose that f (x) is a one-to-one function with inverse function g(x). Then g(x) is also a one-to-one correspondence, with the same pairing of the domain and range as provided by f (x), but with the corresponding pairs reversed. When we apply f (x) then g(x), it is as if nothing has happened, that is, g f (x) = x, for all x in the domain of f (x). And because g(x) sends each number back where it came from, its inverse is f (x), and the other composite f g(x) is also an identity function, f g(x) = x, for all x in the domain of g(x). When we restrict discussion to functions, these two conditions can be taken as an alternative definition of an inverse function: 22 Alternative definition of inverse function using composition • The function g(x) is the inverse function of a function f (x) if and only if: g f (x) = x, for all x in the domain of f (x), and f g(x) = x, for all x in the domain of g(x). That is, if and only if g f (x) and f g(x) are both identity functions. • An identity function is a function I(x) such that: I(x) = x, for all x in the domain of I(x). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6G Inverse function notation 227 Inverse function notation Suppose that f (x) is a one-to-one function, that is, its inverse is also a function. Then that inverse function is written as f −1 (x). The index −1 used here means ‘inverse function’ and must not be confused with its more common use for the reciprocal of a number. To return to the original example in the last section, √ 1 Be careful: f (x) −1 = If f (x) = x3 , then f −1 (x) = 3 x. f (x) As we have seen, the inverse function f −1 (x) is also one-to-one, with inverse f (x), and if the function and the inverse function are applied successively in either order, the result is the original number. Using the example above, √3 √3 √3 3 and f f −1 (8) = 8 = 23 = 8. f −1 f (8) = 83 = 512 = 8 23 Inverse function notation • If a function f (x) is one-to-one, then its inverse relation is also a one-to-one function, and is written as f −1 (x). • The composite of the function and its inverse, in either order, sends every number for which it is defined back to itself: f −1 f (x) = x, for all x in the domain of f (x) f f −1 (x) = x, for all x in the domain of f −1 (x) . • To find the formula for f −1 (x) from the formula for f (x): Convert to y = . . . notation to generate the inverse relation. Write y for x and x for y in all the equations and restrictions. Solve for y, and convert to the notation f −1 (x) = . . . Examples 21–22 demonstrate the method described in the final dotpoint. Example 21 Verifying the formal definition of an inverse function Find the equation of f −1 (x) for each function, then verify that and f f −1 (x) = x . f −1 f (x) = x a f (x) = x3 + 2 b f (x) = 6 − 2x, where x > 0 Solution y = x3 + 2. a Let Then the inverse has equation and solving for y, Hence x = y3 + 2 √3 y = x − 2. √3 f −1 (x) = x − 2 . 3 Verifying, f −1 f (x) = (x3 + 2) − 2 √3 = x3 =x and (the key step) 3 √3 f f −1 (x) = x − 2 + 2 = (x − 2) + 2 =x Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 228 6G Chapter 6 Further graphs b Let Then the inverse has equation The restriction y > 0 means y = 6 − 2x, where x > 0. x = 6 − 2y, where y > 0 y = 3 − 12 x, where y > 0. (the key step) 3 − 12 x > 0 x < 6, −1 so f (x) = 3 − 12 x, where x < 6. Verifying, f −1 f (x) = f −1 (6 − 2x) and f f −1 (x) = f (3 − 12 x) Example 22 = 3 − 12 (6 − 2x) = 6 − 2(3 − 12 x) =3−3+x =6−6+x =x =x Testing whether a function has an inverse function Find the inverse relation of each function. If the inverse is a function, find an expression for f −1 (x), and verify that f −1 f (x) = x and f f −1 (x) = x. 1−x a f (x) = b f (x) = x2 − 9 1+x What is surprising about the result of part a? Solution 1−x . 1+x 1−y x= 1+y y= a Let Then the inverse has equation × (1 + y) (the key step) x + xy = 1 − y Hence y + xy = 1 − x (terms in y on one side) y(1 + x) = 1 − x 1−x . y= 1+x 1−x . f −1 (x) = 1+x (now y occurs only once) Notice that this function f (x) and its inverse f −1 (x) are identical, so that if the function f (x) is applied twice, each number is sent back to itself. 1−x 1 − 1+x and in general, f f (x) = Thus f f (2) = f − 13 1 + 1−x 1+x 1 2 = 13 ÷ 3 (1 + x) − (1 − x) = =2 (1 + x) + (1 − x) =x b The function f (x) = x − 9 fails the horizontal line test. For example, f (3) = f (−3) = 0, which means that the x-axis meets the graph twice. Hence the inverse relation of f (x) is not a function. Alternatively, the inverse relation is x = y2 − 9, which on solving for y gives √ √ y = x + 9 or − x + 9, 2 which is not unique, so the inverse relation is not a function. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6G Inverse function notation 229 Restricting the domain so the inverse is a function When a function is not one–to-one, that is, its inverse is not a function, it is often convenient to restrict the domain of the function so that this new restricted function has an inverse function. The clearest example is the squaring function f (x) = x2 , whose inverse relation is not a function because, for example, 49 has two square roots, 7 and −7. If, however, we restrict the domain of f (x) = x2 to x ≥ 0 and define a new restricted function g(x) = x2 , y y = g( x ) where x ≥ 0, then the new restricted function g(x) is one-to-one, and thus has an inverse √ function. This inverse function has equation g−1 (x) = x, where as explained √ means ‘take the positive square root (or zero)’. earlier, the symbol y = g −1 ( x ) 1 1 x To the right are the graphs of the restricted function and its inverse function, with the unrestricted function and its inverse relation shown dotted. These ideas will be developed a great deal further in Year 12, when inverse of the tirgonometric functions are developed. Exercise 6G 1 FOUNDATION Let f (x) = 2x − 8 and g(x) = 12 x + 4. a Verify by substitution that: i g f (5) = 5 ii f g(5) = 5 iii g f (x) = x iv f g(x) = x b What do you conclude about the functions f (x) and g(x)? 2 3 Each pair of functions f (x) and g(x) are known to be mutual inverses. Show in each case that f g(2) = 2 and g f (2) = 2, and that f g(x) = x and g f (x) = x. a f (x) = x + 13 and g(x) = x − 13 b f (x) = 7x and g(x) = 17 x c f (x) = 2x + 6 and g(x) = 12 (x − 6) d f (x) = x3 − 6 and g(x) = √3 x+6 a Find the inverse function f −1 (x) of f (x) = 2x + 5. Begin ‘Let y = 2x + 5’, then swap x and y to find the inverse, then solve for y, then write down the equation of f −1 (x). b Check your answer by calculating f −1 f (x) and f f −1 (x) . c Similarly find the inverse functions of each function, and check each answer. i f (x) = 4 − 3x ii f (x) = x3 − 2 iii f (x) = 1 x−5 DEVELOPMENT 4 Explain whether the inverse relation is a function by testing whether it is one-to-one. If it is a function, find f −1 (x), specifying its domain. Then verify the two identities f −1 f (x) = x and f f −1 (x) = x. √ a f (x) = x2 b f (x) = x c f (x) = x4 d f (x) = x3 + 1 e f (x) = 9 − x2 f f (x) = 9 − x2 , x ≥ 0 1−x 2 g f (x) = 3−x h f (x) = i f (x) = x2 , x ≤ 0 3+x x+1 j f (x) = x2 − 2x, x ≥ 1 k f (x) = x2 − 2x, x ≤ 1 l f (x) = x−1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 230 6G Chapter 6 Further graphs 5 Let f (x) be the restricted function f (x) = 3x − 2, where 1 ≤ x ≤ 4. a Find the inverse function f −1 (x), being careful to add its restriction. b Show that f −1 f (x) and f f −1 (x) are both identity functions, and find their respective domains. A sketch may make the situation clearer. 6 a What is the gradient of the line y = ax + b? b Write down the inverse relation of y = ax + b. c What are the conditions for this inverse relation to be a function? d When the inverse is a function, solve it for y, find its gradient, and explain why the gradients of the function and its inverse both have the same sign. e Give an argument using reflection in the line y = x for your answers in part c. 7 Sketch on separate graphs: b y = −x2 , for x ≥ 0 a y = −x2 Draw the inverse of each on the same graph, then comment on the similarities and differences between parts a and b. 8 The parabola y = f (x), where f (x) = (x − 3)2 + 1, has its vertex at (3, 1). The inverse of f (x) is not a function. a i Janine says that when she applies the restriction x ≥ a to f (x) the inverse is a function. What is the least value of a? ii Find the equation of the inverse in this case. b i Jacob says that when he applies the restriction x ≤ b to f (x) the inverse is a function. What is the greatest value of b? ii Find the equation of the inverse in this case. 9 a Let f (x) = x2 − 2x − 3 with the restriction x ≥ a. It is known that the inverse of f (x) is a function. Using the previous question as a guide, find the least value of a and find f −1 (x) in that case. b Do the same for the function f (x) = 5 − 4x − x2 with the restriction x ≤ a. 10 a Let f (x) and g(x) be one-to-one functions. That is, both pass the horizontal line test. Let h(x) = g f (x) . 11 Suggest restrictions on the domains of each function to produce a new function whose inverse is also a function (there may be more than one answer). Draw the restricted function and its inverse. √ √ 1 a y = − 4 − x2 b y= 2 c y = x3 − x d y = x2 x Show that the inverse function of h(x) is h−1 (x) = f −1 g−1 (x) . 1 b i Find the inverse function of h(x) = . x−3 ii Express h(x) as the composition of the reciprocal function and a linear function, and hence use part a to find its inverse function. ENRICHMENT 12 a Let f (x) = ax + b and g(x) = αx + β. Find g f (x) , and hence prove that the condition for f (x) and g(x) to be mutually inverse functions is α = a−1 and β = −a−1 b . b Find three linear functions f (x), g(x) and h(x), none of whose graphs pass through the origin, with no two graphs parallel, such that h g f (x) is the identity function. 13 Does the empty function have an inverse function? If so then what is it? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6H Defining functions and relations parametrically 231 6H Defining functions and relations parametrically Learning intentions • Deal with a function or relation defined parametrically. • Eliminate the parameter to obtain the Cartesian equation of the curve. There is an ingenious way of handling curves by making each coordinate a function of a single variable, called a parameter. Each point on the curve is then specified by a single number, rather than by a pair of coordinates. The section uses some trigonometry that is only reviewed in Chapter 7. Angles of any magnitude and the Pythagorean identities are needed. Readers may prefer to delay studying these examples and questions until Chapter 7 is completed. An example of parameters SDB hits a six at the Sydney Cricket Ground. • A cameraman in a distant stand, exactly behind the path of the ball, sees the ball rise and fall, with its height y in metres given by y = −5t2 + 25t, where t is time in seconds after the strike. • A drone filming the shot from high above the Cricket Ground sees the ball move across the ground, with distance x from the batsman given by x = 16t. The cameraman and the drone together have a complete record of the ball’s flight. (Ignore parallax here, and regard both observers as ‘distant’). From their results, we can draw up a table of values of the (x, y) position of the ball at each time t: 0 1 2 2 12 3 4 5 x 0 16 32 40 48 64 80 30 31 14 30 20 0 t y 0 20 The resulting (x, y)-graph shows the path of the ball, and each point on the graph can be labelled with the corresponding value of time t. This variable t is called a parameter, and the equations x = 16t, y 30 20 t=2 t=3 t=1 t=4 10 t=0 40 t=5 80 x y = −5t2 + 25t are called parametric equations of the curve. It is possible to eliminate the parameter t from these two equations. Solving the first equation for t, t = 16x , then substituting into the second, 5 2 x + 25 y = − 256 16 x , −5x2 + 400x 256 5x(80 − x) y= . 256 y= and factoring displays the zeroes, CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 232 6H Chapter 6 Further graphs 24 Parameters • A curve in the (x, y)-plane may be parametrically defined, meaning that x and y are given as functions of a third variable t called a parameter. These two equations for x and y in terms of t are called parametric equations of the curve. • In many situations, the parameter t may be eliminated to give a single equation in x and y for the curve. The single equation in x and y is called the Cartesian equation of the curve. The letter t is often used for the parameter because it stands for ‘time’. Other letters are often used, however, particularly θ and ϕ for an angle, and p and q. Examples of parametrisation — a parabola We can reverse the process of eliminating the parameter, and parametrise familiar curves. Some straightforward parametrisations are given below of a parabola, a circle, and a rectangular hyperbola. The parabola x2 = 4y can be parametrised by the pair of equations x = 2t y=t and y t = −2 2 because elimination of t gives x2 = 4y. The variable point (2t, t2 ) now runs along the whole curve as the parameter t takes all the real numbers as its values: −2 −1 − 12 0 1 2 1 2 x −4 −2 −1 0 1 2 4 1 1 4 0 1 4 1 4 t y 4 4 t = −1 −4 −2 −1 t=0 t=2 1 t=1 4 x 1 2 The sketch shows the curve with the seven plotted points labelled by their parameter. The curve can be regarded as a ‘bent and stretched number line.’ A parametrisation of the circle The circle x2 + y2 = r2 can be parametrised using trigonometric functions by x = r cos θ and θ = 90º, y = r sin θ. y θ = −270º θ = 45º This parametrisation uses the Pythagorean identity, because squaring the two equations and adding them: x2 + y2 = r2 cos2 θ + r2 sin2 θ θ = 180º, r θ = −180º θ = 0º, θ = 360º x = r2 (cos2 θ + sin2 θ) = r2 , because cos2 θ + sin2 θ = 1. θ = 270º, θ = −90º Notice from the table of values below that with these equations, each parameter corresponds to just one point, but each point corresponds to infinitely many different values of the parameter, all differing by multiples of 360◦ : θ −360◦ −270◦ −180◦ −90◦ 0 x r 0 −r 0 r y 0 r 0 −r 0 45◦ √ 1 r 2 2 √ 1 2r 2 90◦ 180◦ 270◦ 360◦ 0 −r 0 r r 0 −r 0 In our previous examples, the map from parameters to points was always one-to-one, but in this parametrisation of the circle, the map is many-to-one. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6H Defining functions and relations parametrically 233 The circle is a relation, but not a function. The great advantage of parametrising the circle is that it is now described by a pair of functions, which in many situations are easier to handle than the original relation. A parametrisation of the rectangular hyperbola The rectangular hyperbola xy = 1 can be parametrised algebraically by 1 x=t and y= . t y t There is a one-to-one correspondence between the points on the curve and the real numbers, with the one exception that t = 0 does not correspond to any point: t −2 −1 x −2 −1 − 12 − 12 y − 12 −1 −2 Example 23 1 2 0 1 2 1 2 1 2 ∗ 2 1 1 2 0 t t t t t 1 2 x 12 Using trig identities to find the Cartesian equaton Find the Cartesian equations of the curves defined by the parametric equations: a x = 4p, y = p2 + 1 b x = sec θ, y = sin θ Describe part a geometrically. Solution a From the first, p = 14 x, and substituting into the b Squaring, x2 = sec2 θ, and y2 = sin2 θ so = 1 − cos2 θ, 1 y2 = 1 − 2 x 2 2 x (1 − y ) = 1. second, 1 2 x + 1, y = 16 which is a parabola with vertex (0, 1) and concave up. Exercise 6H FOUNDATION Note: Some questions in this exercise use trigonometry reviewed in Chapter 7, in particular, angles of any magnitude and the Pythagorean identities: sin2 θ + cos2 θ = 1 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = cosec2 θ . Consider the parametric equations x = t − 2 and y = 2t − 1. a Complete the table to the right. t b Explain from the table why the graph is a line. x c From the table, find the y-intercept and the gradient. d Eliminate t to find the Cartesian equation, and check it from part c. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 −2 −1 0 1 2 y © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 234 6H Chapter 6 Further graphs 2 a The parametric equations x = 2t − 3 and y = t + 1 represent a line. i Find the points A and B with parameters t = 0 and t = 1, and hence find the gradient of the line. ii Find the value of t that makes x = 0, and hence find the y-intercept. iii Check your answers by eliminating t to form a Cartesian equation. b Repeat the steps in part a for these lines: i x = 2t − 3 and y = 6t − 5 ii x = 2t − 3 and y = 3t − 2 3 a Complete the table below for the curve x = 4t, y = 2t2 and sketch its graph. t −6 −4 −2 −1 0 1 2 4 6 x y b Eliminate the parameter to find the Cartesian equation of the curve. c The curve is a parabola. What value of t gives the coordinates of the vertex? 4 5 Repeat the previous question for the curve x = t, y = 12 t2 . c a Show that the point cp, lies on the hyperbola xy = c2 , where c is a constant. p 2 b Complete the table of values below for x = 2p, y = and sketch the graph. p p −3 −2 −1 − 12 − 14 0 1 4 1 2 1 2 3 x y c Explain what happens as p → ∞, p → −∞, p → 0+ and p → 0− . 6 x 2 y2 + = 1. a2 b2 b i Complete a table of values for the curve x = 4 cos θ, y = 3 sin θ, taking the values θ = 0◦ , 30◦ , 60◦ , 90◦ , 120◦ , . . . , 360◦ . ii Sketch the curve and state its Cartesian equation. a Show that the point (a cos θ, b sin θ) lies on the ellipse DEVELOPMENT 7 8 Eliminate the parameter and hence find the Cartesian equation of the curve. a x = 3 − p, y = 2p + 1 b x = 1 + 2 tan θ, y = 3 sec θ − 4 1 1 c x = p + , y = p2 + 2 p p d x = cos θ + sin θ, y = cos θ − sin θ a Show that x = a + r cos θ and y = b + r sin θ define a circle with centre (a, b) and radius r. b Hence sketch a graph of the curve x = 1 + 2 cos θ, y = −3 + 2 sin θ. 9 t2 − 1 2t Show by elimination that x = 2 and y = 2 almost represent the unit circle x2 + y2 = 1. What point t +1 t +1 is missing? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 6H Defining functions and relations parametrically 10 235 [Parameters and Curve Orientation] Let A = (1, 2) and B = (2, 1). a i Show that x = 1 + t, y = 2 − t, 0 ≤ t ≤ 1 parameterises the line segment AB. ii Describe how the point P(1 + t, 2 − t) moves as t increases from 0 to 1. b i Show that x = 2 − u, y = 1 + u, 0 ≤ u ≤ 1 also parameterises the line segment AB. ii Describe how the point P(2 − u, 1 + u) moves as u increases from 0 to 1. c The points A and B also lie on the circle with centre (1, 1) and radius 1. Consider the points P(1 + cos t, 1 + sin t) and Q(1 + sin t, 1 + cos t), where 0◦ ≤ t ≤ 90◦ in both cases. Explain the difference between the curves traced out by the points P and Q. 11 Different parametric representations may result in the same Cartesian equation. The graphical representation, however, may differ due to restrictions in the domain or range. a Find the Cartesian equation of the curve x = 2 − t, y = t − 1 and sketch its graph. b Find the Cartesian equation of the curve (sin2 t, cos2 t). Explain why 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 and sketch a graph of the curve. c Find the Cartesian equation of the curve x = 4 − t2 , y = t2 − 3. Explain why x ≤ 4 and y ≥ −3 and sketch a graph of the curve. 12 A certain graph has parametric equations x = at + b and y = ct + d, where the constants a, b, c and d are real numbers. a Eliminate t from these equations and hence show that they represent a straight line. Express your answer in gradient-intercept form. b Are there any restrictions that need to be placed on the answer to part a? c How could the answer to part a be written in order to avoid these restrictions? Explain your answer. 13 a Show that the parametric equations x = c(sec θ − tan θ) and y = c(sec θ + tan θ) with −90◦ < θ < 90◦ represents the portion of the hyperbola xy = c2 in the first quadrant. b What restriction on θ is needed to get the portion in the third quadrant? 14 Find the Cartesian equation of the curve x = 3 + r cos θ, y = −2 + r sin θ, and describe it geometrically if: a r is constant and θ is variable, b θ is constant and r is variable. 15 [Parameters and Transformations] a The circle with centre the origin and radius 2 has parametric equations x = 2 cos θ and y = 2 sin θ with −180◦ < θ ≤ 180◦ . i By considering translations, write down the parametric equations when this circle is shifted right 1 and up 3. ii By considering dilations, write down the parametric equations when the original circle is stretched horizontally by 2 and vertically by 12 . b A certain curve has parametric equations x = f (t) and y = g(t). Ignoring any possible restrictions on t, answer the following. i Describe the curve generated by x = f (t) + h and y = g(t) + k. ii Describe the curve generated by x = a f (t) and y = b g(t). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 236 6H Chapter 6 Further graphs ENRICHMENT 16 2t + 1 2t2 + 2t Show by elimination that x = 2 and y = 2 almost represent the unit circle x2 + y2 = 1. 2t + 2t + 1 2t + 2t + 1 What point is missing? 17 a Explain why the parametric equations x = cos t, y = sin t, z = t describe a spiral. 18 a Show that the point (a sec θ, b tan θ) lies on the curve 19 After finding the Cartesian equation, sketch the curve whose parametric equations are x2 y2 − = 1. a2 b2 b Complete a table of values for the curve x = 4 sec θ, y = 3 tan θ, where 0◦ ≤ θ ≤ 360◦ . What happens when θ = 90◦ and θ = 270◦ ? c Sketch the curve (it has two asymptotes) and state its Cartesian equation. x = 12 (2t + 2−t ) 20 and y = 12 (2t − 2−t ) . A relation is defined parametrically by x = f (t) and y = g(t). a What transformation of the relation occurs when t is replaced by −t if: i f (t) and g(t) are both even, ii f (t) and g(t) are both odd, iii f (t) is even and g(t) is odd, iv f (t) is odd and g(t) is even. b What is the relationship between this relation and the relation defined by x = g(t) and y = f (t)? c Where is the graph of the relation x = | f (t)| and y = |g(t)| located? d Where is the graph of the relation x = f (t) and y = f (t) located? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 6 review 237 Chapter 6 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 6 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Chapter Review Exercise 1 Solve each inequation. a |x| < 3 b |x + 2| ≥ 4 c |2x − 5| ≤ 11 2 Solve each inequation by multiplying both sides by the square of the denominator. 5 3 x−2 >1 ≤1 ≥4 a b c x x−3 x+1 3 Consider the function f (x) = x(x + 2)(x − 3). Review Note: Graphing software could be very helpful in this exercise. a Write down the zeroes of the function, and draw up a table of signs. b Copy and complete: ‘ f (x) is positive for . . . , and negative for . . . ’ c Write down the solution of the inequation x(x + 2)(x − 3) ≤ 0. d Sketch the graph of the function to confirm these results. 4 Consider the function y = (1 − x)(x − 3)2 . a Write down the zeroes of the function and draw up a table of signs. b Hence solve the inequation (1 − x)(x − 3)2 ≥ 0. c Confirm the solution by sketching a graph of the function. 5 Solve each inequation in Question 4 using the table-of-signs method. First move everything to the left-hand side, then make the LHS into a single fraction, then identify its zeroes and discontinuities, then draw up a table of signs, then read the solution from the table. 6 Consider the linear function f (x) = x − 2. a Sketch y = f (x), clearly indicating the x- and y-intercepts. b Also show on your sketch the points where y = 1 and y = −1. 1 on the same number plane. f (x) 1 , then copy and complete the sentence, ‘As d Write down the equation of the vertical asymptote of y = f (x) x → 2− , y → . . . , and as x → 2+ , y → . . . ’ c Hence sketch the graph of y = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 238 Chapter 6 Further graphs 7 Consider the quadratic function f (x) = 3 − x2 . Review a Sketch y = f (x), clearly indicating the x- and y-intercepts. b Also show on your sketch the points where y = 1 and y = −1. 1 on the same number plane. f (x) c Hence sketch the graph of y = 8 Sketch the reciprocal of each function graphed below, showing all the important features. y a 1 3 - 32 -4 -3 (-2, -1) 3 4 -1 - 32 -1 9 y b x 1 2 x a Find the equations of the vertical asymptotes of each function. 2x + 1 4x 2 ii y = iii y = 2 x+1 x−2 x − 25 b In part iii above, identify the zeroes and discontinuities and draw up a table of signs. Then describe the behaviour of the curve near each vertical asymptote by copying and completing, ‘As x → 5− , y → . . . , and as x → 5+ , y → . . . ’ (and similarly for −5). i y= 10 Consider the function y = 2x x2 − 1 . a Show that it is an odd function. b Find the zeroes and discontinuities, and draw up a table of signs. c Identify any vertical and horizontal asymptotes. d Hence sketch the graph. 11 The graphs of y = f (x) and y = g(x) are sketched to the right. On separate number planes, sketch: y y = f(x) 2 a y = f (x) + g(x) 1 y = g(x) b y = f (x) − g(x) -1 12 The graphs of y = f (x) and y = g(x) are sketched to the right. y a On separate number planes, sketch: 4 i y = f (x) + g(x) ii y = f (x) − g(x) b What symmetry should be evident in your sketches? 13 1 -2 -1 The graph of y = f (x) is sketched to the right. On three separate number planes, sketch: a y = | f (x)| b y = f (|x|) c y = | f (|x|)| CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x 1 -2 y = f(x) y = g(x) 1 2 x 3 x y -1 1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 6 review 14 y y = f (x) a y b y=x 1 y=x -1 -1 x x 1 -1 -1 y c 1 y=x 15 16 1 1 y=x y = f (x) x x Find the equation of the inverse function for each function. 5 5x a y = 5 − 3x b y= c y= x−3 x−3 d y = x3 + 5 Find the inverse function f −1 (x) of each function, and then confirm algebraically that f f −1 (x) = x and f −1 f (x) = x . a f (x) = 12 x + 4 17 y = f (x) y d y = f (x) -1 1 Review Copy each diagram below, then sketch the inverse relation of the function. Also state whether or not the inverse relation is a function. 239 b f (x) = (x + 2)3 c f (x) = 3 −6 x A line is defined parametrically by the equations x = t + 2 and y = 2t + 6. a Copy and complete the table below. t −5 −4 −3 −2 −1 0 1 x y b Use the table to sketch the line, marking each point with its t-value. c Eliminate the parameter t to find the Cartesian equation of the line. 18 A parabola is defined parametrically by the equations x = 12 t and y = 14 t2 . a Copy and complete the table below. t −6 −4 −2 −1 0 1 2 4 6 x y b Use the table to sketch the parabola, marking each point with its t-value. c Eliminate the parameter t to find the Cartesian equation of the parabola. 19 A curve is defined parametrically by the equations x = cos θ − 1 and y = sin θ + 1. a Use the identity cos2 θ + sin2 θ = 1 to find the Cartesian equation of the curve. b Describe the curve, and then sketch it. 20 1 A curve is represented parametrically by the equations x = 2t and y = . t+1 Find the Cartesian equation of the curve and hence sketch it. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7 Trigonometry Chapter introduction Trigonometry is important in modern science principally because the graphs of the sine and cosine functions are waves. Waves appear everywhere in the natural world, for example as water waves, as sound waves, or as the electromagnetic waves that are responsible for radio, heat, light, ultraviolet radiation, X-rays and gamma rays. In quantum mechanics, a wave is associated with every particle. Trigonometry began in classical times, however, in practical situations such as building, surveying, navigation and astronomy. Trigonometry uses the relationships between the angles and the side lengths in a triangle, and its name comes from the Greek words trigonon, ‘triangle’, and metron, ‘measure’. This chapter develops the trigonometric functions and their graphs from the geometry of triangles and circles, and applies the trigonometric functions in practical problems. Some of this chapter will be new to most readers, in particular the extension of the trigonometric functions to angles of any magnitude, the graphs of these functions, and trigonometric identities and equations. The graphs of y = cosec x, y = sec x and y = cot x are included here, in Section 7C, rather than in Chapter 6, so that they take their place amongst the other three trig graphs. Some understanding of graphs of reciprocals — presented in Chapter 6 — is helpful, but readers who have not yet studied Chapter 6 should have little difficulty following the presentation here.. Radian measure is essential later for the calculus of the trigonometric functions. This chapter, however, is already long, and we have placed radian measure a little later in Chapter 11, after differentiation, when its relevance can be explained. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7A Trigonometry with right-angled triangles 241 7A Trigonometry with right-angled triangles Learning intentions • Understand the role of Pythagoras’ theorem and similarity in trigonometry. • Know the definitions of the six trigonometric functions for acute angles. • Evaluate the trigonometric functions of the special angles 30◦ , 45◦ and 60◦ . • Find sides and angles of a right-angled triangle, given sufficient information. This section and the next will review the definitions of the trigonometric functions for acute angles, and apply them to problems involving right-angled triangles. Pythagoras’ theorem You will know from earlier years that the trigonometry of triangles begins with two fundamental ideas — Pythagoras’ theorem and similarity–congruence. Pythagoras’ theorem tells us how to find the third side of a right-angled triangle. The theorem is the best-known theorem in all of mathematics, and has been mentioned several times already in earlier chapters. Here is what it says: The square on the hypotenuse of a right-angled triangle is the sum of the squares on the other two sides. The diagrams below provide a very simple proof. Can you work out how the four shaded triangles have been pushed around inside the square to prove the theorem? b c b A a c B A a C B C Similarity and congruence Similarity is required to define the trigonometric functions, because each function is defined as the ratio of two sides of a triangle. • Two figures are called congruent if one can be obtained from the other by translations, rotations and reflections. • They are called similar if enlargements are allowed as well. A P 6 3 12 6 Q B 10 5 R C In two similar figures: matching angles are equal and CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 matching sides are in ratio. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 242 7A Chapter 7 Trigonometry The trigonometric functions for acute angles Let θ be any acute angle, that is, 0◦ < θ < 90◦ . Construct a right-angled triangle with an acute angle θ, and label the sides: hyp opp θ hyp — the hypotenuse, the side opposite the right angle, opp — the side opposite the angle θ, adj — the third side, adjacent to θ but not the hypotenuse. 1 adj The trigonometric functions for an acute angle θ sin θ = opp hyp cos θ = adj hyp tan θ = opp adj cosec θ = hyp opp sec θ = hyp adj cot θ = adj opp Any two triangles with angles of 90◦ and θ are similar, by the AA similarity test. Hence the values of the six trigonometric functions are the same, whatever the size of the triangle. The full names of the six trigonometric functions are: sine, cosine, tangent, cosecant, secant, cotangent. Question 19 in the next exercise gives some clues about these names, and the way in which the functions were originally defined. Special angles The values of the trigonometric functions for the three acute angles 30◦ , 45◦ and 60◦ can be calculated exactly, using half a square and half an equilateral triangle, and applying Pythagoras’ theorem. A Ö2 C P 45º 45º 1 1 3 B Q Take half a square with side length 1 . The resulting right-angled triangle ABC has two angles of 45◦ . By Pythagoras’ theorem, the hypotenuse AC has √ length 2 . 30º 2 60º 1 R Take half an equilateral triangle with side length 2 by dropping an altitude. The resulting right-angled PQR has angles of 60◦ and 30◦ . √ By Pythagoras’ theorem, PQ = 3 . Applying the definitions on the previous page gives the values in the following table. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7A Trigonometry with right-angled triangles 2 243 A table of exact values θ 30◦ 45◦ 60◦ sin θ cos θ tan θ cosec θ sec θ cot θ √ √ 1 3 1 2 2 3 √ √ 2 2 3 3 √ √ 1 1 1 2 2 1 √ √ 2 2 √ √ 3 1 2 1 3 2 √ √ 2 2 3 3 Trigonometric functions of other angles The calculator is usually used to approximate trigonometric functions of other angles. Alway check first that the calculator is in degrees mode — there is usually a key labelled mode or DRG or something similar. Later, you will be swapping between degrees mode and radian mode (ignore the ‘grads’ unit). Make sure also that you can enter angles in degrees, minutes and seconds, and that you can convert decimal output to degrees, minutes and seconds — there is usually a key labelled ◦ ’ ” or DMS or something similar. Check that you can perform these two procedures: sin 53◦ 47 0.8068 and sin θ = 58 , so θ 38◦ 41 . The reciprocal trigonometric functions The functions cosecant, secant and cotangent can mostly be avoided in simple trigonometric problems by using the sine, cosine and tangent functions. 3 Avoiding the reciprocal trigonometric functions The three reciprocal trigonometric functions can mostly be avoided because 1 1 1 , sec θ = , cot θ = . cosec θ = sin θ cos θ tan θ Finding an unknown side of a right-angled triangle Calculators only have the sine, cosine and tangent functions, so it is best to use only these three functions in problems. 4 To find an unknown side of a right-angled triangle unknown side = . . . . . . (place the unknown at top left) known side 2 Complete the RHS with sin, cos or tan, or the reciprocal of one of these. 1 Start by writing CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 244 7A Chapter 7 Trigonometry Example 1 Finding an unknown side of a right-angled triangle Find the side marked with a pronumeral in each triangle. Give the answer in exact form if possible, or else correct to five significant figures. a b 5 70º 5 x y 60º Solution x = sin 60◦ 5 a opposite hypotenuse b x = 5 sin 60◦ √ 5 3 . = 2 ×5 ×5 y 1 = 5 sin 70◦ 5 y= sin 70◦ 5.3209 . hypotenuse opposite Finding an unknown angle of a right-angled triangle As before, use only sin, cos and tan. 5 Finding an unknown angle, given two sides of a right-angled triangle Work out from the known sides which one of cos θ, sin θ or tan θ to use. Example 2 Finding an unknown angle of a right-angled triangle Find θ in the triangle drawn to the right. Solution The given sides are the opposite and the adjacent sides, so tan θ is known. opposite 12 tan θ = 7 adjacent θ 59◦ 45 . 7 12 Exercise 7A 1 θ FOUNDATION From the diagram to the right, write down the values of: a cos α b tan β c sin α d cos β e sin β f tan α CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7A Trigonometry with right-angled triangles 2 3 4 5 6 7 245 Use your calculator to find, correct to four decimal places: a sin 24◦ b cos 61◦ c tan 35◦ d sin 87◦ e tan 2◦ f cos 33◦ g sin 1◦ h cos 3◦ Use your calculator to find, correct to four decimal places: a tan 57◦ 30 b cos 32◦ 24 c tan 78◦ 40 d cos 16◦ 51 e sin 43◦ 6 f sin 5◦ 50 g sin 8 h tan 57 Use your calculator to find the acute angle θ, correct to the nearest degree, if: a tan θ = 4 b cos θ = 0.7 e cos θ = 2 cos θ = 79 f c sin θ = 15 g d sin θ = 0.456 tan θ = 1 34 h sin θ = 1.1 Use your calculator to find the acute angle α, correct to the nearest minute, if: a cos α = 34 b tan α = 2 c sin α = 0.1 d tan α = 0.3 e sin α = 0.7251 7 f cos α = 13 Find, correct to the nearest whole number, the value of each pronumeral. a b c d Find, correct to the nearest degree, the size of each angle marked with a pronumeral. a b c d CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 246 7A Chapter 7 Trigonometry DEVELOPMENT 8 From the diagram opposite, write down the values of: a sin α b tan β c sec β d cot α e cosec α f sec α β 13 12 α 5 9 a Use Pythagoras’ theorem to find the unknown side in each of the two 15 right-angled triangles in the diagram opposite. x y b Write down the values of: 10 i cos y ii sin x iii cot x iv cosec y v sec x vi cot y Draw the two special triangles containing the acute angles 30◦ , 60◦ and 45◦ . Hence write down the exact values of: a sin 60◦ 11 12 8 10 b tan 30◦ c cos 45◦ d sec 60◦ e cosec 45◦ f cot 30◦ Find, correct to one decimal place, the lengths of the sides marked with pronumerals. a b c d Find, correct to two decimal places, the lengths of the sides marked with pronumerals. a b c d CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7A Trigonometry with right-angled triangles 13 247 Find θ, correct to the nearest minute, in each diagram below. a b c d e f √ 14 It is given that α is an acute angle and that tan α = 25 . a Draw a right-angled triangle (it may be of any size), one of whose angles is α, and show this information. b Use Pythagoras’ theorem to find the length of the unknown side. c Hence write down the exact values of sin α and cos α. d Show that sin2 α + cos2 α = 1. 15 Suppose that β is an acute angle and sec β = √ 11 3 . a Find the exact values of: ii cot β i cosec β b Show that cosec2 β − cot2 β = 1. 16 Find, without using a calculator, the value of: a sin 45◦ cos 45◦ + sin 30◦ b sin 60◦ cos 30◦ − cos 60◦ sin 30◦ c 1 + tan2 60◦ d cosec2 30◦ − cot2 30◦ 17 Without using a calculator, show that: a 1 + tan2 45◦ = sec2 45◦ b 2 sin 30◦ cos 30◦ = sin 60◦ c cos2 60◦ − cos2 30◦ = − 12 d 2 tan 30◦ = tan 60◦ 1 − tan2 30◦ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 248 7A Chapter 7 Trigonometry 18 In the diagram to the right, PQS is a right triangle, and PR is the altitude to the hypotenuse QS . a Explain why ∠RPS = θ. b Find two expressions for tan θ. c Hence show that ab = h2 . 19 a [An earlier definition of the sine function] The value sin θ used to be defined as the length of the semichord subtending an angle θ at the centre of a circle of radius 1. In the diagram to the right, the chord AB of a circle of radius 1 subtends an angle 2θ at the centre O. We know that the perpendicular bisector of the chord 1 passes through the centre O. Prove that sin θ = AB. 2 O TT 1 A B M b [The origin of the notations tan θ and sec θ] The word ‘tangent’ comes from T the Latin tangens meaning ‘touching’, and a tangent to a circle is a line touching it at one point. The word ‘secant’ comes from the Latin secans meaning ‘cutting’, and a secant to a circle is a line cutting it at two points. In the diagram to the right, P lies outside a circle of centre O and radius 1, and a tangent PT and secant PO have been drawn. Let the tangent PT subtend θ at the centre. We know that the radius OT is perpendicular to the tangent. Prove that 1 PT = tan θ 1 P θ O PO = sec θ. and ENRICHMENT 20 An equilateral triangle PQR has side length x, and PS is the perpendicular from P to QR. PS is produced to T so that PT = x. P a Show that ∠PQT = 75◦ and hence that ∠S QT = 15◦ . b c Show that QS = 12 x and that PS = 12 x √ Show that S T = 12 x(2 − 3 ). √ ◦ d Hence show that tan 15 = 2 − 21 √ x x 3. S Q 3. R T [The regular pentagon and the exact value of sin 18◦ ] The regular pentagon ABCDE has sides of length 1 unit. The diagonals AD, BD and AC have been drawn, and the diagonals BD and AC meet at P. a Find the size of each interior angle of the pentagon. ◦ D ◦ b Show that ∠DAB = 72 and ∠DAP = ∠BAP = 36 . c Show that the triangles DAB and ABP are similar. 1 d Let BP = x, and show that AB = AP = DP = 1 and DA = . x √ e Show that AD = 12 5 + 1 . √ f Hence show that sin 18◦ = cos 72◦ = 14 5 − 1 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 E P A C B © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7B Problems involving right-angled triangles 249 7B Problems involving right-angled triangles Learning intentions • Solve problems involving right-angled triangles. • Work with compass bearings and with angles of elevation and depression. Trigonometry developed in the ancient world solving practical problems with right-angled triangles, typically involving compass bearings and angles of elevation or depression. It was also used extensively in ancient astronomy. Angles of elevation and depression Angles of elevation and depression are always measured from the horizontal, and are always acute angles. Sun Observer 25º 80º Boat Observer The angle of elevation of the sun in the diagram above is 80◦ , because the angle at the observer between the sun and the horizontal is 80◦ . Example 3 For an observer on top of the cliff, the angle of depression of the boat is 25◦ , because the angle at the observer between boat and horizontal is 25◦ . Angle of depression From a plane flying at 9000 metres above level ground, I can see a church at an angle of depression of 35◦ from the cabin of the plane. Find how far the church is from the plane, correct to the nearest 100 metres: a measured along the ground, b measured along the line of sight. Solution The situation is illustrated in the diagram by PGC. PC 1 GC a b = tan 55◦ = 9000 9000 cos 55◦ ◦ 9000 GC = 9000 tan 55 PC = cos 55◦ 12 900 metres. 15 700 metres. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 35º 55º P 9000 m 35º C G © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 250 7B Chapter 7 Trigonometry Example 4 Angle of elevation A walker on level ground is 1 kilometre from the base of a 300-metre vertical cliff. a Find, correct to the nearest minute, the angle of elevation of the top. b Find, correct to the nearest metre, the line-of-sight distance to the top. Solution The situation is illustrated by CWB in the diagram to the right. CB a tan θ = b Using Pythagoras’ theorem, WB CW 2 = CB2 + WB2 300 = = 10002 + 3002 1000 3 = 10 = 1 090 000 ◦ θ 16 42 . q C 300 m q W 1000 m B CW 1044 metres. Compass bearings and true bearings Compass bearings are based on north, south, east and west. Any other direction is specified by the deviation from north or south towards the east or west. The diagram to the left below gives four examples. Note that S45◦ W can also be written simply as SW (that is, south west). True bearings are measured clockwise from north (not anti-clockwise as in the coordinate plane). The diagram to the right below gives the same four directions expressed as true bearings. Three digits are used, even for angles less than 100◦ . N20ºW N 20º N30ºE 340ºT 030ºT 30º W E 45º 70º 270ºT 090ºT S70ºE 110ºT S45ºW 225ºT S Example 5 000ºT 180ºT Compass bearings and true bearings A plane flies at 400 km per hour, and flies from A to B in the direction S30◦ E for 15 minutes. The plane then turns sharply to fly due east for 30 minutes to C. a Find how far south and east of A the point B is. A 100 km 30º P B 200 km C b Find the true bearing of C from A, correct to the nearest degree. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7B Problems involving right-angled triangles 251 Solution a The distances AB and BC are PAC, PC tan ∠PAC = AP 50 + 200 = √ 50 3 5 = √ 3 ∠PAC 71◦ . Hence the bearing of C from A is about 109◦ T. b Using opposite over adjacent in 100 km and 200 km respectively. Working in PAB, PB = sin 30◦ 100 PB = 100 sin 30◦ = 50 km, and AP = 100 cos 30◦ √ = 50 3 km. Exercise 7B FOUNDATION 1 A ladder of length 3 metres is leaning against a wall and is inclined at 62◦ to the ground. How far does it reach up the wall? (Answer in metres correct to two decimal places.) 2 Determine, correct to the nearest degree, the angle of elevation of the top T of a 6 metre flagpole FT from a point P on level ground 3 metres from F. 3 Ben cycles from P to Q to R and then back to P in a road-race. Find, correct to the nearest kilometre, the distance he has ridden. 4 A ship sails 78 nautical miles due north from X to Y, then 61 nautical miles due east from Y to Z. Find θ, the bearing of Z from X, correct to the nearest degree. 5 A tree snapped into two sections AB and BC in high winds and then fell. The section BA is inclined at 51◦ 38 to the horizontal and AC is 9.4 metres long. Find the height of the original tree, in metres correct to one decimal place. 6 A ladder makes an angle of 36◦ 42 with a wall, and its foot is 1.5 metres out from the base of the wall. Find the length of the ladder, in metres correct to one decimal place. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 252 7B Chapter 7 Trigonometry 7 Eleni drives 120 km on a bearing of 130◦ T. She then drives due west until she is due south of her starting point. How far is she from her starting point, correct to the nearest kilometre? 8 John is looking out a window W at a car C parked on the street below. If the angle of depression of C from W is 73◦ and the car is 7 metres from the base B of the building, find the height WB of the window, correct to the nearest metre. DEVELOPMENT 9 A ladder of length 5 metres is placed on level ground against a vertical wall. If the foot of the ladder is 1.5 metres from the base of the wall, find, correct to the nearest degree, the angle at which the ladder is inclined to the ground. 10 Find, correct to the nearest tenth of a metre, the height of a tower, if the angle of elevation of the top of the tower is 64◦ 48 from a point on horizontal ground 10 metres from the base of the tower. 11 A boat is 200 metres out to sea from a vertical cliff of height 40 metres. Find, correct to the nearest degree, the angle of depression of the boat from the top of the cliff. 12 Port Q is 45 nautical miles from port P on a bearing of 055◦ T. Port R is 65 nautical miles from port P, and ∠PQR = 90◦ . a Find ∠QPR to the nearest degree. b Hence find the bearing of R from P, correct to the nearest degree. 13 The bearings of towns Y and Z from town X are 060◦ T and 330◦ T respectively. a Show that ∠ZXY = 90◦ . b Given that town Z is 80 km from town X and that ∠XYZ = 50◦ , find, correct to the nearest kilometre, how far town Y is from town X. 14 A ship leaves port P and travels 150 nautical miles to port Q on a bearing of 110◦ T. It then travels 120 nautical miles to port R on a bearing of 200◦ T. a Explain why ∠PQR = 90◦ . 110º P Q 200º b Find, correct to the nearest degree, the bearing of port R from port P. R CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7B Problems involving right-angled triangles 15 a Show that AC = 7 tan 50◦ and BC = 7 tan 25◦ , and hence find the length AB, correct to 1 mm. b Show that AP = 20 sin 56◦ , and c Show that PR = 18 cos 40◦ , find hence find the length of PC, giving your answer correct to 1 cm. A an expression for PQ, and hence find the angle α, correct to the nearest minute. A P 20 cm α D 16 R 18 cm 56º B C √ a Show that x = 10( 3 − 1). 46º P b Show that x = 10 3 (3 − x C √ 3 ). Q √ c Show that x = 20 3 3. x 15º x 15º 30º 30º 10 30º 45º 10 17 8 cm 40º S B 25º 25º 7 cm 253 10 From the ends of a straight horizontal road 1 km long, a balloon directly above the road is observed to have angles of elevation of 57◦ and 33◦ respectively. Find, correct to the nearest metre, the height of the balloon above the road. 57º 33º 1 km ENRICHMENT 18 a Write down two equations involving x and y. b By solving the equations simultaneously, show that y 39º 7 . x= tan 64◦ − tan 39◦ 7 64º x 19 In the diagram, O is the centre of the semicircle ACB, and P is the foot of the perpendicular from C to the diameter AB. Let ∠OAC = θ. C a Show that ∠POC = 2θ and that ∠PCB = θ. b Using the two triangles APC and ABC, show that sin θ cos θ = c Hence show that 2 sin θ cos θ = sin 2θ. 20 PC . AB T A O P B Using the same diagram as the previous question: a Explain why AP − PB = 2 × OP. AP AC PB CB × − × . AC AB CB AB 2 c Hence show that cos2 θ − sin θ = cos 2θ. b Show that cos2 θ − sin2 θ = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 254 7C Chapter 7 Trigonometry 7C Trigonometric functions of a general angle Learning intentions • Use circles to extend the six trigonometric functions to angles of any size. • Work with boundary angles, and determine the domains of the trig functions. • Sketch the six trig functions using their circle definitions. • Identify domain, range, period, amplitude and asymptotes from their graphs. The definitions of the trigonometric functions in Section 7A only apply to acute angles, because in a right-angled triangle, both other angles are acute angles. This section introduces more general definitions based on circles in the coordinate plane (whose equations are Pythagoras’ theorem, as we saw in Section 3G). The new definitions will apply to any angle, but will, of course, give the same values as the previous definitions for acute angles. As always, the graphs of these new functions are drawn as soon as reasonable. Putting a general angle on the coordinate plane Let θ be any angle — possibly negative, possibly obtuse or reflex, possibly greater than 360◦ . We shall associate with θ a ray with vertex at the origin. 6 The ray corresponding to θ • The positive direction of the x-axis is the ray representing the angle 0◦ . • For all other angles, rotate this ray anti-clockwise through an angle θ. If the angle is negative, the ray is rotated backwards, which means clockwise. Here are some angles and corresponding rays. The angles are written at the ends of the arrows representing the rays. y 100º −320º, 40º, 400º Notice that one ray can correspond to many angles. For example, all the following angles have the same ray as 40◦ : . . . , −680◦ , −320◦ , 40◦ , 400◦ , 760◦ , . . . A given ray thus corresponds to infinitely many angles, all differing by multiples of 360◦ . 7 x −160º, 200º −40º, 320º Corresponding angles and rays • For each angle, there is exactly one ray. • For each ray, there are infinitely many angles, all differing from each other by multiples of 360◦ . Defining the trigonometric functions for general angles Construct a circle with centre the origin and any positive radius r. Let the ray corresponding to θ meet the circle at the point P(x, y). y θ P(x,y) r x The six trigonometric functions are now defined in terms of x, y and r as follows: CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7C Trigonometric functions of a general angle 8 255 Definitions of the six trigonometric functions x r r sec θ = x y r r cosec θ = y y x x cot θ = y cos θ = sin θ = tan θ = Note: We chose r to be ‘any positive radius’. If a different radius were chosen, the two figures would be similar, so the lengths x, y and r would stay in the same ratio, so the values of the trigonometric functions would not change. In particular, one may use a circle of radius 1 in the definitions. When this is done, however, we lose the intuition that a trigonometric function is not a length, but is the ratio of two lengths. Agreement with the earlier definition Let θ be an acute angle, and construct the ray corresponding to θ. Let the perpendicular from P meet the x-axis at M. Then θ = ∠POM, so relating the sides to the angle θ, hyp = OP = r, opp = PM = y, adj = OM = x. Hence the old and the new definitions are in agreement. Note: Most people find that the diagram above is the easiest way to learn the new definitions of the trigonometric functions. Take the old definitions in terms of hypotenuse, opposite and adjacent sides, and make the replacements hyp ←→ r, opp ←→ y, adj ←→ x. Boundary angles Integer multiples of 90◦ , that is . . . , −90◦ , 0◦ , 90◦ , 180◦ , 270◦ , 360◦ , 450◦ , . . . are called boundary angles because they lie on the boundaries between quadrants. The values of the trigonometric functions at these boundary angles are not always defined, and are 0, 1 or −1 when they are defined. The diagram to the right can be used to calculate them, and the results are shown in the table below. 9 The boundary angles θ 0◦ 90◦ 180◦ 270◦ x r 0 −r 0 y 0 r 0 −r r r r r r sin θ 0 1 0 −1 cos θ 1 0 −1 0 tan θ 0 undefined 0 undefined 1 undefined −1 cosec θ undefined sec θ 1 undefined −1 undefined cot θ undefined 0 undefined 0 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 256 7C Chapter 7 Trigonometry In practice, read the values of the trigonometric functions at boundary angles off the graphs of the sine, cosine and tangent functions. These graphs need to be known very well indeed. A warning about asymptotes and zeroes You may have concluded from the definitions tan θ = cot θ = 1 tan θ and tan θ = y x and cot θ = that x y 1 . cot θ This is true everywhere except at boundary angles, because: • When θ = 0◦ , tan θ = 0 and cot θ is undefined. • When θ = 90◦ , tan θ is undefined and cot θ = 0. Always be aware that the number 0 has no reciprocal. Also ‘infinity’ is a mathematical idea, but not a number, so it cannot have a reciprocal either. The domains of the six trigonometric functions The trig functions are defined everywhere except where the denominator is zero. 10 Domains of the trigonometric functions • sin θ and cos θ are defined for all angles θ: Their domains are all angles θ. • tan θ and sec θ are undefined when x = 0: Their domains are θ . . . , −270◦ , −90◦ , 90◦ , 270◦ , 450◦ , . . . . • cot θ and cosec θ are undefined when y = 0: Their domains are θ . . . , −360◦ , −180◦ , 0◦ , 180◦ , 360◦ , . . . . The graphs of the six trigonometric functions The diagrams on the next page show the graphs of all six trigonometric functions over an interval extending beyond −360◦ ≤ x ≤ 360◦ . This extended interval shows how the graphs are built up by infinite repetition of a simple element. But readers need to construct these graphs by themselves from the definitions. • Question 9 generates the graphs of y = sin x, y = cos x, and y = tan x using approximations from the circle diagram drawn on graph paper. • Question 14 generates the graphs of y = cosec x, y = sec x, and y = cot x. Period and amplitude The graphs of y = sin x and y = cos x repeat every 360◦ , so the graphs are called periodic. The period is 360◦ , the length of the smallest repeating piece. The periods of the other four graphs will be identified in the exercise. The graphs of y = sin x and y = cos x are waves. They each rise to a maximum height of 1 above the x-axis, then fall to a minimum of −1 below the x-axis, which is the mean position. This difference of 1 is called the amplitude. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7C Trigonometric functions of a general angle 257 11 Period and amplitude • The graphs of y = sin x and y = cos x each consist of a small piece infinitely repeated, so the functions are called periodic. They each have period 360◦ , the length of the smallest repeating piece. • The graphs of y = sin x and y = cos x are waves, rising and falling 1 unit above and below the mean position, so these functions are said to have amplitude 1. The wave graphs of y = sin x and y = cos x are the basis of all mathematics that deals with waves. The later trigonometry in this course will mostly concern these wave properties. y 1 y = sin x −360º −270º −180º −90º 90º 180º 270º 360º x −1 y = cos x y 1 −360º −270º −180º −90º 90º 180º 45º 90º 180º 270º 360º x −1 y = tan x y 1 −45º −360º −270º −180º −90º 270º 360º x −1 y = cosec x y 1 −360º −270º −180º −90º 90º 180º 270º 360º x −1 y = sec x y 1 −360º −270º −180º −90º 90º 180º 270º 360º x −1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 258 7C Chapter 7 Trigonometry y = cot x y 45º 360º 270º 180º 90º 45º 90º 180º 270º 360º x Exercise 7C 1 2 3 4 5 FOUNDATION On a number plane, draw rays representing the following angles. a 40◦ b 110◦ c 190◦ d 290◦ e 420◦ f 500◦ On another number plane, draw rays representing the following angles. a −50◦ b −130◦ c −250◦ d −350◦ e −440◦ f −550◦ For each of the angles in Question 1, write down the size of the negative angle between −360◦ and 0◦ that is represented by the same ray. (b) (c) For each of the angles in Question 2, write down the size of the positive angle between 0◦ and 360◦ that is represented by the same ray. 10º 40º 20º (a) 20º 20º (d) Write down two positive angles between 0◦ and 720◦ and two negative angles between −720◦ and 0◦ that are represented by each of the rays in the diagram to the right. (f) 30º (e) DEVELOPMENT 6 Use the definitions x y y and cos θ = and tan θ = sin θ = r r x to write down the values of the six trigonometric ratios of the angle θ in each diagram. a T (3,4) b T c d 13 5 5 5 T 7 T Find the coordinates of the point P in each diagram. 60º a P b c 150º d P 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 2 2 3 P 315º íR © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7C Trigonometric functions of a general angle 8 259 Find each angle θ, correct to the nearest minute where necessary, given that 0◦ < θ < 360◦ . T (3,4) a b T ( − 5 ,2) c d 3 5 2 (1, 3 ) T T 9 íí [The graphs of sin θ, cos θ and tan θ] The diagram shows angles from 0◦ to 360◦ at 30◦ intervals. The circle has radius 4 units. 90º 120º 4 60º 3 150º 30º 2 1 180º 0º 1 2 3 4 210º 330º 240º 300º 270º a Use the diagram and the definitions of the three trigonometric ratios to copy and complete the following table. Measure the values of x and y correct to two decimal places, and use your calculator only to perform the necessary divisions. θ −30◦ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ 390◦ x y r sin θ cos θ tan θ b Use your calculator to check the accuracy of the values of sin θ, cos θ and tan θ that you obtained in part a. c Using the table of values in part a, graph the curves y = sin θ, y = cos θ and y = tan θ as accurately as possible on graph paper. Use the following scales: On the horizontal axis: let 2 mm represent 10◦ . On the vertical axis: let 2 cm represent 1 unit. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 260 7C Chapter 7 Trigonometry 10 By referring to the graph of the function y = sin x drawn in degrees in the text, write down its: a domain 11 c period d amplitude By referring to the graph of the function y = cos x, write down its: a domain 12 b range b range c period d amplitude By referring to the graph of the function y = tan x, write down: b its range a its domain ◦ c its period ◦ d the equations of its asymptotes for −360 ≤ x ≤ 360 13 a Read off the diagram, correct to two decimal places where necessary, the values of: i cos 60◦ ii sin 210◦ iii sin 72◦ iv cos 18◦ v sin 144◦ vi cos 36◦ vii cos 153◦ viii sin 27◦ ix sin 234◦ x cos 306◦ b Find from the graphs two values of x between 0◦ and 360◦ for which: i sin x = 0.5 ii cos x = −0.5 iii sin x = 0.9 iv cos x = 0.6 v sin x = 0.8 vi cos x = −0.8 vii sin x = −0.4 viii cos x = −0.3 ◦ ◦ c Find two values of x between 0 and 360 for which sin x = cos x. 14 a Copy and complete the following table. Obtain your values at boundary angles using the table constructed in this Section 7C. For other angles, use the values that you obtained in Question 9, together with the fact that, when x is not a boundary angle: 1 1 1 and sec x = and cot x = . cosecx = sin x cos x tan x x −30◦ 0◦ 30◦ 60◦ 90◦ 120◦ 150◦ 180◦ 210◦ 240◦ 270◦ 300◦ 330◦ 360◦ 390◦ cosecx sec x cot x b Using the table of values in part a, graph the curves y = cosec x, y = sec x and y = cot x as accurately as possible on graph paper. Use the same scales as before: On the horizontal axis: let 2 mm represent 10◦ . On the vertical axis: let 2 cm represent 1 unit. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7C Trigonometric functions of a general angle 15 261 a If you have already studied Chapter 6: Further graphs, look at the relationships between the graphs of a function and its reciprocal in Section 6C, and explain how to construct the graphs of cosec x, sec x, and cot x from the graphs of sin x, cos x, and tan x. b Explain why — strictly speaking — cot x is not the reciprocal of tan x. 16 By referring to the graph of the function y = cosec x, write down: a its domain b its range c its period d the equations of its asymptotes for −360◦ ≤ x ≤ 360◦ . 17 By referring to the graph of the function y = sec x, write down: a its domain b its range c its period d the equations of its asymptotes for −360◦ ≤ x ≤ 360◦ . 18 By referring to the graph of the function y = cot x, write down: a its domain b its range c its period d the equations of its asymptotes for −360◦ ≤ x ≤ 360◦ . 19 Classify each of the six trigonometric functions as even, odd, or neither. ENRICHMENT 20 If sin θ = k and θ is obtuse, find an expression for tan(θ + 90◦ ). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 262 7D Chapter 7 Trigonometry 7D Quadrant, sign, and related acute angle Learning intentions • Recognise the quadrant and the related angle of a general angle. • Understand how to find the sign of a trig function from the ASTC diagram. • Use quadrant and related angle to find the trig functions of a general angle. Symmetry is an essential aspect of trigonometric functions. This section uses symmetry to express the values of the trigonometric functions of any angle in terms of trigonometric functions of acute angles. 2nd quadrant 1st quadrant The diagram shows the conventional anti-clockwise numbering of the four quadrants of the coordinate plane. Acute angles are in the first quadrant, obtuse angles are in the second, and reflex angles are in the third and fourth. 3rd quadrant 4th quadrant The quadrant and the related acute angle y 150º 30º The diagram to the right shows the four rays corresponding to the four angles 30º 30º 30◦ , 150◦ , 210◦ and 330◦ . These four rays lie in the four quadrants of the plane, and they all make the same acute angle of 30◦ with the x-axis. They are thus the reflections of each other in the two axes. 30º 30º 210º x 330º 12 Quadrant and related acute angle Let θ be any angle. • The quadrant of θ is the quadrant (1, 2, 3 or 4) in which the ray lies. • The related acute angle of θ is the acute angle between the ray and the x-axis. Each of the four angles above has the same related acute angle 30◦ . Notice that θ and its related angle are the same only when θ is an acute angle. The signs of the trigonometric functions The signs of the trigonometric functions depend only on the signs of x and y. (The radius r is always positive.) The signs of x and y depend in turn only on the quadrant in which the ray lies. Thus we can easily compute the signs of the trigonometric functions from the accompanying diagram and the definitions. quadrant 1st 2nd 3rd 4th x + − − + y + + − − r + + + + sin θ + + − − cos θ + − − + tan θ + − + − cosec θ + + − − (same as sin θ) sec θ + − − + (same as cos θ) cot θ + − + − (same as tan θ) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y (−,+) (+,+) x (−,−) (+,−) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7D Quadrant, sign, and related acute angle 263 In NSW, these results are usually remembered by the phrase: 13 Signs of the trigonometric functions ‘ All Stations To Central’ indicating that the four letters A, S, T and C are placed successively in the four quadrants as shown. The significance of the letters is: A means All six functions are positive, S means only Sine (and cosecant) are positive, T means only Tangent (and cotangent) are positive, C means only Cosine (and secant) are positive. S A T C The graphs of y = sin θ, y = cos θ, and y = tan θ were constructed in the previous exercise, and all six trigonometric functions were drawn on the one page at the end of Section 7C. Study each of the six graphs to see how the table of signs above and the ASTC rule agree with your observations about when the graph is above the x-axis and when it is below. The angle and the related acute angle In the diagram to the right, a circle of radius r has been added to the earlier diagram showing the four angles 30◦ , 150◦ , 210◦ and 330◦ . The four points P, Q, R and S where the four rays meet the circle are all reflections of each other in the x-axis and y-axis. Because of this symmetry, the coordinates of these four points are identical, apart from their signs. Hence the trigonometric functions of these angles will all be the same too, except that the signs may be different. y 150º Q 30º 30º 30º 30º R 210º 30º P x S 330º 14 The angle and the related acute angle • The trigonometric functions of any angle θ are the same as the trigonometric functions of its related acute angle, apart from a possible change of sign. • The sign is best found using the ASTC diagram. Evaluating the trigonometric functions of any angle This gives a two-step way to evaluate the trigonometric functions of any angle. 15 Trigonometric functions of any angle Draw a quadrants diagram, then: • Place the ray in the correct quadrant, and use the ASTC rule to work out the sign of the answer. • Find the related acute angle, and work out the value of the trigonometric function at this related angle. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 264 7D Chapter 7 Trigonometry Example 6 Using the ASTC diagram and the related angle Find the exact value of: a tan 300◦ b sin(−210◦ ) c cos 570◦ b −210◦ is in quadrant 2, c 570◦ is in quadrant 3, the Solution a 300◦ is in quadrant 4, the related angle is 60◦ , so tan 300◦ = − tan 60◦ √ = − 3. the related angle is 30◦ , so sin(−210◦ ) = + sin 30◦ related angle is 30◦ , so cos 570◦ = − cos 30◦ √ = − 12 3 . = 12 . −210º 30º 30º 60º 570º 300º Note: The calculator will give approximate values of the trigonometric functions without any need to find the related acute angle. It will not give exact values, however, when these values involve surds. General angles with pronumerals This quadrants-diagram method generates formulae for expressions such as: sin(180◦ + A) or cot(360◦ − A). The trick is to place A on the quadrants diagram as if it were acute. 16 Some formulae with general angles sin(180◦ − A) = sin A sin(180◦ + A) = − sin A sin(360◦ − A) = − sin A cos(180◦ − A) = − cos A cos(180◦ + A) = − cos A cos(360◦ − A) = cos A tan(180◦ − A) = − tan A tan(180◦ + A) = tan A tan(360◦ − A) = − tan A Some people prefer to learn this list of identities to evaluate trigonometric functions, but this seems unnecessary when the quadrants-diagram method is so clear. Specifying a point in terms of r and θ The definitions of sin θ and cos θ for a general angle θ are: x y sin θ = and cos θ = . r r Rewriting these with x and y as the subject, we obtain: y T P(x,y) r x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7D Quadrant, sign, and related acute angle 265 17 Recovering the coordinates of a point: x = r cos θ y = r sin θ This means that if a point P in the coordinate plane is specified in terms of its distance OP from the origin and the angle of the ray OP, then the x and y coordinates of P can be recovered by means of these formulae. Example 7 Recovering the coordinates of a point The circle x2 + y2 = 36 meets the positive direction of the x-axis at A. Find the coordinates of the points P on the circle such that ∠AOP = 60◦ . Solution The circle has radius 6, so r = 6, and the ray OP has angle 60◦ or −60◦ , so the coordinates (x, y) of P are x = 6 cos 60◦ x = 6 cos(−60◦ ) or = 3 = 3 ◦ y = 6 sin 60 √ = 3 3, √ So P = (3, 3 3) y = 6 sin(−60◦ ) √ = −3 3. √ P = (3, −3 3). or or Exercise 7D 1 Use the ASTC rule to determine the sign (+ or −) of each trigonometric ratio. a sin 20◦ b cos 50◦ c cos 100◦ d tan 140◦ e tan 250◦ f sin 310◦ g sin 200◦ h cos 280◦ i sin 340◦ j cos 350◦ k tan 290◦ l cos 190◦ ◦ ◦ m tan 170 2 3 4 FOUNDATION ◦ n sin 110 p cos 170◦ o tan 80 Find the related acute angle of each angle. a 10◦ b 150◦ c 310◦ d 200◦ e 80◦ f 250◦ g 290◦ h 100◦ i 350◦ j 160◦ Write each trigonometric ratio as the ratio of an acute angle using the correct sign. a tan 130◦ b cos 310◦ c sin 220◦ d tan 260◦ e cos 170◦ f sin 320◦ g cos 185◦ h sin 125◦ i tan 325◦ j sin 85◦ k cos 95◦ l tan 205◦ Use the trigonometric graphs to find the values, if they exist, of these trigonometric ratios of boundary angles. a sin 0◦ b cos 180◦ c cos 90◦ d tan 0◦ e sin 90◦ f cos 0◦ g sin 270◦ h tan 270◦ i sin 180◦ j cos 270◦ k tan 90◦ l tan 180◦ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 266 7D Chapter 7 Trigonometry 5 Find the exact value of: a sin 60◦ b sin 120◦ c sin 240◦ d sin 300◦ e cos 45◦ f cos 135◦ g cos 225◦ h cos 315◦ i tan 30◦ j tan 150◦ k tan 210◦ l tan 330◦ DEVELOPMENT 6 7 8 Find the exact value of: a cos 120◦ b tan 225◦ c sin 330◦ d sin 135◦ e tan 240◦ f cos 210◦ g tan 315◦ h cos 300◦ i sin 225◦ j cos 150◦ k sin 210◦ l tan 300◦ Find the exact value of: a cosec 150◦ b sec 225◦ c cot 120◦ d cot 210◦ e sec 330◦ f cosec 300◦ a If θ is acute, use the ASTC diagram to write as trigonometric functions of θ: ii cos(−θ) iii tan(−θ) i sin(−60◦ ) ii cos(−45◦ ) iii tan(−30◦ ) iv sin(−30◦ ) v cos(−60◦ ) vi tan(−45◦ ) i sin(−θ) b Hence find the exact values of: 9 10 11 12 Use the trigonometric graphs to find, if they exist: a sec 0◦ b cosec 270◦ c sec 90◦ d cosec 180◦ e cot 90◦ f cot 180◦ Find the related acute angle of each angle. a −60◦ b −200◦ c −150◦ d −300◦ e 430◦ f 530◦ g 590◦ h 680◦ Find the exact value of: a cos(−60◦ ) b sin(−120◦ ) c tan(−120◦ ) d sin(−315◦ ) e tan(−210◦ ) f cos(−225◦ ) g tan 420◦ h cos 510◦ i sin 495◦ j sin 690◦ k cos 600◦ l tan 585◦ Given that sin 25◦ 0.42 and cos 25◦ 0.91, write down approximate values, without using a calculator, for: a sin 155◦ b cos 205◦ c cos 335◦ d sin 335◦ e sin 205◦ − cos 155◦ f cos 385◦ − sin 515◦ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7D Quadrant, sign, and related acute angle 13 14 267 Given that tan 35◦ 0.70 and sec 35◦ 1.22, write down approximate values, without using a calculator, for: a tan 145◦ b sec 215◦ c tan 325◦ d tan 215◦ + sec 145◦ e sec 325◦ + tan 395◦ f sec(−145)◦ − tan(−215)◦ Show by substitution into LHS and RHS that each trigonometric identity is satisfied by the given values of the angles. a Show that sin 2θ = 2 sin θ cos θ, when θ = 150◦ . b Show that cos 3θ = 4 cos3 θ − 3 cos θ, when θ = 225◦ . c Show that sin(A + B) = sin A cos B + cos A sin B, when A = 300◦ and B = 240◦ . 15 Write each expression as a trigonometric ratio of θ, using the correct sign: a sin(−θ) b cos(−θ) c tan(−θ) d sec(−θ) ◦ e sin(180 − θ) f sin(360◦ − θ) g cos(180◦ − θ) h tan(180◦ + θ) ENRICHMENT 16 Show that: a sin 330◦ cos 150◦ − cos 390◦ sin 390◦ = 0 √ ◦ ◦ ◦ ◦ b sin 420 cos 405 + cos 420 sin 405 = 3+1 √ 2 2 √ sin 135◦ − cos 120◦ =3+2 2 ◦ ◦ sin 135 + cos 120 d (sin 150◦ + cos 270◦ + tan 315◦ )2 = sin2 135◦ cos2 225◦ sin 120◦ cos 240◦ e − = tan2 240◦ − cosec2 330◦ tan 300◦ cot 315◦ c CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 268 7E Chapter 7 Trigonometry 7E Given one trigonometric function, find another Learning intentions • Find, given a value of a trigonometric function, the exact values of other trigonometric functions. When the exact value of one trigonometric function is known for an angle, the exact values of the other trigonometric functions can easily be found using the circle diagram and Pythagoras’ theorem. Given one trigonometric function, find another 18 Given one trigonometric function, find another • Place a ray or rays on a circle diagram in the quadrants allowed in the question. • Complete the triangle and use Pythagoras’ theorem to find whichever of x, y and r is missing. Example 8 Given one trigonometric function, find another It is known that sin θ = 15 . a Find the possible values of cos θ. b Find cos θ if it is also known that tan θ is negative. Solution a The angle must be in quadrant 1 or 2, because sin θ is positive. y 1 = , we can take y = 1 and r = 5, r 5 √ √ so by Pythagoras’ theorem, x = 24 or − 24 , √ √ 2 6 2 6 so cos θ = or − . 5 5 b Because tan θ is negative, θ can only be in quadrant 2, √ 2 6 so cos θ is negative, cos θ = − . 5 Because sin θ = Exercise 7E FOUNDATION Note: Diagrams have been drawn for Questions 1–4, and similar diagrams should be drawn for the subsequent questions. Many answers will involve surds, but it is not important to rationalise denominators. Do not use the calculator at all in this exercise, because you are looking for exact values, not approximations. 1 Write down the values of sin θ, cos θ and tan θ for each part. a b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c d © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7E Given one trigonometric function, find another 2 3 In each part use Pythagoras’ theorem to find whichever of x, y or r is unknown. Then write down the values of sin α, cos α and tan α. a b c d a b Let cos θ = 35 , where 270◦ < θ < 360◦ . 4 5 Let tan θ = − 12 , where θ is obtuse. i Find sin θ. i Find sin θ. ii Find tan θ. ii Find cos θ. a b √ Suppose that cos θ = − 14 . Suppose that sin θ = − 47 . i Find the possible values of cos θ. ii Find the possible values of tan θ. 5 269 i Find the possible values of sin θ. ii Find the possible values of tan θ. Draw similar diagrams for this question. a If cos θ = 13 and θ is acute, find tan θ. b If cos θ = − 45 and θ is obtuse, find tan θ. c If cos θ = 12 and θ is reflex, find sin θ. d Find cos θ if tan θ = − 23 and θ is reflex. ◦ ◦ e Find sin θ, given that cos θ = − 40 41 and 0 ≤ θ ≤ 180 . f Find tan θ, given that sin θ = √1 and −90◦ ≤ θ ≤ 90◦ . 5 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 270 7E Chapter 7 Trigonometry DEVELOPMENT 6 In this question, each part has two possible answers. a If tan α = 13 , find sin α. c b If cos θ = √2 , find sin θ. 5 If sin θ = 35 , find cos θ. d Find tan θ, given that cos θ = − 23 . e Find tan θ, given that sin θ = − 12 13 . 7 f Find cos θ, given that tan θ = − √2 . 3 a If cos α = 45 and sin α < 0, find tan α. 8 b If tan θ = − 15 and sin θ > 0, find cos θ. c Find cos θ, given that sin θ = 14 and tan θ < 0. d Find sin θ, given that tan θ = 35 12 and cos θ > 0. e Find tan θ, given that sin θ = − 21 29 and cos θ > 0. f Find sin θ, given that cos θ = − 56 and tan θ < 0. 8 a Find sec θ, given that sin θ = √1 . 2 b Find tan θ, given that sec θ = − 17 8 . √ c If sec C = − √7 , find cot C. d If cot D = 9 3 11 5 , find cosec D. √ a Find sec θ, given that cosec θ = 32 and θ is obtuse. 9 b Find sec θ, given that cot θ = 40 and θ is reflex. ◦ ◦ c Find tan θ, given that sec θ = − 17 8 and 0 ≤ θ ≤ 180 . d Find cosec θ, given that cot θ = √2 and −90◦ ≤ θ ≤ 90◦ . 3 10 a If sin A = − 13 and tan A < 0, find sec A. b If cosec B = 73 and cos B < 0, find tan B. √ c Find cot θ, given that sec θ = − 2 and cosec θ < 0. d Find cos θ, given that cosec θ = − 13 5 and cot θ < 0. p , with θ obtuse and p and q both positive, find cos θ and tan θ. q 11 Given that sin θ = 12 If tan α = k, where k > 0, find the possible values of sin α and sec α. 13 a Prove the algebraic identity (1 − t2 )2 + (2t)2 = (1 + t2 )2 . b If cos x = 1 − t2 , where x is acute and t is positive, find expressions for sin x and tan x. 1 + t2 ENRICHMENT 14 If a > 0 and sec θ = a + 1 1 , prove that sec θ + tan θ = 2a or . 4a 2a CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7F Trigonometric identities 271 7F Trigonometric identities Learning intentions • Prove the reciprocal identities, ratio identities, and Pythagorean identities. • Prove the complementary angle identities, and interpret the function names. • Use these identities to prove further identities. Working with the trigonometric functions requires knowledge of a number of formulae called trigonometric identities, which relate trigonometric functions to each other. This section introduces eleven trigonometric identities in four groups: • the three reciprocal identities, • the two ratio identities, • the three Pythagorean identities, • the three identities concerning complementary angles. The three reciprocal identities It follows immediately from the definitions of the trigonometric functions in terms of x, y and r that: 19 The reciprocal identities For any angle θ: 1 sin θ 1 sec θ = cos θ 1 cot θ = tan θ cosec θ = (provided that sin θ 0) (provided that cos θ 0) (provided that tan θ 0 and cot θ 0) Note: We cannot use a calculator to find cot 90◦ or cot 270◦ by first finding tan 90◦ or tan 270◦ , because both are undefined. We already know, however, that cot 90◦ = cot 270◦ = 0. The two ratio identities Again using the definitions of the trigonometric functions: 20 The ratio identities For any angle θ: sin θ tan θ = cos θ cos θ cot θ = sin θ (provided that cos θ 0) (provided that sin θ 0) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 272 7F Chapter 7 Trigonometry y The three Pythagorean identities The point P(x, y) lies on the circle with centre O and radius r, so its coordinates satisfy x 2 + y2 = r 2 . θ P(x,y) r x x 2 y2 Dividing through by r2 , + = 1, r2 r2 that is, sin2 θ + cos2 θ = 1. Dividing through by cos2 θ and using the ratio and reciprocal identities, tan2 θ + 1 = sec2 θ, Dividing instead by sin θ, 1 + cot θ = cosec θ, 2 2 2 provided that cos θ 0. provided that sin θ 0. These identities are called the Pythagorean identities because they rely on the circle equation x2 + y2 = r2 , which is a restatement of Pythagoras’ theorem. 21 The Pythagorean identities For any angle θ: sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ (provided that cos θ 0) cot2 θ + 1 = cosec2 θ (provided that sin θ 0) The three identities for complementary angles The angles θ and 90◦ − θ are called complementary angles because they add to a right angle. Three trigonometric identities relate the values of the trigonometric functions at an angle θ and at the complementary angle 90◦ − θ. 22 The complementary angle identities For any angle θ: cos(90◦ − θ) = sin θ cot(90◦ − θ) = tan θ ◦ cosec(90 − θ) = sec θ For example, (provided that tan θ is defined) (provided that sec θ is defined) cos 20◦ = sin 70◦ , cosec 20◦ = sec 70◦ , cot 20◦ = tan 70◦ . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7F Trigonometric identities 273 Proof The triangle to the right shows that when a right-angled triangle is viewed from 90◦ − θ instead of from θ, then the opposite side and the adjacent side are exchanged. Hence a cos(90◦ − θ) = = sin θ, c a ◦ cot(90 − θ) = = tan θ, b c ◦ cosec(90 − θ) = = sec θ. b B [General angles] For general angles, we take the full circle diagram, and reflect it in the diagonal line y = x. Let P be the image of P under this reflection. 90º− θ A [Acute angles] 1 The image OP of the ray OP corresponds to the angle 90◦ − θ. 2 The image P of P(x, y) has coordinates P (y, x), because reflection in the line y = x reverses the coordinates of each point. c a θ b θ P(x,y) y y=x x P'(y,x) 90º− θ Applying the definitions of the trigonometric functions to the angle 90◦ − θ: y cos(90◦ − θ) = = sin θ, r y ◦ cot(90 − θ) = = tan θ, provided that x 0, x r provided that x 0. cosec(90◦ − θ) = = sec θ, x Cosine, cosecant and cotangent The complementary identities are the origin of the names ‘cosine’, ‘cosecant’ and ‘cotangent’ — the prefix ‘co-’ has the same meaning as the prefix ‘com-’ of ‘complementary’ angle. 23 Cosine, cosecant and cotangent cosine θ = sine (complement of θ) cotangent θ = tangent (complement of θ) cosecant θ = secant (complement of θ) Proving identities An identity is a statement that is true for all values of θ for which both sides are defined. An identity needs to be proven — it is quite different from an equation, which needs to be solved and to have its solutions listed. 24 Proving a trigonometric identity • Work separately on the LHS and the RHS until they are the same. • Use the four sets of identities in Boxes 19–22. • Never treat it as an equation, moving terms from one side to the other. Mostly it is only necessary to work on one of the two sides. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 274 7F Chapter 7 Trigonometry Example 9 Proving an identity using the Pythagorean identities Prove that (1 − cos θ)(1 + cos θ) = sin2 θ. Solution LHS = 1 − cos2 θ (use the difference of squares identity, from algebra) = sin2 θ (use the Pythagorean identities in Box 21) = RHS. Example 10 Using the reciprocal and ratio identities Prove that sin A sec A = tan A. Solution LHS = sin A × 1 cos A (use the reciprocal identities in Box 19) = tan A (use the ratio identities in Box 20) = RHS. Example 11 Prove that Using the Pythagorean and reciprocal identities 1 1 + = sec2 θ cosec2 θ. 2 sin θ cos2 θ Solution 1 1 + 2 2θ cos sin θ 2 2 cos θ + sin θ = sin2 θ cos2 θ 1 = 2 sin θ cos2 θ = sec2 θ cosec2 θ LHS = (use a common denominator) (use the Pythagorean identities in Box 21) (use the reciprocal identities in Box 19) = RHS. Exercise 7F 1 2 3 FOUNDATION Use your calculator to verify that: a sin 16◦ = cos 74◦ b tan 63◦ = cot 27◦ c sec 7◦ = cosec 83◦ d sin2 23◦ + cos2 23◦ = 1 e 1 + tan2 55◦ = sec2 55◦ f cosec2 32◦ − 1 = cot2 32◦ Simplify: 1 a sin θ b 1 tan α c sin β cos β d cos ϕ sin ϕ Simplify: a sin α cosec α CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b cot β tan β c cos θ sec θ © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7F Trigonometric identities 4 Prove: b cot α sin α = cos α a tan θ cos θ = sin θ 5 b sec(90◦ − α) 9 b 1 − cos2 β b 1 + cot2 ϕ Use the reciprocal and ratio identities to simplify: 1 sin2 β a b sec2 θ cos2 β d cos(90◦ − ϕ) sin(90◦ − ϕ) c 1 + tan2 ϕ d sec2 x − tan2 x c cosec2 A − 1 d cot2 θ − cosec2 θ cos2 A sin2 A d sin2 α cosec2 α c Use the reciprocal and ratio identities to prove: b cosec x cos x tan x = 1 a cos A cosec A = cot A 10 1 cot(90◦ − β) Use the Pythagorean identities to simplify: a 1 − sin2 β 8 c Use the Pythagorean identities to simplify: a sin2 α + cos2 α 7 c sin β sec β = tan β Use the complementary identities to simplify: a sin(90◦ − θ) 6 275 Use the reciprocal and ratio identities to simplify: cos α sin α a b sec α cosec α c c sin y cot y sec y = 1 tan A sec A d cot A cosec A DEVELOPMENT 11 12 Prove the following identities: a (1 − sin θ)(1 + sin θ) = cos2 θ b (1 + tan2 α) cos2 α = 1 2 c (sin A + cos A) = 1 + 2 sin A cos A d cos2 x − sin2 x = 1 − 2 sin2 x e tan2 ϕ cos2 ϕ + cot2 ϕ sin2 ϕ = 1 f 3 cos2 θ − 2 = 1 − 3 sin2 θ g 2 tan2 A − 1 = 2 sec2 A − 3 h 1 − tan2 α + sec2 α = 2 i cos4 x + cos2 x sin2 x = cos2 x j cot θ(sec2 θ − 1) = tan θ Prove the following identities: b cot β sec β = cosec β a tan α cosec α = sec α c cosec γ + sec γ = cosec γ sec γ d tan δ + cot δ = cosec δ sec δ e cosec ϕ − sin ϕ = cos ϕ cot ϕ f sec θ − cos θ = tan θ sin θ 2 13 2 2 Prove the following identities: a sin θ cos θ cosec2 θ = cot θ b (cos ϕ + cot ϕ) sec ϕ = 1 + cosec ϕ c sin4 A − cos4 A = sin2 A − cos2 A d sin β + cot β cos β = cosec β 1 + tan x = tan2 x 1 + cot2 x 1 1 + = 2 sec2 θ g 1 + sin θ 1 − sin θ 1 + cot x = cot x 1 + tan x 1 1 − = 2 tan ϕ h sec ϕ − tan ϕ sec ϕ + tan ϕ 2 e 14 2 f Prove that each expression is independent of θ by simplifying it. cos2 θ cos2 θ a b tan θ(1 − cot2 θ) + cot θ(1 − tan2 θ) + 1 + sin θ 1 − sin θ tan θ + cot θ tan θ + 1 cot θ + 1 c d − sec θcosecθ sec θ cosecθ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 276 7F Chapter 7 Trigonometry ENRICHMENT 15 Prove the identities: 2 cos3 θ − cos θ = cot θ a sin θ cos2 θ − sin3 θ 1 + sin y b sec y + tan y + cot y = sin y cos y c cos A − tan A sin A = 1 − 2 sin2 A cos A + tan A sin A d (sin ϕ + cos ϕ)(sec ϕ + cosecϕ) = 2 + tan ϕ + cot ϕ 1 1 cos4 x − = 1 + tan2 x 1 + sec2 x 1 + cos2 x cos θ sin θ + = sin θ + cos θ f 1 − tan θ 1 − cot θ e sec α cosecα + 2 cosec α sec2 α 1 cos θ h = sec θ − tan θ = sec θ + tan θ 1 + sin θ g (tan α + cot α − 1)(sin α + cos α) = i sin2 x(1 + n cot2 x) + cos2 x(1 + n tan2 x) = n + 1 = sin2 x(n + cot2 x) + cos2 x(n + tan2 x) j 16 1 + cosec2 A tan2 C 1 + cot2 A sin2 C = 1 + cosec2 B tan2 C 1 + cot2 B sin2 C a If b ab a = , show that sin A cos A = 2 . sin A cos A a + b2 b If a−b a2 − b2 a+b = , show that cosecx cot x = . cosecx cot x 4ab CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7G Trigonometric equations 277 7G Trigonometric equations Learning intentions • Solve various types of trigonometric equations. It is important to be able to confidently solve trignometric equations, as many problems result in a trigonometric equation that has to be solved. There are many small details to be wary of when solving trigonometric equations. Pay attention to any restriction on the angle To begin with a warning, before any other details: 25 Restrictions on the angle Pay careful attention to any restriction on the interval in which the angle can lie. Equations involving boundary angles Boundary angles are a special case because they do not lie in any quadrant. 26 The boundary angles If the solutions are boundary angles, read the solutions off a sketch of the graph. Example 12 Solving trig equations involving boundary angles a Solve sin x = −1, for 0◦ ≤ x ≤ 720◦ . b Solve sin x = 0, for 0◦ ≤ x ≤ 720◦ . Solution a The graph of y = sin x is drawn to the right. Examine where the curve touches the line y = −1, and read off the x-coordinates of these points. The solution is x = 270◦ or 630◦ . b Examine where the graph crosses the x-axis. The solution is x = 0◦ , 180◦ , 360◦ , 540◦ or 720◦ . y 1 270º 630º 360º x 720º −1 The standard method — quadrants and the related acute angle Most trigonometric equations eventually come down to one or more equations such as sin x = − 12 , where −180◦ ≤ x ≤ 180◦ . Provided that the angle is not a boundary angle, the method is: CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 278 7G Chapter 7 Trigonometry 27 The quadrants-and-related-angle method 1 Draw a quadrant diagram, then draw a ray in each quadrant that the angle could be in. 2 Find the related acute angle — only work with positive numbers here: a using special angles, or b using the calculator to find an approximation. Never enter a negative number into the calculator at this point. 3 Mark the angles on the ends of the rays, taking account of any restrictions on x, and write a conclusion. Example 13 Solving trig equations — the standard method a Solve the equation sin x = − 12 , for −180◦ ≤ x ≤ 180◦ . b Solve tan x = −3, for 0◦ ≤ x ≤ 360◦ , correct to the nearest degree. Solution a Here sin x = − 12 , where −180◦ ≤ x ≤ 180◦ . Because sin x is negative, x is in quadrant 3 or 4. The sine of the related acute angle is + 12 , so the related angle is 30◦ . Hence x = −150◦ or −30◦ . b Here tan x = −3, where 0◦ ≤ x ≤ 360◦ . Because tan x is negative, x is in quadrant 2 or 4. The tangent of the related acute angle is +3, so the related angle is about 72◦ . Hence x 108◦ or 288◦ . 108º 72º 72º 288º Note: When using the calculator, never enter a negative number and take an inverse trigonometric function of it. In the example above, the calculator was used to find the related acute angle whose tan is 3, which is 71◦ 34 , correct to the nearest minute. The positive number 3 was entered, not −3. The three reciprocal functions The calculator doesn’t have specific keys for secant, cosecant and cotangent. Convert these functions to sine, cosine and tangent as quickly as possible. 28 The reciprocal functions Take reciprocals to convert the three reciprocal functions secant, cosecant and cotangent to the three more common functions. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7G Trigonometric equations Example 14 279 Solving trigonometric equations with the three reciprocal functions a Solve cosec x = −2, for −180◦ ≤ x ≤ 180◦ . b Solve sec x = 0.7, for −180◦ ≤ x ≤ 180◦ . Solution a Taking reciprocals of both sides gives sin x = − 12 , which was solved in the previous example, so x = −150◦ or −30◦ . b Taking reciprocals of both sides gives cos x = 10 7 , which has no solutions, because cos θ can never be greater than 1. Equations with compound angles In some equations, the angle is a function of x rather than simply x itself. For example, √ tan 2x = 3, where 0◦ ≤ x ≤ 360◦ , or √ 3 ◦ , where 0◦ ≤ x ≤ 360◦ . sin(x − 250 ) = 2 These equations are really trigonometric equations with the compound angles 2x and (x − 250◦ ) respectively. The secret lies in solving for the compound angle, and in first calculating any restrictions for that compound angle. 29 Equations with compound angles 1 Let u be the compound angle. 2 Find any restrictions on u from the given restrictions on x. 3 Solve the trigonometric equation for u. 4 Hence solve for x. Example 15 Solving trigonometric equations with compound angles √ ◦ ◦ 3, where √ 0 ≤ x ≤ 360 . 3 b Solve sin(x − 250◦ ) = , where 0◦ ≤ x ≤ 360◦ . 2 a Solve tan 2x = Solution a u = 2x. √ tan u = 3 . Let Then The restriction on x is ×2 and replacing 2x by u, 60º,420º 0◦ ≤ x ≤ 360◦ ◦ ◦ 0 ≤ 2x ≤ 720 60º 60º 0◦ ≤ u ≤ 720◦ . (The restriction on u is the key step here.) Hence from the diagram, u = 60◦ , 240◦ , 420◦ or 600◦ . Because x = 12 u, x = 30◦ , 120◦ , 210◦ or 300◦ . 240º,600º Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 280 7G Chapter 7 Trigonometry b u = x − 250◦ . √ 3 . sin u = 2 0◦ ≤ x ≤ 360◦ Let Then The restriction on x is − 250◦ 60º −240º 60º 60º −250◦ ≤ x − 250◦ ≤ 110◦ and replacing x − 250◦ by u, −250◦ ≤ u ≤ 110◦ . (Again, the restriction on u is the key step here.) Hence from the diagram, u = −240◦ or 60◦ . Because x = u + 250◦ , x = 10◦ or 310◦ . Equations with more than one trigonometric function Some trigonometric equations involve more than one trigonometric function, as in √ sin x + 3 cos x = 0. The general approach is to use trigonometric identities to produce an equation in only one trigonometric function. Example 16 Trigonometric equations with more than one trig function √ Solve sin x + 3 cos x = 0, where 0◦ ≤ x ≤ 360◦ . Solution ÷ cos x √ sin x + 3 cos x = 0 √ tan x + 3 = 0 √ tan x = − 3 , where 0◦ ≤ x ≤ 360◦ . Because tan x is negative, x is in quadrants 2 or 4. √ The tan of the related acute angle is 3 , so the related angle is 60◦ . Hence x = 120◦ or 300◦ . Exercise 7G FOUNDATION 1 Solve each equation for 0◦ ≤ θ ≤ 360◦ . (Each related acute angle is 30◦ , 45◦ or 60◦ .) √ √ 3 1 a sin θ = b sin θ = c tan θ = 1 d tan θ = 3 2 2 √ √ 3 1 1 e cos θ = − √ f tan θ = − 3 g sin θ = − h cos θ = − 2 2 2 2 Solve each equation for 0◦ ≤ θ ≤ 360◦ . (The trigonometric graphs are helpful here.) 3 a sin θ = 1 b cos θ = 1 c cos θ = 0 d cos θ = −1 e tan θ = 0 f sin θ = −1 Solve each equation for 0◦ ≤ x ≤ 360◦ . Use your calculator to find the related acute angle in each case, and give solutions correct to the nearest degree. a cos x = 37 d sin x = − 23 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b sin x = 0.1234 e tan x = − 20 9 c tan x = 7 f cos x = −0.77 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7G Trigonometric equations 4 281 Solve for 0◦ ≤ α ≤ 360◦ . Give solutions correct to the nearest minute where necessary. a sin α = 0.1 b cos α = −0.1 c tan α = −1 d cosec α = −1 e sin α = 3 f sec α = −2 g √ 3 tan α + 1 = 0 h cot α = 3 DEVELOPMENT 5 Solve for −180◦ ≤ x ≤ 180◦ . Give solutions correct to the nearest minute where necessary. b cos x = 0 a tan x = −0.3 √ c sec x = 2 6 7 d sin x = −0.7 Solve each equation for 0◦ ≤ θ ≤ 720◦ . a 2 cos θ − 1 = 0 b cot θ = 0 c cosec θ + 2 = 0 d tan θ = Solve each equation for 0◦ ≤ x ≤ 360◦ . (Let u = 2x.) c cos 2x = − √1 2 9 √ 3 d sin 2x = −1 b tan 2x = a sin 2x = 12 8 √ 2−1 Solve each equation for 0◦ ≤ α ≤ 360◦ . (Let u be the compound angle.) √ a tan(α − 45◦ ) = √1 3 b sin(α + 30◦ ) = − 23 c cos(α + 60◦ ) = 1 d cos(α − 75◦ ) = − √1 2 Solve each equation for 0◦ ≤ θ ≤ 360◦ . a sin θ = cos θ √ c sin θ = 3 cos θ b sin θ + cos θ = 0 d √ 3 sin θ + cos θ = 0 ENRICHMENT 10 11 Solve for 0◦ ≤ θ ≤ 360◦ , giving solutions correct to the nearest minute where necessary: a sin2 θ − sin θ = 0 b 2 cos2 θ = cos θ c 2 sin θ cos θ = sin θ d 2 sin2 θ + sin θ = 1 e sec2 θ + 2 sec θ = 8 f 3 cos2 θ + 5 cos θ = 2 g 4 cosec2 θ − 4 cosec θ − 15 = 0 h 4 sin3 θ = 3 sin θ Use the quadratic formula to solve 4 cos2 θ + 2 sin θ = 3 for 0◦ ≤ θ ≤ 360◦ . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 282 7H Chapter 7 Trigonometry 7H The sine rule and the area formula Learning intentions • Prove the sine rule and the area formula. • Use the sine rule and the area formula to solve non-right-angled triangles. • Work with the ambiguities with these formulae in the SSA situation. These last three sections review the sine rule, the area formula and the cosine rule. These three rules extend trigonometry to non-right-angled triangles. The three formulae are closely related to the standard congruence tests: SSS, SAS, AAS, RHS from Euclidean geometry, and to the spurious ‘SSA non-test’ and its ambiguities. A The usual statement of all three rules uses the convention shown in the diagram to the right. The vertices are named with upper-case letters, then each side takes its name from the lower-case letter of the opposite vertex. b c a C B Statement of the sine rule The sine rule states that the ratio of each side of a triangle to the sine of its opposite angle is constant for the triangle: 30 The sine rule In any triangle ABC, b c a = = . sin A sin B sin C ‘The ratio of each side to the sine of the opposite angle is constant.’ Proving the sine rule by constructing an altitude Any proof of the sine rule must involve a construction with a right angle, because trigonometry so far has been restricted to right-angled triangles. Thus we construct an altitude, which is the perpendicular from one vertex to the opposite side. This breaks the triangle into two right-angled triangles, where previous methods can be used. Given: Let ABC be any triangle. There are three cases, depending on whether ∠A is an acute angle, a right angle, or an obtuse angle. C a b h A C C M B c Case 1: ∠A is acute Aim: To prove that A a a b c B Case 2: ∠A = 90◦ h b M A B c Case 3: ∠A is obtuse b a = . sin A sin B CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7H The sine rule and the area formula 283 b , so the result is clear. a Construction: In the remaining cases 1 and 3, construct the altitude from C, meeting AB (produced if necessary) at M. Let h be the length of CM. In case 2, sin A = sin 90◦ = 1, and sin B = Proof Case 1 — Suppose that ∠A is acute. h = sin A In ACM, b h = b sin A. ×b h = sin B In BCM, a ×a h = a sin B. Equating these, b sin A = a sin B a b = . sin B sin A Case 3 — Suppose that ∠A is obtuse. h In ACM, = sin(180◦ − A), b and because sin(180◦ − A) = sin A, ×b In h = b sin A. h = sin B a h = a sin B. BCM, ×a Equating these, b sin A = a sin B a b = . sin B sin A Using the sine rule to find a side — the AAS congruence situation When using the sine rule to find a side, one side and two angles must be known. This is the situation described by the AAS congruence test from geometry, so we know that there will only be one solution. 31 Using the sine rule to find a side In the AAS congruence situation: known side unknown side = . sine of its opposite angle sine of its opposite angle Always place the unknown side at the top left of the equation. If two angles of a triangle are known, so is the third, because the angles add to 180◦ . Example 17 Using the sine rule to find a side length Find the length of the side x in the triangle drawn to the right. Solution Using the sine rule, and placing the unknown at the top left 12 x = ◦ sin 30 sin 135◦ 12 sin 30◦ × sin 30◦ x= . sin 135◦ 1 Using special angles, sin 30◦ = , 2 (sine is positive for obtuse angles) and sin 135◦ = + sin 45◦ 1 = √ . (the related acute angle is 45◦ ) 2 √ 2 1 Hence x = 12 × × 1 √ 2 = 6 2. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x 12 30º 135º © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 284 7H Chapter 7 Trigonometry Using the sine rule to find an angle — the ambiguous SSA situation The SAS congruence test requires that the angle be included between the two sides. When two sides and a non-included angle are known, the resulting triangle may not be determined by congruence, and two different triangles may be possible. This situation may be referred to as ‘the ambiguous SSA situation’. When the sine rule is applied in the ambiguous SSA situation, there is only one answer for the sine of an angle. There may be two possible solutions for the angle itself, however, one acute and one obtuse. 32 Using the sine rule to find an angle In the ambiguous SSA situation, where two sides and a non-included angle of the triangle are known, sine of unknown angle sine of known angle = . its opposite side its opposite side Always check the angle sum to see whether both answers are possible. Example 18 Using the sine rule in the ambiguous SSA situation Find the angle θ in the triangle drawn to the right. Solution sin θ sin 45◦ (always place the unknown at the top left) √ = 14 7 6 √ 1 1 1 × √ , because sin 45◦ = √ , sin θ = 7 × 6 × 14 2 2 √ 3 sin θ = 2 θ = 60◦ or 120◦ . √ 3 , one of them acute and the other obtuse. Note: There are two angles whose sine is 2 Moreover, 120◦ + 45◦ = 165◦ , leaving just 15◦ for the third angle in the obtuse case, so it all seems to work. Opposite is the ruler-and-compasses construction of the triangle, showing how two different triangles can be produced from the same given SSA measurements. In many examples, however, the obtuse angle solution can be excluded using the fact that the angle sum of a triangle cannot exceed 180◦ . Example 19 Finding only one result in the ambiguous SSA situation Find the angle θ in the triangle drawn to the right. Solution sin θ sin 80◦ = (always place the unknown at the top left) 4 7 ◦ 4 sin 80 sin θ = 7 ◦ θ 34 15 or 145◦ 45 4 80º 7 θ But θ 145◦ 45 is impossible, because the angle sum would then exceed 180◦ , so θ 34◦ 15 is the only solution. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7H The sine rule and the area formula 285 The area formula The standard formula for the area of a triangle is well known: area = 12 × base × height. This can be developed to a formula involving two sides and the included angle. 33 The area formula for a triangle In any triangle ABC, area ABC = 12 bc sin A . ‘The area of a triangle is half the product of any two sides times the sine of the included angle.’ Proof Use the same three diagrams as in the proof of the sine rule. In case 2, ∠A = 90◦ and sin A = 1, so area = 12 bc = 12 bc sin A, as required. Otherwise, area = 12 × base × height = 12 × AB × h = 12 × c × b sin A, because we proved before that h = b sin A. Using the area formula — the SAS congruence situation The area formula requires the SAS congruence situation in which two sides and the included angle are known. 34 Using the area formula In the SAS congruence situation: area = (half the product of two sides) × (sine of the included angle) . Example 20 Finding the area of a triangle Find the area of the triangle drawn to the right. 3 135º 4 Solution Using the formula, Because 1 × 3 × 4 × sin 135◦ . 2 1 (as in the previous worked exercise) sin 135◦ = √ , 2 1 area = 6 × √ 2 √ 1 2 =6× √ × √ (rationalise the denominator) 2 2 √ = 3 2 square units. area = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 286 7H Chapter 7 Trigonometry Using the area formula to find a side Substituting a known area into the area formula may allow an unknown side to be found. Example 21 Using the area formula to find a side Find x, correct to four significant figures, given that the triangle to the right has area 72 m2 . Solution Substituting into the area formula, x 72 = 12 × 24 × x × sin 67◦ 72 = 12 × x × sin 67 6 x= sin 67◦ 6.518 metres. ÷ 12 sin 67◦ ◦ 67º 24 m Using the area formula to find an angle Substituting a known area into the area formula when two sides are known, will allow sin θ to be found uniquely, where θ is the included angle. But unless θ = 90◦ , there will be two answers for θ, supplementary to each other. Example 22 Using the area formula to find an angle Find θ in the triangle to the right, correct to the nearest minute, given that the triangle has area 60 cm2 . Solution Substituting into the area formula, 60 = 12 × 13 × 12 × sin θ 60 = 6 × 13 sin θ ÷ (6 × 13) Hence sin θ = 10 13 . θ 50◦ 17 or 129◦ 43 . Notice that the second angle is the supplement of the first. 35 Two ambiguous situations • When finding an angle in the SSA situation, sin θ will be unique, but there may be two possible answers for θ, supplementary to each other. • When finding an angle using the area formula, sin θ will be unique, but unless θ = 90◦ , there will be two possible answers for θ, supplementary to each other. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7H The sine rule and the area formula Exercise 7H 1 287 FOUNDATION Find x in each triangle, correct to one decimal place. a b c 4 85º 70º x x 50º 55º 10 d 2 4 f Find the value of the pronumeral in each triangle, correct to two decimal places. a 3 e b c 15 48º c 112º Find θ in each triangle, correct to the nearest degree. a b c d e f Find the area of each triangle, correct to the nearest square centimetre. a CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b c © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 288 7H Chapter 7 Trigonometry ABC in which A = 43◦ , B = 101◦ and a = 7.5 cm. b Find b and c, in cm correct to two decimal places. 5 a Sketch 6 Sketch ABC in which a = 2.8 cm, b = 2.7 cm and A = 52◦ 21 . a Find B, correct to the nearest minute. b Hence find C, correct to the nearest minute. c Hence find the area of ABC in cm2 , correct to two decimal places. DEVELOPMENT 7 8 9 There are two triangles that have sides 9 cm and 5 cm, and in which the angle opposite the 5 cm side is 22◦ . Find, in each case, the size of the angle opposite the 9 cm side, correct to the nearest degree. Two triangles are shown, with sides 6 cm and 4 cm, in which the angle opposite the 4 cm side is 36◦ . Find, in each case, the angle opposite the 6 cm side, correct to the nearest degree. Sketch PQR in which p = 7 cm, q = 15 cm and ∠P = 25◦ 50 . a Find the two possible sizes of ∠Q, correct to the nearest minute. b For each possible size of ∠Q, find r in cm, correct to one decimal place. 10 A travelling salesman drove from town A to town B, then to town C, and finally directly home to town A. Town B is 67 km north of town A, and the bearings of town C from towns A and B are 039◦ T and 063◦ T respectively. Find, correct to the nearest kilometre, how far the salesman drove. 11 Melissa is standing at A on a path that leads to the base B of a vertical flagpole. The path is inclined at 12◦ to the horizontal and the angle of elevation of the top T of the flagpole from A is 34◦ . a Explain why ∠T AB = 22◦ and ∠ABT = 102◦ . b Given that AB = 20 metres, find the height of the flagpole, correct to the nearest metre. 12 ABC, sin A = 14 , sin B = 23 and a = 12. Find the value of b. b In PQR, p = 25, q = 21 and sin Q = 35 . Find the value of sin P. a In CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7H The sine rule and the area formula 13 289 Substitute into the area formula to find the side length x, given that each triangle has area 48 m2 . Give your answers in exact form, or correct to the nearest centimetre. a b c P 130º 36 m x 9m x x 70º 14 Substitute into the area formula to find the angle θ, given that each triangle has area 72 cm2 . Give answers correct to the nearest minute, where appropriate. a 15 b Find the exact value of x in each diagram. a b 6 60º 16 17 c x 45º c 6 x 30º 45º d 6 x 6 x 45º 60º 45º 30º The diagram to the right shows an isosceles triangle in which the apex angle is 35◦ . Its area is 35 cm2 . Find the length of the equal sides, correct to the nearest millimetre. The summit S of a mountain is observed by climbers at two points P and Q that are 250 metres apart. The point Q is uphill and due north from P, and the summit is uphill and due north from Q — thus the whole diagram to the right all lies in one vertical plane. Construct perpendicular lines PM and QR to the vertical line through S . The line segment PQ is inclined at 18◦ to the horizontal, and the respective angles of elevation of S from P and Q are 42◦ and 54◦ . a Explain why ∠PS Q = 12◦ and ∠PQS = 144◦ . 35º S 250 m 54º Q 42º 18º P M 250 sin 144◦ . sin 12◦ c Hence find the vertical height S M, correct to the nearest metre. b Show that S P = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 290 7H Chapter 7 Trigonometry ENRICHMENT 18 a Suppose that we are using the sine rule in an ASS situation to find the angle θ in a triangle known to exist. Using the known angle β and the two sides, we have found the value of sin θ, and from that we have found the related angle α. Explain why the obtuse value θ = 180◦ − α is a solution if and only if the related angle α is greater than the known angle β. b Wayan used the sine rule in the ASS situation to find ∠A in ABC where a = 3, b = 2 and ∠B = 150◦ . He found that sin A = 34 , and discarding the obtuse solution, he concluded that ∠A 49◦ . What is wrong, and why did it go wrong? 19 b c a , and are equal. Each is a length over [The circumcircle] By the sine rule, the three ratios sin A sin B sin C a ratio, so is a length, and in this question, you will show that this common length is the diameter of the circumcircle that passes through all three vertices. C A D D M A A C 180ºD M D O O O B C B B a is the sin A diameter BM of the circumcircle. The proof requires a circle theorem that some will have proven in earlier years, that ∠A = ∠M because they both stand on the same arc BC. b Prove the result in the middle diagram, where the circumcentre lies on BC. c Prove the result in the right-hand diagram, where the circumcentre lies outside the triangle. Here you will need the theorem that the opposite angles of the quadrilateral ABMC whose vertices lie on the circle are supplementary. a In the left-hand diagram, the circumcentre lies inside the triangle ABC. Show that 20 b c a , and are each equal to the [The circumcircle] We saw in the previous question that sin A sin B sin C diameter of the circumcircle of ABC. a In a particular triangle ABC, ∠A = 60◦ and BC = 12. Find the diameter DC of the circumcircle. b A triangle 21 PQR with ∠RPQ = 150◦ is inscribed inside a circle of diameter DC . Find the ratio DC : RQ. x sin α sin β in the diagram to the right. sin(α − β) h h b Use the fact that tan α = and tan β = to show that y−x y x tan α tan β h= . tan α − tan β c Combine the expressions in parts a and b to show that sin(α − β) = sin α cos β − cos α sin β. d Hence find the exact value of sin 15◦ . a Show that h = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 D h A E x D B y C © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7I The cosine rule 291 7I The cosine rule Learning intentions • Prove the cosine rule using Pythagoras’ theorem. • Understand the cosine rule as Pythagoras’ theorem with correction term. • Use the cosine rule to find a side or an angle of a triangle. The cosine rule is a generalisation of Pythagoras’ theorem to non-right-angled triangles. It gives a formula for the square of any side in terms of the squares of the other two sides and the cosine of the opposite angle. The cosine rule and Pythagoras’ theoprem 36 The cosine rule In any triangle ABC, a2 = b2 + c2 − 2bc cos A. ‘The square of any side of a triangle equals: the sum of the squares of the other two sides, minus twice the product of those sides and the cosine of their included angle.’ The formula should be understood as Pythagoras’ theorem with the term −2bc cos A, which disappears when ∠A = 90◦ . 37 The cosine rule and Pythagoras’ theorem • When ∠A = 90◦ , then cos A = 0 and the cosine rule is Pythagoras’ theorem. • The last term is thus a correction to Pythagoras’ theorem when ∠A 90◦ . • When ∠A < 90◦ , then cos A is positive, so a2 < b2 + c2 . • When ∠A > 90◦ , then cos A is negative, so a2 > b2 + c2 . Proof of the cosine rule The proof is based on Pythagoras’ theorem, and again begins with the construction of an altitude. Given: Let ABC be any triangle. Again, there are three cases, according as to whether ∠A is acute, obtuse, or a right angle. B B B c a A x M a c h C b Case 1: ∠A is acute A b h C Case 2: ∠A = 90◦ c a M x A C b Case 3: ∠A is obtuse Aim: To prove that a2 = b2 + c2 − 2bc cos A. In case 2, cos A = 0, and this is just Pythagoras’ theorem. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 292 7I Chapter 7 Trigonometry Construction: In the remaining cases 1 and 3, construct the altitude from B, meeting AC, produced if necessary, at M. Let BM = h and AM = x. Proof Case 1 — Suppose that ∠A is acute. By Pythagoras’ theorem in BMC, a2 = h2 + (b − x)2 . By Pythagoras’ theorem in BMA, h2 = c2 − x2 , so a2 = c2 − x2 + (b − x)2 Case 3 — Suppose that ∠A is obtuse. By Pythagoras’ theorem in BMC, a2 = h2 + (b + x)2 . By Pythagoras’ theorem in BMA, h2 = c2 − x2 , so a2 = c2 − x2 + (b + x)2 = c2 − x2 + b2 − 2bx + x2 = b2 + c2 − 2bx. Using trigonometry in ABM, x = c cos A. So = c2 − x2 + b2 + 2bx + x2 (∗) = b2 + c2 + 2bx. Using trigonometry in ABM, x = c cos(180◦ − A) a2 = b2 + c2 − 2bc cos A. (∗) = −c cos A. So a = b2 + c2 − 2bc cos A. 2 Note: The identity cos(180◦ − A) = − cos A is the key step in Case 3 of the proof. The cosine rule appears in Euclid’s geometry book, but without any mention of the cosine ratio — the form given there is approximately the two statements in the proof marked with (∗). Using the cosine rule to find a side — the SAS situation For the cosine rule to be applied to find a side, the other two sides and their included angle must be known. This is the SAS congruence situation. 38 Using the cosine rule to find a side In the SAS congruence situation: (square of any side) = (sum of squares of other two sides) − (twice the product of those sides) × (cosine of their included angle). Example 23 Using the cosine rule to find a side Find x in the triangle drawn to the right. x 12 110º 30 Solution Using the cosine rule: x2 = 122 + 302 − 2 × 12 × 30 × cos 110◦ = 144 + 900 − 720 cos 110◦ = 1044 − 720 cos 110◦ , and because cos 110◦ = − cos 70◦ , (cosine is negative in the second quadrant) x2 = 1044 + 720 cos 70◦ (until this point, all calculations have been exact) Using the calculator to approximate x2 , and then to take the square root, x 35.92 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7I The cosine rule 293 Using the cosine rule to find an angle — the SSS situation To use the cosine rule to find an angle, all three sides need to be known, which is the SSS congruence situation. Finding the angle is done most straightforwardly by substituting into the usual form of the cosine rule: 39 Using the cosine rule to find an angle In the SSS congruence situation: • Substitute into the cosine rule and solve for cos θ. There is an alternative approach. Solving the cosine rule for cos A gives a formula for cos A. Some readers may prefer to remember and apply this second form of the cosine rule — but the triangle may then need to be relabelled. 40 The cosine rule with cos A as subject In any triangle ABC, cos A = b2 + c2 − a2 . 2bc Notice that cos θ is positive when θ is acute, and is negative when θ is obtuse. Hence there is only ever one solution for the unknown angle, unlike the situation for the sine rule, where there are often two possible angles. Example 24 Using the cosine rule to find an angle Find θ in the triangle drawn to the right. 6 3 θ 4 Solution Substituting into the cosine rule: 6 = 3 + 4 − 2 × 3 × 4 × cos θ 2 2 2 24 cos θ = −11 11 cos θ = − 24 θ 117◦ 17 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 OR Using the boxed formula: 32 + 42 − 62 cos θ = 2×3×4 −11 = 24 θ 117◦ 17 . © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 294 7I Chapter 7 Trigonometry Exercise 7I 1 FOUNDATION Find x in each triangle, correct to one decimal place. a b c d x 10 115º 8 e 2 f Find θ in each triangle, correct to the nearest degree. a b c d e f CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7I The cosine rule 3 Using the fact that cos 60◦ = 12 and cos 120◦ = − 12 , find x as a surd in each triangle. a b 3 x 60º 4 295 x 4 1 a If cos A = 14 , find the exact value of a. C a 120º b If cos C = 23 , find the exact value of c. A c B 2 5 B 2 3 6 A C DEVELOPMENT 5 a Find the smallest angle of the triangle, correct to the nearest minute. 5 6 b Find the largest angle of the c Find the value of cos θ. triangle, correct to the nearest minute. 8 10 14 7 6 In the diagram to the right, ship A is 120 nautical miles from a lighthouse L on a bearing of 072◦ T, while ship B is 180 nautical miles from L on a bearing of 136◦ T. Calculate the distance between the two ships, correct to the nearest nautical mile. 72º 120 nm A L 136º 180 nm B 7 A golfer at G wishes to hit a shot between two trees P and Q, as shown in the diagram opposite. The trees are 31 metres apart, and the golfer is 74 metres from P and 88 metres from Q. Find the angle within which the golfer must play the shot, correct to the nearest degree. P 31 m Q 88 m 74 m G 8 The sides of a triangle are in the ratio 5 : 16 : 19. Find the smallest and largest angles of the triangle, correct to the nearest minute where necessary. 9 In ABC, a = 31 units, b = 24 units and cos C = 59 62 . a Show that c = 11 units. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b Show that A = 120◦ . © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 296 7I Chapter 7 Trigonometry 10 √ In PQR, p = 5 3 cm, q = 11 cm and R = 150◦ . b Find cos P. a Find r. 11 A ship sails 50 km from port A to port B on a bearing of 63◦ , then sails 130 km from port B to port C on a bearing of 296◦ . C a Show that ∠ABC = 53◦ . 130 km b Find, to the nearest km, the distance of port A from port C. B 63R c Use the cosine rule to find ∠ACB, and hence find the bearing of port A from port C, correct to the nearest degree. 296R 50 km A ENRICHMENT 12 In a parallelogram ABCD, ∠ADC = 60◦ , AB = 9 cm and AD = 3 cm. The point P lies on DC such that DP = 3 cm. ADP is equilateral, and hence find AP. b Use the cosine rule in BCP to find √ BP. 7 . c Let ∠APB = x. Show that cos x = − 14 a Explain why 13 9 cm A B 3 cm x 60º D 3 cm P C Use the cosine rule to find the two possible values of x in the diagram to the right. 7 x 60º 14 The diagram shows ABC in which ∠A = 30◦ , AB = 6 cm and BC = 4 cm. Let AC = x cm. √ a Use the cosine rule to show that x2 − 6 3 x + 20 = 0. √ √ b Use the quadratic formula to show that AC has length 3 3 + 7 cm or √ √ 3 3 − 7 cm. c Copy the diagram and indicate on it (approximately) the other possible position of the point C. A 8 6 cm 30º x cm B 4 cm C 15 The sides of a triangle are n2 + n + 1, 2n + 1 and n2 − 1, where n > 1. Find the largest angle of the triangle. 16 Two identical rods BA and CA are hinged at A. When BC = 8 cm, ∠BAC = 30◦ and when BC = 4 cm, ∠BAC = α. Show that √ 6+ 3 . cos α = 8 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7J Problems involving general triangles 297 7J Problems involving general triangles Learning intentions • Work with the relationship between congruence tests. A triangle has three lengths and three angles, and most triangle problems involve using three of these six measurements to calculate some of the others. The key to deciding which formula to use is to see which congruence situation applies. Trigonometry and the congruence tests There are four standard congruence tests — RHS, SSS, SAS and AAS. These tests can also be seen as theorems about constructing triangles from given data. Thus if three measurements of a triangle are known, including at least one length, then apart from the ambiguous SSA situation, any two triangles with these three measurements are congruent, and the other sides and angles can be calculated. 41 The standard congruence tests, and the sine, cosine, and area formulae Use simple trigonometry and Pythagoras’ theorem. SSS: Use the cosine rule to find any angle. SAS: Use the cosine rule to find the third side, then the sine rule will find any angle. Use the area formula to find the area. AAS: Use the sine rule to find each of the other two sides. [SSA]: [Not a congruence test — this is the ambiguous situation] When using the sine rule or the area formula to find an unknown angle, sin θ will be unique, but there may be two possible solutions for θ. RHS: In the ambiguous SSA situation, it is also possible to use the cosine rule to find the third side. See the last two questions in the previous exercise. Problems requiring two steps Various situations with non-right-angled triangles require two steps for their solution. For example, two steps are needed to find the other two angles in an SAS situation, or to find the area in an AAS, SSA, or SSS situations. Example 25 Solving a two-step problem A boat sails 6 km due north from the harbour H to A, and a second boat sails 10 km from H to B on a bearing of 120◦ T. a What is the distance AB? b What is the bearing of B from A, correct to the nearest minute? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 298 7J Chapter 7 Trigonometry Solution a This is an SAS situation, so we use the cosine rule to find AB: A AB2 = 62 + 102 − 2 × 6 × 10 × cos 120◦ 6 km = 36 + 100 − 120 × (− 12 ) 120º = 196 H AB = 14 km. 10 km B b Because AB is now known, this is an SSS situation, so we use the cosine rule in reverse to find ∠A: 102 = 142 + 62 − 2 × 14 × 6 × cos A 12 × 14 × cos A = 196 + 36 − 100 132 cos A = 12 × 14 11 = 14 A 38◦ 13 , and the bearing of B from A is about 141◦ 47 T. Exercise 7J 1 FOUNDATION Use right-angled triangle trigonometry, the sine rule or the cosine rule to find x in each triangle, correct to one decimal place. a b x 23 43º 57º c 52º 13.5 x d x 11 62º 13 e f x 7.4 9.2 98º CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7J Problems involving general triangles 2 3 Use right-angled triangle trigonometry, the sine rule or the cosine rule to find θ in each triangle, correct to the nearest degree. a b c d e f [This question is designed to show that the sine and cosine rules work in right-angled triangles, but are not the most efficient methods.] In each part find the pronumeral (correct to the nearest cm or to the nearest degree), using either the sine rule or the cosine rule. Then check your answer using right-angled triangle trigonometry. a b c 73º x d x 27 cm 4 35º 40 cm In PQR, ∠Q = 53◦ , ∠R = 55◦ and QR = 40 metres. The point T lies on QR such that PT ⊥ QR. 40 sin 55◦ a Use the sine rule in PQR to show that PQ = . sin 72◦ b Use PQT to find PT , correct to the nearest metre. P 53º Q 5 299 55º T 40 m In ABC, ∠B = 90◦ and ∠A = 31◦ . The point P lies on AB such that AP = 20 cm and ∠CPB = 68◦ . R C a Explain why ∠ACP = 37◦ . 20 sin 31◦ . sin 37◦ c Hence find PB, correct to the nearest centimetre. b Use the sine rule to show that PC = 31º A 68º P B 20 cm CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 300 7J Chapter 7 Trigonometry 6 In the diagram to the right, AB = 6.7 cm, AD = 8.3 cm and DC = 9.2 cm. Also, ∠A = 101◦ and ∠C = 73◦ . a Use the cosine rule to find the diagonal BD, correct to the nearest millimetre. A 6.7 cm B 101º 8.3 cm b Hence use the sine rule to find ∠CBD, correct to the nearest degree. 73º C 9.2 cm D 7 In ABC, AB = 4 cm, BC = 7 cm and CA = 5 cm. A a Use the cosine rule to find ∠ABC, correct to the nearest minute. b Hence calculate the area of 4 cm B 8 In a triangle XYZ, ∠Y = 72◦ and ∠Y XZ = 66◦ . XP ⊥ YZ and XP = 25 cm. C 7 cm X 66º PXY to show that XY 26.3 cm. b Hence use the sine rule in XYZ to find YZ, correct to the nearest centimetre. c Check your answer to part b by using the tangent ratio in triangles PXY and PXZ to find PY and PZ. a Use the sine ratio in 9 5 cm ABC, correct to the nearest square centimetre. 72º 25 cm Y P Z A triangle has sides 13 cm, 14 cm and 15 cm. Use the cosine rule to find one of its angles, and hence show that its area is 84 cm2 . DEVELOPMENT 10 A ship sails 53 nautical miles from P to Q on a bearing of 026◦ T. It then sails 78 nautical miles due east from Q to R. Q 26º 53 a Explain why ∠PQR = 116◦ . 78 R 10 m B b How far apart are P and R, correct to the nearest nautical mile? P 11 A golfer at G, 60 metres from the hole H, played a shot that landed at B, 10 metres from the hole. The direction of the shot was 7◦ away from the direct line between G and H. H a Find, correct to the nearest minute, the two possible sizes of ∠GBH. b Hence find the two possible distances the ball has travelled. (Answer in metres 60 m B correct to one decimal place.) 7º G 12 Two towers AB and PQ stand on level ground. The angles of elevation of the top of the taller tower from the top and bottom of the shorter tower are 5◦ and 20◦ respectively. The height of the taller tower is 70 metres. a Explain why ∠APJ = 15◦ . 5º A J ◦ BP sin 15 . sin 95◦ 70 c Show that BP = . sin 20◦ d Hence find the height of the shorter tower, correct to the nearest metre. b Show that AB = P CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 K 70 m 20º B Q © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 7J Problems involving general triangles 13 14 The diagram shows three straight roads, AB, BC and CA, where AB = 8.3 km, AC = 15.2 km, and the roads AB and AC intersect at 57◦ . Two cars, P1 and P2 , leave A at the same instant. Car P1 travels along AB and then BC at 80 km/h while P2 travels along AC at 50 km/h. Which car reaches C first, and by how many minutes? (Answer correct to one decimal place.) B C 8.3 km In the diagram to the right, ABCD is a trapezium in which AB DC. The diagonals AC and BD meet at P. Also, AB = AD = 4 cm, DC = 7 cm and ∠ADC = 62◦ . 4 cm a Find ∠ACD, correct to the nearest minute. D 301 57º 15.2 km A A 4 cm B P 62º b Explain why ∠PDC = 12 ∠ADC. C 7 cm c Hence find, to the nearest minute, the acute angle between the diagonals of the trapezium. 15 A bridge spans a river, and the two identical sections of the bridge, each of length x metres, can be raised to allow tall boats to pass. When the two sections are fully raised, they are each inclined at 50◦ to the horizontal, and there is an 18-metre gap between them, as shown in the diagram. Calculate the width of the river in metres, correct to one decimal place. 18 m x 50º x 50º river ENRICHMENT 16 Let ABC be a triangle and let D be the midpoint of AC. Let BD = m and ∠ADB = θ. A a Simplify cos(180◦ − θ). c 4m + b − 4c . Write down a similar expression for b Show that cos θ = 4mb ◦ cos(180 − θ) in terms of a, b and m. c Hence show that a2 + c2 = 2m2 + 12 b2 . 2 2 m b T D 2 17 In ABC, a cos A = b cos B. Use the cosine rule to prove that the triangle is either isosceles or right-angled. 18 [Heron’s formula for the area of a triangle in terms of the side lengths] B a C a By repeated application of factoring by the difference of squares, prove the identity (2ab)2 − (a2 + b2 − c2 )2 = (a + b + c)(a + b − c)(a − b + c)(−a + b + c). b Let ABC be any triangle, and let s = 12 (a + b + c) be the semiperimeter. Prove that (a + b + c)(a + b − c)(a − b + c)(−a + b + c) = 16s(s − a)(s − b)(s − c). c Write down the formula for cos C in terms of the sides a, b and c, then use parts a and b and the √ 2 s(s − a)(s − b)(s − c) . Pythagorean identities to prove that sin C = ab √ d Hence show that the area A of the triangle is A = s(s − a)(s − b)(s − c) . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 302 Chapter 7 Trigonometry Chapter 7 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 7 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Review Chapter Review Exercise 1 2 Find, correct to four decimal places: a cos 73◦ b tan 42◦ c sin 38◦ 24 d cos 7◦ 56 Find the acute angle θ, correct to the nearest minute, given that: a sin θ = 0.3 b tan θ = 2.36 cos θ = 14 d tan θ = 1 13 c 3 Find, correct to two decimal places, the side marked x in each triangle below. a b 7.2 x 16 x 36º 42º c 4 d Find, correct to the nearest minute, the angle θ in each triangle below. a b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c d © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 7 review 5 Use the special triangles to find the exact values of: b sin 45◦ c cos 30◦ d cot 45◦ e sec 60◦ f cosec 60◦ A vertical pole stands on level ground. From a point on the ground 8 metres from its base, the angle of elevation of the top of the pole is 38◦ . Find the height of the pole, correct to the nearest centimetre. 7 At what angle, correct to the nearest degree, is a 6-metre ladder inclined to the ground if its foot is 2.5 metres out from the wall? 8 A motorist drove 70 km from town A to town B on a bearing of 056◦ T, and then drove 90 km from town B to town C on a bearing of 146◦ T. B 146º a Explain why ∠ABC = 90◦ . c Find ∠BAC, and hence find the bearing of town C from town A, correct to the 56º A 70 km km 90 b How far apart are the towns A and C, correct to the nearest kilometre? nearest degree. 11 12 13 b y = cos x c y = tan x Write each trigonometric ratio as the ratio of its related acute angle, using the correct sign. a cos 125◦ b sin 312◦ c tan 244◦ d sin 173◦ Find the exact value of: a tan 240◦ b sin 315◦ c cos 330◦ d tan 150◦ Use the graphs of the trigonometric functions to find these values, if they exist. a sin 180◦ b cos 180◦ c tan 90◦ d sin 270◦ Use Pythagoras’ theorem to find whichever of x, y or r is unknown. Then write down the values of sin θ, cos θ and tan θ. a 14 C Sketch each graph for 0◦ ≤ x ≤ 360◦ . a y = sin x 10 Review a tan 60◦ 6 9 303 b a If tan α = 12 5 and α is acute, find the values of sin α and cos α. √ b If sin β = 2 7 6 and β is acute, find the values of cos β and tan β. 9 c If tan α = − 40 and 270◦ < α < 360◦ , find the values of sin α and cos α. √ d If sin β = 2 7 6 and 90◦ < β < 180◦ , find the values of cos β and tan β. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 304 Chapter 7 Trigonometry Review 15 16 17 Simplify: 1 a cos θ d 1 − sin2 θ c Prove the following trigonometric identities. a cos θ sec θ = 1 b tan θ cosec θ = sec θ d 2 cos2 θ − 1 = 1 − 2 sin2 θ e 4 sec2 θ − 3 = 1 + 4 tan2 θ cot θ = cosec θ cos θ f cos θ + tan θ sin θ = sec θ c Solve each trigonometric equation for 0◦ ≤ x ≤ 360◦ . a cos x = 12 b sin x = 1 c tan x = −1 d cos x = 0 3 tan x = 1 √ h 2 cos x + 3 = 0 k cos(x − 75◦ ) = 12 f tan x = 0 e √ 2 sin x + 1 = 0 j cos 2x = 12 g 18 sin θ cos θ f cosec2 θ − 1 1 cot θ e sec2 θ − tan2 θ b √ i cos2 x = 12 √ l sin x = − 3 cos x Use the sine rule or the cosine rule in each triangle to find x, correct to one decimal place. a b x 47º 8 7 63º 75º c 9 d 112º 15 9.3 x 18º30¢ 40º x 19 x 17.2 Calculate the area of each triangle, correct to the nearest cm2 . a b 8 cm 42º 20 10 cm Use the sine rule or the cosine rule in each triangle to find θ, correct to the nearest minute. a b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c d © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 7 review A triangle has sides 7 cm, 8 cm and 10 cm. Use the cosine rule to find one of its angles, and hence find the area of the triangle, correct to the nearest cm2 . 22 a Find the side a in 23 A helicopter H is hovering above a straight, horizontal road AB of length 600 metres. The angles of elevation of H from A and B are 7◦ and 13◦ respectively. The point C lies on the road directly below H. 600 sin 7◦ a Use the sine rule to show that HB = . sin 160◦ b Hence find the height CH of the helicopter above the road, correct to the nearest metre. 24 25 Review 21 ABC, where ∠C = 60◦ , b = 24 cm and the area is 30 cm2 . b Find the size of ∠B in ABC, where a = 9 cm, c = 8 cm and the area is 18 cm2 . H A 7º 13º C 600 m 305 B T A man is sitting in a boat at P, where the angle of elevation of the top T of a vertical cliff BT is 15◦ . He then rows 50 metres directly towards the cliff to Q, where the angle of elevation of T is 25◦ . 50 sin 15◦ a Show that T Q = . sin 10◦ b Hence find the height h of the cliff, correct to the nearest tenth of a metre. h 15º 25º P 50 m Q A ship sailed 140 nautical miles from port P to port Q on a bearing of 050◦ T. It then sailed 260 nautical miles from port Q to port R on a bearing of 130◦ T. Q 130º a Explain why ∠PQR = 100◦ . 50º 0 14 P b Find the distance between ports R and P, correct to the nearest nautical mile. c Find the bearing of port R from port P, correct to the nearest degree. B 26 0 R 26 In each diagram, find CD correct to the nearest centimetre: a D C 122º 35º 40º A 8m B 27 b B C 6m c 37º 6m 7m 85º A D A 4m 65º P 6m 7m B D C Two towers AB and PQ stand on level ground. Tower AB is 12 metres taller than tower PQ. From A, the angles of depression of P and Q are 28◦ and 64◦ respectively. AKP to show that KP = BQ = 12 tan 62◦ . b Use ABQ to show that AB = 12 tan 62◦ tan 64◦ . c Hence find the height of the shorter tower, correct to the nearest metre. d Solve the problem again by finding AP using AKP and then using the sine rule in APQ. A 12 m K 64º 28º P a Use CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 B Q © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 306 Chapter 7 Trigonometry Review 28 Two trees T 1 and T 2 on one bank of a river are 86 metres apart. A sign S on the opposite bank is between the trees and the angles S T 1 T 2 and S T 2 T 1 are 53◦ 30 and 60◦ 45 respectively. S a Find S T 1 in exact form. T1 b Hence find the width of the river, correct to the nearest metre. 29 In the diagram to the right, ∠ACD = α, ∠BCD = β, AB = h and BD = x. h cos α . a Show that BC = sin(α + β) h sin β cos α . b Hence show that x = sin(α + β) T2 A D E C D h x B 30 In the diagram to the right, ABC is isosceles with BA = BC = x units and ∠BAC = 36◦ . The point P is on AC such that PA = BA. Let BP = y units. y . a Use the cosine rule in ABP to show that cos 72◦ = 2x b Show that BCP is isosceles. c Hence show that cos 36◦ cos 72◦ = 14 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 B x C y P x 36° A © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8 Lines in the coordinate plane Chapter introduction The earlier chapters have used graphs to turn functions into geometric objects within the coordinate plane, thus allowing them to be visualised and studied by geometric as well as algebraic methods. In the coordinate plane: Points are represented by pairs of numbers. Lines are represented by linear equations. Circles, parabolas and other curves are represented by non-linear equations. Points, lines and line segments are the main concern of this chapter. Its purpose is to prepare for Chapter 10, where the new topic of calculus will be introduced in the coordinate plane, heavily based on geometric ideas of tangents and areas. This material may all have been covered in earlier years. Readers should do as much or as little of it as they need in order to master the skills. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 308 8A Chapter 8 Lines in the coordinate plane 8A Lengths and midpoints of line segments Learning intentions • Define a line segment, and recognise the unique line containing it. • Prove and use the distance formula for a line segment. • Prove and use the midpoint formula for a line segment. • Review the definitions of and tests for special quadrilaterals. Many of the familiar objects of geometry — triangles, squares, parallelograms, hexagons — are made up of line segments, and we need to know how to deal with line segments in the coordinate plane. Line segments A line segment is completely determined by specifying its two distinct endpoints. 1 Line segment We know that two distinct points P and Q determine a line . • The line segment PQ is the part of the line contained between these two distinct points, including the two endpoints P and Q. • The line is the only line containing the line segment PQ. The word ‘interval’ used to be used for both intervals and line segments, but the two closely related ideas now have separate words, which is more satisfactory. The symbol PQ now has three distinct meanings. It can mean; the line PQ, or the line segment PQ, or the distance PQ. In practice, this rarely causes problems, but if it does, add the preceding noun. There are simple formulae for the length and the midpoint of a line segment. The distance formula The formula for the length of a line segment PQ in the coordinate plane is just Pythagoras’ theorem in different notation. Let P(x1 , y1 ) and Q(x2 , y2 ) be two points in the plane. Construct the right-angled triangle PQA, where A(x2 , y1 ) lies level with P and vertically above or below Q. Then PA = |x2 − x1 | and QA = |y2 − y1 |, so by Pythagoras’ theorem in PQA, Q(x2,y2) y2 y1 P(x1,y1) x1 PQ2 = (x2 − x1 )2 + (y2 − y1 )2 . 2 y A x2 x Distance formula Let P(x1 , y1 ) and Q(x2 , y2 ) be two points in the plane. Then PQ2 = (x2 − x1 )2 + (y2 − y1 )2 . First find the square PQ2 of the distance, then take the square root. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8A Lengths and midpoints of line segments Example 1 309 Using the distance formula Find the lengths of the sides AB and AC of the triangle with vertices A(1, −2), B(−4, 2), and C(5, −7), and hence say why ABC is isosceles. Solution First, AB2 = (x2 − x1 )2 + (y2 − y1 )2 = − 4 − 1 2 + 2 − (−2) 2 Secondly, AC 2 = (x2 − x1 )2 + (y2 − y1 )2 = 5 − 1 2 + − 7 − (−2) 2 = (−5)2 + 42 so = 42 + (−5)2 = 41, √ AB = 41 . so = 41, √ AC = 41 . Because the two sides AB and AC are equal, the triangle is isosceles. The midpoint formula The midpoint of a line segment is found by taking the averages of the coordinates of the two points. Congruence is the basis of the proof below. Let P(x1 , y1 ) and Q(x2 , y2 ) be two points in the plane, and let M(x, y) be the midpoint of PQ. Construct S (x, y1 ) and T (x2 , y), as shown. (AAS). Then PMS ≡ MQT Hence PS = MT (matching sides of congruent triangles). y Q(x2,y2) M(x,y) P(x1,y1) T S x − x1 = x2 − x 2x = x1 + x2 x1 + x2 , x= 2 x1 x x2 x which is the average of x1 and x2 . The calculation of the y-coordinate of M is similar. 3 Midpoint formula Let M(x, y) be the midpoint of the line segment joining P(x1 , y1 ) and Q(x2 , y2 ). y1 + y2 x1 + x 2 and y = (take the averages of the coordinates) Then x = 2 2 Example 2 Using the midpoint formula and the distance formula The line segment joining A(3, −1) and B(−7, 5) is a diameter of a circle. Find: a the centre M CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b the radius c the equation. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 310 8A Chapter 8 Lines in the coordinate plane Solution a The centre of the circle is the midpoint M(x, y) of the line segment AB. x1 + x2 2 3−7 = 2 = −2 Using the midpoint formula, x = and y1 + y2 2 −1 + 5 = 2 =2 y= so the centre is M(−2, 2). b Using the distance formula, AM 2 = (x2 − x1 )2 + (y2 − y1 )2 = − 2 − 3 2 + 2 − (−1) 2 = 34, √ AM = 34 . √ Hence the circle has radius 34 . c The equation of the circle is therefore (x + 2)2 + (y − 2)2 = 34. Testing for special quadrilaterals Special quadrilaterals are not the study of this chapter. But as soon as there are line segments, triangles and quadrilaterals appear, and the methods of this chapter illustrate many of the results obtained geometrically in previous years. Some of the questions in this chapter ask for a proof that a quadrilateral is of a particular type. The most obvious way is to test the definition itself. 4 Definitions of the special quadrilaterals • A trapezium is a quadrilateral with one pair of opposite sides parallel. • A parallelogram is a quadrilateral with both pairs of opposite sides parallel. • A rhombus is a parallelogram with a pair of adjacent sides equal. • A rectangle is a parallelogram with one angle a right angle. • A square is both a rectangle and a rhombus. There are, however, several further standard tests that the exercises assume, and that were developed in earlier years. (Tests involving angles are omitted here, being irrelevant in the chapter.) 5 Further standard tests for special quadrilaterals A quadrilateral is a parallelogram: • if the opposite sides are equal, or • if one pair of opposite sides are equal and parallel, or • if the diagonals bisect each other. A quadrilateral is a rhombus: • if all sides are equal, or • if the diagonals bisect each other at right angles. A quadrilateral is a rectangle: • if the diagonals are equal and bisect each other. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8A Lengths and midpoints of line segments 311 These definitions and tests are important ideas underlying future Extension 1 and 2 work on vectors, and are also needed for complex numbers in Extension 2. Exercise 8A FOUNDATION Note: Diagrams should be drawn wherever possible. 1 Find the midpoint of each line segment AB. Use x = a A(3, 5) and B(1, 9) c A(−4, 7) and B(8, −11) e A(0, −8) and B(−11, −12) 2 3 y1 + y2 x1 + x2 and y = . 2 2 b A(4, 8) and B(6, 4) d A(−3, 6) and B(3, 1) f A(4, −7) and B(4, 7) Find the length of each line segment. Use AB2 = (x2 − x1 )2 + (y2 − y1 )2 , then find AB. a A(1, 4), B(5, 1) b A(−2, 7), B(3, −5) c A(−5, −2), B(3, 4) d A(3, 6), B(5, 4) e A(−4, −1), B(4, 3) f A(5, −12), B(0, 0) a Find the midpoint M of the line segment joining P(−2, 1) and Q(4, 9). b Find the lengths PM and MQ, and verify that PM = MQ. 4 a Find the length of each side of the triangle formed by P(0, 3), Q(1, 7) and R(5, 8). b Hence show that 5 PQR is isosceles. The vertices of ABC are given by A(0, 0), B(9, 12) and C(25, 0). a Find the lengths of the three sides AB, BC, AC. b By checking that AB2 + BC 2 = AC 2 , show that 6 ABC is right-angled. ABC, where A = (0, 5), B = (3, −2) and C = (−3, 4). b Find the midpoint of each side of this triangle ABC. a Find the length of each side of DEVELOPMENT 7 a A circle with centre O(0, 0) passes through A(5, 12). What is its radius? b A circle with centre B(4, 5) passes through the origin. What is its radius? c Find the centre of the circle with diameter CD, where C = (2, 1) and D = (8, −7). d Show that E(−12, −5) lies on the circle with centre the origin and radius 13. 8 In previous studies of special quadrilaterals, you should have seen the result that A quadrilateral whose diagonals bisect is a parallelogram, that is, a quadrilateral with opposite sides parallel. a Find the midpoint of the line segment joining A(4, 9) and C(−2, 3). b Find the midpoint of the line segment joining B(0, 4) and D(2, 8). c Explain why the quadrilateral ABCD is a parallelogram. 9 A quadrilateral with four sides of equal lengths is a rhombus. The points A(3, 1), B(10, 2), C(5, 7) and D(−2, 6) are the vertices of a quadrilateral. a Find the lengths of all four sides. b What can you conclude about ABCD? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 312 8A Chapter 8 Lines in the coordinate plane 10 a Find the side lengths of the triangle with vertices X(0, −4), Y(4, 2) and Z(−2, 6). b Show that XYZ is a right-angled isosceles triangle by showing that its side lengths satisfy Pythagoras’ theorem. c Hence find the area of XYZ. 11 √ √ a Find the distance of each point A(1, 4), B(2, 13 ), C(3, 2 2 ) and D(4, 1) from the origin O. Hence explain why the four points lie on a circle with its centre at the origin. b What are the radius, diameter, circumference and area of this circle? 12 The circle with centre (h, k) and radius r has equation (x − h)2 + (y − k)2 = r2 . By identifying the centre and radius, find the equations of: a the circle with centre (5, −2) and passing through (−1, 1), b the circle with K(5, 7) and L(−9, −3) as endpoints of a diameter. 13 The point M(3, 7) is the midpoint of the line segment joining A(1, 12) and B(x2 , y2 ). Find the coordinates x2 and y2 of B by substituting into the formulae y1 + y2 x1 + x2 and y= . x= 2 2 14 Solve each problem using the same methods as in the previous question. a If A(−1, 2) is the midpoint of S (x, y) and T (3, 6), find the coordinates of S . b The midpoint of the line segment PQ is M(2, −7). Find the coordinates of P if: ii Q = (5, 3) i Q = (0, 0) iii Q = (−3, −7) c Find B, if AB is a diameter of a circle with centre Q(4, 5), and A = (8, 3). d Given that P(4, 7) is one vertex of the square PQRS , and that the centre of the square is M(8, −1), find the coordinates of the opposite vertex R. 15 Each set of three points given below forms a triangle of one of these types: A isosceles, B equilateral, C right-angled, D none of these. Find the side lengths of each triangle below and hence determine its type. √ a A(−1, 0), B(1, 0), C(0, 3) b P(−1, 1), Q(0, −1), R(3, 3) c D(1, 1), E(2, −2), F(−3, 0) d X(−3, −1), Y(0, 0), Z(−2, 2) 16 a Given the point A(7, 8), find the coordinates of three points P with integer coordinates such that √ AP = 5. √ b Given that the distance from U(3, 7) to V(1, y) is 13, find the two possible values of y. √ c Find a, if the distance from A(a, 0) to B(1, 4) is 18 units. ENRICHMENT 17 a Given that C(x, y) is equidistant from each of the points P(1, 5), Q(−5, −3) and R(2, −2), use the distance formula to form two equations in x and y and solve them simultaneously to find the coordinates of C. b Find the coordinates of the point M(x, y) that is equidistant from each of the points P(4, 3) and Q(3, 2), and is also equidistant from R(6, 1) and S (4, 0). 18 Suppose that A, B and P are the points (0, 0), (3a, 0) and (x, y) respectively. Use the distance formula to form an equation in x and y for the point P, and describe the curve so found if: a PA = PB, CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b PA = 2PB. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8B Gradients of line segments and lines 313 8B Gradients of line segments and lines Learning intentions • Develop and apply the theory of the gradient of line segments and lines. • Develop and apply the conditions for lines to be parallel or perpendicular. • Define and apply the angle of inclination of a line. Gradient is the key idea that will be used in the next section to bring lines and their equations into the coordinate plane. The gradient of a line segment Let P(x1 , y1 ) and Q(x2 , y2 ) be two points in the plane. The gradient of the line segment PQ is a measure of its steepness, as someone walks along it from P to Q. The rise is the vertical difference y2 − y1 , and the run is the horizontal difference x2 − x1 . The gradient of the line segment PQ is defined to be the ratio of the rise and the run: rise gradient of PQ = run y2 − y1 . = x2 − x 1 6 y Q(x2,y2) y2 rise y1 P(x1,y1) x1 run x2 x The gradient of a line segment Let P(x1 , y1 ) and Q(x2 , y2 ) be any two points. Then rise y2 − y1 gradient of PQ = , that is, gradient of PQ = . run x2 − x 1 • Horizontal line segments have gradient zero, because the rise is always zero. • Vertical line segments don’t have a gradient — the run is always zero, so the fraction is undefined. Positive and negative gradients If the rise and the run have the same sign, then the gradient is positive, as in the first diagram below. In this case the line segment slopes upwards as we move from left to right. If the rise and run have opposite signs, then the gradient is negative, as in the second diagram. The line segment slopes downwards as we move from left to right. y y Q P P Q x x If the points P and Q are interchanged, in either diagram, then the rise and the run both change sign, but the gradient remains the same. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 314 8B Chapter 8 Lines in the coordinate plane Example 3 Finding gradients Find the gradients of the sides of XYZ, where X = (2, 5), Y = (5, −2) and Z = (−3, 4). Solution y2 − y1 x2 − x1 −2 − 5 = 5−2 −7 = 3 7 =− 3 gradient of XY = y 2 − y1 x2 − x 1 4 − (−2) = −3 − 5 4+2 = −8 3 =− 4 gradient of YZ = y2 − y1 x2 − x 1 5−4 = 2 − (−3) 1 = 2+3 1 = 5 gradient of ZX = The gradient of a line The gradient of a line is defined to be the gradient of any line segment within the line. This definition makes sense because any two line segments on the same line always have the same gradient. y B Q To prove this, suppose that PQ and AB are two line segments on the same line . Construct right triangles PQR and ABC underneath the line segments, with sides parallel to the axes. Because these two triangles are similar (by the AA similarity test), the ratios of their heights and bases are the same, which means that the two line segments AB and PQ have the same gradient. 7 P A C R x The Gradient of a Line • The gradient of a line is found by taking any two distinct points P and Q on the line and finding the gradient of the line segment PQ. • It doesn’t matter which two points on the line are taken, because the ratio of rise over run will be same, as seen using similar triangles. A condition for two lines to be parallel Any two vertical lines are parallel. Otherwise, lines are parallel when their gradients are equal. 8 Parallel lines Two lines are parallel if and only if: • they have the same gradient OR they are both vertical. y The phrase ‘if and only if’ means that Q • if the condition holds, then the lines are parallel, and l1 B l2 • if the lines are parallel, then the condition holds, and the two statements are called converses of each other. ‘If and only if’ is a very convenient abbreviation, but its logic needs attention. P CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 R A C x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8B Gradients of line segments and lines 315 Proof (assuming neither is vertical) The two lines 1 and 2 meet the x-axis at P and A. Construct PR = AC, then construct QR and BC perpendicular to the x-axis. A If the lines 1 and 2 are parallel, then Hence ∠QPR = ∠BAC (corresponding angles on parallel lines). PQR ≡ (AAS) ABC QR = BC (matching sides of congruent triangles) QR BC = (because PR = AC). PR AC gradient 1 = gradient 2 . B Conversely, if the two gradients are equal, then QR BC = PR AC QR = BC (because PR = AC). so PQR ≡ Hence 1 2 so Example 4 ABC (SAS) (corresponding angles are equal). Testing whether lines are parallel Given the four points A(3, 6), B(7, −2), C(4, −5) and D(−1, 5), show that the quadrilateral ABCD is a trapezium with AB CD. Solution 5 − (−5) −1 − 4 10 = −5 = −2. −2 − 6 7−3 −8 = 4 = −2, gradient of CD = gradient of AB = Hence AB CD because their gradients are equal, so ABCD is a trapezium. Testing for collinear points Three points are called collinear if they all lie on one line. 9 Testing for collinear points • To test whether three points A, B and C are collinear, test whether AB and BC are parallel. (Are they both vertical, or do they have the same gradient?) • If AB and BC are parallel, then the three points are collinear, because then AB and BC are parallel lines passing through a common point B. Example 5 Testing whether three points are collinear Test whether the three points A(−2, 5), B(1, 3) and C(7, −1) are collinear. Solution −1 − 3 3−5 gradient of BC = 7−1 1+2 2 2 =− . =− , 3 3 Because the gradients are equal, the points A, B and C are collinear. gradient of AB = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 316 8B Chapter 8 Lines in the coordinate plane Gradient and the angle of inclination The angle of inclination of a line is the angle between the upward direction of the line and the positive direction of the x-axis. y y P P O D D M x M x O The two diagrams above show that lines with positive gradients have acute angles of inclination, and lines with negative gradients have obtuse angles of inclination. They also illustrate the trigonometric relationship between the gradient and the angle of inclination α. 10 Angle of inclination Suppose that a line has angle of inclination α. Then: • gradient of line = tan α OR the line is vertical and α = 90◦ . Proof A When α is acute, as in the first diagram, then the rise MP and the run OM are the opposite and adjacent sides of the triangle POM, so MP tan α = = gradient of OP. OM B When α is obtuse, as in the second diagram, then ∠POM = 180◦ − α, so MP tan α = − tan ∠POM = − = gradient of OP. OM Example 6 Finding angles of inclination a Given the points A(−3, 5), B(−6, 0) and O(0, 0), find the angles of y A inclination of the line segments AB and AO. b What sort of triangle is ABO? B −6 −3 5 O x Solution 5 0−5 = , −6 + 3 3 5 and using a calculator to solve tan α = , 3 angle of inclination of AB 59◦ . 0−5 5 Secondly, gradient of AO = =− , 0+3 3 5 and using a calculator to solve tan α = − , 3 angle of inclination of AO 121◦ . a First, gradient of AB = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8B Gradients of line segments and lines 317 ∠AOB = 59◦ (straight angle). Thus the base angles of AOB are equal, so the triangle is isosceles. b Hence A condition for lines to be perpendicular A vertical and a horizontal line are perpendicular. Otherwise, the condition is that the product of their gradients is −1: 11 Perpendicular lines Two lines are perpendicular if and only if: • m1 m2 = −1, where m1 and m2 are the gradients of the two lines. OR • one is vertical and the other is horizontal. 1 . The condition m1 m2 = −1 can also be written as m2 = − m1 Proof (assuming that neither is vertical) Shift each line sideways, without rotating it, so that the lines intersect at the origin. We can assume that one line has a positive gradient and the other a negative gradient, otherwise one of the angles between them would be acute. So let 1 be a line with positive gradient through the origin, and let 2 be a line with negative gradient through the origin. Construct the two triangles POQ and AOB as shown in the diagram, with the run OQ of OB QP QP 1 equal to the rise OB of 2 . Then m1 × m2 = × − =− , OQ AB AB because OQ = OB. A If the lines are perpendicular, then ∠AOB = ∠POQ. Hence AOB ≡ so QP = AB so m1 × m2 = −1. POQ y l1 P l2 B A O Q x (adjacent angles at O) (AAS) (matching sides of congruent triangles) B Conversely, if m1 × m2 = −1, then QP = AB. Hence AOB ≡ POQ (SAS) so ∠AOB = ∠POQ, (matching angles of congruent triangles) and so 1 and 2 are perpendicular. Example 7 Finding perpendicular gradients 2 What is the gradient of a line perpendicular to a line with gradient ? 3 Solution Perpendicular gradient = − 3 2 (take the negative of the reciprocal of 23 ) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 318 8B Chapter 8 Lines in the coordinate plane Example 8 Testing whether line segments are perpendicular Show that the diagonals of the quadrilateral ABCD are perpendicular, where the vertices are A(3, 7), B(−1, 6), C(−2, −3) and D(11, 0). Solution y2 − y 1 x2 − x 1 −3 − 7 = −2 − 3 = 2, y 2 − y1 x2 − x1 0−6 = 11 + 1 1 =− . 2 gradient of BD = Gradient of AC = Hence AC ⊥ BD, because the product of the gradients of AC and BD is −1. Example 9 Using the condition for lines to be perpendicular The line segment joining the points C(−6, 0) and D(−1, a) is perpendicular to a line with gradient 10. Find the value of a. Solution a−0 −1 + 6 a = . 5 Because the line segment CD is perpendicular to a line with gradient 10, a 10 × = −1 (the product of the gradients is −1) 5 1 a × 2 = −1 1 a=− . ÷2 2 The line segment CD has gradient = Exercise 8B FOUNDATION Note: Diagrams should be drawn wherever possible. 1 a Write down the gradient of a line parallel to a line with gradient: i 2 3 4 iii −1 12 ii b Find the gradient of a line perpendicular to a line with gradient: i 2 3 4 iii −1 12 ii CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8B Gradients of line segments and lines 2 3 4 319 y2 − y 1 to find the gradient of each line segment AB below. Then find the x2 − x 1 gradient of a line perpendicular to it. Use the formula gradient = a A(1, 4), B(5, 0) b A(−2, −7), B(3, 3) c A(−5, −2), B(3, 2) d A(3, 6), B(5, 5) e A(−1, −2), B(1, 4) f A(−5, 7), B(15, −7) Each of the lines AB in this question are either horizontal (with gradient zero), vertical (with gradient undefined) or neither. State which in each case. a A(1, 2), B(1, 3) b A(4, 3), B(7, 3) c A(5, 1), B(4, 5) d A(6, −3), B(1, −3) e A(−2, 2), B(2, −2) f A(0, 4), B(0, −4) The points A(2, 5), B(4, 11), C(12, 15) and D(10, 9) form a quadrilateral. We can show this by demonstrating that the opposite sides are parallel. a Find the gradients of AB and DC, and hence show that AB DC. b Find the gradients of BC and AD, and hence show that BC AD. c What type of quadrilateral is ABCD? d Calculate the lengths AB, CD and BC, DA. e What result have you verified? 5 a Show that A(−2, −6), B(0, −5), C(10, −7) and D(8, −8) form a parallelogram. b Show that A(2, 5), B(3, 7), C(−4, −1) and D(−5, 2) do not form a parallelogram. 6 Use the formula gradient = tan α to find the gradient, correct to two decimal places where necessary, of a line with angle of inclination: a 15◦ 7 b 135◦ c 22 12 ◦ d 72◦ Use the formula gradient = tan α to find the angle of inclination, correct to the nearest degree where necessary, of a line with gradient: √ 1 a 1 b − 3 c 4 d √ 3 DEVELOPMENT 8 To show that a parallelogram is a rectangle, it is enough to show that one internal angle is 90◦ . To show that a rectangle is a square, it is enough to show that one pair of adjacent sides are equal in length. Consider the quadrilateral ABCD, with vertices A(−1, 1), B(3, −1), C(5, 3) and D(1, 5). a Show that the opposite sides are parallel, and hence that ABCD is a parallelogram. b Show that AB ⊥ BC. What can you conclude about ABCD? c Show that AB = BC. What can you conclude about ABCD now? 9 Use gradients to show that each quadrilateral ABCD below is a parallelogram. Then use the methods of the previous question to show that it is: a a rhombus, for the vertices A(2, 1), B(−1, 3), C(1, 0) and D(4, −2), b a rectangle, for the vertices A(4, 0), B(−2, 3), C(−3, 1) and D(3, −2), c a square, for the vertices A(3, 3), B(−1, 2), C(0, −2) and D(4, −1). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 320 8B Chapter 8 Lines in the coordinate plane 10 Find the gradients of PQ and QR, and hence determine whether P, Q and R are collinear. a P(−2, 7), Q(1, 1), R(4, −6) b P(−5, −4), Q(−2, −2), R(1, 0) 11 Show that the four points A(2, 5), B(5, 6), C(11, 8) and D(−16, −1) are collinear. 12 The triangle ABC has vertices A(−1, 0), B(3, 2) and C(4, 0). Calculate the gradient of each side, and hence show that ABC is a right-angled triangle. 13 Similarly, show that each triangle below is right-angled. Then find the lengths of the sides enclosing the right angle, and calculate the area of each triangle. a P(2, −1), Q(3, 3), R(−1, 4) b X(−1, −3), Y(2, 4), Z(−3, 2) 14 The line segment PQ has gradient −3. A second line passes through A(−2, 4) and B(1, k). a Find k if AB is parallel to PQ. b Find k if AB is perpendicular to PQ. 15 Find the points A and B where each line below meets the x-axis and y-axis respectively. Hence find the gradient of AB and its angle of inclination α (correct to the nearest degree). a y = 3x + 6 b y = − 12 x + 1 c 3x + 4y + 12 = 0 d 16 x y − =1 3 2 The quadrilateral ABCD has vertices A(1, −4), B(3, 2), C(−5, 6) and D(−1, −2). a Find the midpoints P of AB, Q of BC, R of CD, and S of DA. b Prove that PQRS is a parallelogram by showing that PQ RS and PS QR. 17 a Show that the points A(−5, 0), B(5, 0) and C(3, 4) all lie on the circle x2 + y2 = 25. b Explain why AB is a diameter of the circle. c Show that AC ⊥ BC. 18 Given the points X(−1, 0), Y(1, a) and Z(a, 2), find a if ∠Y XZ = 90◦ . 19 For the four points P(k, 1), Q(−2, −3), R(2, 3) and S (1, k), it is known that PQ is parallel to RS . Find the possible values of k. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8B Gradients of line segments and lines 321 ENRICHMENT 20 a The points P(x1 , y1 ), Q(x2 , y2 ) and R(x, y) are collinear. Use the gradient formula to show that (x − x1 )(y − y2 ) = (y − y1 )(x − x2 ). b If AB is the diameter of a circle and P another point on the circumference, then Euclidean geometry tells us that ∠APB = 90◦ . Use this fact to show that the equation of the circle whose diameter has endpoints A(x1 , y1 ) and B(x2 , y2 ) is (x − x1 )(x − x2 ) + (y − y1 )(y − y2 ) = 0. 21 a Three points A1 (a1 , b1 ), A2 (a2 , b2 ), A3 (a3 , b3 ) form a triangle. By dropping perpendiculars to the x-axis and taking the areas of the resulting trapeziums, show that the area A of the triangle A1 A2 A3 is A = 21 |a1 b2 − a2 b1 + a2 b3 − a3 b2 + a3 b1 − a1 b3 |, with the expression inside the absolute value sign positive if and only if the vertices A1 , A2 and A3 are in anti-clockwise order. b Use part a to generate a test for A1 , A2 and A3 to be collinear. c Generate the same test by putting gradient A1 A2 = gradient A2 A3 . 22 Consider the points P(2p, p2 ), Q(− 2p , p12 ) and T (x, −1). Find the x-coordinate of T if: a the three points are collinear, b PT and QT are perpendicular. 23 The points P(p, 1/p), Q(q, 1/q), R(r, 1/r) and S (s, 1/s) lie on the curve xy = 1. a If PQ RS , show that pq = rs. b Show that PQ ⊥ RS if and only if pqrs = −1. c Use part b to conclude that if a triangle is drawn with its vertices on the rectangular hyperbola xy = 1, then the altitudes of the triangle intersect at a common point which also lies on the hyperbola (an altitude of a triangle is the perpendicular from a vertex to the opposite side). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 322 8C Chapter 8 Lines in the coordinate plane 8C Equations of lines Learning intentions • Develop equations for horizontal and vertical lines. • Develop gradient–intercept form and general form equations for lines. • Develop tests for lines to be parallel and perpendicular. • Develop and apply the various forms of the equation of a line. In the coordinate plane, a line is represented by a linear equation in x and y. This section and the next develop various useful forms for the equation of a line. Horizontal and vertical lines All the points on a horizontal line through P(a, b) have the same x-coordinate, but the y-coordinate can take any value. Hence the equation of the line is y = b. All the points on a vertical line through P(a, b) have the same x-coordinate a, but the y-coordinate can take any value. Hence the equation of the line is x = a. 12 Horizontal and vertical lines The horizontal line through the point P(a, b) has the equation y = b. The vertical line through the point P(a, b) has the equation x = a. y y x=a y=b P(a,b) b x a P(a,b) x Gradient–intercept form There is a simple form of the equation of a line whose gradient and y-intercept are known. Let have gradient m and y-intercept b, passing through the point B(0, b). Let Q(x, y) be any other point in the plane. Then the condition for Q to lie on the line is that is, gradient of BQ = m, y−b =m x−0 y − b = mx. y Q(x,y) B(0,b) (this is the formula for gradient) x y = mx + b. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8C Equations of lines 323 13 Gradient–intercept form The line with gradient m and y-intercept b is y = mx + b. The gradient–intercept formula is often written using c for the y-intercept as: y = mx + c, but such a change of pronumeral makes no difference whatsoever. Example 10 Using gradient–intercept form a Write down the gradient and the y-intercept of the line : y = 3x − 2. b Find the equation of the line that goes through B(0, 5) and is parallel to . c Find the equation of the line that goes through B(0, 5) and is perpendicular to . Solution a The line : y = 3x − 2 has gradient 3 and y-intercept −2. b The line through B(0, 5) parallel to has gradient 3 and y-intercept 5, so its equation is y = 3x + 5. c The line through B(0, 5) perpendicular to has gradient − 13 and y-intercept 5, so its equation is y = − 13 x + 5. General form It is often useful to write the equation of a line so that all the terms are on the LHS and only zero is on the RHS. This is called general form. 14 General form The equation of a line is said to be in general form if it is ax + by + c = 0, where a, b and c are constants. When an equation is given in general form, it should usually be simplified by multiplying out all fractions and dividing through by all common factors. Example 11 Converting from general form a Put the equation of the line 3x + 4y + 5 = 0 into gradient–intercept form. b Hence write down the gradient and y-intercept of the line 3x + 4y + 5 = 0. Solution a Solving the equation for y, 4y = −3x − 5 5 3 y = − x − , which is gradient–intercept form. 4 4 b Hence the line has gradient − 34 and y-intercept −1 14 . ÷4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 324 8C Chapter 8 Lines in the coordinate plane Example 12 Converting to general form Find, in general form, the equation of the line passing through B(0, −2) and: a perpendicular to a line with gradient 23 , b having angle of inclination 60◦ . Solution a The line through B perpendicular to has gradient − 32 and y-intercept −2, so its equation is y = − 32 x − 2 ×2 2y = −3x − 4 3x + 2y + 4 = 0. (this is gradient–intercept form) √ 3, (this is gradient–intercept form) b The line through B with angle of inclination 60◦ has gradient tan 60◦ = √ y= x 3−2 so its equation is √ x 3 − y − 2 = 0. Exercise 8C 1 FOUNDATION Determine, by substitution, whether the point A(3, −2) lies on the line: a y = 4x − 10 b 8x + 10y − 4 = 0 c x=3 2 Find the x-intercept and y-intercept of each line. a 3x + 4y = 12 b y = 4x − 6 c x − 2y = 8 3 Write down the coordinates of any three points on the line x + 3y = 24. 4 Write down the equations of the vertical and horizontal lines through: a (1, 2) b (0, −4) c (5, 0) 5 6 Write down the gradient and y-intercept of each line. a y = 4x − 2 b y = 15 x − 3 c y=2−x d y = − 57 x Use the formula y = mx + b to write down the equation of the line with a gradient of −3 and: a y-intercept 5, 7 b y-intercept − 23 , c y-intercept 0. Use the formula y = mx + b to write down the equation of the line with y-intercept −4 and: a gradient 5, CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b gradient − 23 , c gradient 0. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8C Equations of lines 8 Use the formula y = mx + b to write down the equation of each line, giving your answer in general form. a line with a gradient of 1 and y-intercept 3 c 9 10 325 line with a gradient of 15 and y-intercept −1 b line with a gradient of −2 and y-intercept 5 d line with a gradient of − 12 and y-intercept 3. Solve each equation for y and hence write down its gradient m and y-intercept b. a x−y+3=0 b y+x−2=0 c x − 3y = 0 d 3x + 4y = 5 Write down the gradient m of each line. Then use the formula gradient = tan α to find its angle of inclination α, correct to the nearest minute where appropriate. a y= x+3 b y = −x − 16 c y = 2x d y = − 34 x DEVELOPMENT 11 Substitute y = 0 and x = 0 into the equation of each line below to find the points A and B where the line crosses the x-axis and y-axis respectively. Hence sketch the line. a 5x + 3y − 15 = 0 b 2x − y + 6 = 0 c 3x − 5y + 12 = 0 12 Find the gradient of the line through each pair of given points. Then find its equation, using gradient– intercept form. Give your final answer in general form. a (0, 4), (2, 8) b (0, 0), (1, −1) c (−9, −1), (0, −4) 13 Find the gradient of each line below. Hence find, in gradient–intercept form, the equation of a line passing through A(0, 3) and: ii perpendicular to it. i parallel to it a 2x + y + 3 = 0 14 b 5x − 2y − 1 = 0 c 3x + 4y − 5 = 0 In each part, find the gradients of the four lines, and hence state what sort of special quadrilateral they enclose. a 3x + y + 7 = 0, x − 2y − 1 = 0, 3x + y + 11 = 0, x − 2y + 12 = 0 b 4x − 3y + 10 = 0, 3x + 4y + 7 = 0, 4x − 3y − 7 = 0, 3x + 4y + 1 = 0 15 Find the gradients of the three lines 5x − 7y + 5 = 0, 2x − 5y + 7 = 0 and 7x + 5y + 2 = 0. Hence show that they enclose a right-angled triangle. 16 Using a sketch, find the equations of the sides of: a the rectangle with vertices P(3, −7), Q(0, −7), R(0, −2) and S (3, −2), b the triangle with vertices F(3, 0), G(−6, 0) and H(0, 12). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 326 8C Chapter 8 Lines in the coordinate plane 17 In each part below, the angle of inclination α and the y-intercept A of a line are given. Use the formula gradient = tan α to find the gradient of each line, then find its equation in general form. a α = 45◦ , A = (0, 3) b α = 60◦ , A = (0, −1) c α = 30◦ , A = (0, −2) d α = 135◦ , A = (0, 1) 18 A triangle is formed by the x-axis and the lines 5y = 9x and 5y + 9x = 45. a Find (correct to the nearest degree) the angles of inclination of the two lines. b What sort of triangle has been formed? 19 Consider the two lines 1 : 3x − y + 4 = 0 and 2 : x + ky + = 0. Find the value of k if: a 1 is parallel to 2 , b 1 is perpendicular to 2 . 20 Show by substitution that the line y = mx + b passes through A(0, b) and B(1, m + b). Then show that the gradient of AB, and hence of the line, is m. ENRICHMENT 21 Find the equations of the four circles which are tangent to the x-axis, the y-axis, and the line x + y = 2. 22 a Show that the four lines y = 2x − 1, y = 2x + 1, y = 3 − 12 x and y = k − 12 x enclose a rectangle. b Find the possible values of k if they enclose a square. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8D Further equations of lines 327 8D Further equations of lines Learning intentions • Develop and apply point–gradient form of the equation of a line. • Find the line through two given points. • Test three lines for concurrency. This section introduces another standard form of the equation of a line, called point–gradient form. It also deals with lines through two given points, and the point of intersection of two lines. Point–gradient form Point–gradient form gives the equation of a line with gradient m passing through a particular point P(x1 , y1 ). Let Q(x, y) be any other point in the plane. y Then the condition for Q to lie on the line is Q(x,y) gradient of PQ = m, y − y1 =m (this is the formula for gradient) that is, x − x1 P(x1,y1) y − y1 = m(x − x1 ). x The proof doesn’t work when the point Q coincides with P, that is, when x = x1 , first because PP does not have a gradient, and secondly because you can’t divide by x − x1 = 0. But (x1 , y1 ) is still a point on the line y − y1 = m(x − x1 ), as is obvious by direct substitution. 15 Point–gradient form The line with gradient m through the point (x1 , y1 ) is y − y1 = m(x − x1 ). If a question doesn’t specify which form to express the equation of a line in, then there are two forms that are acceptable in this situation: • Gradient–intercept form: • General form: y = 3x + 7 3x − y + 7 = 0 If general form is used, then in most circumstances it should be simplified by multiplying out any fractions, and dividing through by any common factor. Example 13 Using point–gradient form a Find the equation of the line through (−2, −5) and parallel to y = 3x + 2. b Express the answer in gradient–intercept form, and hence find its y-intercept. Solution a The line y = 3x + 2 has a gradient of 3. Hence the required line is y − y1 = m(x − x1 ) (this is point–gradient form) y + 5 = 3(x + 2) y + 5 = 3x + 6 y = 3x + 1. (this is gradient–intercept form) b Hence the new line has y-intercept 1. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 328 8D Chapter 8 Lines in the coordinate plane The line through two given points Given two distinct points, there is just one line passing through them both. Its equation is best found by a two-step approach. 16 The line through two given points • First find the gradient of the line, using the gradient formula. • Then find the equation of the line, using point–gradient form. Example 14 Finding the line through two given points Find the equation of the line passing through A(1, 5) and B(4, −1). Solution First, using the gradient formula: −1 − 5 gradient of AB = 4−1 = −2. Then, using point–gradient form for a line with gradient −2 through A(1, 5), the line AB is y − y1 = m(x − x1 ) (this is point–gradient form) y − 5 = −2(x − 1) (the coordinates of A(1, 5) have been used) y − 5 = −2x + 2 y = −2x + 7. Note: Using the coordinates of B(4, −1) rather than A(1, 5) would give the same equation. Example 15 Finding the perpendicular bisector Given the points A(6, 0) and B(0, 9), find, in general form, the equation of the perpendicular bisector of AB. Solution y2 − y1 x2 − x 1 9−0 = 0−6 3 =− , 2 so any line perpendicular to AB has gradient of 23 . 6+0 0+9 , Secondly, the midpoint of AB = 2 2 gradient of AB = First, = (3, 4 12 ). Hence, using point–gradient form, the perpendicular bisector of AB is y − y1 = m(x − x1 ) y − 4 12 = 23 (x − 3) ×6 6y − 27 = 4x − 12 4x − 6y + 15 = 0. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8D Further equations of lines 329 Intersection of lines — concurrent lines The point where two distinct lines intersect can be found using simultaneous equations, as discussed in Sections 1E and 3C. Three distinct lines are called concurrent if they all pass through the same point. 17 Testing for concurrent lines To test whether three lines are concurrent: • First find the point of intersection of two of them. • Then test by substitution whether this point lies on the third line. Example 16 Testing whether three lines are concurrent Test whether these three lines are concurrent. 1 : 5x − y − 10 = 0, 2 : x + y − 8 = 0, 3 : 2x − 3y + 9 = 0. Solution A Solve 1 and 2 simultaneously. Adding 1 and 2 , 6x − 18 = 0 x = 3, and substituting into 2 , 3 + y − 8 = 0 y=5 so the lines 1 and 2 intersect at (3, 5). B Substituting the point (3, 5) into the third line 3 , LHS = 6 − 15 + 9 =0 = RHS, so the three lines are concurrent, meeting at (3, 5). Exercise 8D 1 FOUNDATION Find, in general form, the equation of the line: a through the point (1, 1) with a gradient of 2 b with a gradient of −1 through the point (3, 1) c through the point (0, 0) with a gradient of −5 1 3 e with a gradient of − 45 through the point (3, −4). d through the point (−1, 3) with a gradient of − 2 Find, in gradient–intercept form, the equation of: a the line through the point (2, 5) and parallel to y = 2x + 5 b the line through the point (2, 5) and perpendicular to y = 2x + 5 c the line through the point (5, −7) and perpendicular to y = −5x d the line through the point (−7, 6) and parallel to y = 37 x − 8 e the line through the point (−4, 0) and perpendicular to y = − 25 x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 330 8D Chapter 8 Lines in the coordinate plane 3 a Find the gradient of the line through the points A(4, 7) and B(6, 13). b Hence use point–gradient form to find, in general form, the equation of the line AB. 4 5 Find the gradient of the line through each pair of points, and hence find its equation. a (3, 4), (5, 8) b (−1, 3), (1, −1) c (5, 6), (−1, 4) d (−1, 0), (0, 2) e (0, −1), (−4, 0) f (0, −3), (3, 0) a Find the gradient of the line through the points A(1, −2) and B(−3, 4). b Hence find, in general form, the equation of: ii the line through A and perpendicular to AB. i the line AB 6 Find the equation of the line parallel to 2x − 3y + 1 = 0 and: b passing through (3, −1). a passing through (2, 2) 7 Find the equation of the line perpendicular to 3x + 4y − 3 = 0 and: a passing through (−1, −4) b passing through (−2, 1). DEVELOPMENT 8 a Find the point M of intersection of the lines 1 : x + y = 2 and 2 : 4x − y = 13. b Show that M lies on 3 : 2x − 5y = 11, and hence that 1 , 2 and 3 are concurrent. c Use the same method to test whether each set of lines is concurrent. ii 6x − y = 26, 5x − 4y = 9 and x + y = 9 i 2x + y = −1, x − 2y = −18 and x + 3y = 15 9 Put the equation of each line into gradient–intercept form and hence write down the gradient. Then find, in gradient–intercept form, the equation of the line that is: i parallel to it and goes through A(3, −1), a 2x + y + 3 = 0 10 11 ii perpendicular to it and goes through B(−2, 5). b 5x − 2y − 1 = 0 c 4x + 3y − 5 = 0 The angle of inclination α and a point A on a line are given below. Use the formula gradient = tan α to find the gradient of each line, then find its equation in general form. a α = 45◦ , A = (1, 0) b α = 120◦ , A = (−1, 0) c α = 30◦ , A = (4, −3) d α = 150◦ , A = (−2, −5) Determine, in general form, the equation of each straight line sketched below. a y b x (ii) (3,−1) c y (i) d y (i) 3 y x −6 4 (ii) 2 x 45º 120º x 30º 12 Explain why the four lines 1 : y = x + 1, 2 : y = x − 3, 3 : y = 3x + 5 and 4 : y = 3x − 5 enclose a parallelogram. Then find the vertices of this parallelogram. 13 Show that the triangle with vertices A(1, 0), B(6, 5) and C(0, 2) is right-angled. Then find the equation of each side. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8D Further equations of lines 14 The three points A(1, 0), B(0, 8) and C(7, 4) form a triangle. Let θ be the angle between AC and the x-axis. 331 y B(0,8) a Find the gradient of the line segment AC and hence determine θ, correct to the C(7,4) nearest degree. b Derive the equation of the line segment AC. c Find the coordinates of the midpoint D of AC. d Show that AC is perpendicular to BD. e What type of triangle is ABC? f Find the area of this triangle. D θ x A(1,0) g Write down the coordinates of a point E such that the parallelogram ABCE is a rhombus. 15 The line crosses the x- and y-axes at L(−4, 0) and M(0, 3). The point N lies on , and P is the point P(0, 8). y P(0,8) a Copy the sketch and find the equation of . N b Find the lengths of ML and MP and hence show that LMP is an isosceles triangle. c If M is the midpoint of LN, find the coordinates of N. d Show that ∠NPL = 90◦ . e Write down the equation of the circle through N, P and L. M(0,3) L(−4,0) 16 Find k if the lines 1 : x + 3y + 13 = 0, 2 : 4x + y − 3 = 0 and 3 : kx − y − 10 = 0 are concurrent. (Hint: Find the point of intersection of 1 and 2 and substitute into 3 .) 17 Consider the two lines 1 : 3x + 2y + 4 = 0 and 2 : 6x + μy + λ = 0. Write down the value of μ if: a 1 is parallel to 2 , 18 x b 1 is perpendicular to 2 . Write down the equation of the line with the intercepts (a, 0) and (0, b), then rewrite it in general form. ENRICHMENT 19 The line passing through M(a, b) intersects the x-axis at A and the y-axis at B. Find the equation of the line, given that M bisects AB. 20 The tangent to a circle is perpendicular to the radius at the point of contact. Use this fact to show that the tangent to x2 + y2 = r2 at the point (a, b) has equation ax + by = r2 . 21 [Perpendicular form of a line] Consider the line with equation ax + by = c where, for the sake of convenience, the values of a, b and c are positive. Suppose that this line makes an acute angle θ with the y-axis as shown. a b a Show that cos θ = √ and sin θ = √ . y 2 2 2 a +b a + b2 b The perpendicular form of the line is c b a b c x+ √ y= √ . √ θ a2 + b 2 a2 + b 2 a2 + b2 Use part a to help show that the RHS of this equation is the perpendicular c a distance from the line to the origin. x c Write these lines in perpendicular form and hence find their perpendicular distances from the origin: i 3x + 4y = 5 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii 3x − 2y = 1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 332 8E Chapter 8 Lines in the coordinate plane 8E Using pronumerals in place of numbers Learning intentions • Work with the various formulae using pronumerals in place of numbers. • Use these approaches to prove some general results about geometric figures. Most problems of this chapter so far have used numbers for the coordinates of points, and for the coefficients of x and y. When constant pronumerals are used, however, far more general results can be obtained. For example, Chapter 3 used pronumeral constants in the study of quadratics and their parabolic graphs. This final section consolidates the methods of the last four sections, but uses pronumerals, rather than numbers, for points and coefficients. Because of the generality, most of the questions have a geometric interpretation as a result. When the French mathematician and philosopher René Descartes introduced the coordinate plane in the 17th century, he intended it as an alternative approach to geometry, in which all the ancient theorems of Euclid would be proven again using algebraic rather than geometric arguments. In our course, however, Descartes’ methods are used with the reverse purpose — we are turning functions that have been defined algebraically into geometric objects that can be visualised and interpreted using the methods of geometry. Questions here, and throughout the course, should be interpreted both algebraically and geometrically. Example 17 Applying formulae with pronumerals rather than numbers a A triangle has vertices at A(1, −3), B(3, 3) and C(−3, 1). i Find the coordinates of the midpoint P of AB and the midpoint Q of BC. ii Show that PQ AC and that PQ = 12 AC. b Repeat this procedure for a triangle with vertices A(2a, 0), B(2b, 2c) and C(0, 0), where a > 0. Solution a i Using the midpoint formula, −3 + 3 1+3 y= For P, x = 2 2 =2 = 0. 3−3 3+1 For Q, x = y= 2 2 =0 = 2. Hence P = (2, 0) and Q = (0, 2). C y 3 2Q 1 -3 -3 B P 1 23 x A ii The lines PQ and AC are parallel because they both have a gradient of −1: 1+3 2−0 gradient AC = 0−2 −3 − 1 = −1 = −1. Using the distance formula, PQ2 = (0 − 2)2 + (2 − 0)2 AC 2 = (−3 − 1)2 + (1 + 3)2 gradient PQ = =8 √ so PQ = 2 2 and √ = 32 AC = 4 2 = 2PQ. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8E Using pronumerals in place of numbers b 333 y i Using the midpoint formula, 0 + 2c 2a + 2b y= For P, x = 2 2 =a+b = c. 2c + 0 2b + 0 y= For Q, x = 2 2 = c. =b Hence P = (a + b, c) and Q = (b, c). B(2b, 2c) Q C P A(2a, 0) x ii The lines PQ and AC are parallel because they are both horizontal. Using the distance formula, PQ2 = (b − a − b)2 + (c − c)2 = a2 so PQ = a and AC 2 = (0 − 2a)2 + (0 − 0)2 AC = 2a = 2PQ. = 4a2 Any triangle can be translated and rotated into the position of the triangle in part b, so this example demonstrates that the line segment joining the midpoints of two sides of a triangle is parallel to the base and half its length. Exercise 8E FOUNDATION Note: This exercise involves lines with equations defined in terms of various constants k, m, b, etc. Without further definition, these are taken to be some constant real value. 1 Find k for each of the following situations: a The line y = 2x + k passes though (3, 2). b The line y = (2k − 1)x + 4 has gradient 3. c The line y = kx + (3 − k) passes through the origin. 2 The two lines y = kx + 3 and y = (2 − 3k)x + 7 are defined in terms of a constant k. Find the value of k if: a The lines are parallel. b The lines are perpendicular. 3 Find the area of the triangle bounded by the line y = mx, the x-axis and the line x = 4. 4 Find the area of the triangle bounded by the line y = mx, the y-axis and the line y = 6. 5 The triangle OAB has vertices O(0, 0), A(k + 2, 0) and B(k, 3). It is known to have an area of 6 units squared. Find k. 6 The line y = 4 − kx cuts the x-axis at A and the y-axis at B. Find the area of triangle OAB, where O is the origin O(0, 0). 7 a A triangle has vertices at A(3, 1), B(7, 3) and C(1, −1). i Find the coordinates of the midpoint P of AB and the midpoint Q of BC. ii Show that the equation of the line through P parallel to AC is y = x − 3. iii Show that Q lies on the line found in part ii. That is, show that bisects BC. iv Show that PQ = 12 AC. b Consider a triangle by placing its vertices at A(2a, 0), B(2b, 2c) and C(0, 0), where a > 0, and repeat the steps in part a. In this question you have shown that the line segment through the midpoint of a side and parallel to the base bisects the third side and is half the length of the base. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 334 8E Chapter 8 Lines in the coordinate plane DEVELOPMENT 8 The line y = mx − 4 is a tangent to the parabola y = x2 . a By solving the equations simultaneously, find a quadratic involving m. b Use the discriminant, and the fact the quadratic has one solution, to find m. 9 The line y = 2x + b is a tangent to the parabola y = x2 . a Find a quadratic involving b for their point of intersection. b Use the discriminant to find b. c Where do the line and parabola intersect? 10 a These three lines are concurrent: y = 2x + 1 y = 3x − 4 y = mx + 7 Find m. b These three lines lines are concurrent. y= x+k x=4 y = − 12 x + 3k Find their common point of intersection. 11 Consider the points A(2, 0), B(4, 2) and a point P(x, y). It is known that P is the same distance from A and B (we say P is equidistant from A and B). a Use the distance formula to write expressions for square distances PA2 and PB2 in terms of x and y. b Hence find the equation of the line on which P sits. c What can you state about triangle ABP and the midpoint of AB? 12 Consider the points A(2, 0) and C(0, −4) and a point P(x, y). It is known that P is equidistant from A and C. a Use the methods of the previous question to find the equation of the line on which P sits. b Which point in the plane is equidistant from points A, C, and also from point B in the previous question? 13 a On a number plane, plot the points O(0, 0), A(6, 0), B(6, 6) and C(0, 6), which form a square. i Find the gradients of the diagonals OB and AC. ii Hence show that the diagonals OB and AC are perpendicular. b Let the vertices of a square be O(0, 0), A(a, 0), B(a, a) and C(0, a). i Find the gradients of the diagonals OB and AC. ii Hence show that the diagonals OB and AC are perpendicular. In this question you have shown that the diagonals of a square are perpendicular. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 8E Using pronumerals in place of numbers 14 a The points O(0, 0), P(8, 0) and Q(0, 10) form a right-angled triangle. Let M be 335 y the midpoint of PQ. Q(0,2q) i Find the coordinates of M. M ii Find the distances OM, PM and QM, and show that M is equidistant from each of the vertices. iii Explain why a circle with centre M can be drawn through the three vertices O, P and Q x P(2p,0 ) O b Consider a right-angled triangle with vertices at O(0, 0), P(2p, 0) and Q(0, 2q) and repeat the steps of part a. In this question you have shown that the midpoint of the hypotenuse of a right-angled triangle is the centre of a circle through all three vertices. 15 The points A(a, 0) and Q(q, 0) are points on the positive x-axis, and the points B(0, b) and P(0, p) lie on the positive y-axis. Show that AB2 − AP2 = QB2 − QP2 . 16 The diagram to the right shows the points A, B, C and D on the number plane. y a Show that ΔABC is equilateral. B a 3 b Show that ΔABD is isosceles, with AB = AD. c Show that AB2 = 13 BD2 . 17 C −a A a Dx 3a Place three vertices of a parallelogram at A(0, 0), B(2a, 2b) and D(2c, 2d). a Use gradients to show that with C = (2a + 2c, 2b + 2d), the quadrilateral ABCD is a parallelogram. b Find the midpoints of the diagonals AC and BD. c Explain why this proves that the diagonals of a parallelogram bisect each other. ENRICHMENT 18 Two circles x2 + y2 = 4 and (x − 5)2 + y2 = 16 intersect at P and Q. Solve the equations simultaneously to find the points of intersection of the two circles, and hence find the equation of the secant PQ. Note: The secant is the line through two distinct points on a curve. 19 The point P(a, a2 ) lies on the parabola y = x2 . a Write down the equation of the line through P with a gradient of m. b Substitute the parabola into the line to produce a quadratic equation in x (that is, begin to solve simultaneously the equations of the line and the parabola). c Find the discriminant of this quadratic equation, and write the discriminant as a quadratic function of m. d If is known to be a tangent to y = x2 , use this discriminant to find m in terms of a. e Complete this sentence: The tangent to f (x) = x2 at P(a, a2 ) has gradient g(a) = . . .. (The function g is called the gradient function for f (x) and we shall investigate this more closely with better techniques in later chapters.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 336 8E Chapter 8 Lines in the coordinate plane 20 a The points A(1, −2), B(5, 6) and C(−3, 2) are the vertices of a triangle, and P, Q and R are the midpoints of BC, CA and AB respectively. i Find the equations of the three medians BQ, CR and AP. ii Find the intersection of BQ and CR, and show that it lies on the third median AP. b Now consider the more general situation of a triangle with vertices at A(6a, 6b), B(−6a, −6b) and C(0, 6c), and follow these steps. i Find the midpoints P, Q and R of BC, CA and AB respectively. ii Show that the median through C is x = 0 and find the equations of the other two medians. iii Find the point where the median through C meets the median through A, and show that this point lies on the median through B. In this question you have shown that the three medians of a triangle are concurrent. (Their point of intersection is called the centroid.) 21 Let the circle x2 + y2 = r2 intersect the radius y = mx at P. Assume m > 0 and the point of intersection is in quadrant 1. Let the line through P perpendicular to the radius have equation x + my = mc, where c is the y-intercept of . √ a i Use algebra to show that mc = r 1 + m2 . ii Confirm your result using similarity. b Use simultaneous equations to show that the intersection points of and the circle are determined by the quadratic (1 + m2 )y2 − 2mcy + c2 − r2 = 0. c Show that the line perpendicular to the radius at point of contact is a tangent, i.e. it only intersects the circle at one point. You may have seen the theorem that a tangent is perpendicular to the radius at the point of contact. This is the converse result. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 8 review 337 Chapter 8 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 8 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise Let X = (2, 9) and Y = (14, 4). Use the standard formulae to find: b the gradient of XY a the midpoint of XY 2 A triangle has vertices A(1, 4), B(−3, 1) and C(−2, 0). a Find the lengths of all three sides of b What sort of triangle is 3 c the length of XY. Review 1 ABC. ABC? A quadrilateral has vertices A(2, 5), B(4, 9), C(8, 1) and D(−2, −7). a Find the midpoints P of AB, Q of BC, R of CD and S of DA. b Find the gradients of PQ, QR, RS and S P. c What sort of quadrilateral is PQRS ? 4 A circle has a diameter AB, where A = (2, −5) and B = (−4, 7). a Find the centre C and radius r of the circle. b Use the distance formula to test whether P(6, −1) lies on the circle. LMN, given L(3, 9), M(8, −1) and N(−1, 7). b Explain why LMN is a right-angled triangle. 5 a Find the gradients of the sides of 6 a Find the gradient of the line segment AB, where A = (3, 0) and B = (5, −2). b Find a if AP ⊥ AB, where P = (a, 5). c Find the point Q(b, c) if B is the midpoint of AQ. d Find d if the line segment AD has length 5, where D = (6, d). 7 Find, in general form, the equation of the line: a with a gradient of −2 and a y-intercept of 5, 2 and goes through the point A(3, 5), 3 c that goes through the origin and is perpendicular to y = 7x − 5, d that goes through B(−5, 7) and is parallel to y = 4 − 3x, e with a y-intercept of −2 and an angle of inclination 60◦ . b with a gradient of CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 338 Chapter 8 Lines in the coordinate plane Review 8 Put the equation of each line into gradient–intercept form, and hence find its y-intercept b, its gradient m and its angle of inclination α (correct to the nearest minute when necessary). b 4x + 4y − 3 = 0 a 5x − 6y − 7 = 0 9 Find the gradient of each line AB, then find its equation in general form. b A(5, −2) and B(7, −7) a A(3, 0) and B(4, 8) 10 a Are the points L(7, 4), M(13, 2) and N(25, −3) collinear? b Are the lines 2x + 5y − 29 = 0, 4x − 2y + 2 = 0 and 7x − 3y + 1 = 0 concurrent? 11 a Determine whether the lines 8x + 7y + 6 = 0, 6x − 4y + 3 = 0 and 2x + 3y + 9 = 0 enclose a right-angled triangle. b Determine what sort of figure the lines 4x + 8y + 3 = 0, 5x − 2y + 7 = 0, x + 2y − 6 = 0 and 9x − 3y = 0 enclose. 12 a Find the points where the line 5x + 4y − 30 = 0 meets the x-axis and y-axis. b Hence find the area of the triangle formed by the line, the x-axis and the y-axis. 13 A sketch is essential in this question. a Find the gradient, length and midpoint M of the line segment joining A(10, 2) and B(2, 8). b Show that the perpendicular bisector of the line segment AB has the equation 4x − 3y − 9 = 0. c Find the point C where the perpendicular bisector meets the line x − y + 2 = 0. d Use the distance formula to show that C is equidistant from A and B. e Show that CM = 15 and hence find the area of f ABC. Let θ = ∠ACB. Use the area formula area = 12 × AC × BC × sin θ to find θ, correct to the nearest minute. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9 Exponential and logarithmic functions Chapter introduction Suppose that the amount of mould on a piece of cheese is doubling every day. In one day, the mould increases by a factor of 2. In two days, the mould increases by a factor of 22 = 4. In three days, the mould increases by a factor of 23 = 8. In four days, the mould increases by a factor of 24 = 16. And so on, until the cheese is thrown out. Applying mathematics to situations such as this requires indices and logarithms, and their graphs. Logarithms may be quite unfamiliar to many readers. The application questions throughout, and particularly the final exercise, indicate some of the extraordinary variety of situations where exponential and logarithmic functions are needed. Readers who have studied this material well in earlier years may be selective with the earlier questions in the exercises. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 340 9A Chapter 9 Exponential and logarithmic functions 9A Indices Learning intentions • Use the terms power, base, and index correctly. • Develop the index laws for powers whose indices are positive whole numbers. • Extend to powers whose indices are zero or a negative integer. This section reviews powers such as 23 and 5−2 , whose indices are integers. Then the various index laws for working with them are also reviewed. Power, base and index These words ‘power’, ‘base’, and ‘index’ have three different very exact meanings. 1 Power, base and index • An expression an is called a power. • The number a is called the base. • The number n is called the index or exponent. Thus 23 is a power with base 2 and index 3. The words exponent and index (plural indices) mean exactly the same thing. Powers whose indices are positive whole numbers Powers are defined differently depending on what sort of number the index is. When the index is a positive integer, the definition is quite straightforward. 2 Powers whose indices are positive whole numbers For any real number a, a1 = a, a2 = a × a, a3 = a × a × a, ... n factors In general, an = a × a × a × · · · × a, for all positive whole numbers n. For example, 71 = 7 34 = 3 × 3 × 3 × 3 95 = 9 × 9 × 9 × 9 × 9 = 81 = 59 049 although with a large number such as 95 , the index form may be more convenient. Scientific calculators can help to evaluate or approximate powers. The button for powers is labelled xy or ∧ , depending on the calculator. For example, 226 = 67 108 864 350 7.179 × 1023 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 π4 97.41 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9A Indices 341 Index laws — combining powers with the same base The first three index laws show how to combine powers when the base is fixed. 3 Index laws — products, quotients and powers of powers • To multiply powers with the same base, add the indices: am × an = am+n • To divide powers with the same base, subtract the indices: am am ÷ an = am−n also written as n = am−n a • To raise a power to a power, multiply the indices: (am )n = amn Demonstrating the results when m = 5 and n = 3 should make these laws clear. • a5 × a3 = (a × a × a × a × a) × (a × a × a) a×a×a×a×a a×a×a = a2 • a5 ÷ a3 = = a8 = a5+3 = a5−3 • (a5 )3 = a5 × a5 × a5 = a5+5+5 (to multiply powers of the same base, add the indices) = a5×3 Example 1 Using the index laws Use the index laws above to simplify each expression. a x3 × x7 b 3 x × 35x c w12 ÷ w3 d 10a+b ÷ 10b e (y6 )9 f (23x )2y a x3 × x7 = x10 b 3 x × 35x = 36x c w12 ÷ w3 = w9 d 10a+b ÷ 10b = 10a e (y6 )9 = y54 f (23x )2y = 26xy Solution Index laws — powers of products and quotients The next two index laws show how to work with powers of products and powers of quotients. 4 Index laws — powers of products and quotients • The power of a product is the product of the powers: (ab)n = an × bn • The power of a quotient is the quotient of the powers: a n an = n b b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 342 9A Chapter 9 Exponential and logarithmic functions Demonstrating the results when n = 3 should make the general results obvious. a 3 a a a • (ab)3 = ab × ab × ab • = × × b b b b =a×a×a×b×b×b a3 = 3 b = a3 b3 Example 2 Powers of products and quotients Expand the brackets in each expression. x 4 a (10a)3 b 3 2 c (3x y) Solution x4 3 34 x4 = 81 2 3 2 a3 a d = 2 3b 3 × b2 a6 = 2 9b a (10a)3 = 103 × a3 b = 1000 a3 3 c (3x2 y)3 = 33 × x2 x 4 3 2 a d 3b 3 × y3 = 27x6 y3 = Zero and negative indices All the indices were positive whole numbers in the examples above. Powers with zero and negative indices are defined so that these laws also hold for them. Zero and negative indices involve division by the base a, so a cannot be zero. a3 ÷ a3 = 1. 1 We know that If we use the index laws, however, a3 ÷ a3 = a3−3 = a0 . a0 = 1. 1 2 We know that a2 ÷ a 3 = . a If we used the index laws, however, a2 ÷ a3 = a2−3 = a−1 . 1 Hence it is convenient to define a−1 = . a 1 1 3 Similarly, we shall define a−2 = 2 , and a−3 = 3 , and so on. a a Hence it is convenient to define 5 Zero and negative indices Let the base a be any non-zero real number. • Define a0 = 1. 1 • Define a−1 = , a define a−2 = • In general, define a−n = 1 , a2 define a−3 = 1 , a3 ... 1 , for all whole numbers n. an CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9A Indices 343 Thus the negative sign in the index means, ‘take the reciprocal’. Always do this first, and always write the 1 reciprocal of 23 as 32 , never as 2 . For example, 3 −3 3 2 3 27 , where we have dealt with the negative sign first. = = 3 2 8 Example 3 Converting powers with negative indices to fractions Write each expression using fractions instead of negative indices, then simplify it. a 12−1 b 7−2 c ( 23 )−1 d ( 23 )−4 Solution 1 72 1 a 12−1 = 12 b 7−2 = c ( 23 )−1 = 32 1 = 49 4 d ( 23 )−4 = 32 (take the reciprocal first) (the negative index means, ‘take the reciprocal’) Example 4 = 81 16 (then take the fourth power) Converting fractions to powers with negative indices Write each expression using negative indices instead of fractions. 1 1 5 a 20 b 3 c 7 2 x a d w2 v4 Solution a 1 = 2−20 220 Example 5 b 1 = x−3 x3 c 1 5 =5× 7 a7 a = 5a−7 d 1 w2 = w2 × 4 v4 v = w2 v−4 Simplifying expressions with indices Use the index laws, extended to negative indices, to simplify these expressions. Give your answers in fraction form without negative indices. . a x4 × x−10 b 12a2 b4 ÷ 6a3 b4 c (2n )−3 2−3 4−4 b 12a2 b4 ÷ 6a3 b4 = 12 b 6 a c (2n )−3 = 2n×(−3) Solution a x4 × x−10 = x4−10 = x−6 1 = 6 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 = 2 a−1 b0 = 2 a−1 2 = a = 2−3n 1 = 3n 2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 344 9A Chapter 9 Exponential and logarithmic functions Exercise 9A FOUNDATION Note: Do not use a calculator in this exercise, except for the very last question. 1 The town of Goldhope had a small population in 1970, but the population has been increasing by a factor of 3 every decade since then. a By what factor had the population increased from its 1970 value in 1980, 1990, 2000, 2010 and 2020? b If the town’s population was 10 000 in 1970, during which decade did the population pass 1 000 000? 2 Preparation: The answers to this question will be needed for all the exercises in this chapter. Keep the results in a place where you can refer to them easily. a Write down all the powers of 2 from 20 = 1 to 212 = 4096. b In the same way, write down: i the powers of 3 to 36 = 729 ii the powers of 5 to 55 = 3125 iii the powers of 6 to 63 = 216 iv the powers of 7 to 73 = 343 c From your list of powers of 2, read off: ii powers of 8. i powers of 4 3 Simplify: 6 a 2 1 5 f 9 k 10−2 2 2 b 3 3 2 c 3 4 d 2 4 e 7 g 70 h 5−1 i 11−1 j 6−2 l 3−3 m 5−3 n 2−5 o 10−6 3 10 4 Simplify. Remember that with fractions, a negative index means, ‘take the reciprocal,’ −1 −1 −1 −1 2 7 10 1 a b c d e (0.1)−1 11 7 2 23 −3 −4 1 1 −1 −1 −2 f (0.01) g (0.02) h 5 i j 5 2 −6 −2 −4 −2 0 1 2 3 2 3 k l m n o 10 3 2 5 7 5 Simplify, leaving your answers in index form: a 29 × 25 b a8 × a7 c 72 × 7−10 d x7 × x−5 e 96 × 9−6 f a5 × a−5 g 5 × 5−4 h 84 × 8 × 8−4 i 78 ÷ 7 3 j a5 ÷ a7 k x10 ÷ x−2 l x−10 ÷ x2 m 28 ÷ 2−8 n 28 ÷ 28 o y12 ÷ y p y ÷ y12 q x5 × x5 × x5 u 6 a−2 3 3 r x5 −7 v 54 7 s z2 −5 w y−2 t a−2 × a−2 × a−2 x 2−4 −4 Expand the brackets in each expression. a (3x)2 5 1 f x b (5a)3 2 3 g x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c (2c)6 h y 2 5 d (3st)4 e (7xyz)2 7a 5 3x 2y i 2 j 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9A Indices 7 8 345 Write each expression as a fraction without negative indices or brackets. a 9−1 b x−1 c b−2 d −a−4 e (7x)−1 f 7x−1 g −9x−1 h (3a)−2 i 3a−2 j 4x−3 First change each mixed numeral or decimal to a fraction, then simplify the expression. −1 −1 −2 −1 1 2 1 1 a 1 b 2 c 2 d 2 2 3 3 2 −3 −2 1 2 e 3 f 6 g 0.2−1 h 2.4−1 3 3 i 2.25−1 j 2.5−2 k 2.5−3 l 0.05−2 DEVELOPMENT 9 A huge desert rock is entirely made up of coarse grains of sandstone, each about 1 mm3 in size, and is very roughly a rectangular block 3 km long, 2 km wide, and 500 m high. a Find the approximate volume of the rock in cubic kilometres. b How many cubic millimetres are in a cubic kilometre? Answer in index notation. c Find the approximate number of grains of sandstone in the rock. 10 11 12 Write each fraction in index form. 1 1 a b − 2 x x 12 7 f 5 g 3 x x a 7 x = 17 b 2 x = 18 e 9x = 1 f i x2 = 100 169 j x m 32x−8 = 81 n 47−x = 14 x 5 8 −2 1 x3 1 j − 2 4x 9 x2 1 i 6x Write down the solutions of these index equations. c x 1 3 x d e − =3 d ( 56 ) x = 65 = 85 h = 25 64 g = 14 = 27 x+1 o 13 = 27 5 8 −3 k x x 3 5 −2 = 25 9 = 64 81 x+1 x−1 1 1 p 5 = 125 l x Simplify each expression, giving the answer without negative indices. a 2 x × 23 b 3x × 3 c 74x × 7−5x d 52x ÷ 53 e 102x 3 g (6 x )4 × 62x 13 12 x 6 h − x c − 5 f 54x −2 h (2 x )3 ÷ 24 Simplify each expression, giving the answer without negative indices. a x2 y × x4 y3 b x−5 y3 × x3 y−2 c 3ax−2 × 7a2 x d e 56x2 y6 ÷ 8xy8 g 3 3 2 3x2 y3 × xy4 3 2 k 5st2 ÷ 5s−1 t3 i s2 y−3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 −2 3 s t × 5st−5 15 f 35a−2 b4 ÷ 28a4 b−6 −1 h 5c−2 d 3 −2 3 j 2a−2 y3 × 2ay−3 3 2 l 10x2 y−2 ÷ 2x−1 y3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 346 9A Chapter 9 Exponential and logarithmic functions 14 15 Expand and simplify, answering without using negative indices: 2 2 a x + x−1 b x − x−1 Write as a single fraction, without negative indices, and simplify: 1 − y−1 a a−1 − b−1 b 1 − y−2 −1 −1 a +b d −2 e x−2 y−2 (x2 y−1 − y2 x−1 ) a − b−2 c x2 − x−2 2 c (x−2 − y−2 )−1 f (a2 − 1)−1 (a − 1)−1 16 Explain why 12n = (22 × 3)n = 22n × 3n . Using similar methods, write these expressions with prime bases and simplify them. 9n+2 × 3n+1 a 2n × 4n × 8n b 3 × 27n x 12 × 18 x c 6x × 4x ÷ 3x d x 3 × 2x 24 x+1 × 8−1 e 1002n−1 × 25−1 × 8−1 f 62x 17 Explain why 3n + 3n+1 = 3n (1 + 3) = 4 × 3n . Then use similar methods to simplify: 3n+3 − 3n a 7n+2 + 7n b 3n 7n + 7n+2 c 5n − 5n−3 d n−1 7 + 7n+1 22n − 2n−1 e 22n+2 − 22n−1 f 2n − 2−1 18 Use the methods of the last two questions to simplify: 6n + 3 n 12 x + 1 a n+1 b 2x 2 +2 6 + 3x n n 12 − 18 c 3n − 2n ENRICHMENT 19 A calculator is required in this question — give your answers in scientific notation. a The mass of a neutron is about 1.675 × 10−27 kg. About how many neutrons are there in 1 kg of neutrons? b The radius of a neutron is about 1.11 × 10−15 m. Use the formula for the volume of a sphere V = 4 3 πr to 3 find its approximate volume. mass to find its approximate density in kg/m3 . c Use the formula density = volume Note: This question makes several extremely naive assumptions about what a neutron actually is, but the extraordinarily high density that this calculation gives is close to the density of neutron stars, which are the densest objects in the universe. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9B Fractional indices 347 9B Fractional indices Learning intentions • Define and work with powers with fractional indices. In this section, the definition of powers is extended to powers such as 4 2 and 27− 3 , where the index is a positive or negative fraction. Again, the definitions are made so that the index laws work for all indices. 1 2 Numerical answers can be checked on the calculator using xy or ∧ . Fractional indices with numerator 1 Let the base a be any real number a ≥ 0. √ 2 1 We know that a = a. 1 2 If we use the index laws, however, a 2 = a1 = a. √ 1 Hence it is convenient to define a2 = a . √ (Remember that a means the non-negative square root of a.) √3 3 2 We know that a = a. 1 3 If we use the index laws, however, a 3 = a1 = a. √ 1 Hence it is convenient to define a3 = 3 a . Remember that we have restricted the base a to non-negative numbers. 6 Fractional indices with numerator 1 Let the base a be any real number a ≥ 0. √ √ √ 1 1 1 • Define a 2 = a , define a 3 = 3 a , define a 4 = 4 a , . . . √ 1 • In general, define a n = n a , for all positive whole numbers n, where in every case, the non-negative root of a is to be taken. Example 6 Converting powers with fractional indices to surds Write each expression using surd notation, then simplify it. 1 1 Solution 1 1 b 27 3 a 64 2 a 64 2 = √ 64 1 b 27 3 = =8 1 c 10 000 4 √3 =3 27 1 d 32 5 c 10 000 4 = √4 10 000 = 10 1 d 32 5 = √5 32 =2 General fractional indices Now let the base a be any positive real number, so that we can also take reciprocals. If the index laws are to apply 2 to a power such as a 3 , then we must be able to write 1 2 1 2 2 a3 = a3 and (a2 ) 3 = a 3 , because to raise a power to a power, we multiply the indices. 1 2 √ 2 √3 1 2 Hence we shall define a 3 as a 3 = 3 a , which is the same as (a2 ) 3 = a2 . As before, the negative says take the reciprocal — define a− 3 = 1/a 3 . 2 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 348 9B Chapter 9 Exponential and logarithmic functions 7 General fractional indices Let the base a be any positive real number, and let m and n be whole numbers with n ≥ 1. √ m √n m • Define a n = n a , which is the same as am . 1 m • Define a− n = m . an When there are negatives and fractions in the index, first deal with the negative sign, then deal with the fractional index: 8 Deal with complicated indices in this order 1 If the index is negative, take the reciprocal. 2 Then if the index has a denominator, take the corresponding root. 3 Finally, take the power indicated by the numerator of the index. − 32 32 3 4 9 3 27 For example, . = = = 9 4 2 8 Example 7 Using fractional indices Simplify each power. First take the root indicated by the denominator. 2 3 a 125 3 b 100 2 c Solution 2 3 a 125 3 = 52 b 100 2 = 103 = 25 = 1000 Example 8 c 34 16 81 34 16 81 = 3 2 3 8 = 27 Using negative fractional indices Simplify each power. First take the reciprocal as indicated by the negative index. − 43 2 3 8 a 25− 2 b 1000− 3 c 27 Solution a 25− 2 = 3 = 32 1 25 b 1000− 3 = 1 5 = 2 3 1 = 125 Example 9 2 3 1 1000 2 1 10 3 x3 = x 2 8 27 = 43 27 8 4 3 2 = 81 16 Converting surds to fractional indices c Solution √ − 43 = 1 = 100 Write each expression using a fractional index. √ 1 a x3 b √ x3 a c 1 b √ x3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 = x− 2 3 c √3 √3 1 d √3 x2 2 x2 = x 3 x2 1 d √3 x2 = x− 3 2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9B Fractional indices 349 Negative and zero bases, and irrational indices Negative numbers do not have square roots, so negative bases are impossible if a square root or fourth root or any 1 1 even root is involved. For example, it may be reasonable to define (−64) 3 = −4, but (−64) 2 is undefined. Powers of zero are only defined for positive indices x, in which case 0 x = 0 — for example, 03 = 0. If the index x is negative, then 0 x is undefined, being the reciprocal of zero — for example, 0−3 is undefined. If the index is zero, then 00 is also undefined. We can only define irrational indices informally. To define 2π , for example, take the sequence 3, 3.1, 3.14, 3.141, 3.1415, . . . of truncated decimals for π, then form the sequence of rational powers 23 , 23.1 , 23.14 , 23.141 , 23.1415 , . . . . Everything then works as expected. Exercise 9B FOUNDATION Note: Do not use a calculator in this exercise unless the question asks for it. Make sure that you can refer easily to the the list of powers of 2, 3, 5, . . . from the previous exercise. 1 Simplify these powers. First take the root indicated by the denominator. 1 1 1 b 64 3 a 36 2 3 c 1 000 000 6 2 d 25 2 4 g 27 3 3 e 27 3 f 16 4 2 i 42.5 h 64 3 j 320.6 2 Simplify these powers. First convert mixed numerals and decimals to fractions. 13 12 13 1 25 27 a b c 8 49 8 43 23 27 8 7 d e f 0.25 2 8 27 3 3 g 0.09 2 3 h 0.04 2 Simplify these powers. First take the reciprocal as indicated by the negative index. a 25− 2 b 100− 2 c 125− 3 d 16− 4 3 e 27− 3 f 81− 4 − 14 − 13 1 125 − 32 1 k 2 4 1 g j 4 1 16 16 81 1 1 2 3 h − 34 i l 1 16 − 34 125 8 − 23 Use the index laws to simplify these expressions, leaving answers in index form. 1 1 4 12 3 12 a x2 2 × x3 2 d x ÷x g (x− 3 )6 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b x4 × x− 2 e x−7 ÷ x 2 h (x9 ) 3 1 c 3x−2 y × 5x4 y−2 −2 12 f 14a4 b−5 ÷ 2a5 b−4 i 9s−4 t5 1.5 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 350 9B Chapter 9 Exponential and logarithmic functions 5 Use the index laws to simplify these expressions, giving answers as integers or fractions. 1 b 2 2 × 2− 2 1 1 a 22 × 22 1 1 6 f 73 ÷ 73 g 33 × 34 ÷ 310 h 3− 3 1 1 2 6 i 1 2 14 98 Write down the solutions of these index equations. a 9x = 3 b 121 x = 11 d 64 = 2 1 25 e x c 81 x = 3 x 1 1 f = 8 2 1 = 5 Rewrite each expression using surds instead of fractional indices. 1 1 d (7x) 2 c 7x 2 3 1 1 1 b x3 5 f x2 e 15x 4 9 1 e 25 ÷ 25 2 a x2 8 c 2 2 × 22 2 d 34 2 ÷ 3 5 2 x 7 1 4 h x3 g 6x 2 Rewrite each expression using fractional indices instead of surds. √ √ √ a x b 3 x c 3x √ √ √ e 9 6x f x3 g x9 √ √5 d 12 3 x h 25 x6 Rewrite each expression as a single power of x using fractional indices. √ √3 √ x a x2 x b 3 c x3 x2 x x d √3 x2 DEVELOPMENT 10 Use the button labelled xy or ∧ on your calculator to approximate these powers. Express your answers in scientific notation correct to four significant figures. b 565 5 √ e 2π 11 c 5−0.12 2 a 78 f 10 2 g √ 7 d (0.001)0.7 −√7 π h 0.001− 2 The cost of installing a standard swimming pool is rising exponentially according to the formula C = $6000 × (1.03)n , where n is the number of years after 1st January 2015. a What was the original cost on 1st January 2015? b What was the cost on 1st January 2016? c Use your calculator to find, correct to the nearest $10, the cost of installing a pool on: ii 1st July 2015 i 1st January 2020 12 Given that x = 16 and y = 25, evaluate: 1 1 1 a x2 + y2 1 1 b x4 − y2 d (y − x) 2 × (4y)− 2 1 c x − 2 − y− 2 13 iii 1st July 2018. 1 1 Simplify each expression, giving the answer in index form. 1 b 5a 3 b 3 × 7a− 3 b 3 d x 2 y3 ÷ x 2 y 2 1 e a 2 b 2 ÷ a− 2 b 2 g h 1 a 3x 2 y × 3x 2 y2 1 1 8x3 y−6 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 1 2 1 1 3 10 p 5 q− 5 1 1 1 1 1 21 s 2 × 24s−2 8 1 f a−2 b4 2 3 4 4 3 i x4 × x3 c © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9B Fractional indices 14 15 16 17 Rewrite these expressions, using fractional and negative indices. 1 12 5 a √ b √ c −√ x x x √ 4 5 e − √3 f x x g √ x x x2 15 x d √3 √ h 8x2 x Write down the solutions of these index equations. a 49 x = 17 b 81 x = 13 c 8x = 4 d 8 x = 14 f 4 x = 18 g 81 x = 27 1 h 27 x = 81 i x 1 25 Expand and simplify, giving answers in index form: 1 1 1 2 1 2 a x 2 + x− 2 b x 2 − x− 2 c Expand and simplify, answering without using negative indices: 2 2 a x + 5x−1 b x2 − 7x−2 c e 4x = 8 =5 j 8 x = 25 125 4 5 5 2 x 2 − x− 2 3x 2 − 2x− 2 1 1 2 18 Explain why if 33x−1 = 9, then 3x − 1 = 2, so x = 1. Similarly, by reducing both sides to powers of the same base, solve: √ a 125 x = 15 b 25 x = 5 c 8 x = 14 x √ d 64 x = 32 e 8 x+1 = 2 × 4 x−1 f 19 = 34 19 By taking appropriate powers of both sides, solve: 1 a b 3 = 17 20 b n−2 = 121 c x− 4 = 27 b 8 x ÷ 4y = 4 c 13 x+4y = 1 1 11y−x = 11 25 x+5y = 5 3 Solve simultaneously: a 72x−y = 49 2 x+y = 128 21 351 1 1 a If x = 2 3 + 4 3 , show that x3 = 6(1 + x). √ x2 + x−2 = 3. x − x−1 pq−1 − p−1 q pq c Show that 2 −2 = 2 . p q − p−2 q2 p + q2 b If x = 12 + 12 5, show that ENRICHMENT 22 1 1 1 a 3 3 and 2 2 23 1 1 b 2 2 and 5 5 3 c 7 2 and 20 1 1 d 5 5 and 3 3 Find (and check on the calculator) the smallest whole numbers m and n for which: a 12 < 2m/n < 13 24 1 By taking 6th powers of both sides, show that 11 3 < 5 2 . Using similar methods (followed perhaps by a check on the calculator), compare: b 13 < 2m/n < 14 Find lim x→0+ 0 x and lim x→0 x0 . Explain what these two limits have to do with the remark made in the notes that 00 is undefined. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 352 9C Chapter 9 Exponential and logarithmic functions 9C Logarithms Learning intentions • Understand what a logarithm and its base are. • Convert a log equation to an index equation. • Use a calculator to approximate powers and base 10 logs. • Locate a log between two integers. The introduction to this chapter mentioned that mould growing on some cheese was doubling every day. Suppose now that someone asks the question backwards: ’How many days does it takes for the mould to increase by a factor of 8?’ The answer is 3 days, because 8 = 23 , so in 3 days, the mould doubles three times. This index 3 is called the logarithm base 2 of 8, and is written as log2 8 = 3: log2 8 = 3 means that 8 = 23 . Read this as, ‘The log base 2 of 8’, mentioning the base first, and then the number. Logarithms The logarithm is the index. More precisely: 9 The logarithm is the index • The logarithm base a of a positive number x is the index, when the number x is expressed as a power of the base a: y = loga x x = ay . means that • The base a must always be positive, and not equal to 1. The basic skill with logarithms is converting between statements about indices and statements about logarithms. The next box should be learnt off by heart. 10 A sentence to commit to memory This one sentence should fix most problems: ‘ log2 8 = 3 8 = 23 .’ means that Examine the sentence and notice that: • The base of the log is the base of the power (in this case 2). • The log is the index (in this case 3). Example 10 Rewriting an index statement in log form Rewrite each index statement in logarithmic form. a 103 = 1000 b 34 = 81 Solution a 103 = 1000 log10 1000 = 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b 34 = 81 log3 81 = 4 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9C Logarithms Example 11 353 Rewriting a log statement in index form Rewrite each logarithmic statement in index form. Then state whether it is true or false. a log2 16 = 4 b log3 27 = 4 Solution a log2 16 = 4 16 = 2 , 4 b log3 27 = 4 27 = 34 , which is true. which is false. Finding logarithms To find a logarithm, it can be easier to rewrite it in index form. You will usually need to introduce a pronumeral. Example 12 Convert a log equation to an index equation. a Find log2 32. b Find log10 1 000 000. Solution x = log2 32. a Let x = log10 1 000 000. b Let Then x 2 = 32. Then Hence x = 5. Hence 10 x = 1 000 000. x = 6. Negative and fractional indices A logarithmic equation can involve a negative or a fractional index. Example 13 Convert a log equation to an index equation Solve each logarithmic equation by changing it to an index equation. a x = log7 17 b 1 3 = log8 x c log x 9 = −1 b 1 3 = log8 x c log x 9 = −1 Solution a x = log7 17 x 1 7 =7 x=8 x = −1 =2 1 3 x−1 = 9 x = 19 Logarithms on the calculator Most scientific calculators only allow direct calculations of logarithms base 10, using the button labelled log . (They also allow calculation of logarithms using the important mathematical constant e as the base — this is the purpose of the button marked ln , as explained later in Chapter 12.) The function 10x is usually on the same key as log , and is reached by pressing shift followed by log . These two functions log and 10x are called inverses of each other — when used one after the other, the original number returns. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 354 9C Chapter 9 Exponential and logarithmic functions Example 14 Solve an index equation by converting it to logs Write each statement below in logarithmic form, then use the function labelled log on your calculator to approximate x, correct to four significant figures. Then check your calculation using the button labelled 10x . a 10 x = 750 b 10 x = 0.003 Solution a 10 x = 750 10 x = 0.003 b x = log10 750 2.875, x = log10 0.003 −2.523, using log . x using log . Check this using the 10 button: Check this using the 10x button: 102.875 750. 10−2.523 0.003. Locating a logarithm between two integers Consider again the cheese with a mould that doubles in amount every day. Suppose that someone asks how many days it takes for the amount of mould to increase by a factor of 10. The answer is log2 10 days, but this is an awkward question, because the answer is not an integer. – In three days the mould increases 8-fold. – In four days it increases 16-fold. Thus we can at least say that the answer lies between 3 days and 4 days. Example 15 Locating a logarithm between two integers Use a list of powers of 5 to explain why log5 100 is between 2 and 3. Solution We know that 52 = 25 and 53 = 125, so 52 < 100 < 53 . Hence taking logarithms base 5, 2 < log5 100 < 3. Note: The change-of-base formula in the next section will allow log2 10 to be approximated on a calculator. Why can’t the base a of a logarithm be 1? We restricted discussion of powers to bases a > 0. With logarithms, we must also exclude a = 1, because all powers of 1 are 1. For example, log1 4 would be undefined because 1 x = 4 has no solutions. And log1 1 could be any number. Exercise 9C FOUNDATION Note: Do not use a calculator in this exercise unless the question asks for it. Make sure that you can refer easily to the list of powers of 2, 3, 5, . . . from Exercise 8A. 1 Copy and complete each sentence. a ‘log2 8 = 3 because . . . ’ b ‘log5 25 = 2 because . . . ’ c ‘log10 1000 = 3 because . . . ’ d ‘72 = 49, so log7 49 = . . .’ e ‘34 = 81, so . . . ’ f ‘105 = 100 000, so . . . ’ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9C Logarithms 2 Copy and complete the following statements of the meaning of logarithms. b ‘y = a x means that . . . ’ a ‘y = loga x means that . . . ’ 3 4 5 6 355 Rewrite each equation in index form, then solve it for x. a x = log10 1000 b x = log10 10 c x = log10 1 1 d x = log10 100 e x = log3 9 f x = log5 125 g x = log2 64 h x = log4 64 i x = log8 64 j 1 m x = log6 36 n x = log7 17 1 x = log4 64 k o 1 x = log12 12 1 x = log8 64 1 l x = log11 121 1 p x = log2 64 Rewrite each equation in index form, then solve it for x. a log7 x = 2 b log5 x = 3 c log2 x = 5 d log100 x = 3 e log7 x = 1 f log11 x = 0 g log13 x = −1 h log12 x = −2 i log5 x = −3 j log7 x = −3 k log2 x = −5 l log3 x = −4 Rewrite each equation in index form, then solve for x. Remember that a base must be positive and not equal to 1. a log x 49 = 2 b log x 8 = 3 c log x 27 = 3 d log x 10 000 = 4 e log x 10 000 = 2 f log x 64 = 6 g log x 64 = 2 h log x 125 = 1 i log x 11 = 1 1 = −1 j log x 17 1 m log x = −2 9 1 p log x = −2 81 1 k log x = −1 6 1 n log x = −2 49 1 = −1 7 1 o log x = −3 8 l log x Use the calculator buttons log or 10x to approximate each expression. Give answers correct to three significant figures, then check each answer using the other button. a log10 2 b log10 20 c 100.301 d 101.301 e 100.5 f 101.5 g log10 3.16 h log10 31.6 i log10 0.7 j log10 0.007 k 10 −0.155 l 10−2.15 DEVELOPMENT 7 Rewrite each equation in index form and then solve for x (where a is a constant). a x = loga a b loga x = 1 c log x a = 1 1 a g x = loga 1 e loga x = −1 f log x d x = loga 8 9 1 = −1 a i log x 1 = 0 h loga x = 0 Given that a is a positive real number not equal to 1, evaluate: 1 a loga a b loga c loga a3 a √ 1 1 e loga 5 f loga a g loga √ a a d loga 1 a2 h loga 1 Use the list of powers of 2 prepared in Exercise 8A to find which two integers each expression lies between. a log2 3 e log2 50 b log2 7 c log2 1.8 d log2 13 f log2 1000 4 g log2 5 h log2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 356 9C Chapter 9 Exponential and logarithmic functions 10 11 12 Use the tables of powers of 2, 3, 5, . . . prepared in Exercise 8A to find which two integers each expression lies between. a log10 35 b log10 6 c log10 2000 d log3 2 e log3 50 f log3 100 g log5 100 h log7 20 i log10 0.4 j log10 0.05 Solve each equation for x. These questions involve fractional indices. √ √ 1 a x = log7 7 b x = log11 11 c log9 x = 2 √3 1 1 e log x 3 = f log x 13 = g x = log6 6 2 2 1 1 1 i log64 x = j log16 x = k log x 2 = 3 4 3 1 m x = log8 2 n x = log125 5 o log7 x = 2 1 1 1 1 1 q log x = − r log x s x = log4 =− 7 2 20 2 2 1 1 1 1 u log121 x = − v log81 x = − w log x = − 2 4 2 4 d log144 x = 1 2 h x = log9 3 l log x 2 = 1 6 1 2 1 t x = log27 3 1 x log x 2 = − 4 p log7 x = − Problems arise when a logarithm is written down with too few significant figures. a Find log10 45 correct to two significant figures. b Then find 10 to that index, again correct to two significant figures. c To how many significant figures should log10 45 be recorded for the inverse procedure to yield 45.00, that is, to be correct to four significant figures? ENRICHMENT 13 A googol is 10100 — the company name ‘Google’ was an accidental misspelling of ‘googol’. A googolplex is 10googol . A googol is about 100 billion times greater than the number of elementary particles in the observable universe, and a googolplex is very big. a Find: i log10 googol ii log100 googol iii log10 googolplex iv log100 googolplex v loggoogol googol vi loggoogol googolplex b Which whole numbers does log2 googol lie between? Do not use log or ln . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9D The laws for logarithms 357 9D The laws for logarithms Learning intentions • Use the fact that a log function is the inverse function of an exponential function to solve problems. • Use the log laws in expressions and equations. Logarithms are indices, so the laws for manipulating indices can be rewritten as laws for manipulating logarithms. As always, the base a of a logarithm must be a positive number not equal to 1. Logarithmic and exponential functions are inverse functions We have already seen that the functions log and 10x are inverse functions. That is, when the two functions are applied to a number one after the other, in any order, the original number returns. Check, for example, using the buttons log and 10x on your calculator, that log10 103 = 3 and 10log10 3 = 3. These relationships are true whatever the base of the logarithm, simply because of the definition of logarithms. For example: log2 26 = 6, because 6 is the index, when 26 is written as a power of 2. 2log2 64 = 64, because log2 64 is the index, when 64 is written as a power of 2. The exponential and logarithmic functions to any base a are therefore called inverse functions. 11 The functions y = a x and y = log a x are inverse functions Let the base a be any positive real number not equal to 1. Then • loga a x = x, • aloga x = x, Example 16 for all real x, and for all real x > 0. Simplify expressions using the inverse property a Simplify log7 712 . b Simplify 5log5 11 . Solution a Using the first identity above, 12 log7 7 = 12 Example 17 b Using the second identity above, 5log5 11 = 11 Write any number as power or as a logarithm a Write 3 as a power of 10. b Write 7 as a logarithm base 2. Solution a Using the second identity above, 3 = 10 log10 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b Using the first identity above, 7 = log2 27 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 358 Chapter 9 Exponential and logarithmic functions 9D Three laws for logarithms Three laws are particularly useful when working with logarithms: 12 Three laws for logarithms • The log of a product is the sum of the logs: loga xy = loga x + loga y • The log of a quotient is the difference of the logs: x loga = loga x − loga y y • The log of a power is the multiple of the log: loga xn = n loga x The simple reason for these three laws is that logarithms are indices, and these laws therefore mirror the index laws. • When multiplying powers, add the indices. • When dividing powers, subtract the indices. • And when raising a power to a power, multiply the indices. Here is a more formal proof. Proof To prove each law, show that aLHS = aRHS , by using the second identity in the previous paragraph and the index laws for powers. aloga x + loga y = aloga x × aloga y (to multiply powers, add the indices) = xy = aloga (xy) aloga x − loga y = aloga x ÷ aloga y (to divide powers, subtract the indices) = x÷y = yx x = aloga y n an(loga x) = aloga x (to raise a power to a power, multiply the indices) = xn n = aloga (x ) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9D The laws for logarithms Example 18 359 Working with approximations to logarithms Suppose that it has been found that log3 2 0.63 and log3 5 1.46. Use the log laws to find approximations for: b log3 25 a log3 10 c log3 32 d log3 18 Solution a Because 10 = 2 × 5, b log3 25 = log3 2 − log3 5 log3 10 = log3 2 + log3 5 0.63 − 1.46 0.63 + 1.46 −0.83. 2.09. c Because 32 = 25 , d Because 18 = 2 × 32 , log3 32 = 5 log3 2 log3 18 = log3 2 + 2 log3 3. 3.15. = log3 2 + 2, because log3 3 = 1, 2.63. Some particular values of logarithmic functions Some particular values of logarithmic functions occur very often, and are worth committing to memory. 13 Some particular values and identities of logarithmic functions loga 1 = 0, because 1 = a0 . loga a = 1, √ loga a = 12 , 1 loga = −1, a 1 loga = − loga x, x Example 19 because a = a1 . √ 1 because a = a 2 . 1 because = a−1 . a 1 because loga = loga x−1 = − loga x. x Simplifying expressions involving logarithms Use the log laws to expand: a log3 7x3 b log5 5 x Solution a log3 7x3 = log3 7 + log3 x3 = log3 7 + 3 log3 x b log5 5 = log5 5 − log5 x x = 1 − log5 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (the log of a product is the sum of the logs) (the log of a power is the multiple of the log) (the log of a quotient is the difference of the logs) (log5 5 = 1, because 5 = 51 ) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 360 9D Chapter 9 Exponential and logarithmic functions Exercise 9D FOUNDATION Note: Do not use a calculator in this exercise. Make sure that you can refer easily to the list of powers of 2, 3, 5, . . . from Exercise 8A. 1 2 Use the three log laws in Box 12 to evaluate (each answer is a whole number): a log6 9 + log6 4 b log5 75 − log5 3 c log15 3 + log15 5 d log12 72 + log12 2 e log10 50 + log10 20 f log3 15 − log3 5 g log2 24 − log2 3 h log3 810 − log3 10 i log2 12 + log2 6 − log2 9 j log30 2 + log30 3 + log30 5 k log12 9 + log12 8 + log12 2 l log2 12 + log2 6 − log2 72 b log2 6 − log2 48 c log3 8 − log3 6 − log3 12 Use the log laws to simplify: a log5 2 − log5 50 d log5 12 − log5 20 − log5 15 3 e loga 1 8 e log2 18 e log2 3 2 c loga 64 1 32 g loga 2 f loga √ 1 2 1 h loga √ 2 d loga b log2 25 c log2 6 d log2 10 f log2 20 2 g log2 3 h log2 2 1 2 b log2 9 f log2 3 5 c log2 10 d log2 50 2 3 h log2 75 g log2 Use the identity loga a x = x to simplify: a log10 103 7 b loga 16 Given that log2 3 1.58 and log2 5 2.32, use the log laws to find approximations for: a log2 15 6 f log7 19 + log7 63 Express each logarithm in terms of log2 3 and log2 5. Remember that log2 2 = 1. a log2 9 5 e Use the log law loga xn = n loga x to write each expression in terms of loga 2. a loga 8 4 log6 13 − log6 12 b log7 75 c log12 121.3 d log8 8n c 4log4 3.6 d 6log6 y Use the identity aloga x = x to simplify: a 10log10 100 b 3log3 7 DEVELOPMENT 8 9 Use the log law loga xn = n loga x and the identity loga a = 1 to simplify: 1 a loga a2 b 5 loga a3 c loga a Use the log law loga xn = n loga x to write each expression in terms of loga x. √ 1 a loga x3 b loga c loga x x 1 3 5 e loga x − loga x f loga x4 + loga 2 g 2 loga a4 − loga x8 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 √ d 12 loga a 1 x2 √ 1 h loga √ + 3 loga x x d loga © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9D The laws for logarithms 10 11 Write these expressions in terms of loga x, loga y and loga z. z b loga c loga y4 a loga yz y x2 y √ e loga xy3 f loga 3 g loga y z 1 x2 √ h loga xz d loga Using the identities x = aloga x and x = loga a x , express: a 10 as a power of 3 b 3 as a power of 10 c 0.1 as a power of 2 d 2 as a logarithm base 10 e −4 as a logarithm base 3 f g 8 as a power of a h 15 as a logarithm base a 1 as a logarithm base 7 2 i y as a logarithm base 5 12 Given that log10 2 0.30 and log10 3 0.48, use the fact that log10 5 = log10 10 − log10 2 to find an approximation for log10 5. Then use the log laws to find approximations for: √ a log10 20 b log10 0.2 c log10 360 d log10 2 √ √ 1 1 e log10 8 f log10 √ g log10 12 h log10 √ 10 5 13 If x = loga 2, y = loga 3 and z = loga 5, simplify: 1 a loga 64 b loga 30 18 e loga 1.5 f loga 25a 14 15 16 361 100 a 8 h loga 15a2 c loga 27a5 d loga g loga 0.04 Give exact values of: a log25 45 − log25 9 b log81 12 − log81 36 c log8 3 − log8 96 d log 1 5 − log 1 10 32 32 Simplify these expressions. a 5− log5 2 b 122 log12 7 c 2log2 3+log2 5 d an loga x e 7− log7 x f 5 x+log5 x g 2 x log2 x h 3 x log3 x Rewrite these relations in index form (that is, without using logarithms). a loga (x + y) = loga x + loga y b log10 x = 3 + log10 y c log3 x = 4 log3 y d 2 log2 x + 3 log2 y − 4 log2 z = 0 e x loga 2 = loga y f loga x − loga y = n loga z g 1 1 2 log2 x = 3 log2 y − 1 h 2 log3 (2x + 1) = 3 log3 (2x − 1) ENRICHMENT 17 Prove by contradiction that log2 3 is irrational. Begin with the sentence, ‘Suppose by way of contradiction a that log2 3 = , where a and b are positive whole numbers.’ b CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 362 9E Chapter 9 Exponential and logarithmic functions 9E Equations involving logarithms and indices Learning intentions • Apply the change-of-base formula. • Solve index and logarithmic equations to any base. • Solve exponential and logarithmic inequations. The calculator allows approximations to logarithms base 10. This section explains how to obtain approximations to any other base. Index equations can then be solved approximately, whatever the base. As always, every base must be positive and not equal to 1. The change-of-base formula Finding approximations of logarithms to bases not on the calculator requires a formula that converts logarithms from one base to another. 14 The change-of-base formula Let the bases a and b be positive real numbers not equal to 1. • To write log base a in terms of log base b, use the formula logb x . loga x = logb a Remember this as ‘the log of the number over the log of the base’. • The calculator button log gives approximations for logs base 10, so to find logarithms to another base a, put b = 10 and use the formula log10 x loga x = . log10 a y = loga x. Proof To prove this formula, let x = ay Then by the definition of logs, and taking logs base b of both sides, logb x = logb ay . Using the log laws, logb x = y logb a logb x y= , logb a and rearranging, Example 20 as required. Finding approximations of logs to other bases Find, correct to four significant figures: a log2 5 b log3 0.02. Then check your approximations using the button labelled xy or ∧ . Solution In each part, use the change-of-base-formula to change to logs base 10. log10 5 log10 0.02 a log2 5 = b log3 0.02 = log10 2 log10 3 2.322 Checking, 22.322 5. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 −3.561 Checking, 3−3.561 0.02. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9E Equations involving logarithms and indices 363 Solving index equations An index equation such as 5 x = 18 is solved, in exact form, by rewriting it in terms of logarithms, then changing to base 10 allows approximation by calculator: 5 x = 18 x = log5 18 log10 18 = log10 5 (the exact solution) 1.796 (correct to four significant figures). Example 21 Solving index equations Solve these equations correct to four significant figures. Then check your approximations using the button labelled xy or ∧ . a 2x = 7 b 3 x = 0.05 Solution 2x = 7 a Solving for x, Changing to base 10, 3 x = 0.05 b x = log2 7 log10 7 x= log10 2 Solving for x, Changing to base 10, 2.807 Checking, 2 Example 22 2.807 −2.727 7. Checking, 3 x = log3 0.05 log10 0.05 x= log10 3 −2.727 0.05. A problem with rabbits The rabbit population on Kanin Island is doubling every year. A few years ago, there were 500 rabbits there. How many years later will the rabbit population be 10 000? Answer in exact form, then correct to the nearest tenth of a year. Solution Let P be the rabbit population n years after the first estimate. Then Put P = 10 000, then P = 500 × 2n . 10 000 = 500 × 2n 2n = 20 n = log2 20 log10 20 = log10 2 (the exact answer) 4.3 years CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (correct to 0.1 years). © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 364 9E Chapter 9 Exponential and logarithmic functions Exponential equations reducible to quadratics Some exponential equations can be reduced to quadratics with a substitution. Example 23 Solving an exponential equation using a substitution Use the substitution u = 2 x to solve the equation 4 x − 7 × 2 x + 12 = 0. Solution Writing 4 x = (2 x )2 , the equation becomes and returning to x, 2 x = 4 or 2 x = 3 (2 x )2 − 7 × 2 x + 12 = 0. Substitute u = 2 x , x = 2 or log2 3. u2 − 7u + 12 = 0 (u − 4)(u − 3) = 0 u = 4 or 3, Solving exponential and logarithmic inequations Provided that the base a is greater than 1, the exponential function y = a x and the logarithmic function y = loga x both increase as x increases, so there are no difficulties solving inequations such as 2 x ≤ 64 and log10 x > 4. Example 24 Solving exponential and logarithmic inequations Solve: a 2 x ≤ 64 b log10 x ≥ 4 Solution a 2 x ≤ 64 b log10 x ≥ 4 2 ≤2 x ≥ 104 x≤6 x ≥ 10 000 x 6 Taking logarithms of both sides When an approximate solution of an index equation is required, it is often quicker to take the logarithm base 10 of both sides. This is particularly useful when powers with different bases are involved. Example 25 Taking logarithms of both sides Solve these equations correct to four significant figures by taking the logarithm base 10 of both sides. a 53x = 3 × 5 x+1 b 23x = 3 x+1 Solution a 53x = 3 × 5 x+1 b Taking log base 10 of both sides, log10 5 = log10 3 + log10 5 3x x+1 3x log10 5 = log10 3 + (x + 1) log10 5 2x log10 5 = log10 3 + log10 5 log10 3 + log10 5 x= 2 log10 5 0.8413 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 23x = 3 x+1 Taking log base 10 of both sides, 3x log10 2 = (x + 1) log10 3 x(3 log10 2 − log10 3) = log10 3 log10 3 x= 3 log10 2 − log10 3 1.1201. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9E Equations involving logarithms and indices 365 Making both sides the indices of powers with the same base This is particularly useful in equations involving logarithms with the same base. Example 26 Both sides become indices of powers with the same base Solve log3 x − log3 (x − 6) = 1 Solution Using the base 3, and writing LHS and RHS of the equation as indices, 3log3 x−log3 (x−6) = 31 and using the log laws, From the log identities, 3log3 x ÷ 3log3 (x−6) = 3 x =3 x−6 x = 3x − 18 x = 9. Exercise 9E 1 Use the change-of-base formula, loga x = a log2 8 = 3 2 3 FOUNDATION log10 x , and your calculator, to verify that: log10 a b log3 81 = 4 c log100 10 000 = 2 log10 x , to approximate these logarithms, correct to four log10 a significant figures. Check each answer using the button labelled xy or ∧ . Use the change-of-base formula, loga x = a log2 7 b log2 26 c log2 0.07 d log3 4690 e log5 2 f log7 31 g log6 3 h log12 2 i log3 0.1 j log3 0.0004 k log11 1000 l log0.8 0.2 m log0.03 0.89 n log0.99 0.003 o log0.99 1000 Rewrite each equation with x as the subject, using logarithms. Then use the change-of-base formula to solve it, giving your answers correct to four significant figures. Check each answer using the button labelled xy or ∧ . a 2 x = 15 b 2x = 5 c 2 x = 1.45 d 2 x = 0.1 e 2 x = 0.0007 f 3 x = 10 g 3 x = 0.01 h 5 x = 10 i 12 x = 150 7 9 m 0.7 x = 0.1 k 6 x = 1.4 l 30 x = 2 n 0.98 x = 0.03 o 0.99 x = 0.01 j 8x = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 366 9E Chapter 9 Exponential and logarithmic functions 4 Give exact solutions to these inequations. Do not use a calculator. a 2 x > 32 b 2 x ≤ 32 c 2 x < 64 e 5x > 5 f 4x ≤ 1 g 2x < d 3 x ≥ 81 1 2 h 10 x ≤ 0.001 DEVELOPMENT 5 a Use the substitution u = 2 x to solve 4 x − 9 × 2 x + 14 = 0. b Use the substitution u = 3 x to solve 32x − 8 × 3 x − 9 = 0. c Use similar substitutions to solve: 6 7 8 i 25 x − 26 × 5 x + 25 = 0 ii 9 x − 5 × 3 x + 4 = 0 iii 32x − 3 x − 20 = 0 iv 72x + 7 x + 1 = 0 v 22x − 64 = 0 vi 2 x − 3 × 2 2 x + 2 = 0 1 Give exact solutions to these inequations. Do not use a calculator. a log2 x < 3 b log2 x ≥ 3 c log10 x > 3 d log10 x ≥ 1 e log5 x > 0 f log6 x < 1 g log5 x ≤ 3 h log6 x > 2 Rewrite each inequation in terms of logarithms, with x as the subject. Then use the change-of-base formula to solve it, giving your answer correct to three significant figures. a 2 x > 12 b 2 x < 100 c 2 x < 0.02 d 2 x > 0.1 e 5 x < 100 f 3 x < 0.007 g 1.2 x > 10 h 1.001 x > 100 A few years ago, Rahul and Fiona were interested in building a new garage, whose price is rising with inflation at 5% per annum. Its price then was $12 000. a Explain why the cost C in dollars is C = 12 000 × 1.05n , where n is the years since. b Substitute into the formula to find, correct to the nearest 0.1 years, how many years later the cost will be $18 000. 9 Use the change-of-base formula loga x = a log8 x = 10 1 log2 x 3 b logan x = 1 loga x n Solve these equation and inequations, correct to three significant figures if necessary. a 2 x+1 = 16 x−3 1 e =8 2 11 logb x to prove that: logb a b 32x = 81 f 1 10 5x = 1 10 1 c 5 x−1 < 1 d 10 3 x ≤ 1 000 g 2 x−2 < 7 h 3 x+5 > 10 Solve these equations by making both sides indices of powers with the same base. Check your solutions to make sure that LHS and RHS are defined when they are substituted. a log2 (x + 1) = log2 (11 − x) + 1 b log3 (x + 3) + log3 (x − 3) = log3 (4x + 3) c log2 x + log2 (x − 3) − log2 (x + 2) = 2 12 a Solve 2 x < 1010 . How many positive integer powers of 2 are less than 1010 ? b Solve 3 x < 1050 . How many positive integer powers of 3 are less than 1050 ? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9E Equations involving logarithms and indices 13 367 a Explain why, if a > 1, then loga x is: ii negative for 0 < x < 1. i positive for x > 1, b Explain why, if 0 < a < 1, then loga x is: ii positive for 0 < x < 1. i negative for x > 1, c Explain why loga x is never defined for: ii x = 0. i x < 0, d Explain why the base of logarithms ii cannot be 1. i cannot be 0, 14 Use a substitution such as u = 4 x to solve each equation. Give each solution as a rational number, or approximate to three decimal places. 2x x a 24x − 7 × 22x + 12 = 0 b 100 x − 10 x − 1 = 0 c 15 − 7 × 15 + 10 = 0 15 By taking logarithms base 10 of both sides, solve each index equation correct to four significant figures. 16 a 3 x−4 = 47 b 2 x−5 = 5 × 22x c 52x = 6 x+1 d 71−x = 6 × 5 x−3 a Explain why log10 300 lies between 2 and 3. b If x is a two-digit number, what two integers does log10 x lie between? c If log10 x = 4.7, how many digits does x have to the left of the decimal point? d Find log10 2520 , and hence find the number of digits in 2520 . e Find log10 21000 , and hence find the number of digits in 21000 . ENRICHMENT 17 Let S = 12 (2 x + 2−x ) and D = 12 (2 x − 2−x ). a Simplify S D, S + D, S − D and S 2 − D2 . b Rewrite the formulae for S and D as quadratic equations in 2 x . Hence express x in terms of S , and in terms of D, in the case where x > 1. 1+y , where y = DS −1 . c Show that x = 12 log2 1−y CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 368 9F Chapter 9 Exponential and logarithmic functions 9F Exponential and logarithmic graphs Learning intentions • Sketch and describe graphs of exponential and logarithmic functions. • Apply the standard transformations to exponential and logarithmic graphs. • Develop the interaction of transformations with the index laws and log laws. The function y = a x is an exponential function, because the variable x is in the exponent or index. The function y = loga x is a logarithmic function. The graphs of y = 2 x and y = log2 x These two graphs will demonstrate the characteristic features of all exponential and logarithmic graphs. Here are their tables of values: y = 2x y = log2 x y 2 y = 2x 1 x −2 −1 0 1 2 y 1 4 1 2 1 2 4 1 2 x 1 4 1 2 1 2 4 y −2 −1 y = log2 x 0 1 2 x Here are some of the properties of the two graphs: • The two graphs are reflections of each other in the diagonal line y = x. This is because they are inverse functions of each other, so their tables of values are the same, except that the x-values and y-values have been swapped. • For y = 2 x , the domain is all real x, also written as (−∞, ∞), and the range is y > 0, also written as (0, ∞). For y = log2 x, the domain is x > 0, also written as (0, ∞), and the range is all real y, also written as (−∞, ∞). • For y = 2 x , the x-axis is a horizontal asymptote — as x → −∞, y → 0. For y = log2 x, the y-axis is a vertical asymptote — as x → 0+ , y → −∞. Notation: The symbol x → 0+ here means, ‘As x approaches 0 from the right’. Elsewhere, we will use symbol x → 0− to mean, ‘As x approaches 0 from the left’. • As x → ∞, 2 x → ∞, and y = 2 x gets steeper all the time. As x → ∞, log2 x → ∞, but y = log2 x gets flatter all the time. • The graph of y = 2 x is concave up, but y = log2 x is concave down. 15 Exponential and logarithmic functions are inverse functions Let the base a be any positive number not equal to 1. • The functions y = a x and y = loga x are inverse functions, meaning that loga (a x ) = x and aloga x = x. • The graphs of y = a x and y = loga x are reflections of each other in the diagonal line y = x. • The domains, the asymptotes, the steepness, and the concavity, of the two graphs are clearly seen from their graphs and this reflection property. • We find base 2 logarithms by reading the graph of y = 2 x backwards. Input the number on the y-axis, go horizontally across to the graph of y = 2 x , go vertically down to the x-axis, and read the output off the x-axis. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9F Exponential and logarithmic graphs 369 Reflections of exponential and logarithmic functions Shifting, reflecting and dilating apply to exponential and logarithmic functions. Transformations of some exponential functions were done earlier, but logarithmic functions were only introduced in this chapter. Example 27 Reflections in the x-axis and y-axis For each pair of functions: i Draw up tables of values. ii Sketch both functions on one set of axes. iii Explain what transformation transforms one graph onto the other, and why. iV State their domain and range. a y = 3 x and y = 3−x b y = 3 x and y = −3 x Solution a i x −2 −1 0 1 2 3x 1 9 1 3 1 3 9 3−x 9 3 1 1 3 1 9 3 iii The graphs are reflections of each other in the i x −2 −1 0 1 2 3x 1 9 − 19 1 3 − 13 1 3 9 −3 x x 1 y 3 −1 −3 −9 y = 3-x 1 3 -1 ii y = 3x 1 iii The graphs are reflections of each other in the x-axis, because y has been replaced by −y. iv Both functions have domain all real x. For y = 3 x , the range is y > 0. For y = −3 x , the range is y < 0. 1 y = 3x y-axis, because x has been replaced by −x. iv For both functions, the domain is all real x, and the range is y > 0. b y ii 1 3 -1 - 31 1 x -1 -3 y = -3x Exponential functions with bases between 0 and 1 are reflections In the introduction, the mould on the cheese was doubling every day, so its growth is proportional to the function y = 2 x . But the decay of radioactive element is specified by its half-life — for example, approximately half a sample of thorium-231 decays every day — so its decay is proportional to the exponential function y = ( 12 ) x , whose base is between 0 and 1. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 370 9F Chapter 9 Exponential and logarithmic functions x The surprising result here is that the decay function y = 12 and the growth function y = 2 x are reflections of each other in the y-axis, by successive applications of the index laws: x is the same as y = 21x , which is the same as y = 2−x . y = 12 Because of this reflection, we can see immediately that the domains and ranges of the two functions are the same, and the x-axis is horizontal asymptote for both. But for 1 x y = 2 , the limiting behaviours as x → −∞ and as x → ∞ are the reverse of y = 2 x . Logarithmic functions with bases between 0 and 1 are reflections Now let us look at the two logarithmic functions y = log 1 x and y = log2 x. We start this time by applying the log 2 laws to y = log 1 x: 2 y = log 1 x The original function is 2 y= using the change-of-base formula, writing the denominator as a power of 2, using the definition of logs, so y = log 1 x is the same as 2 log2 x log2 12 log2 x y= log2 2−1 log2 x y= −1 y = − log2 x. The result is that this time y = log 1 x is the reflection of y = log2 x in the x-axis rather than the y-axis. The reader 2 may have expected this because of the symmetry of the inverse functions y = 2 x and y = log2 x in the line y = x. Once again, the domains and ranges of the two functions are the same, and now the y-axis is a horizontal asymptote for both. And again, for y = log 1 x, the limiting behaviours as x → 0+ and as x → ∞ are the reverse of 2 y = log2 x. 16 A function with base between 0 and 1 can be understood as a reflection Let a > 1 be a base greater than 1. Exponential functions: x • The function y = 1a is y = a−x , so is y = a x reflected in the y-axis. x Thus y = 1a has the same domain, range, and asymptotes as y = a x , but its behaviours as x → −∞ and as x → ∞ are the reverse of y = a x . Logarithmic functions: • The function y = log 1 is y = − loga x, so is y = loga x reflected in the x-axis. a Thus y = log 1 has the same domain, range, and asymptotes as loga x, but its behaviours as x → 0+ a and as x → ∞ are the reverse of y = a x . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9F Exponential and logarithmic graphs 371 Translations of exponential and logarithmic functions Translations of exponential functions are in Section 5A, but not log functions. Example 28 Translations horizontally and vertically Repeat the steps of Example 27 and state the asymptotes for: a y = log2 x and y = log2 (x − 1) b y = log2 x and y = 3 + log2 x Solution a i 1 4 x 1 2 ii 1 2 4 log2 x −2 −1 0 1 2 x 1 14 1 12 y 2 y = log2 x 1 1 2 2 3 5 1 21 1 log2 (x − 1) −2 −1 0 1 2 2 3 4 y = log2(x - 1) -1 iii The second graph is the first shifted right by 5 x 1 unit, because x has been replaced by x − 1. iv Both functions have range all real y. For y = log2 x, the domain is x > 0, and the y-axis is a vertical asymptote. For y = log2 (x − 1), the domain: x > 1, (and x = 1 is a vertical asymptote). b i x log2 x 3 + log2 x 1 4 1 2 ii 1 2 4 −2 −1 0 1 2 1 2 y 5 4 3 4 5 iii The second graph is the first shifted up by 3 units, because y has been replaced by y − 3 (or because 3 has been added to each value of y). iv Both have domain x > 0, and range all real y. y = log2 x + 3 3 2 y = log2 x 1 1 2 1 2 4 x -1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 372 9F Chapter 9 Exponential and logarithmic functions Dilations of exponential and logarithmic functions As with translations, we did not cover log functions when discussing dilations in Chapter 5. Example 29 Dilations horizontally and vertically Repeat the steps of Example 27, and state the asymptotes, for: a y = log2 x and y = log2 4x b y = log2 x and y = 4 log2 x Solution a i x log2 x 1 8 1 4 1 2 1 2 4 ii −3 −2 −1 0 1 2 log2 4x −1 0 1 2 3 4 iii The second is the first dilated horizontally factor 14 , because x has been replaced by 4x = x/ 14 . iv Both functions have domain x > 0, and range all real y. Both functions have the y-axis as an asymptote. b i x 1 8 1 4 1 2 log2 x −3 −2 −1 0 1 2 1 2 4 ii 4 log2 x −12 −8 −4 0 4 8 iii The second is the first dilated vertically factor 4, because y has been replaced by y/4. iv Both functions have domain x > 0, and range all real y. Both functions have the y-axis as an asymptote. Dilations with negative factors When a dilation with a negative factor is applied to an exponential or log function, it is usually better to regard it as the composition of a dilation with positive factor, and a reflection — in either order. In particular, the behaviour of the transformed function near asymptotes and for large positive and negative values of x often reverses with a reflection, and the situation is usually far more straightforward to follow if it is done in two steps. The following worked example should be enough to illustrate this point. There are four cases to cover, as is done in the challenging Question 12 in Exercise 9F. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9F Exponential and logarithmic graphs Example 30 373 Applying a dilation with a negative factor a Interpret y = −2 × 3 x as the composition of a dilation and a reflection. b Sketch y = 3 x , y = 2 × 3 x , and y = −2 × 3 x on one set of axes. c Detail the domain and range, the asymptotes, and the behaviour as x → ∞ and as x → −∞, of the three functions. Solution a A vertical dilation with factor 2 transforms b y = 3 x to y = 2 × 3 x . A reflection in the x-axis then transforms y = 2 × 3 x to y = −2 × 3 x . c For all three functions, the x-axis is an asymptote, the domain is all real x, and y → 0 as x → −∞. For y = 3 x and y = 2 × 3 x , the range is y > 0, and y → ∞ as x → ∞. For y = −2 × 3 x , the range is y < 0, and y → −∞ as x → ∞. The word ‘logarithm’ The Scottish mathematician John Napier (1550–1617) formed his new word ‘log | arithm’ from the Greek words ‘logos’, meaning ‘ratio’, and ‘arithmos’, meaning ‘number’. Until the invention of calculators, the routine method for difficult calculations in arithmetic used tables of logarithms, invented by Napier, to convert products to sums, quotients to differences, and powers to multiples. Exercise 9F FOUNDATION Technology: All these graphs can be confirmed and further developed using graphing software, after which experimentation with further graphs can easily be done. 1 a Use the calculator button log to complete the following table of values, giving each entry correct to two significant figures where appropriate. x 0.1 0.25 0.5 0.75 1 2 3 4 5 6 7 8 9 10 log10 x b Hence sketch the graph of y = log10 x. Use a large scale, with the same scale on both axes. Ideally use graph paper so that you can see the shape of the curve. c Copy and complete: i ‘As x → ∞, log10 x → . . .’ CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii ‘As x → 0+ , log10 x → . . .’ © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 374 9F Chapter 9 Exponential and logarithmic functions 2 a Copy and complete these two tables of values. i x 2 −3 −2 −1 0 1 2 3 ii x −3 x 2 −2 −1 0 1 2 3 −x −x b Hence sketch the graphs of y = 2 and y = 2 x on one set of axes. c How are the two tables of values related to each other? d What symmetry does the diagram of the two graphs display, and why? e Write down the domain and range of: ii y = 2−x . i y = 2x f Write down the equations of the asymptotes of: ii y = 2−x . i y = 2x g Copy and complete: i ‘As x → −∞, 2 x → . . .’ ii ‘As x → ∞, 2 x → . . .’ h Copy and complete: i ‘As x → −∞, 2−x → . . .’ 3 ii ‘As x → ∞, 2−x → . . .’ a Copy and complete these two tables of values. i x −2 −1 0 1 2 ii 3x x 1 3 1 9 1 3 9 log3 x b Hence sketch the graphs of y = 3 and y = log3 x on one set of axes. x c How are the two tables of values related to each other? d What symmetry does the diagram of the two graphs display, and why? e Write down the domain and range of: ii y = log3 x. i y = 3x Both domains and both ranges are intervals — use both notations for them. f Write down the equations of the asymptotes of: ii y = log3 x. i y = 3x g Copy and complete: ii ‘As x → 0+ , log3 x → . . .’ i ‘As x → −∞, 3 x → . . .’ DEVELOPMENT 4 a Sketch on one set of axes the graphs of y = 3 x and y = 3−x . b Sketch on one set of axes the graphs of y = 10 x and y = 10−x . 5 Sketch the four graphs below on one set of axes. a y = 2x 6 c y = 2−x b y = −2 x d y = −2−x Sketch each set of graphs on one set of axes, clearly indicating the asymptotes, the y-intercepts, and the x-intercepts if they exist. Use shifting of the graphs in the previous questions, but also use a table of values to confirm your diagram. a y = 2x b y = −2 x c y = 2−x y = 2x + 3 y = 2 − 2x y = 2−x + 1 y = 2x − 1 y = −2 − 2 x y = 2−x − 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9F Exponential and logarithmic graphs 7 a Use reflections in the x- and y-axes to sketch the four graphs below on one set of axes. y = − log2 x y = log2 x b Copy and complete: y = log2 (−x) Use shifting to sketch each set of graphs on one set of axes, clearly indicating the asymptote and the intercepts with the axes. b y = − log2 x a y = log2 x 9 y = − log2 (−x) ii ‘As x → 0− , log2 (−x) → . . .’ i ‘As x → −∞, log2 (−x) → . . .’ 8 375 y = log2 x + 1 y = 2 − log2 x y = log2 x − 1 y = −2 − log2 x a Use dilations to sketch these graphs on the one set of axes: y = 2 × 3x y = 32x y = 3x b Use dilations to sketch these graphs on the one set of axes: y = 2 log3 x y = log3 2x y = log3 x 10 y 3 2 1 −1 1 0 x The diagram above shows the graph of y = 2 x . Use the graph to answer the questions below, giving your answers correct to no more than two decimal places. a Read from the graph the values of: i 22 ii 2−2 iii 21.5 iv 20.4 v 2−0.6 b Find x from the graph if: i 2x = 2 ii 2 x = 3 iii 2 x = 1.2 iv 2 x = 0.4 iii 1.5 ≤ 2 x ≤ 3 iv 0.5 ≤ 2 x ≤ 2 iii log2 1.4 iv log2 0.8 c Find the values of x for which: i 1 ≤ 2x ≤ 4 ii 1 ≤ 2 x ≤ 2 d Read the graph backwards to find: i log2 4 ii log2 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 376 9F Chapter 9 Exponential and logarithmic functions 11 Use the change-of-base formula to show that log 13 x = − log3 x, and hence sketch these graphs on one set of axes: y = 2 log 13 x y = log 13 2x y = log 13 x 12 Each function below is either y = 5 x or y = log5 x transformed by a horizontal or vertical dilation with negative factor. For each of the four functions: i Interpret the function as the original function transformed by a dilation with positive factor, then by a reflection. Give the equation of the middle stage. ii Then sketch all three stages of the transformation on one set of axes. iii From the sketch, describe, for all three stages, the domain and range, the asymptotes, and the behaviour as x → ∞ and as x → −∞. a y = −2 × 5 x 13 14 b y = 5−2x c y = −2 log5 x d y = log5 (−2x) Characterise these graphs as single or as multiple transformations of y = log5 x: a y = 4 log5 x b y = − 23 log5 x c y = log5 (3x) d y = 3 log5 (−x) e y = log5 (5x + 8) f y = −2 log5 (3x − 7) These questions will required the index laws and the log laws. a Describe two different transformations that send y = 3 x to y = 3 x+4 . b Describe two different transformations that send y = log2 x to y = log2 4x. c Describe two different transformations that send y = log2 x to y = log2 7x. d Describe two different transformations that send y = 6 × 5 x to y = 5 x . 15 Sketch, on separate axes, the graphs of: a y = 2 x−2 d y = log2 (x − 1) 16 b y = 2 x+1 e y = 12 (2 x + 2−x ) c y = log2 (x + 1) f y = 12 (2 x − 2−x ) Sketch, on separate axes, the graphs of: a y = | log2 x| |x| d y=2 b y = |2 x | c y = log2 |x| e y = | log2 |x| | f y = |2|x| | ENRICHMENT 17 Look at the graphs of y = 2 x and y = log2 x at the beginning of this section. Let the line y = 2 − x meet the √ curves at A and B. Explain why the distance AB is less than 2. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9G Applications of these functions 377 9G Applications of these functions Learning intentions • Solve practical problems involving exponential and logarithmic functions. Some applications of these functions have been scattered through the preceding exercises — the applications in this exercise are longer and more sustained. The intention of all the questions is to indicate how diverse the applications of exponential and logarithmic functions are in various sciences and technologies. The questions require conversion between a statement in exponential form and a statement in logarithmic form. The pattern for this is 23 = 8 means that 3 = log2 8 . The change-of-base formula is also needed to convert logs to base 10, log10 x . loga x = log10 a The later questions in this exercise are more difficult than in other exercises, but they are included because of their vital interest to readers studying computing, physics, geology and chemistry. Many of them could easily be adapted to projects that would go into the subjects in more detail. Exercise 9G FOUNDATION Technology: Graphing software will display these functions, and give opportunities to answer many further questions about the situations described in the exercise. 1 t A quantity Q is varying over time t according to the formula Q = 5 × 10 2 . Give answers correct to four significant figures. a Find Q when t = 6, and when t = 5.43. b Rewrite the formula with t as the subject. c Find t when Q = 500, and when Q = 256. 2 A quantity Q is is varying over time t according to the formula t = 20 log2 2Q. a Find t when Q = 4, and when Q = 6. b Rewrite the formula with Q as the subject. c Find Q when t = 40, and when t = 45. DEVELOPMENT 3 The population of a country is doubling every 30 years, and was 3 000 000 at the last census. n a Explain why the population P after another n years is P = 3 000 000 × 2 30 . b Find the population (nearest million) after: i 90 years ii 100 years. c By substituting into the equation, find (nearest year) when the population will be: i 48 000 000 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii 60 000 000. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 378 9G Chapter 9 Exponential and logarithmic functions 4 The price of a particular metal has been rising with inflation at 5% per annum, from a base price of $100 per kilogram in 1900. Let $P be the price n years since 1900, so that P = 100 × (1.05)n . a Copy and complete the following table of values, giving values correct to the nearest whole number. n 0 20 40 60 80 100 P 100 b Sketch the graph of the function. c Now copy and complete this table for log10 P, correct to two decimal places. n 0 20 40 60 80 100 log10 P 2 d Draw a graph with n on the horizontal axis and log P on the vertical axis. e Use the log laws to prove that log10 P = 2 + n log10 (1.05), and hence explain the shape of the second graph. 5 [Moore’s law] Gordon Moore predicted in 1965 that over the next few decades, the number of transistors within a computer chip would very roughly double every two years — this is called Moore’s law. Let D0 be the density in 1975. n a Explain why the density D after n more years is predicted to be D = D0 2 2 . b What was the predicted density for 2015? c Substitute into the formula to find, correct to the nearest year, the prediction of the year when the density increases by a factor of 10 000 000. 6 [Newton’s law of cooling] A container of water, originally just below boiling point at 96◦ C, is placed in a fridge whose temperature is 0◦ C. The container is known to cool in such a way that its temperature halves every 20 minutes. 3n a Explain why the temperature T ◦ C after n hours in the fridge is given by the function T = 96 × 12 . b Draw up a table of values of the function and sketch its graph. c What is the temperature after 2 hours? d Write the equation with the time n in hours as the subject. e How long does it take, correct to the nearest minute, for the temperature to fall to 1◦C? 7 [Decay of Uranium-235] Uranium-235 is a naturally-occurring isotope of uranium. It is particularly important because it can be made to fission, when it releases vast amounts of energy in a nuclear power station or a nuclear bomb. It has a half-life of about 700 million years, which means that if you store a mass of uranium-235 for 700 million years, half of it will then have decayed into other elements. a Explain why the mass of uranium-235 in the Earth n years after the present is given by n M = M0 × 12 700 000 000 , where M0 is the mass in the Earth now. b The Andromeda Galaxy will collide with our Milky Way galaxy in about 4 billion years. What percentage of the present uranium-235 will still be present in the Earth? c The Earth itself is about 4.5 billion years old. How many times more uranium-235 was present in the Earth when it was formed? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 9G Applications of these functions 8 379 [Decibels] Noise intensity I is measured in units of watts per square metre (W/m2 ). The more common (absolute) decibel scale for noise intensity uses a reference level of Ir = 10−12 W/m2 , which is the auditory threshold of someone with perfect hearing. The (absolute) decibel level n of noise with intensity I is then defined as the ratio I n = 10 log10 dB. Ir (One decibel is one-tenth of a bel, which explains the initial 10 in the formula.) a What decibel level corresponds to 4 × 10−3 W/m2 (nearest decibel)? b Write the formula with I as the subject, and find what intensity corresponds to 75 decibels (animated conversation). c The sound level rises from 72 dB to 108 dB (rock concert level). By what multiple is the power increased? d What does a decibel level of 70 dB fall to if the power decreases by a factor of 1600? 9 [The Richter scale] Earthquake strengths are usually reported on the Richter scale, which has several minor variants. They are all based on the log base 10 of the ‘shaking amplitude’ of the earthquake wave at some standardised distance from the epicentre. (An earthquake occurs kilometres underground — the epicentre is the point on the surface above the place where the earthquake occurs.) a An earthquake of strength 4.0–4.9 is classified as ‘light’, and will cause only minimal damage. An earthquake of strength 7.0 or above is classified as ‘major’, and will cause damage or total collapse to most buildings. What is the ratio of the shaking amplitudes of the smallest light quake and the smallest major quake? b The energy released by the quake is proportional to the 32 th power of the shaking amplitude. What is the ratio of the energies released by the quakes in part a? c An earthquake of magnitude 9.0 will cause total destruction. What are the ratios of the shaking amplitudes and the energies released of such an earthquake and the smallest light quake? ENRICHMENT 10 [pH of a solution] The pH of a liquid is traditionally defined on a logarithmic scale as pH = − log10 [H+ ], where [H+ ] is the hydrogen ion concentration in units of moles per litre (mol/L). It is not necessary to understand the units, except to know that the greater the concentration of hydrogen ions, the greater the acidity. a Rewrite the formula with [H+ ] as the subject. b Pure water has a pH of about 7. What is its hydrogen ion concentration? c Lemon juice typically has a pH of about 2.0. What is its hydrogen ion concentration? How many times more acidic is it than pure water? d Sea water typically has a pH of about 8.1. What is its hydrogen ion concentration? How many times more alkaline (meaning ‘less acidic’) is it than pure water? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 380 Chapter 9 Exponential and logarithmic functions 11 9G [Keyboard tuning] When a keyboard is tuned to modern concert pitch, the note A below middle C vibrates at 220 Hz (1 hertz is 1 vibration per second), and the note A above middle C vibrates at twice that frequency, which is 440 Hz (these names are not standard, but are convenient for this question). When modern equal temperament is used for the twelve semitones of the scale from A and A , the frequency 1 ratio of each semitone is 2 12 , so that on a logarithmic scale base 10, the log of all the semitones ratios is 1 12 log10 2. a Find, correct to two decimal places, the frequencies of C, C and E, if the interval A C consists of three semitones, A C of four semitones, and A E of seven semitones. b Unfortunately, intervals sound best when their ratios consist of small whole numbers, as most of them are in meantone temperament, the most common temperament from say 1500 to 1800. Using that temperament, find the frequencies of C, C and E, if the interval A C is in the ratio 6 : 5 (meaning frequency of C : frequency of A = 6 : 5), A C is in ratio 5 : 4, and A E is in ratio 3 : 2. c When two frequencies close together are sounded together, there is an ugly beat frequency, which is the difference between the frequencies. Calculate the beat frequencies between the equal-tempered and meantone versions of C, C and E. d Meantone temperament is actually very complicated. Suppose that a careless tuner tuned A at 220 Hz, then tuned three major thirds so that A C was in ratio 5 : 4, C F was in ratio 5 : 4, and F A was in ratio 5 : 4. What would the resulting frequency of A be, and how would it beat with A tuned to 440 Hz on another instrument? e Meantone temperament gets its name from the fact that B between A and C is tuned so that the log of its frequency is the mean of the logs of the frequencies of A and C (B is two semitones above A, and two semitones below C). Find the meantone frequency of B, and the equal-temperament frequency of B. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 9 review 381 Chapter 9 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 9 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise Write each expression as an integer or fraction. a 53 b 28 e 9−2 f 2−3 3 2 i 3 m 27 q 2 3 4 j 1 3 14 59 7 12 n 8 0 r 2 3 9 25 Simplify: 5 a x4 b 3 a3 c 109 d 17−1 g 3−4 h 270 −2 5 k 6 −1 o 9 − 12 s Write each expression in index form. 7 1 a b 2 x x √ 4 e 5 36x f √ x 4 c 1 4 2 p 49 − 32 9 t 100 5 2 2 8 3 27 c − g 1 l 36 2 y x 1 2x 25x d √ x √ h 2y x 6 12 d 1 8r3 3 t6 Simplify each expression, leaving the answer in index form. a x 2 y × y2 x b 15xyz × 4y2 z4 c 3x−2 y × 6xy−3 d 4a−2 bc × a5 b2 c−2 e x3 y ÷ xy3 f 14x−2 y−1 ÷ 7xy−2 g m 2 n− 2 × m1 2 n− 2 h (2st)3 × 3s3 1 5 Review 1 1 1 1 2 i 4x2 y−2 3 3 ÷ 2xy−1 Solve each equation for x. 1 7 a 3 x = 81 b 5 x = 25 c 7x = e f g 25 x = 5 x 1 1 = 3 9 x 8 2 = 3 27 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 d 2x = 1 32 h 8x = 2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 382 Chapter 9 Exponential and logarithmic functions Review 6 Rewrite each logarithmic equation as an index equation and solve for x. b x = log3 9 c x = log10 10 000 d x = log5 1 49 f x = log13 1 g x = log9 3 h x = log2 2 i 2 = log7 x j −1 = log11 x m 2 = log x 36 n 3 = log x 1000 1 = log16 x 2 1 o −1 = log x 7 1 = log27 x 3 1 = log x 4 p 2 e x = log7 7 9 10 11 12 k l Use the log laws and identities to simplify: a log22 2 + log22 11 8 1 5 √ a x = log2 8 b log10 25 + log10 4 c log7 98 − log7 2 1 d log5 6 − log5 150 e log3 54 + log3 6 Write each expression in terms of loga x, loga y and loga z. a loga xyz b loga yx e loga x2 y5 f loga xz2 f log12 d loga z12 c loga x3 y2 g loga √ 2 7 + log12 7 2 x h loga √ xyz Between what two consecutive integers do the following logarithms lie? a log10 34 b log7 100 c log3 90 d log2 35 e log10 0.4 f log10 0.007 g log2 0.1 h log7 0.1 Use a calculator, and the change-of-base formula if necessary, to approximate these logarithms, correct to four significant figures. a log10 215 b log10 0.0045 c log7 50 d log2 1000 e log5 0.215 f log1.01 2 g log0.5 8 h log0.99 0.001 Use logarithms to solve these index equations, correct to four significant figures. a 2 x = 11 b 2 x = 0.04 c 7 x = 350 d 3 x = 0.67 e 1.01 x = 5 f 1.01 x = 0.2 1 g 0.8 x = 10 1 h 0.99 x = 100 a Sketch on one set of axes y = 3 x , y = 3−x , y = −3 x and y = −3−x . b Sketch on one set of axes y = 2 x and y = log2 x. c Sketch on one set of axes y = 3 x , y = 3 x − 1, and y = 3 x + 2, and write down the asymptotes of the three 13 14 graphs. a Sketch on one set of axes y = log5 x, y = log 1 x, and y = log5 (−x). 5 b Sketch on one set of axes y = log5 x, y = log5 (x − 2), and y = log5 (2 − x), and write down the asymptotes of the three graphs.. c Sketch on one set of axes y = log5 x, y = log5 2x, and y = 2 log5 x − 2. Describe a sequence of transformations that will transform: 1 a y = log6 x to y = −4 log6 (3x − 7). 15 b y = 6 x to y = −6 2 x+9 + 10. and then write down the domain and range of the transformed function, in both notations. The number of bacteria growing in a Petri dish is doubling every four hours. Initially there were 100 bacteria. n a Explain why the formula for the population P after n hours is P = 100 × 2 4 . b How many bacteria are there after: ii 13 hours? i 12 hours c Write the formula with n as the subject. d How long, correct to the nearest hour, before there are 10 000 000 bacteria? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10 Differentiation Chapter introduction The next step in studying functions and their graphs is called calculus. Calculus begins with two processes called differentiation and integration. Differentiation looks at the changing steepness of a curve. Integration looks at the areas of regions bounded by curves. Both processes involve taking limits. They were well known to the Greeks, but it was not until the late 17th century that Gottfried Leibniz in Germany and Sir Isaac Newton in England independently gave systematic accounts of them. Leibniz’s notation is in standard use today. This chapter deals with differentiation, and introduces the derivative as the gradient of the tangent to a curve. Integration will be introduced in Year 12. Section 10J on rates of change begins to show how essential calculus is for science, for economics, and for solving practical problems. The proofs of the basic results here are difficult. Some are given in an Enrichment question or in the Appendix at the end of this chapter, and some are omitted. The Enrichment questions and the Appendix could perhaps be tackled during a second reading. The final two sections are an introduction to motion, which is another form of rate. But motion is far more complicated than the rates covered in Section 10J. Readers may like to delay the treatment of motion until much later in the year. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 384 10A Chapter 10 Differentiation 10A Tangents and the derivative Learning intentions • Understand the derivative as the gradient of a tangent. • Understand the derivative as a measure of instantaneous change. • Find the derivatives of linear and constant functions. The formula for a function y = f (x) tells us how to find the value of a quantity y given a value of a quantity x. But throughout science, engineering, and economics, we are constantly asking, ‘How does the quantity y change as x changes?’. Differentiation answers this question very precisely. Its basic approach is to draw a tangent to the curve, and calculate its gradient. This gradient give us a measure of how y is changing as x changes, at the exact point where the tangent is drawn. An example of how secants and tangents are used to measure how a quantity is changing A bottle of water was taken out of a fridge on a hot day when the air temperature was 40◦ C. The graph y = f (x) above shows how the temperature increased over the next 80 minutes. The horizontal axis gives the time x in minutes, and the vertical axis gives the temperature y◦ C. The water temperature was originally 0◦ C (at O), and 20 minutes later it had risen to 20◦ C (at A). Thus during the first 20 minutes, the temperature was rising at an average rate of 1◦ C per minute. This rate is the gradient of the secant OA. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10A Tangents and the derivative 385 Measuring the instantaneous rate of temperature increase, however, requires a tangent to be drawn. The gradient of the tangent is the instantaneous rate of increase at the time x — such gradients are easy to measure on the rise graph paper by counting little divisions and using the formula gradient = . run The gradient of the tangent to the curve y = f (x) at any point is called the derivative, which is written as f (x). Measuring the gradients at the marked points O, A, B and C gives a table of values of the derivative: x 0 20 40 60 f (x) 1.39 0.69 0.35 0.17 This derivative f (x) is a new function. It has a table of values, and it can be sketched, just as the original function was. You will be sketching this derivative in the first question of the following exercise. Geometric definition of the derivative Here is the definition of the derivative expressed in geometric language. 1 The Derivative f (x) defined geometrically f (x) is the gradient of the tangent to y = f (x) at each point on the curve. The gradient of a tangent to a curve is often called the gradient of the curve at the point. Thus the derivative f (x) is also called the gradient function of f (x). The derivaties of linear and constant functions When a graph is a straight line, the tangent at every point on the graph is just the line itself. The diagram on the left below is the line f (x) = mx + b with gradient m. At every point on the line, the tangent is the same line, and so also has gradient m. Hence the derivative is the constant function f (x) = m. y y y = mx + b y=c b c x x In particular, a horizontal straight line has gradient zero, as in the graph of y = c on the right above. The tangent at any point is this same horizontal line. Hence the derivative is the zero function f (x) = 0. 2 The derivative of linear and constant functions • If f (x) = mx + b is a linear function, then f (x) = m is a constant function — it is the gradient of the line. • In particular, if f (x) = c is a constant function, then f (x) = 0 is the zero function. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 386 10A Chapter 10 Differentiation Example 1 Derivatives of linear functions Write down the derivative f (x) of each linear function. 5 − 2x a f (x) = 3x + 2 b f (x) = 3 c f (x) = −4 Solution In each case the derivative is just the gradient of the line: a f (x) = 3 b f (x) = − 23 c f (x) = 0 The derivatives of semicircle functions The two Enrichment questions at the end of Exercise 10A differentiate semicircle functions using geometric methods. Secants, chords, and tangents These three words were introduced in earlier years only for circles, but are now being used in the context of graphs. y A • A secant (meaning ‘cutting’) is the line through two distinct points A and B on secant B chord Q P the graph. • A chord (meaning ‘cord’, such as a bowstring) is the line segment joining two distinct points P and Q on the graph. It is part of the secant PQ, and has the same gradient. T tangent x • Informally, a tangent (meaning ‘touching’) at a point T on the graph is the line that continues in the direction that the curve is going at the point T — think of the headlights of a car going around a bend. The definition used with circles, ‘meets the curve only at T ’, doesn’t work for graphs. The curve may twist around so that the tangent meets the curve again at one or more other points. Section 10B has a more formal definition. Some preliminary investigations Questions 2–5 of the following exercise consist of four investigations that apply the definition of the derivative directly to several functions. They may be done by photocopying the graphs, or by using graphing software. Section 10B develops these ideas into a more rigorous treatment of the derivative. Exercise 10A 1 INVESTIGATION A question about the graph of water temperature at the start of the chapter. a Look at the graph, and say in words what happens to the temperature of the water bottle as time goes on. b What is the average rate of temperature increase in the first 40 seconds, giving units? c From the table of values of f (x), draw a sketch of the derivative f (x). Label the x-axis Time, and the y-axis Instantaneous rate of temperature increase. d What is the instantaneous rate of temperature increase after 40 seconds, giving units? e From your graph, describe in words the behaviour of the rate of temperature increase. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10A Tangents and the derivative 2 387 [Graph paper] y 4 x2 y 3 2 1 x a Photocopy the sketch above of f (x) = x2 . b At the point P(1, 1), construct the tangent. Place your pencil point on P, bring your ruler to the pencil, then rotate the ruler about P until it seems reasonably like a tangent. rise to measure the gradient of this tangent correct to at most two decimal c Use the definition gradient = run places. Choose the run to be 10 little divisions, and count how many vertical divisions the tangent rises as it runs across the 10 horizontal divisions. d Copy and complete the following table of values of the derivative f (x) by constructing a tangent at each of the nine points on the curve and measuring its gradient. x −2 −1 12 −1 − 12 0 1 2 1 1 12 2 f (x) e On a separate set of axes, use your table of values to sketch the curve y = f (x). f Make a reasonable guess as to what the equation of the derivative f (x) is. 3 [The same investigation using technology] a Use graphing software to sketch the graph of y = x2 . Use the same scale on both axes so that gradients are represented correctly. The values on the y-axis should run from at least y = 0 to y = 4, and on the x-axis from at least x = −2 to x = 2. b Construct the tangent at the point P(1, 1). c Find the gradient of this tangent, either by asking the software, or by producing a graph-paper background and counting little squares as in Question 1. d Copy and complete the table of values of f (x) as in Question 1d. e On a new set of axes, plot these values of f (x), and join them up to form the graph of y = f (x). f Make a reasonable guess as to what the equation of the derivative f (x) is. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 388 10A Chapter 10 Differentiation 4 [Graph paper, but easily adapted to technology] a Photocopy the cubic graph above. b Let P and Q be the points on the curve with x-ccordinates −4 and 4. Find the gradients of the secants OP and OQ, and draw a conclusion about the three points. c The tangent at the origin O(0, 0) has been drawn — notice that it crosses the curve at the origin. Use the rise to measure the gradient of this tangent. definition gradient = run d Copy and complete the following table of values of the derivative f (x) by constructing a tangent at each of the seven points on the curve and measuring its gradient. x −3 −2 −1 0 1 2 3 f (x) e On a separate set of axes, use your table of values to sketch the curve y = f (x). 5 [Technology, but easily done on graph paper] a Use graphing software to sketch the graph of y = x3 , using the same scale on both axes. The values on the y-axis should run from just y = −1 to y = 1, and on the x-axis from x = −1 to x = 1. (Larger values of x will send the graph off the screen.) b Using the software to construct tangents and calculate their gradients, copy and complete the following table of values of the derivative f (x). x −1 −0.9 −0.8 −0.5 0 0.5 0.8 0.9 1 f (x) c On a new set of axes, plot the graph of y = f (x). d Make a reasonable guess as to what the equation of the derivative f (x) is. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10A Tangents and the derivative 389 Conclusions • Either of the first two investigations should demonstrate reasonably clearly that the derivative of the quadratic function f (x) = x2 is the linear function y = 2x. • Either of the last two investigations should demonstrate at least that the derivative of a cubic curve looks very much like a quadratic curve — the last investigation may have indicated that f (x) = x3 has derivative f (x) = 3x2 . DEVELOPMENT 6 Write each function in the form f (x) = mx + b, and hence write down f (x). a f (x) = 2x + 3 b f (x) = 5 − 3x c f (x) = 12 x − 7 d f (x) = −4 e f (x) = ax + b f f (x) = 23 (x + 4) h f (x) = 52 (7 − 43 x) i f (x) = 12 + 13 g f (x) = 3 − 5x 4 7 Write each function in the form f (x) = mx + b, and hence write down the derived function. k − x k + x 5 − 2x 3 + 5x a f (x) = b f (x) = (x + 3)2 − (x − 3)2 c f (x) = − + 2 2 r r 8 Sketch graphs of these functions, draw tangents at the points where x = −2, −1, 0, 1, 2, estimate their gradients, and hence draw a reasonable sketch of the derivative. 1 a f (x) = 4 − x2 b f (x) = c f (x) = 2 x x ENRICHMENT 9 Here is a geometric approach to differentiating a semicircle function. √ 2 Let f (x) = 25 − x be the upper semicircle with centre O and radius 5. Let the point P on the semicircle have x-coordinate x. a Find the gradient of the radius OP. b Use the fact that the tangent at P is perpendicular to the radius at P to find the derivative f (x) of f (x). 10 Use similar methods to find the derivatives of the following. Begin each with a sketch. √ √ √ a f (x) = 1 − x2 b f (x) = − 1 − x2 c f (x) = 4 − x2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 390 10B Chapter 10 Differentiation 10B The derivative as a limit Learning intentions • Understand a tangent as the limit of secants. • Differentiate using the limit formula for the derivative. • Work with tangents and their gradients. • Use a spreadsheet to model the limit definition of the derivative. This section introduces a limiting process to find the gradient of a tangent at a point P on a curve. This will become the formal definition of the derivative. The tangent as the limit of secants For non-linear functions, we look at secants through P that cross the curve again at another point Q near P, and then take the limit as Q moves towards P. The diagram below shows f (x) = x2 and the tangent at P(1, 1) on the curve. Let Q be another point on the curve, and join the secant PQ. Let the x-coordinate of Q be 1 + h, where h 0. y (1 + h)2 Then the y-coordinate of Q is (1 + h) . y2 − y1 Hence gradient PQ = (rise over run) x2 − x 1 (1 + h)2 − 1 = (1 + h) − 1 2h + h2 = h = 2 + h, where we can cancel because h 0. Q 2 1 P 1 (1 + h) x As Q moves along the curve, to the right or left of P, the secant PQ changes. The closer Q is to the point P, the closer the secant PQ is to the tangent at P. The gradient of the secant PQ becomes ‘as close as we like’ to the gradient of the tangent as Q moves sufficiently close to P. This is called ‘taking the limit as Q approaches P’, with notation limQ→P . gradient (tangent at P) = lim (gradient PQ) Q→P = lim (2 + h), because h → 0 as Q → P = 2, because 2 + h → 2 as h → 0. h→0 Thus the tangent at P has gradient 2, which means that f (1) = 2. The point Q cannot coincide with P, because there would then be no line PQ. The point Q may, however, be on the left of P (h < 0) or on the right of P (h > 0). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10B The derivative as a limit 391 The derivative as a limit This process can be applied to any function f (x). Let P x, f (x) be any point on the curve y = f (x). Let Q be any other point on the curve, left or right of P, and let Q have the x-coordinate x + h, where h 0, and y-coordinate f (x + h). y f (x + h) Then using the formula gradient = rise over run, f (x + h) − f (x) (rise over run). gradient of secant PQ = h As h → 0, the point Q moves ‘as close as we like’ to P, and the gradient of secant PQ becomes ‘as close as we like’ to the gradient of the tangent at P. f (x + h) − f (x) , Hence gradient of tangent at P = lim h→0 h f (x + h) − f (x) that is, f (x) = lim . h→0 h f (x) Q P x (x + h) x This last expression is the limit formula for the derivative, and is taken as its definition. 3 The derivative f (x) defined as a limit For each value of x, the derivative is defined as f (x + h) − f (x) f (x) = lim (if the limit exists). h→0 h The limit is two-sided — it involves both positive and negative small values of h. But h cannot be zero, because then the expression is 00 , which is not defined. f (x + h) − f (x) is sometimes called a difference quotient, because the numerator f (x + h) − f (x) is h the difference between the heights at P and Q, and the denominator h is the difference between the x-coordinates of P and Q. Geometrically, the expression is the gradient of the secant PQ. The expression Using the definition of the derivative — first-principles differentiation First-principles differentiation is using the limit formula to find the derivative. Example 2 First-principles differentiation a For f (x) = x2 , use first-principles differentiation to show that f (5) = 10. b i Use first-principles differentiation to find the derivative f (x). ii Substitute x = 5 to confirm that f (5) = 10, as in part a. Solution a For all h 0, f (5 + h) − f (5) (5 + h)2 − 52 = h h 25 + 10h + h2 − 25 = h 10h + h2 = h = 10 + h, because h 0. Taking the limit as h → 0, f (5) = 10. Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 392 10B Chapter 10 Differentiation b For all h 0, f (x + h) − f (x) (x + h)2 − x2 = h h x2 + 2xh + h2 − x2 = h 2xh + h2 = h = 2x + h, because h 0. Taking the limit as h → 0, f (x) = 2x. c Substituting x = 5, f (5) = 10, as established in part a. Finding points on a curve with a given gradient Once the derivative has been found, we can find the points on a curve where the tangent has a particular gradient. 4 Finding points on a curve with a given gradient • To find the points where the tangent has a given gradient of m, solve the equation f (x) = m. • To find the y-coordinates of the points, substitute back into f (x). The points on the curve where the tangent is horizontal are particularly important. Example 3 Finding points with a given gradient Find the point on f (x) = 6x − x2 where the tangent is horizontal, then sketch the curve. Solution Here f (x) = 6x − x2 , so f (x) = 6 − 2x. Put f (x) = 0 Then 6 − 2x = 0 (because the tangent is horizontal). x = 3. Substituting, f (3) = 18 − 9 = 9, so the tangent is horizontal at (3, 9). (This is, of course, the vertex of the parabola.) Differentiable (or smooth) at a point Two issues may raise questions, and need explanation. y First, the graph f (x) = |x| sketched to the right is continuous at the origin, but there is clearly a problem differentiating at the origin O, because the curve is not smooth there. 2 When we substitute a = 0 into the limit formula, there is no limit f (0 + h) − f (0) limh→0 , because all the secants on the right of O have gradient 1, and h on the left they all have gradient −1. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 −2 2 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10B The derivative as a limit 393 We say that y = |x| is not differentiable (or smooth ) at x = 0. If a curve is not continuous at a point, it is not differentiable there. What is a tangent Secondly, the term ‘tangent’ was introduced in Section 10A without a formal definition. We can now use the limit formula for the derivative to define a tangent (provided that it is not vertical) as the line through the point with gradient f (x). 5 Differentiable at a point, and the definition of a non-vertical tangent • A function f (x) is called differentiable or smooth at x = a if the limit f (a + h) − f (a) f (a) = lim exists. h→0 h If f (x) is not continuous at x = a, it is not differentiable there. • If f (x) is differentiable at x = a, the tangent at the point P where x = a is the line through P with gradient f (a). Vertical tangents need further discussion in Year 12. As with continuity at a point, we make a second standard assumption: 6 Assumption about differentiability at a point The functions in this course are differentiable (smooth) for every value of x in their domains, except where there is an obvious problem. Exercise 10B FOUNDATION Note: The questions in this exercise use the formula for the derivative as a limit: f (x + h) − f (x) . h→0 h f (x) = lim 1 Consider the function f (x) = 5x2 . a Show that f (1) = 5 and f (1 + h) = 5 + 10h + 5h2 . f (1 + h) − f (1) = 10 + 5h. h c Take the limit as h → 0 to show that f (1) = 10. b Show that 2 Consider again the function f (x) = 5x2 . a Show that f (x + h) = 5x2 + 10xh + 5h2 . f (x + h) − f (x) = 10x + 5h. h c Take the limit as h → 0 to show that f (x) = 10x. d Substitute x = 1 into f (x) to confirm that f (1) = 10, as found in Question 1. b Show that CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 394 10B Chapter 10 Differentiation 3 Consider the function f (x) = x2 − 4x. f (x + h) − f (x) a Show that = 2x + h − 4. h b Show that f (x) = 2x − 4 by taking the limit as h → 0. c Evaluate f (1) to find the gradient of the tangent at A(1, −3). d Similarly, find the gradients of the tangents at B(3, −3) and C(2, −4). e The function f (x) = x2 − 4x is graphed above. Place your ruler on the curve at A, B and C to check the reasonableness of the results obtained above. y 0 1 −1 2 3 4 x y = x2 − 4x −2 −3 A −4 B C 4 In Section 10A, you found by geometry that a linear function f (x) = mx + b has derivative f (x) = m. This question confirms that the limit formula for f (x) gives the same answer. f (x + h) − f (x) a Let f (x) = 3x + 7. Find , and hence show that f (x) = 3. h f (x + h) − f (x) , and hence show that f (x) = m. b Let f (x) = mx + b. Find h f (x + h) − f (x) c Let f (x) = c. Find , and hence show that f (x) = 0. h 5 a f (x + h) − f (x) for the function f (x) = x2 + 10, then find f (x). h ii Find the gradient of the tangent at the point P on the curve where x = 2. iii Find the coordinates of any points on the curve where the tangent is horizontal. b Repeat the steps of part a for f (x) = x2 + 6x + 2. c Repeat the steps of part a for f (x) = 2x2 − 20x. d Repeat the steps of part a for f (x) = 9 − 4x2 . i Simplify DEVELOPMENT 6 a Sketch f (x) = x2 − 10x, then show that the derivative is f (x) = 2x − 10. b Hence find the gradient of the tangent at the points where: i x=0 ii x = 10 iii x = 5 iv x = 4 21 v x = 5 12 c What is the angle between the tangents in part iv and v? 7 a Show that the derivative of g(x) = x2 − 5x is g (x) = 2x − 5. b Hence find the coordinates of the points on y = x2 − 5x where the tangent has gradient: i 1 ii −1 iii 5 iv −5 v 0 c Then sketch the curve and the tangents. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10B The derivative as a limit 8 395 a Sketch f (x) = x2 − 7x + 6, then show that the derivative is f (x) = 2x − 7. b Find the two x-intercepts, find the gradients of the tangents there, and show that they are opposites. c Find the coordinate of the y-intercept A and the gradient m at A, then find the point B on the curve where the tangent has the opposite gradient −m. d Find the coordinates of the point where the tangent is horizontal. 9 a Show that the derivative of the quadratic f (x) = ax2 + bx + c is f (x) = 2ax + b. b Find the x-coordinate of the point V on y = f (x) where the tangent is horizontal. c What geometric significance does the result of part b have? 10 The two binomial expansions in this question will be studied more systematically in Chapter 15 on binomial expansions. a i Prove that (x + h)3 = x3 + 3x2 h + 3xh2 + h3 by writing (x + h)3 = (x + h) × (x + h)2 and then expanding brackets. ii Hence find the derivative of f (x) = x3 . iii Show that the tangents to y = x3 at the points A(a, a3 ) and B(−a, −a3 ) are parallel. b i Similarly prove that (x + h)4 = x4 + 4x3 h + 6x2 h2 + 4xh3 + h4 . ii Hence find the derivative of f (x) = x4 . iii Show that the tangents to y = x4 at the points A(a, a4 ) and B(−a, a4 ) have opposite gradients. 1 f (x + h) − f (x) −1 , show that = . x h x(x + h) 1 b Hence show that f (x) = − 2 . x c Show from this formula, and from sketching the curve, that all tangents slope downwards (that is, have negative gradient). What happens to the slopes of the tangents as x → 0+ and as x → 0− ? What happens as x → ∞ and as x → −∞? 11 a For f (x) = 12 a If f (x) = √ f (x + h) − f (x) . h b Use the method of rationalising the numerator: √ √ √ √ √ √ x+h− x x+h+ x x+h− x = √ √ h h x+h+ x x, where x ≥ 0, write down to show that 1 f (x + h) − f (x) = √ √ . h x+h + x c Hence find f (x). d Show from this formula, and from sketching the curve, that all tangents slope upwards. What happens to the slopes of the tangents as the point of contact moves towards the origin? What happens as x → ∞? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 396 10B Chapter 10 Differentiation 13 [Spreadsheet technology (written for Excel)] Consider the function f (x) = x3 . f (2 + h) − f (2) a Use the formula f (2) = limh→0 to prove that f (2) = 12. You may want to use the h standard expansion (A + B)3 = A3 + 3A2 B + 3AB2 + B3 . b In the top row of a spreadsheet, in cell A1 enter the letter h in cell A2 enter the formula f (2 + h) – f (2) and in cell A3 enter the word gradient c In the second row, in cell B1 enter the number 1, in cell B2 enter the string =(2+A2)∧3 – 2∧3 and in cell B3 enter =B2/A2 d In the first column, in cell A3 enter 0.5, in cell A4 enter 0.1, and in cell A5 enter 0.05. e Select cell B2, then hold the Shift key and use the arrows to highlight the cells that are in rows 2, 3, 4, and 5 and in columns B and C. f With these cells highlighted, press Ctrl+D to Fill Down the formulae into the rows below. g Continue down column A with ever smaller numbers, using Fill Down each time. h Explain what this spreadsheet procedure has to do with part a. ENRICHMENT 14 Using the methods of Questions 11 and 12, find f (x + h) − f (x) to show that: h 2 1 , then f (x) = − 3 . 2 x x 1 1 b If f (x) = √ , then f (x) = − √ . x 2x x a If f (x) = 15 [Algebraic differentiation of x2 ] Let P(a, a2 ) be any point on the curve y = x2 , then the line through P with gradient m has equation y − a2 = m(x − a). Show that the x-coordinates of the points where meets the curve are x = a and x = m − a. Find the value of the gradient m for which these two points coincide, and explain why it follows that the derivative of x2 is 2x. 16 [An alternative algebraic approach] Find the x-coordinates of the points where the line : y = mx + b meets the curve y = x2 , and hence deduce that the derivative of x2 is 2x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10C A rule for differentiating powers of x 397 10C A rule for differentiating powers of x Learning intentions • Find and use the derivative of xn for whole-number indices where n > 0. • Work with tangents, normals, and angles of inclinations. From this limit definition, we will develop formulae for differentiation for each type of function. This section concerns a formula that applies to all powers of x. A formula for differentiating powers of x The long calculations of f (x) in the previous exercise had quite simple answers, as you will probably have noticed. Here is the pattern: f (x) 1 f (x) 0 x 1 x2 x3 x4 x5 ··· 2x 2 3 4 ··· 3x 4x 5x These results are examples of a simple rule for differentiating any power of x: 7 The derivative of any power of x Let f (x) = xn , where n is any real number. Then the derivative is f (x) = nxn−1 . This rule is usually memorised as: ‘Take the index as a factor, and reduce the index by 1.’ The general proof is tricky, with several stages — here we prove the result only for whole-number indices n > 0, then the proofs for negative and rational indices are completed in Sections 10F & 10G. First, four examples using the theorem. Example 4 Differentiating powers of x Differentiate: a f (x) = x8 b f (x) = x100 c f (x) = x−4 2 d f (x) = x 3 Solution a f (x) = x8 f (x) = 8x7 b f (x) = x100 f (x) = 100x99 c f (x) = x−4 f (x) = −4x−5 2 d f (x) = x 3 2 1 f (x) = x− 3 3 Proof when n is a whole number Begin by looking carefully at the first two terms in the expansion of (x + h)4 : (x + h)4 = (x + h)(x + h)(x + h)(x + h) = x4 + (xxxh + xxhx + xhxx + hxxx) + (terms with at least two hs) = x4 + 4x3 h + (terms in h2 and above). Every term in the expansion has four factors. The terms in hx3 are xxxh, xxhx, xhxx and hxxx, being the four factors h, x, x and x in all four possible orders. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 398 10C Chapter 10 Differentiation The general case of the expansion of (x + h)n is similar: (x + h)n = (x + h)(x + h) · · · (x + h)(x + h) (there are n factors) = xn + (xn−1 h + xn−2 hx + xn−3 hx2 + · · · + hxn−1 ) + (terms with at least two hs) = xn + nxn−1 h + (terms in h2 and above). Every term in the expansion has n factors. The n factors of each term in hxn−1 consist of one h, and n − 1 copies of x, in any one of the n possible orders. (x + h)n − xn Hence f (x) = lim h→0 h nxn−1 h + (terms in h2 and above) = lim h→0 h n−1 = lim nx + (terms in h and above) h→0 n−1 = nx . Later in the course, the binomial theorem, and mathematical induction, will each provide proofs that are more concise than what is possible now. Linear combinations of functions Functions formed by taking sums and multiples of simpler functions can be differentiated in the obvious way, one term at a time. The proofs here are reasonably straightforward. 8 Linear combinations of functions • If f (x) = u(x) + v(x), then f (x) = u (x) + v (x). • If f (x) = au(x), then f (x) = au (x) where a is a constant. Proof For the first: f (x + h) − f (x) f (x) = lim h→0 h u(x + h) + v(x + h) − u(x) − v(x) = lim h→0 h u(x + h) − u(x) v(x + h) − v(x) + lim = lim h→0 h→0 h h = u (x) + v (x). Example 5 For the second: f (x + h) − f (x) f (x) = lim h→0 h au(x + h) − au(x) = lim h→0 h u(x + h) − u(x) = a lim h→0 h = au (x). Differentiating linear combinations of powers of x Differentiate each function. a f (x) = x2 + 3x + 5 b f (x) = 4x2 − 3x + 2 c f (x) = 3x10 + 4x9 d f (x) = 12 x6 − 16 x3 Solution a f (x) = x2 + 3x + 5 b f (x) = 4x2 − 3x + 2 f (x) = 2x + 3 + 0 f (x) = 4 × 2x − 3 + 0 = 2x + 3 = 8x − 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10C A rule for differentiating powers of x c f (x) = 3x10 + 4x9 d 399 f (x) = 12 x6 − 16 x3 f (x) = 3 × 10x9 + 4 × 9x8 f (x) = 12 × 6x5 − 16 × 3x2 = 30x9 + 36x8 = 3x5 − 12 x2 Expanding products Sometimes a product needs to be expanded before differentiating the function. Example 6 Expanding a product before differentiating Differentiate each function after first expanding the bracket, factoring if possible. a f (x) = x3 (x − 10) b f (x) = (x + 2)(2x + 3) Solution a f (x) = x3 (x − 10) = x − 10x 4 b f (x) = (x + 2)(2x + 3) = 2x2 + 7x + 6 3 f (x) = 4x3 − 30x2 = 2x2 (2x − 15) f (x) = 4x + 7 Gradient of a curve, and angle of inclination of a tangent The gradient of a curve at a particular point P on the curve is defined to be the gradient of the tangent at P — the gradient of the curve at P is thus the value of the derivative at P. The steepness of a tangent can be expressed either by giving its gradient, or by giving its angle of inclination. As we saw in Chapter 8, negative gradients correspond to obtuse angles of inclination. 9 Gradient of a curve, and angle of inclination of a tangent • The gradient of a curve at a point P on it is the gradient of the tangent at P. • The steepness of a tangent can also be expressed by its angle of inclination. Example 7 Differentiating to find the angle of inclination a Differentiate f (x) = x2 + 2x. b Find the gradient and the angle of inclination of the curve at the origin. c Find the gradient and the angle of inclination of the curve at the point A(−2, 0). d Sketch the curve and the tangents, marking their angles of inclination. Solution a Differentiating f (x) = x2 + 2x gives f (x) = 2x + 2. b At the origin O(0, 0), gradient of tangent = f (0) = 2. Let α be the angle of inclination of the tangent. Then tan α = 2, where 0◦ ≤ α < 180◦ , so α 63◦ 26 . Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 400 10C Chapter 10 Differentiation c At A(−2, 0), gradient of tangent = f (−2) = −4 + 2 = −2. Let β be the angle of inclination of the tangent. Then tan β = −2, where 0◦ ≤ β < 180◦ , β 180◦ − 63◦ 26 so 116◦ 34 . d Method: Enter 2 into the calculator, take tan−1 2, then find its supplement. Finding points on a curve with a given gradient or angle of inclination When the angle of inclination is given, use the tan θ function to find the gradient. Example 8 Finding points with a given angle of inclination Find the points on the graph of f (x) = x2 − 5x + 4 where: a the tangent has a gradient of −3, b the tangent has an angle of inclination of 45◦ . Solution Here f (x) = x2 − 5x + 4, so f (x) = 2x − 5. a First, tan 45◦ = 1, 2x − 5 = −3 so put f (x) = 1. 2x = 2 Then 2x − 5 = 1 Put f (x) = −3. Then b x = 1. 2x = 6 Substituting x = 1 into the function, x = 3. f (1) = 1 − 5 + 4 Substituting x = 3 into the function, = 0, so the tangent has gradient −3 at the point (1, 0). f (3) = 9 − 15 + 4 = −2, so the tangent has angle of inclination 45◦ at the point (3, −2). 10 Finding points on a curve with a given angle θ of inclination dy = tan θ, dx because the gradient of the tangent is tan θ. • Solve CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10C A rule for differentiating powers of x 401 Tangents and normals to a curve The normal to a curve at a point P on the curve is the line through P that is perpendicular to the tangent at P. Provided that the tangent and normal are not parallel to the axes, the product of their gradients is −1, so their equations are easily found. 11 Tangents and normals to a curve at a point P on the curve • The tangent at P is the line through P whose gradient is the derivative at P. • The normal at P is the line through P that is perpendicular to the tangent at P. • If they are not parallel to the axes, their gradients m1 and m2 are related by 1 or equivalently, m2 = − . m1 m2 = −1, m1 • Their equations are then easily found using point–gradient form: y − y1 = m(x − x1 ). Example 9 Finding tangents, normals, and areas For the curve y = f (x), where f (x) = (x + 1)2 : a Find the equation and angle of inclination of the tangent at A(0, 1). b Find the equation and angle of inclination of the normal at A, and draw a sketch. c Find the x-intercepts of the tangent and normal by finding their equations. d Find the area of the triangle formed by the tangent, normal and x-axis. Solution a Expanding, so f (x) = x2 + 2x + 1, f (x) = 2x + 2. At A(0, 1) f (0) = 2, so the tangent at A(0, 1) is y − 1 = 2(x − 0) y = 2x + 1. Solving tan α = 2, where 0◦ ≤ α < 180◦ , its angle of inclination is about 63◦ 26 . b The normal has gradient − 12 (take the opposite of the reciprocal), so the normal at A(0, 1) is y − 1 = − 12 (x − 0) y = − 12 x + 1. Solving tan β = − 12 , where 0◦ ≤ β < 180◦ , its angle of inclination is about 153◦ 26 . (The two angles differ by 90◦ , by the exterior-angle-of-a-triangle theorem.) c The tangent is y = 2x + 1 (using gradient–intercept form) and substituting y = 0 gives x = − 12 . The normal is y = − 12 x + 1 and substituting y = 0 gives x = 2. d Hence area of triangle = 12 × 2 12 × 1 = 1 14 square units. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 402 10C Chapter 10 Differentiation Exercise 10C FOUNDATION Note: The next seven exercises, Exercise 10C–10I, contain exhaustive practice of differentiation skills, which are essential for most of what follows in the course. As is the case throughout the book, the exercises are there to support readers learning the relevant skills — once those skills are mastered, it is not necessary to do all the questions. 1 Use the rule for differentiating xn to differentiate these functions. a f (x) = x7 b f (x) = 9x5 c f (x) = 13 x6 d f (x) = 3x2 − 5x e f (x) = x4 + x3 + x2 + x + 1 f f (x) = 2 − 3x − 5x3 g 2 f (x) = 13 x6 − 12 x4 + x2 − 2 h f (x) = 14 x4 + 13 x3 + 12 x2 + x + 1 Differentiate these functions by first expanding the products. a f (x) = 3x(4 − 2x) b f (x) = x(x2 + 1) c f (x) = x2 (3 − 2x − 4x2 ) d f (x) = (x + 4)(x − 2) e f (x) = (2x + 1)(2x − 1) f f (x) = (x − 7)2 g f (x) = (x2 + 3)2 h f (x) = x(7 − x)2 i f (x) = (x2 + 3)(x − 5) 3 4 Differentiate these functions, given that a, b, c, k and are constants. a f (x) = ax4 − bx2 + c b f (x) = (ax − 5)2 c f (x) = k(ax + b)(ax − b) d f (x) = x e f (x) = x5a+1 f f (x) = bx3b Find the gradients of the tangent and normal at the point on y = f (x) where x = 3. a f (x) = x2 − 5x + 2 b f (x) = x3 − 3x2 − 10x c f (x) = 2x2 − 18x Then find the angles of inclination of these tangents and normals. 5 6 7 Find the equations of the tangent and normal to the graph of f (x) = x2 − 8x + 15 at: a A(1, 8) b B(6, 3) c the y-intercept d C(4, −1) Find any points on the graph of each function where the tangent is parallel to the x-axis, and write down the equation of each horizontal tangent. a f (x) = 4 + 4x − x2 b f (x) = x3 − 12x + 24 c f (x) = 3x + 2 d f (x) = 4ax − x2 e f (x) = x4 − 2x2 f f (x) = x5 + x Differentiate f (x) = x3 . Hence show that the tangents to y = x3 have positive gradient everywhere except at the origin, and show that the tangent there is horizontal. Explain the situation using a sketch. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10C A rule for differentiating powers of x 403 DEVELOPMENT 8 Show that the line y = 3 meets the parabola y = 4 − x2 at D(1, 3) and E(−1, 3). Find the equations of the tangents to y = 4 − x2 at D and E, and find the point where these tangents intersect. Sketch the situation. 9 Find the equation of the tangent to f (x) = 10x − x3 at the point P(2, 12). Then find the points A and B where the tangent meets the x-axis and y-axis respectively, and find the length of AB and the area of ΔOAB. 10 The tangent and normal to f (x) = 9 − x2 at the point K(1, 8) meet the x-axis at A and B respectively. Sketch the situation, find the equations of the tangent and normal, find the coordinates of A and B, and hence find the length AB and the area of ΔAKB. 11 The tangent and normal to the cubic f (x) = x3 at the point U(1, 1) meet the y-axis at P and Q respectively. Sketch the situation and find the equations of the tangent and normal. Find the coordinates of P and Q, and the area of ΔQUP. 12 a Show that the tangents at the x-intercepts of f (x) = x2 − 4x − 45 have opposite gradients. b Find the vertex of the parabola, and use symmetry to explain geometrically why the result in part a has occurred. 13 Find the derivative of the cubic f (x) = x3 + ax + b, and hence find the x-coordinates of the points where the tangent is horizontal. For what values of a and b do such points exist? 14 [Change of pronumeral] a Find G (t) and G (3), if G(t) = t3 − 4t2 + 6t − 27. b Given that (h) = 5h4 , find (h) and (2). c If Q(k) = ak2 − a2 k, where a is a constant, find: i Q (k) 15 ii Q (a) iii Q (0) iv |Q (0) − Q (a)| Sketch the graph of f (x) = x2 − 6x and find the gradient of the tangent and normal at the point A(a, a2 − 6a) on the curve. Hence find the value of a if: a the tangent has gradient 2, b the normal has gradient 4, c the tangent has angle of inclination 135◦ , d the normal has angle of inclination 30◦ , e the tangent is parallel to 2x − 3y + 4 = 0, f the normal is parallel to 2x − 3y + 4 = 0. 16 Show that the derivative of f (x) = x2 − ax is f (x) = 2x − a, and find the value of a when: a the tangent at the origin has gradient 7, b y = f (x) has a horizontal tangent at x = 3, c the tangent at the point where x = 1 has an angle of inclination of 45◦ , d the tangent at the non-zero x-intercept has gradient 5, e the tangent at the vertex has y-intercept −9. 17 a The tangents to y = x2 at two points A(a, a2 ) and B(b, b2 ) on the curve meet at K. Find the coordinates of K. b When is K above the x-axis? Explain algebraically and geometrically. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 404 10C Chapter 10 Differentiation ENRICHMENT 18 Let n be the number of horizontal tangents to the cubic y = ax3 + bx2 + cx + d. Find the condition on a, b, c and d for: a n=2 b n=1 c n = 0. 19 f (x + h) − f (x − h) . Draw a diagram to justify this 2h 2 3 formula, then use it to find the derivatives of x and x . 20 a Write down the equation of the tangent to the parabola y = ax2 + bx + c (where a 0) at the point P Another formula for the derivative is f (x) = limh→0 where x = t, and show that the condition for the tangent at P to pass through the origin is at2 − c = 0. Hence find the condition on a, b and c for such tangents to exist, and the equations of these tangents. b Find the points A and B where the tangents from the origin touch the curve, and show that the y-intercept C(0, c) is the midpoint of the line segment joining the origin and the midpoint of the chord AB. Show also that the tangent at the y-intercept C is parallel to the chord AB. c Hence show that the triangle ΔOAB has four times the area of the triangle ΔOCA, and find the area of ΔOAB. 21 At this point, someone inevitably asks, “Can the function f (x) = 5 be differentiated by putting f (x) = 5x0 , and then using the rule to obtain f (x) = 0 × 5x−1 = 0?” One problem, which we try to suppress in this course, is that 00 is undefined. a Write down the domains of f (x) = 5 and g(x) = 5x0 . b Write down the domains of f (x) = 0 and g (x) = 0x−1 . c What is the second error in the initial argument presented in the first sentence above? From now on, you are free to ignore the problem. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10D The notation 10D The notation dy dx dy for the derivative dx 405 for the derivative Learning intentions dy • Use notation for the derivative. dx • Find tangents through other points in the plane. Leibniz’s original notation for the derivative remains the most widely used and best-known notation, particularly dy . The notation is extremely in science. It is even said that Dee Why Beach was named after the derivative dx flexible, and clearly expresses the fact that the derivative behaves very much like a fraction. Small changes in x and in y dy The diagram used to explain is exactly the same as for f (x), but the notation dx is a little different. Let P(x, y) be any point on the graph of a function. y Q y + Dy Let x change by a small amount Δx to x + Δx, and let y change by a corresponding amount Δy to y + Δy. Thus Q(x + Δx, y + Δy) is another point on the curve. Then Δy (rise over run). gradient PQ = Δx y P x x + Dx x Δy When Δx is small, the secant PQ is almost the same as the tangent at P, and the derivative is the limit of as Δx Q gets ‘as close as we like’ to P, that is, as Δx → 0. This is the basis for Leibniz’s notation: 12 An alternative notation for the derivative Let Δy be the small change in y resulting from a small change Δx in x. Then dy Δy = lim . dx Δx→0 Δx (The letter Δ is the upper-case Greek letter delta, symbolising Difference.) The object dx is intuitively understood as an ‘infinitesimal change’ in x, and dy is understood as the corredy is then understood as the ratio of these two infinitesimal sponding ‘infinitesimal change’ in y. The derivative dx changes. ‘Infinitesimal changes’, however, are for the intuition only — the logic of the situation is: 13 Fraction notation and the derivative The derivative dy dx Δy dy is not a fraction, but is the limit of the fraction as Δx → 0. dx Δx rise , and the notation The notation is very clever because the derivative is a gradient, any gradient is a fraction run dy preserves the intuition of fractions. dx CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 406 10D Chapter 10 Differentiation The words ‘differentiation’ and ‘calculus’ The small differences Δx and Δy, and the infinitesimal differences dx and dy, are the origins of the word ‘differentiation’. The symbol Δ is the upper-case Greek letter ‘Delta’, standing for difference. Calculus is a Latin word meaning ‘stone’. An abacus consists of stones sliding on bars and was used routinely in the past to help with arithmetic — this is the origin of the modern word ‘calculate’. The word ‘calculus’ in English can refer to any systematic method of calculation, but is most often used for the twin theories of differentiation and integration studied in this course. Second and higher derivatives The derivative of a function is another function, so it can in turn be differentiated to give the second derivative of d2 y dy the original function. This is written as f (x) in f (x) notation, and as 2 in the new notation. dx dx Thus for a function such as x4 + 5x3 , the two notations are: f (x) = x4 + 5x3 y = x4 + 5x3 dy = 4x3 + 15x2 f (x) = 4x3 + 15x2 dx d2 y = 12x2 + 30x f (x) = 12x2 + 30x dx2 This process can be continued indefinitely to third and higher derivatives: d3 y = 24x + 30 f (x) = 24x + 30 dx3 d4 y = 24 f (4) (x) = 24 dx4 and after that, the fifth and higher derivatives for this function are all zero. Notice that to avoid counting multiple dashes, we usually write f (4) (x) rather than f (x). The second derivative f (x) is the gradient function of the gradient function. Next year, we will interpret the second derivative in terms of curvature, but there are no obvious geometric interpretations of higher derivatives. Operator notation d dy can also be regarded as the operator operating on the function y. This gives an alternative dx dx notation for the derivative: d 2 d 5 and (x ) = 5x4 (x + x − 1) = 2x + 1 . dx dx (An operator acts on a function to give another function. It is thus a function whose domain and range are sets of functions rather than sets of numbers.) The derivative Thus the standard rule for differentiating powers of x can be written as d n (x ) = n xn−1 , for all real numbers n. dx Setting out using dy dx notation The remaining worked examples in this section show how the new notation is used in calculations on the geometry of a curve. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10D The notation Example 10 dy for the derivative dx 407 Find tangents and normals to a curve Find the equations of the tangent and normal to the curve y = 4 − x2 at the point P(1, 3) on the curve. Solution dy = −2x, dx dy = −2 (substitute x = 1 into the derivative), so at P(1, 3), dx Differentiating, 1 so the tangent at P has gradient −2 and the normal has gradient . 2 Hence the tangent is y − 3 = −2(x − 1) y = −2x + 5, 1 (x − 1) 2 1 y = x + 2 12 . 2 and the normal is y − 3 = Finding the tangents passing through some fixed point in the plane Sometimes we want to find the equation of a tangent to a curve y = f (x) passing through a fixed point F in the plane. This problem can be solved by finding the equation of the tangent at a variable point P t, f (t) on the curve, then substituting the coordinates of F into this equation. The method can be applied to other similar problems. Example 11 Finding tangents from a fixed point in the plane a Find the equation of the tangent to y = x2 + x + 1 at the point P on the curve where x = t. b Hence find the equations of the tangents to the curve passing through the origin, and the corresponding point of contact of each tangent. Solution dy = 2x + 1, dx dy so at P, = 2t + 1. dx The coordinates of P are (t, t2 + t + 1), a Differentiating, y 3 so the tangent is y − (t2 + t + 1) = (2t + 1)(x − t) 1 y = (2t + 1)x − t2 + 1. b Substituting (0,0), 0 = −t2 + 1 t = 1 or −1. When t = 1, the tangent is y = 3x and the point of contact is P = (1, 3). When t = −1, the tangent is y = −x and the point of contact is P = (−1, 1). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 −1 1 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 408 10D Chapter 10 Differentiation Exercise 10D Note: You are invited to use FOUNDATION dy notation as well as f (x) notation in this exercise. dx dy dy of each function, and the value of when x = −1. dx dx a y = x3 + 3x2 + 6x + 8 b y = x4 + x2 + 8x c y=7 d y = (2x − 1)(x − 2) 1 Find the derivative 2 a Using dy notation, find the first, second and third derivatives of each function. dx i y = x6 + 2x ii y = 5x2 − x5 iii y = 4x b Using f (x) notation, find the 1st, 2nd, 3rd and 4th derivatives of each function. i f (x) = 10x3 + x c ii f (x) = 2x4 + 2 iii f (x) = 5 i How many times must y = x be differentiated to give the zero function? 5 ii How many times must y = xn be differentiated to give the zero function? 3 Find any points on each graph where the tangent has gradient −1. a y = x4 − 5x 4 b y = x3 − 4x c y = x3 − 4 Find the x-coordinates of any points on each curve where the normal is vertical. a y = 3 − 2x + x2 b y = x4 − 18x2 c y = x3 + x 5 In each function, first divide through by the denominator, then differentiate. Then factor the derivative and state any values of x where the tangent is horizontal. 5x6 + 3x5 3x4 − 9x2 a y= b y= x 3x3 6 Using dy notation, find the tangent and normal to each curve at the point indicated. dx a y = x2 − 6x at O(0, 0) b y = x2 − x4 at J(−1, 0) c y = x3 − 3x + 2 at P(1, 0) DEVELOPMENT 7 Find, correct to the nearest minute, the angles of inclination of the tangents to the parabola y = 1 − x2 at its x-intercepts. 8 Find any points on each curve where the tangent has the given angle of inclination. a y = 13 x3 − 7, 45◦ 9 b y = x2 + 13 x3 , 135◦ c y = x2 + 1, 120◦ a Find the equations of the tangent and normal to y = 9 − x2 at the point K(1, 8). b Find the points A and B where the tangent and normal respectively meet the x-axis. c Sketch the situation, then find the length AB and the area of ΔABK. 10 a Find the equation of the tangent to y = x2 + 9 at the point P(p, p2 + 9). b Hence find the points on the curve where the tangents pass through the origin, and the equation of those tangents. Draw a sketch of the situation. 11 For each curve below, find the equation of the tangent at the point P where x = t. Hence find the equations of any tangents passing through the origin. a y = x2 − 10x + 9 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b y = x2 + 15x + 36 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10D The notation 12 dy for the derivative dx 409 a Find the equation of the tangent to y = x2 + 2x − 8 at the point K on the curve with x-coordinate t. b Hence find the points on the curve where the tangents from H(2, −1) touch the curve. 13 Differentiate y = x2 + bx + c, and hence find b and c given that: a the parabola passes through the origin, and the tangent there has gradient 7. b the parabola has y-intercept −3 and gradient −2 there, c the parabola is tangent to the x-axis at the point (5, 0), d when x = 3 the gradient is 5, and x = 2 is a zero, e the parabola is tangent to 3x + y − 5 = 0 at the point T (3, −4), f the line 3x + y − 5 = 0 is a normal at the point T (3, −4). 14 Find the tangent to the curve y = x4 − 4x3 + 4x2 + x at the origin, and show that this line is also the tangent to the curve at the point (2, 2). 15 Find the points where the line x + 2y = 4 cuts the parabola y = (x − 1)2 . Show that the line is the normal to the curve at one of these points. 16 a The tangent at T (a, a2 ) to y = x2 meets the x-axis at U and the y-axis at V. Find the equation of this tangent, and show that ΔOUV has area | 14 a3 | square units. b Hence find the coordinates of T for which this area is 31 14 . 17 Show that the line x + y + 2 = 0 is a tangent to y = x3 − 4x, and find the point of contact. (Hint: Find the equations of the tangents parallel to x + y + 2 = 0, and show that one of them is this very line.) 18 a Find the points A and B where y = −2x meets the parabola y = (x + 2)(x − 3). b Find the midpoint M of the chord AB, find the point T where the vertical line through M meets the parabola, and show that the tangent at T is parallel to the chord AB. 19 20 dP dP dP , and (assuming that when differentiating with respect to one If P = tx2 + 3tu2 + 3xu + t, find dx du dt variable, the other pronumerals are constant). [For discussion] Sketch the graph of y = x3 . Then choose any point in the plane and check by examining the graph that at least one tangent to the curve passes through every point in the plane. What points in the plane have three tangents to the curve passing through them? This problem can also be solved algebraically, but that is considerably harder. ENRICHMENT 21 a Show that the tangent to y = ax2 + bx + c with gradient m has y-intercept (m − b)2 . 4a b Hence find the equations of any quadratics that pass through the origin and are tangent to both y = −2x − 4 and to y = 8x − 49. c Find also any quadratics that are tangent to y = −5x − 10, to y = −3x − 7 and to y = x − 7. c− 22 Let y = ax3 + bx2 + cx + d be a cubic (so that a 0). a Write down the equation of the tangent at the the point T on the curve where x = t. b Show that every point P(x0 , y0 ) in the plane lies on at least one tangent to the cubic. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 410 10E Chapter 10 Differentiation 10E The chain rule Learning intentions • Use the chain rule to differentiate composite functions. In Section 10C, we easily developed rules for differentiating sums and multiple of functions. Differentiating composite functions, and products or quotients of functions, is a little more elaborate. This section concerns composite functions. Composition of functions — a chain of functions Section 5F introduced composite functions, formed by putting two functions into a chain so that the output of the first function becomes the input of the second. We now want to apply the process in reverse. For example, we can decompose the function y = (x2 − 2)4 into a chain of two functions — ‘square and subtract 2’, followed by ‘raise to the power 4’. x 0 −→ 1 −→ −2 −→ x −→ u Square and subtract 2 y −→ −2 −→ −→ −1 −→ 2 −→ −→ −→ x − 2 −→ 2 Take the fourth power −→ 16 −→ 1 −→ 16 −→ (x − 2)4 2 The middle column is the output of the first function ‘Square and subtract 2’. This output is then the input of the second function ‘Take the fourth power’. The resulting decomposition of the original function y = (x2 − 2)4 into the chain of two functions may be expressed as follows: ‘Let u = x2 − 2. Then y = u4 .’ Our original function has now been decomposed into the composition of two simpler functions, ready to be differentiated using the chain rule. The chain rule — differentiating a composite function Suppose then that y is a function of u, where u is a function of x. Δy dy = lim dx Δx→0 Δx Δy Δu × (multiplying top and bottom by Δu) = lim Δx→0 Δu Δx Δy Δu × lim (because Δu → 0 as Δx → 0) = lim Δu→0 Δu Δx→0 Δx dy du × . = du dx 14 The chain rule Suppose that y is a function of u, where u is a function of x. Then dy dy du = × . dx du dx dy is not a fraction, the usual approach to this rule is that ‘the du’s Although the proof uses limits, and dx cancel out’. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10E The chain rule Example 12 411 Differentiating using the chain rule Use the chain rule to differentiate each function. a (x2 − 2)4 b 7(3x + 4)5 Note: The working in the right-hand column is the strongly recommended setting out of the calculation. Always begin with that working, because the first step is the decomposition of the function into a chain of two functions. Solution a Let Then y = (x2 − 2)4 . dy dy du = × dx du dx = 4(x2 − 2)3 × 2x = 8x(x2 − 2)3 . b Let Then y = 7(3x + 4)5 . dy dy du = × dx du dx = 35(3x + 4)4 × 3 = 105(3x + 4)4 . Let u = x2 − 2. y = u4 . du = 2x Hence dx dy = 4u3 . and du Let u = 3x + 4. Then y = 7u5 . du =3 Hence dx dy = 35u4 . and du Then This is much easier than expanding the brackets first and then differentiating, Powers of a linear function Consider the derivative of (ax + b)n . Let y = (ax + b)n . dy dy du = × dx du dx = n(ax + b)n−1 × a Let u = ax + b. y = un . du =a Hence dx dy and = nun−1 . = an(ax + b)n−1 . du This result occurs so often that it should be remembered as a formula for differentiating any linear function of x raised to a power. Then Then 15 Differentiating a power of a linear function d (ax + b)n = an(ax + b)n−1 dx Example 13 Differentiating a power of a linear function d (ax + b)n = an(ax + b)n−1 to differentiate each function. dx a y = (4x − 1)7 b y = 3(7 − 4x)5 Use the formula CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 412 10E Chapter 10 Differentiation Solution a y = (4x − 1)7 dy = 4 × 7 × (4x − 1)6 dx = 28(4x − 1)6 (a = 4, b = −1 and n = 7) b y = 3(7 − 4x)5 Here a = −4, b = 7 and n = 5. dy = 3 × (−4) × 5 × (7 − 4x)4 dx = −60(7 − 4x)4 The chain rule and tangents The usual methods of dealing with tangents apply when the chain rule is used to find the derivative. Example 14 The chain rule and tangents Differentiate y = (1 − x4 )4 , and hence find the points on the curve where the tangent is horizontal. Solution y = (1 − x4 )4 . dy dy du = × dx du dx = 4(1 − x4 )3 × (−4x3 ) Let u = 1 − x4 . y = u4 . du Hence = −4x3 dx dy = 4u3 . and du Then = −16x3 (1 − x4 )3 , which is zero when x = −1, 0 or 1, so the points are (−1, 0), (0, 1) and (1, 0). A shorter setting-out of the chain rule It is important to practise the full setting-out. • That is the best way to handle many tricky calculations. • The process will be reversed in Year 12 and must be clearly understood. Many people like to shorten the setting-out — if so, write down at least the function u on the right, and take at least one middle step in the working. Example 15 Shorter setting out of the chain rule Differentiate y = (3 − 5x3 )7 with a shorter setting-out. Solution y = (3 − 5x3 )7 dy = 7(3 − 5x3 )6 × (−15x2 ) dx = −105x2 (3 − 5x3 )6 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 Let u = 3 − 5x3 . (further steps if it gets tricky) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10E The chain rule 413 Another approach to the chain rule For a composite function h(x) = f g(x) , it is not difficult to see that the derivative is given by the formula: h (x) = f g(x) g (x) Again, always begin by writing down the functions f (x) and g(x) on the right. Example 16 Using another formula for the chain rule Differentiate y = (3 − 5x3 )7 using the formula above. Solution h(x) = (3 − 5x3 )7 h (x) = f g(x) g (x) = −105x2 (3 − 5x3 )6 Let f (x) = x7 , and g(x) = 3 − 5x3 . Then f (x) = 7x6 , and g (x) = −15x2 . Exercise 10E FOUNDATION d (ax + b)n = an(ax + b)n−1 when the function is a power of a linear function. dx Otherwise use the full setting-out of the chain rule, or use a shorter form if you are quite confident to do so. Note: Use the standard form 1 2 3 Copy and complete the setting out below to differentiate (x2 + 9)5 by the chain rule. Let u = x2 + 9. Let y = (x2 + 9)5 . Then y = u5 . dy dy du Then = × du dx du dx Hence = ··· = ··· × ··· dx dy = ··· . and = ··· . du dy dy du = × to differentiate each function. Use the full setting-out above, first dx du dx identifying u as a function of x, and y as a function of u. Use the chain rule a y = (3x + 7)4 b y = (5 − 4x)7 c y = (x2 + 1)12 d y = 8(7 − x2 )4 e y = (x2 + 3x + 1)9 f y = −3(x3 + x + 1)6 Use the standard form a y = (5x − 7)5 d y = (4 − 3x)7 d (ax + b)n = an(ax + b)n−1 to differentiate: dx b y = (7x + 3)7 4 e y = 6 12 x − 1 c y = 9(5x + 3)4 f y = 23 5 − 13 x 4 Find the first six derivatives of y = (5x − 2)4 . 5 a Differentiate y = (x − 3)2 + 12 by expanding the RHS and differentiating each term. b Differentiate y = (x − 3)2 + 12 using the standard form 6 4 d (ax + b)n = an(ax + b)n−1 . dx Find the x-coordinates of any points on y = (4x − 7)3 where the tangent is: a parallel to y = 108x + 7 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b perpendicular to x + 12y + 6 = 0 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 414 10E Chapter 10 Differentiation 7 Find the equations of the tangent and normal to each curve at the point where x = 1. b y = (x2 + 1)3 a y = (5x − 4)4 8 Differentiate each function, and hence find any points where the tangent is horizontal. a y = 4 + (x − 5)6 b y = 24 − 7(x − 5)2 c y = a(x − h)2 + k d y = (x2 − 1)3 e y = (x2 − 4x)4 f y = (2x + x2 )5 DEVELOPMENT 9 Find the value of a if: a y = (x − a)3 has gradient 12 when x = 6. b y = (5x + a)4 has gradient −160 when x = 3. 10 Find the values of a and b if the parabola y = a(x + b)2 − 8: a has tangent y = 2x at the point P(4, 8), b has a common tangent with y = 2 − x2 at the point A(1, 1). 11 a Find the x-coordinates of the points P and Q on y = (x − 7)2 + 3 such that the tangents at P and Q have gradients 1 and −1 respectively. b Show that the square formed by the tangents and normals at P and Q has area 12 . 12 a Show that the equation of the tangent to y = (x − 2)4 − 16 at the point T where x = t is y = (t − 2)4 + 4(t − 2)3 (x − t) − 16. b Show that if the tangent passes through P( 12 , −16), then (t − 2)4 + 4(t − 2)3 ( 12 − t) = 0. c Solve this equation to find the tangents to the curve through P. d Use translations, and your knowledge of the graph y = x4 , to sketch the curve and the tangents. 13 14 Find the tangent to each curve at the point where t = 3. a x = 5t2 , y = 10t b x = (t − 1)2 , y = (t − 1)3 Use the chain rule to show that: d 23 (x ) = 6x5 a dx b d k (x ) = kxk−1 dx ENRICHMENT 15 a Find the x-coordinates of the points P and Q on y = (x − h)2 + k such that the tangents at P and Q have gradients m and −m respectively. b Find the area of the quadrilateral formed by the tangents and normals at P and Q. 16 a Let T be the point on the parabola y = a(x − h)2 + k where x = α. Show that the tangent at T has equation y = 2a(α − h)x + k − a(α2 − h2 ). b Let the vertical line through the vertex V meet the tangent at the point P. Show that V P is proportional to the square of the distance between A(α, 0) and the axis of symmetry. c Use the equation in part a to find the equations of the tangents to the parabola through the origin, and the x-coordinates of the points of contact. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10F Differentiating powers with negative indices 415 10F Differentiating powers with negative indices Learning intentions • Differentiate powers with negative indices. d n (x ) = nxn−1 to all rational indices n. This section dx 1 deals with powers whose indices are negative integers, allowing us to differentiate functions such as y = and x 1 y = 2. x The chain rule allows us to extend the standard formula Proving the standard formula for n = −1 The first step is to use first principles differentiation to prove the standard formula for the index n = −1, that is, 1 for y = (as done in Question 11 of Exercise 10B). x 1 Let f (x) = . x f (x + h) − f (x) 1 1 1 Then = − h h x+h x x − (x + h) = (with a common denominator) (x + h)xh −h = (x + h)xh 1 =− provided that h 0. (x + h)x 1 Taking the limit as h → 0, f (x) = − 2 . x −2 This can be written as −x , which is the result given by the standard formula. Proving the standard formula for all negative integers The formula can now be proven for all negative integers using the chain rule. The result for n = −1 is used at (*) in the fourth line on the right below. Suppose then that y = x−m , where m ≥ 2 is an integer. dy dy du By the chain rule, = × dx du dx 1 = − 2m × mxm−1 x = −mx−m−1 , as required. Let then So and u = xm , 1 y= . u du = mxm−1 , dx dy 1 =− 2 du u (∗) 16 Differentiating powers with negative indices: d n (x ) = nxn−1 applies for all integers n. For example, dx d −4 (x ) = −4x−5 . dx 1 d 1 = − 2. • In particular, dx x x • The formula CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 416 10F Chapter 10 Differentiation 1 is useful to know because hyperbolas occur so often. Readers may choose whether x or not to memorise it as a special result. The particular derivative of Example 17 Differentiate powers with negative indices Differentiate each function, giving your answer without negative indices. 5 1 a y = −5x−4 b f (x) = 2 c f (x) = 5 x 3x Solution a y = −5x−4 dy = 20x−5 dx 5 x2 = 5x−2 f (x) = b c f (x) = −10x−3 10 =− 3 x 1 3x5 = 13 x−5 f (x) = f (x) = − 53 x−6 5 =− 6 3x The chain rule and negative indices The chain rule can be combined with these results, as in the next worked example. Example 18 Differentiate y = The chain rule with negative integer indices 1 . 4 − x2 Solution a 1 4 − x2 dy dy du = × dx du dx 1 × (−2x) =− (4 − x2 )2 2x . = (4 − x2 )2 y= Let Then Hence and u = 4 − x2 . 1 y= . u du = −2x dx 1 dy = − 2. du u Dividing through by the denominator If the denominator is a single term, divide through by it first, then differentiate. Example 19 Divide through by a denominator that is a single term In each function, first divide through by the denominator, and differentiate. Then find any points on the curve y = f (x) where the tangent is horizontal. 16 − 24x a f (x) = x3 5 − x2 + 5x4 b f (x) = x2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10F Differentiating powers with negative indices 417 Solution a 16 − 24x x3 −3 = 16x − 24x−2 f (x) = f (x) = −48x−4 + 48x−3 48 48 = − 4 + 3 x x −48 + 48x = x4 b f (x) = 5 − x2 + 5x4 x2 = 5x−2 − 1 + 5x2 f (x) = −10x−3 + 10x = −10 + 10x4 . x3 48(x − 1) f (x) = x4 48(x − 1) = 0 x4 x = 1. Factoring, Hence f (x) = 0 when that is, when Substituting, f (1) = 16 − 24 = −8, so the tangent is horizontal at the point (1, −8). 10(x − 1)(x + 1)(x2 + 1) x3 x = 1 or x = −1. f (x) = Factoring, Hence f (x) = 0 when Substituting, f (1) = 9 and f (−1) = 9 so the tangent is horizontal at the point (1, 9) and at the point (−1, 9). (The function is even.) Exercise 10F 1 2 3 FOUNDATION Write down the derivative f (x) of each function. Leave your answers with negative indices. a f (x) = x−1 b f (x) = x−5 c f (x) = 3x−1 d f (x) = 5x−2 e f (x) = − 43 x−3 f f (x) = 2x−2 + 12 x−8 Rewrite each function using a negative power of x, then differentiate it, then convert the answer back to a fraction. 1 1 3 1 a f (x) = b f (x) = 2 c f (x) = 4 d f (x) = x x x x d Use the standard form (ax + b)n = an(ax + b)n−1 to differentiate each function, leaving results in index dx form. 1 4 3 a y = (7x − 5)−6 b y= c y= d y= 3 + 5x 2x − 1 4(5x + 6)7 1 . x 4 Write down, in index form, the first five derivatives of y = 5 Find any points on these curves where the tangent has gradient −1. 1 a y= b y = 12 x−2 x 6 Find the equations of the tangent and normal to each curve at the point where x = 1. 3 3 a y= b y= 4x − 1 (x − 2)3 7 Use the fact that the derivative of f (x) = a f (x) = 3 x 1 1 is f (x) = − 2 to differentiate: x x 1 7 b f (x) = c f (x) = − 3x 3x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 d f (x) = a x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 418 10F Chapter 10 Differentiation DEVELOPMENT 8 Expand brackets, then differentiate: 2 2 2 1 1 1 a y= x+ b y= x+ − x− x x x c y= √ 1 x− √ x 2 9 Use the full setting out of the chain rule to differentiate each function. Then find the coordinates of any points where the tangent is horizontal. −3 1 a y= b y= 4 2 1+x x − 2x2 − 1 10 Consider the function f (x) = x−1 . a Find f (x), f (x), f (x), f (4) (x) and f (5) (x). b Find f (1), f (1), f (1), f (4) (1) and f (5) (1). c Describe the pattern in the results of part b. d Describe the pattern if −1 is substituted rather than 1. 1 at the point where x = 6 has gradient −1. x+a 11 Find a, if the tangent to y = 12 a Find the equation of the tangent to y = 1 at the point L where x = b. x−4 b Hence find the equations of the tangents passing through: ii W(6, 0). i the origin, 13 a Differentiate the hyperbola y = 9/x, and sketch the curve. b Show that the tangent to the hyperbola at the point T (t, 9/t) is 9x + t2 y = 18t. c Find the points A and B where the tangent meets the x-axis and y-axis respectively. d Find the area of ΔOAB, and show that it is constant as T varies. e Show that T bisects AB and that OT = AT = BT . f Hence explain why the rectangle with diagonal OT has a constant area that is half the area of ΔOAB. 14 12 at M(2, 6) and T (−2, −6) are parallel. x b Find the tangents at M(2, 6) and N(6, 2), and their point of intersection. c Sketch the situation in parts a and b, and explain the particular symmetry of the rectangular hyperbola that is illustrated in each part. a Show that the tangents to f (x) = ENRICHMENT 15 This question shows how to differentiate f (x) = x−m , where m is a whole number, using only the definition of the derivative. f (x + h) − f (x) xm − (x + h)m = a Use a common denominator to show that . h hxm (x + h)m b Simplify this expression using the fact, proven in Section 9C, that (x + h)m = xm + mhxm−1 + (terms in h2 and above). c Hence show that f (x) = − mx−m−1 , agreeing with the formula. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10G Differentiating powers with fractional indices 419 10G Differentiating powers with fractional indices Learning intentions • Differentiate powers with fractional indices. • Understand that the same rules apply for powers with real indices. We now extend the standard formula d n (x ) = nxn−1 to all rational indices n. dx Proving the standard formula for the reciprocal of a positive integer Extending the proof to fractional indices requires two steps, each using the chain rule. The first step extends the result to the fractional indices 12 , 13 , 14 , 15 , . . . . 1 where k ≥ 2 is an integer. Consider the function y = xk , Forming the inverse function, x = y. dx = kyk−1 , because k is a positive integer. dy −1 dy 1 dx dy = , , because = dx dy dx kyk−1 1 = k−1 kx k 1 1 −1 x k , as required. = k Differentiating, Taking reciprocals, The statement that k dy dx and are reciprocals will be developed in Year 12. dy dx Proving the standard formula for any rational number m Consider the function y = x k , where m and k are integers and k ≥ 2. By the chain rule, dy dy du = × dx du dx m−1 1 1 = mx k × x k −1 m m −1 k xk . = k 1 Let u = xk , then y = um . So and du dx dy du = 1 1 −1 x k , by the first step, k = mum−1 . 17 Differentiating powers with fractional indices d n (x ) = nxn−1 applies for all rational numbers n. For example, dx 5 8 d −5 d 1 1 −3 x4 = x 4 x 3 = − x− 3 . and dx 4 dx 3 d √ 1 • In particular, x= √ . dx 2 x • The rule CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 420 10G Chapter 10 Differentiation √ The calculation of the derivative of x by the rule is d 1 1 −1 1 d √ x2 = x 2 = √ . x= dx dx 2 2 x This result is useful to know because functions involving square roots occur so often. Again, readers may choose whether to memorise it as a special result. A note on irrational indices √ We do not have a precise definition of powers such as xπ or x 2 with irrational indices, so we cannot provide a rigorous proof of the fact that the derivative of y = xn is indeed y = nxn−1 for irrational values of n. Nevertheless, every irrational is ‘as close as we like’ to a rational number for which the theorem is certainly true, so the result is intuitively clear for irrational indices. and we assume the result. The domain of x n A power xn does not always exist, because zero has no reciprocal, and negative numbers have no square roots. • When n is negative or zero, x cannot be zero. m , where k is even, x cannot be negative. k • When n is irrational, x cannot be negative. • When n is a cancelled fraction Example 20 Differentiating powers with fractional indices Write each function using a fractional index, then differentiate it, then convert the answer back to surd form. √ 1 3 a y= √ b f (x) = x3 x c f (x) = √ x x3 x Solution a 3 f (x) = √ x = 3x b √ f (x) = x3 x c f (x) = 1 = x3 × x 2 − 12 1 1 f (x) = − 12 × 3 × x−1 2 3 =− √ 2x x = x3 2 1 f (x) = 72 x2 2 = 1 √ x3 x 1 1 x3 × x 2 1 = x−3 2 f (x) = − 72 x−4 2 1 Sometimes surds and whole-number powers need to be combined, and the index laws applied, as in the following two more difficult worked exercises. The chain rule and general indices The chain rule can be combined with these results, as in worked Example 20 and 21. Example 21 Combining the chain rule and fractional indices Differentiate these functions. √ a y = 4 − x2 (an upper semicircle) 1 b y= √ 1 + 4x2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10G Differentiating powers with fractional indices 421 Solution a b √ y = 4 − x2 dy dy du = × dx du dx 1 = √ × (−2x) 2 4 − x2 −x =√ . 4 − x2 1 y= √ 1 + 4x2 dy dy du = × dx du dx 1 = − 12 (1 + 4x2 )−1 2 × 8x = u = 4 − x2 . √ y = u. du = −2x dx dy 1 = √ . du 2 u Let −4x 1 . Then Hence and Let u = 1 + 4x2 . Then y = u− 2 . du = 8x dx 1 dy = − 12 u−1 2 . du 1 Hence and (1 + 4x2 )1 2 Dividing through by the denominator If a function’s denominator is a single term, divide through by it first, then differentiate. Example 22 Dividing through by the denominator In each function, divide through by the denominator and differentiate. Then find any values of x where the tangent is horizontal. 3x2 + 4x − 7 12x − 6 a y= b y= √ x x Solution a 4x 7 3x2 + − y= x x x = 3x + 4 − 7x−1 dy = 3 + 7x−2 dx 3x2 + 7 = , x2 which is never zero. b y = 12x1 1 x2 1 = 12x 2 − 6 1 x2 1 − 6x− 2 1 1 dy = 6x− 2 + 3x−1 2 dx 3(2x + 1) = , 1 x1 2 which is never zero (x = − 12 is outside the domain). Exercise 10G 1 Use the rule FOUNDATION d n (x ) = nxn−1 to find the derivative of each function in index notation. dx 1 2 1 a y = x− 2 b y = x1 2 1 d y = 12x− 3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1 c y = 6x 3 1 e y = 4x 4 + 8x− 4 1 f y = 7x2 3 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 422 10G Chapter 10 Differentiation √ 1 dy = √ . dx 2 x 5 √ dy 5 √ b Explain why y = x2 x can be written as y = x 2 , and hence why = x x. dx 2 c Using similar procedures, differentiate each function and write the answer using surds. √ 1 1 i y=x x ii y = √ iii y = √ x x x √ 3 Write down, in index form, the first five derivatives of y = x. 2 4 a Write y = x as a power of x, and use the rule to show that d (ax + b)n = an(ax + b)n−1 to differentiate each function, leaving results in Use the standard form dx index form. 1 1 √ √ a y = (7 + 2x)1 3 b y= x+4 c y = 5 − 3x d y = 4(2 − 5x)−2 4 5 Find any points on these curves where the tangent has gradient −1. √ √ a y=− x b y= x 6 Find the equations of the tangent and normal to each curve at the point T (4, 2). √ √ a y= x b y= 8−x 7 Use the fact that the derivative of √ a f (x) = 3 x √ 1 x is √ to differentiate: 2 x √ √ b f (x) = 10 x c f (x) = 49x d f (x) = √ 7x DEVELOPMENT 8 Find any points on each curve where the tangent has the given angle of inclination. √ √ a y = 2 x, 30◦ b y = 3 3 x, 45◦ 9 Divide each function through by the denominator, then differentiate. √ x3 − 3x + 8 x2 + 6x x + x a y= b y= x x2 10 Use the full setting-out of the chain rule to differentiate each function. Then find the coordinates of any points where the tangent is horizontal. √ √ √ 1 a y = x2 − 2x + 5 b y = x2 − 2x c y = 7 x2 + 1 d y= √ 1+ x 11 Show that the tangent to y = x 3 at T (−1, −1) meets the curve again at (8, 2). √ Find a, if the normal to y = x + a at the point where x = 4 has gradient −6. √ a Differentiate the semicircle y = 169 − x2 , and find the equations of the tangent and the normal at P(12, 5). Show that the normal passes through the centre. b Find the x-intercept and y-intercept of the tangent, and the area of the triangle enclosed by the tangent and the two axes. c Find the perimeter of this triangle. √ 3x − 2 x c y= √ x 1 12 13 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10G Differentiating powers with fractional indices 423 √ √ 25 − x2 , and let Q(4, 95 ) lie on the curve y = 35 25 − x2 (which is half an ellipse). Find the equations of the tangents at P and at Q, and show that they intersect on the x-axis. √ b Find the equation of the tangent at the point P with x-coordinate x0 > 0 on the curve y = λ 25 − x2 (again, half an ellipse). Let the tangent meet the x-axis at T , let the ellipse meet the x-axis at A(5, 0), and let the vertical line through P meet the x-axis at M. Show that the point T is independent of λ, and show that OA is the geometric mean of OM and OT . 14 a Let the point P(4, 3) lie on the semicircle y = 15 a If y = Axn , show that x dy = ny. dx C dy b If y = n , show that x = −ny. x dx √ dy is a constant. c i If y = a x, show that y dx dy ii Conversely, if y = axn and a 0, find y and show that it is constant if and only if n = 12 or 0. dx ENRICHMENT dy , where y is a function of u, dx 1 . u is a function of v, and v is a function of x. Hence differentiate y = √ 1 + 1 − x2 b Generalise the chain rule to n steps. 16 a Develop a three-step chain rule for the derivative 17 a Write down the first four derivatives of f (x) = x− 2 , leaving the numerator and denominator of the 1 coefficient in factored form. b Write down a formula (using dots) for the nth derivative f (n) x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 424 10H Chapter 10 Differentiation 10H The product rule Learning intentions • Use the product rule to differentiate a product of two functions. The product rule extends the methods for differentiation to functions that are products of two simpler functions. For example, y = x(x − 10)4 is the product y = x × (x − 10)4 , and we shall decompose y as y = uv, where u = x and v = (x − 10)4 . The product rule — differentiating the product of two functions The product rule expresses the derivative in terms of the derivatives of the two simpler functions u and v: 18 The product rule Suppose that the function y=u×v is the product of two functions u and v, each of which is a function of x. Then du dv dy =v +u or y = vu + uv . dx dx dx dy The second form uses the convention of the dash to represent differentiation with respect to x. That is, y = dx du dv and v = . and u = dx dx Proof of the product rule Proof Suppose that x changes to x + Δx. Suppose that as a result, u changes to u + Δu and v changes to v + Δv, and that finally y changes to y + Δy. Then y = uv, and y + Δy = (u + Δu)(v + Δv) and expanding y + Δy = uv + v Δu + u Δv + Δu Δv, so subtracting y = uv, Hence, dividing by Δx, and taking limits as Δx → 0, Δy = v Δu + u Δv + Δu Δv. Δy Δu Δv Δu Δv =v + u + × × Δx, Δx Δx Δx Δx Δx du dv dy =v + u + 0, as required. dx dx dx Using the product rule The working in the right-hand column is the recommended setting out. The first step is to decompose the function into the product of two simpler functions. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10H The product rule Example 23 425 Differentiating the product of two functions Differentiate each function, writing the result in fully factored form. Then state for what value(s) of x the derivative is zero. b x2 (3x + 2)3 a x(x − 10)4 Solution a Let Then y = x(x − 10)4 . dy du dv =v +u dx dx dx = (x − 10)4 × 1 + x × 4(x − 10)3 = (x − 10)3 (x − 10 + 4x) = (x − 10)3 (5x − 10) = 5(x − 10)3 (x − 2), so the derivative is zero for x = 10 and for x = 2. b Let y = x2 (3x + 2)3 . Then Let u= x v = (x − 10)4 . du Then =1 dx dv = 4(x − 10)3 and dx (using the chain rule). and Let u = x2 and v = (3x + 2)3 . = (3x + 2)3 × 2x + x2 × 9(3x + 2)2 Then u = 2x = x(3x + 2) (6x + 4 + 9x) and v = 9(3x + 2)2 . y = vu + uv 2 = x(3x + 2)2 (15x + 4), so the derivative is zero for x = 0, x = − 23 and 4 . for x = − 15 Using the chain rule first to differentiate one of the factors The product rule is sometimes difficult to use with the algebraic functions under consideration at present, because the calculations can easily become quite involved. The rule will seem far more straightforward later in the context of exponential and trigonometric functions. This happens particularly when one of the factors u or v needs to be differentiated first using the chain rule. See the Development Questions 5 and 7 on the following page. Exercise 10H 1 FOUNDATION Copy and complete the setting out below to differentiate 5x(x − 2)4 by the product rule. Let y = 5x(x − 2)4 . Let u = 5x dy du dv and v = (x − 2)4 . Then =v +u dx dx dx du = ··· Then = ··· × ··· + ··· × ··· dx dv = 5(x − 2)4 + 20x(x − 2)3 and = ··· . dx = · · · (take out the common factor) = 5(x − 2)3 (5x − 2). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 426 10H Chapter 10 Differentiation 2 Differentiate each function: i by expanding the product and then differentiating each term, ii using the product rule. a y = x3 (x − 2) 3 b y = (2x + 1)(x − 5) Differentiate these functions using the product rule, identifying the factors u and v in each example. Express your answers in fully factored form, and state the values of x for which the derivative is zero. a y = x(3 − 2x)5 b y = x3 (x + 1)4 c y = x (1 − x) d y = (x − 1)(x − 2)3 5 7 e y = 2(x + 1)3 (x + 2)4 4 c y = (x2 − 3)(x2 + 3) f y = (2x − 3)4 (2x + 3)5 Find the tangents and normals to these curves at the indicated points. a y = x(1 − x)6 at the origin b y = (2x − 1)3 (x − 2)4 at A(1, 1) DEVELOPMENT 5 Differentiate each function using the product rule, giving your answer in fully factored form. One of the factors will require the chain rule to differentiate it. a y = x(x2 + 1)5 b y = 2πx3 (1 − x2 )4 c y = −2x(x2 + x + 1)3 6 a Find the first and second derivatives of y = (x2 − 1)4 . b What are the degrees of the three polynomials y, y and y ? c What values of x are zeroes of all three polynomials? 7 8 9 10 Differentiate y = (x2 − 10)3 x4 , using the chain rule to differentiate the first factor. Hence find the points on the curve where the tangent is horizontal. 3 Show that the function y = x3 (1 − x)5 has a horizontal tangent at a point P with x-coordinate . Show that 8 3 3 × 55 . the y-coordinate of P is 224 Differentiate each function using the product rule. Then combine terms using a common denominator and factoring the numerator completely. State the values of x for which the derivative is zero. √ a y = 6x x + 1 √ b y = −4x 1 − 2x √ c y = 10x2 2x − 1 √ a What is the domain of y = x 1 − x2 ? √ b Differentiate y = x 1 − x2 , using the chain rule to differentiate the second factor, then combine the terms using a common denominator. c Find the points on the curve where the tangent is horizontal. d Find the tangent and the normal at the origin. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10H The product rule 11 427 a Use the product rule to differentiate y = a(x − 1)(x − 5), where a > 0 is a constant, then sketch the curve. b Show that the tangents at the x-intercepts have opposite gradients and meet at a point M whose x-coordinate is the average of the x-intercepts. c Find the point V where the tangent is horizontal. Show that M is vertically above or below V and twice as far from the x-axis. Sketch the situation. 12 Show that if a polynomial f (x) can be written as a product f (x) = (x − a)n q(x) of the polynomials (x − a)n and q(x), where n ≥ 2, then f (x) can be written as a multiple of (x − a)n−1 . What does this say about the shape of the curve near x = a? ENRICHMENT 13 a Show that the function y = xr (1 − x) s , where r, s > 1, has a horizontal tangent at a point P whose x-coordinate p lies between 0 and 1. b Show that the vertical line through P divides the line segment joining O(0, 0) and A(1, 0) in the ratio r : s, and find the y-coordinate of P. What are the coordinates of P if s = r? 14 Establish the rule for differentiating a product y = uvw, where u, v and w are functions of x. Hence differentiate each function, and find the zeroes of y . a x5 (x − 1)4 (x − 2)3 √ b x(x − 2)4 2x + 1 15 Establish the rule for differentiating a product y = u1 u2 . . . un of n functions of x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 428 10I Chapter 10 Differentiation 10I The quotient rule Learning intentions • Use the quotient rule to differentiate a quotient of two functions. The quotient rule extends the formulae for differentiation to functions that are quotients of two simpler functions. For example, 2x + 1 is the quotient of the two functions 2x + 1 and 2x − 1. y= 2x − 1 The quotient rule — differentiating the quotient of two functions The quotient rule, like the product rule, expresses the derivative in terms of the derivatives of the two simpler functions. 19 The quotient rule Suppose that the function u y= v is the quotient of two functions u and v, each of which is a function of x. Then du dv vu − uv dy v dx − u dx = or y = . 2 dx v v2 Proof of quotient rule Proof Differentiate uv−1 using the product rule. Let Then y = uv−1 . dU dV dy =V +U dx dx dx dv −1 du =v − uv−2 dx dx du dv v −u = dx 2 dx v Let V = v−1 . dU du = Then dx dx and by the chain rule, dV dV dv = × dx dv dx dv . = −v−2 dx and v2 after multiplying by 2 . v Example 24 U=u Differentiating the quotient of two functions Differentiate each function, stating any values of x where the derivative is zero. 2x + 1 a y= 2x − 1 x b y= 2 x +1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10I The quotient rule 429 Solution a Let Then b Let Then 2x + 1 Let u = 2x + 1 . 2x − 1 du dv and v = 2x − 1. dy v dx − u dx du = Then =2 dx v2 dx 2(2x − 1) − 2(2x + 1) dv = = 2. and (2x − 1)2 dx −4 = , which is never zero. (2x − 1)2 x y= 2 Let u= x . x +1 vu − uv and v = x2 + 1. y = 2 v Then u = 1 (x2 + 1) − 2x2 = and v = 2x. (x2 + 1)2 1 − x2 = 2 (x + 1)2 (1 − x)(1 + x) = , which is zero when x = 1 or x = −1. (x2 + 1)2 y= Note: Both these functions could have been differentiated using the product rule after writing them as (2x + 1)(2x − 1)−1 and x(x2 + 1)−1 . The quotient rule, however, makes the calculations much easier. Exercise 10I 1 FOUNDATION 2x + 3 by the quotient rule. Copy and complete the setting out below to differentiate 3x + 2 2x + 3 . Let u = 2x + 3 Let y= 3x + 2 dy v u − u v and v = 3x + 2. Then = dx v2 Then u = · · · ··· × ··· − ··· × ··· = and v = ··· ··· −5 . = (3x + 2)2 2 Differentiate each function using the quotient rule, taking care to identify u and v first. Express your answer in fully factored form, and state any values of x for which the tangent is horizontal. x+1 2x 3 − 2x x2 a y= b y= c y= d y= x−1 x+2 x+5 1−x x2 − 1 x x3 − 1 x2 − 9 e y= 2 f y= g y = h y = x +1 1 − x2 x3 + 1 x2 − 4 3 Differentiate y = 4 For each curve below, find the equations of the tangent and normal and their angles of inclination at the given point. x x2 − 4 at K(2, −2) at L(4, 4) a y= b y= 5 − 3x x−1 1 in two different ways: 3x − 2 1 a Use the chain rule with u = 3x − 2 and y = . This is the better method. u b Use the quotient rule with u = 1 and v = 3x − 2. This method is longer. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 430 10I Chapter 10 Differentiation DEVELOPMENT 5 6 7 Differentiate these functions, where m, b, a and n are constants. x2 − a mx + b a y= b y= 2 bx + m x −b x2 . x+1 x2 + k b Find the value of k if f (−3) = 1, where f (x) = 2 . x −k x−α a Differentiate y = . x−β b Show that for α > β, all tangents have positive gradient, and for α < β, all tangents have negative gradient. c What happens when α = β? 5 + 2x : 5 − 2x a using the quotient rule (the better method), b using the product rule, with the function in the form y = (5 + 2x)(5 − 2x)−1 . Differentiate y = 9 a Find the normal to the curve x = 10 Differentiate, stating any zeroes of the derivative: √ x+1 a √ x+2 √ √ x+ 2 a Evaluate f (8) if f (x) = √ √ . x− 2 (r − 6)3 − 1 b Evaluate g (5) if g(r) = . (r − 4)3 + 1 12 xn − 3 xn + 3 a Find the value of c if f (c) = −3, where f (x) = 8 11 c y= t t and y = at the point T where t = 2. t+1 t−1 b Eliminate t from the two equations (by solving the first for t and substituting into the second). Then differentiate this equation to find the gradient of the normal at T . a Sketch the hyperbola y = x−3 b √ x+1 x , showing the horizontal and vertical asymptotes, and state its domain x+1 and range. b Show that the tangent at the point P where x = a is x − (a + 1)2 y + a2 = 0. c Let the tangent at A(1, 12 ) meet the x-axis at I, and let G be the point on the x-axis below A. Show that the origin bisects GI. d Let T (c, 0) be any point on the x-axis. i Show that for c > 0, no tangents pass through T . ii Show that for c < 0 and c −1, there are two tangents through T , the x-coordinates of whose points of contact are opposites of each other. For what values of c are these two points of contact on the same and on different branches of the hyperbola? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10I The quotient rule 431 dy du u , where u is a function of x. Show that y + x = . x dx dx x du dy b Suppose that y = , where u is a function of x. Show that y +u = 1. u dx dx 13 a Suppose that y = 14 a Find the first and second derivatives of: 2x + 1 x2 iii y = x−1 x−1 x−1 x+1 2 b Repeat part i by writing = − , and similarly for part ii–iii. x+1 x+1 x+1 i y= 15 x−1 x+1 ii y = Differentiate, using the most appropriate method. Factor each answer completely. x2 + 3x − 2 1 a (3x − 7)4 b c (2x + 3)(2x − 3) d 2 x x −9 1 3−x e x(4 − x)3 f g (x4 − 1)5 h √ 3+x 2−x 2 x + x + 1 x i (x3 + 5)2 j k 23 x2 (x3 − 1) l √ x + 5 2 x 2 2 √ √ √ 1 1 m x x + x2 x n x− o x3 (x − 1)8 p x+ √ x x ENRICHMENT 16 Sketch a point P on a curve y = f (x) where x, f (x) and f (x) are all positive. Let the tangent, normal and vertical at P meet the x-axis at T , N and M respectively. Let the (acute) angle of inclination of the tangent be θ = ∠PT N, so that y = tan θ. a Using trigonometry, show that: i MN = yy iv cosecθ = 1+y 2 y iii sec θ = v PN = y 1 + y 2 b Hence find the four lengths when x = 3 and: i y = x2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 1+y 2 vi PT = y 1 + y 2 y ii T M = y/y ii y = 3x − 1 x+1 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 432 10J Chapter 10 Differentiation 10J Rates of change Learning intentions • Interpret the derivative as an instantaneous rate. • Solve practical problems involving instantaneous and average rates. • Find where a cubic function is stationary, increasing, and decreasing. • Determine when a cubic function of time is increasing, and decreasing. So far, differentiation has been developed geometrically in terms of tangents to curves. We now return to the initial idea in Section 10A that the derivative is a measure of how some quantity y is changing when y is a function of x and x is changing. This section and the next will interpret the derivative as a rate of change of some quantity Q that is a function of time t. Differentiation with respect to time will give the instantaneous rate at which the quantity Q is changing over time. This will allow us to analyse and solve problems in various practical situations. Two important type of rates are velocity and acceleration, but we have left a fuller discussion of motion until Year 12 because motion requires a great deal more theory than we have at the moment. A couple of questions in the next exercise, however, indicate how velocity and acceleration are particular types of rates. A quantity Q that is a cubic function of time is accessible, because the rate is a quadratic function. A few examples of when such a quantity Q is increasing or decreasing break the ice for a general approach to curve sketching next year. Average rates and instantaneous rates Suppose that a quantity Q is given as a function of time t, as in the diagram to the right. There are two types of rates: • An average rate of change corresponds to a secant. In the diagram, the quantity Q takes the value Q1 at time t1 , and the value Q2 at time t2 . The average rate of change from time t1 to time t2 is the usual gradient formula: Q2 − Q 1 . average rate = t2 − t 1 • An instantaneous rate of change corresponds to a tangent. The instantaneous rate of change at time t1 is the value of the derivative Q Q2 Q1 t1 t2 t dQ at time t = t1 : dt dQ , evaluated at t = t1 . dt One can see from the diagram that as the time t2 gets closer to t1 , the average rate from t = t1 to t = t2 gets closer to the instantaneous rate at t1 (provided that the curve is continuous and smooth at t = t1 ). This is exactly the way in which first-principles differentiation was defined using a limit in Section 10B. instantaneous rate = CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10J Rates of change 433 20 Average and instantaneous rates of change Suppose that a quantity Q is a function of time t. • The average rate of change from the time t = t1 , when the value is Q1 , to the time t = t2 , when the value is Q2 , is the gradient of the secant: Q2 − Q 1 average rate = . t2 − t 1 • The instantaneous rate of change at time t = t1 is the gradient of the tangent, that is, the value of the dQ evaluated at t = t1 : derivative dt dQ , evaluated at t = t1 . instantaneous rate = dt • In this course, the unqualified phrase ‘rate of change’ will always mean the ‘instantaneous rate of change’. • Take particular care to use the correct units for the quantity, the time, and the rate, in final answers. In many problems, it is useful to sketch two graphs, one of the quantity Q as a function of time t, the other of the dQ rate of change also as a function of time t . dt The quantity Q is usually replaced by a more convenient pronumeral, such as P for population, or V for volume, or M for mass. Example 25 Interpreting the derivative as a rate A cockroach plague hit the suburb of Berrawong last year, but was gradually brought under control. The council estimated that the cockroach population P in millions, t months after 1st January, was given by P = 7 + 6t − t2 . dP of the cockroach population. dt b Find the cockroach population on 1st January, and the rate at which the population was increasing at that time. Be careful with the units. c Find the cockroach population on 1st April, and the rate at which the population was increasing at that time. d When did the council manage to stop the cockroach population increasing any further, and what was the population then? e When were the cockroaches finally eliminated? f What was the average rate of increase from 1st January to 1st April? dP g Draw the graphs of P and as functions of time t, and state their domains and ranges. Add to your graph dt of P the tangents and secants corresponding to parts b, d and f. a Differentiate to find the rate of change Solution a The population function was Differentiating, the rate of change was P = 7 + 6t − t2 . dP = 6 − 2t. dt Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 434 10J Chapter 10 Differentiation b When t = 0, P = 7, dP = 6. and when t = 0, dt Thus on 1st January, the population was 7 million cockroaches, and the population was increasing at 6 million cockroaches per month. (The time t is in months, so the rate is ‘per month’.) c On 1st April, t = 3, so P = 16 million cockroaches, dP = 0 cockroaches/month. and dt d We found in part b that the population stopped increasing on 1st April. and that the population then was 16 million. e To find the time when the population was zero, make P = 0, that is, 7 + 6t − t2 = 0 × (−1) t2 − 6t − 7 = 0 (t − 7)(t + 1) = 0 t = 7 or − 1. Hence t = 7, and the cockroaches were finally eliminated on 1st August. (The negative solution is irrelevant because council action started at t = 0.) f From part b, the population was 7 million on 1st January, and from part c, the population was 16 million on 1st April. P 2 − P1 Hence average rate of increase = (gradient of chord formula) t2 − t1 16 − 7 = 3 = 3 million cockroaches per month. g h P 6 dP dt 16 7 3 7 3 7 −8 t Domain: 0 ≤ t ≤ 7, range: 0 ≤ P ≤ 16 t Domain: 0 ≤ t ≤ 7, dP range: −8 ≤ ≤6 dt Increasing, decreasing, and stationary We can see from the left-hand graph above that the cockroach population was increasing when 0 < t < 3 — every tangent to the left of t = 3 has positive gradient. For 3 < t < 7, however, the cockroach population was decreasing — every tangent to the right of t = 3 has negative gradient. At t = 3, the cockroach population was stationary, that is, it was neither increasing nor decreasing and the tangent is horizontal. These three words are used throughout calculus to describe functions, whether or not time is involved. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10J Rates of change 435 21 Increasing, decreasing, stationary Suppose that y = f (x) is defined at x = a, with f (a) also defined. • If f (a) is positive (that is, if the tangent slopes upwards), then the curve is called increasing at x = a. • If f (a) is negative (that is, if the tangent slopes downwards), then the curve is called decreasing at x = a. • If f (a) is zero (that is, if the tangent is horizontal), then the curve is called stationary at x = a. In part h of Example 25 is a graph of the rate at which the population is changing. It is a straight line with negative gradient — this means that the rate of change of the cockroach population was decreasing at a constant rate. Where is a cubic stationary, increasing, and decreasing? dQ These problems require first finding the stationary points, then drawing up table of test points dodging the dt zeroes and discontinuities. The principle being used here to solve the quadratic inequations was introduced in Chapter 6: dQ can only change sign at zeroes and discontinuities. A function here dt Alternatively, sketch the graph and read the answer off the graph, as in Question 7 of the following exercise. Example 26 Where is a cubic stationary, increasing and decreasing? Consider the cubic function Q = t3 − 9t2 + 15t + 1. dQ a Find , and hence find the stationary points of the cubic. dt dQ dQ > 0 by taking a table of test points of dodging around the stationary b Solve the quadratic inequality dt dt points. c What mathematical principle is being used in part b? d Hence state where the curve is increasing, and where is it decreasing. Solution dQ = 3t2 − 18t + 15 dt = 3(t − 1)(t − 5). Hence the stationary points are (1, 8) and (5, −24). dQ dQ dQ b We solve the inequations > 0 and < 0 using a table of values of . dt dt dt t 0 1 2 5 6 dQ 15 0 −9 0 15 dt a gradient of Q + 0 − 0 + dQ dQ > 0 for t < 1 and for t > 5, and < 0 for 1 < t < 5. dt dt dQ c The quadratic can only change sign at zeroes.. dt d Hence Q is increasing for t < 1 and for t > 5, and is decreasing for 1 < t < 5. so CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 436 10J Chapter 10 Differentiation Questions with a diagram or a graph instead of an equation In some problems about rates, the graph of some quantity Q over time is known, but its equation is unknown. Such problems require careful reading of the graph — pay attention to zeroes of the quantity and of the rate, and to where the quantity and rates are increasing, decreasing, and stationary. Occasionally, it may be often useful to draw an approximate sketch of the rate of change of the rate of dQ , and is thus the second change. This second rate-of-change graph is the derivative of the rate of change dt d2 Q derivative 2 of the quantity. dt Example 27 Interpreting a graph of a rate with no function formula The graph to the right shows the temperature T ◦ C of a patient suffering from Symond’s syndrome at time t hours after her admission to hospital at midnight. T a When did her temperature reach its maximum? b When was her temperature increasing most rapidly, when was it decreasing most rapidly, and when was it stationary? c What happened to her temperature eventually? dT at which the temperature is changing. d Sketch the graph of the rate dt e When was the rate of change decreasing, and when was it increasing? 3 6 9 t Solution a The maximum temperature occurred when t = 3, that is, at 3:00 am. b Her temperature was increasing most rapidly at the start when t = 0, that is, at midnight. It was decreasing most rapidly at 6:00 am. It was stationary at 3:00 am. c The patient’s temperature eventually stabilised. dT d The graph of is zero at t = 3, and minimum at t = 6. dt dT → 0, so the t-axis is a horizontal asymptote. As t → ∞, dt e The rate of change of temperature was decreasing from midnight to 6:00 am, and increasing from 6:00 am onwards. Exercise 10J FOUNDATION Note: Except in Questions 1–3, be careful to give the correct units in all final answers. dQ for the rate of change of Q. dt dQ when t = 2. b Hence find the quantity Q and the rate dt 1 a If Q = t3 − 10t2 , find the function 2 a If Q = t2 + 6t, find the function dQ for the rate of change of Q. dt dQ b Hence find the quantity Q and the rate when t = 2. dt c When is the quantity Q: i stationary CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii increasing iii decreasing. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10J Rates of change 3 437 a If Q = 8t − t2 , find the values of Q when t = 1 and t = 3. b Hence find the average rate of change from t = 1 to t = 3. c Similarly find the average rate of change from t = 5 to t = 7. 4 Orange juice is being poured at a constant rate into a glass. After t seconds there are V ml of juice in the glass, where V = 60t. a How much juice is in the glass after 3 seconds? b Show that the glass was empty to begin with. c If the glass takes 5 seconds to fill, what is its capacity? d Differentiate to find the rate at which the glass is being filled. e What sort of function is the derivative in part d? 5 The volume V litres of fuel in a tanker, t minutes after it has started to empty, is given by V = 200(400 − t2 ). Initially the tanker was full. a Find the initial volume of fuel in the tanker. b Find the volume of fuel in the tanker after 15 minutes. c Find the time taken for the tanker to empty. d Show that dV = −400t, and hence find the rate at which the tanker is emptying after 5 minutes. dt DEVELOPMENT 6 [When is a cubic stationary, increasing or decreasing?] Consider the cubic function Q = t3 − 6t2 + 9t. dQ in factored form, and hence find the stationary points of the cubic. a Find dt dQ dQ b Solve the quadratic inequality > 0 by taking a table of test points of dodging around the dt dt stationary points. c What principle did you use in part b? d Hence state where the curve is increasing, and where is it decreasing. 7 [When is a cubic stationary, increasing or decreasing?] dQ , and factor it. a For Q = t3 − 3t2 , find the rate of change dt b Find the zeroes of Q, and the rate of change at those zeroes. c Find the stationary points of the function Q. d i Find Q when t = −1 and t = 4 dQ when t = −1 and t = 1. dt e Hence sketch the curve, and from the graph find: ii find i when Q is positive, and when Q is negative? ii when Q is increasing, and when Q is decreasing? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 438 10J Chapter 10 Differentiation 8 The graph to the right shows the share price P in Penn & Penn Stationery Suppliers t months after 1st January. P a When is the price stationary, neither increasing nor decreasing? b When is the price maximum and when is it minimum? c When is the price increasing and when is it decreasing? 2 d When is the share price increasing most rapidly? 4 6 8 t e When is the share price increasing at an increasing rate? f Sketch a possible graph of 9 dP as a function of time t. dt The graph shows the level of pollution in a river between 1995 and 2000. In 1995, the local council implemented a scheme to reduce the level of pollution in the river. Comment briefly on whether this scheme worked and how the level of pollution changed. Include mention of the rate of change, and a sketch of the graph of the dp . rate dt p 1995 2000 t 10 Water surges into a rock pool, and then out again. The mass M in kilograms of water in the pool t seconds after time zero is given by the function M = 10t − t2 . dM as a function of time t. a Find the rate dt dM b Find (with units) the values of M and when t = 4. dt c Find the value of M when t = 2, and hence find the average rate of change over the time interval from t = 2 to t = 4. d At what times is M zero? e For how long was there water in the pool? dM = zero? f How long after time zero is dt g For how long was the water level in the pool rising, and for how long was it falling? dM as functions of t, showing these results. Add the secant corresponding to h Sketch the graphs of M and dt the result in part c and the tangent corresponding to the result in part f. 11 The share price $P of the Eastcom Bank t years after it opened on 1st January 1970 was P = −0.4t2 + 4t + 2. a What was the initial share price? b What was the share price after one year? c At what rate was the share price increasing after two years? d Find when the share price was stationary. e Explain why the maximum share price was $12, at the start of 1975. f Explain why the rate of change of the price decreased at a constant rate. g The directors decided to close the bank when the share price fell back to its initial value. When did this happen? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10J Rates of change 12 439 For a certain brand of medicine, the amount M present in the blood after t hours is given by M = 3t2 − t3 , for 0 ≤ t ≤ 3. a Sketch a graph of M as a function of t. b When is the amount of medicine in the blood a maximum? dM , and find the axis of symmetry yof the resulting parabola. dt d When is the amount of medicine increasing most rapidly? c Differentiate to find the rate 13 A lightning bolt hits the ground with a tremendous noise. The noise spreads out across the suburbs in a circle of area A, startling people. a The speed of sound in air is about 13 km/sec. Show that the area A in which people are startled as a function of the time t in seconds after the strike is A = π9 t2 . b Find the rate at which the area of startled people is increasing. c Find, correct to four significant figures, the time when the area is 5 km2 , and the rate at which the area is increasing at that time. 14 A 3000 cm long water trough has the shape of a triangular prism. Its cross-section is an isosceles triangle with an apex of 90◦ , and the trough is built with the apex at the bottom. a i Find the area of an isosceles triangle with apex angle 90◦ , and height h centimetres. ii Find the volume of water in the water trough when the water height is h cm. b The water trough was initially empty, and then was filled with water at a constant rate of 27 litres per minute. i Find the height h cm of the water after t minutes, and the rate the water was rising. ii Find the water height, and the rate the water was rising, after 25 minutes. iii Find when the water height was 30 cm, and the rate the water was then rising. ENRICHMENT 15 A glass sphere with an opening at the top is being filled slowly with water in a controlled way so that the height rises at a constant rate. Show that the rate at which the surface area of the water is changing is decreasing at a constant rate. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 440 10K Chapter 10 Differentiation 10K Average velocity and average speed Learning intentions • Understand the displacement of a particle moving on a line as a function of time. • Calculate average velocity, and interpret it as the gradient of a secant. • Calculate distance, distance travelled, and average speed. Velocity is also a rate, but the ideas and language of motion are quite different from the rates of the previous section, and readers may like to delay these two sections until later in the year. Motion will be covered far more thoroughly in Year 12. These last two sections set up motion in one dimension, using a function or graph to relate the time and displacement of an object. Average velocity is the gradient of a secant on the graph, and instantaneous velocity is the gradient of a tangent. Motion in one dimension — displacement and time When a particle is moving in one dimension, turn the line that it is moving on into a number line with an origin and a positive direction. Its displacement x at any time t after time zero is then read off the number line. Thus its motion is described by giving x as a function of the time t, or by drawing the graph. Suppose, for example, that a ball is hit vertically upwards from ground level, and lands 8 seconds later in the same place. Its motion can be described approximately by the following equation of motion and table of values, x = 5t(8 − t) 0 2 4 6 8 x 80 x 0 60 80 60 0 60 t 40 Here x is the height in metres of the ball above the ground t seconds after it is thrown. The diagram to the right shows the path of the ball up and down along the same vertical line. 20 • This vertical line has thus been made into a number line, with the ground as the origin, upwards as the positive direction, and metres as the units of displacement. • Time has also become a number line. The origin of time is when the ball is thrown, and the units of time are seconds. Graphing the equation of motion The graph to the right is the resulting graph of the equation of motion x = 5t(8 − t). The horizontal axis is time, and the vertical axis is displacement — the graph must not be mistaken as a picture of the ball’s path. The graph is part of a parabola with a vertex at (4, 80), so the ball reaches maximum height 80 metres after 4 seconds. When t = 8, the height is zero, and the ball returns to the ground. The equation of motion therefore has restricted domain and range: Domain: 0 ≤ t ≤ 8 and x 80 60 B A C 40 20 O 2 4 6 D 8 t Range: 0 ≤ x ≤ 80. Most equations of motion have this sort of restriction on the domain of t. In particular, it is a convention of this course that negative values of time are excluded unless the question specifically allows it. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10K Average velocity and average speed 441 22 Motion in one dimension • Motion in one dimension is specified by making the line into a number line, and giving the displacement x as a function of time t after time zero. • Negative values of time are excluded unless otherwise stated. Example 28 Working with an equation of motion Consider the example above, where x = 5t(8 − t). a Find the height of the ball after 1 second. b At what other time is the ball at this same height above the ground? Solution a When t = 1, x = 5 × 1 × 7 = 35. Hence the ball is 35 metres above the ground after 1 second. b To find when the height is 35 metres, solve the equation x = 35 for t. Substituting into x = 5t(8 − t) gives 5t(8 − t) = 35 t(8 − t) = 7 ÷5 8t − t2 − 7 = 0 × (−1) t2 − 8t + 7 = 0 (t − 1)(t − 7) = 0 t = 1 or 7. Hence the ball is 35 metres high after 1 second, and again after 7 seconds. Average velocity During its ascent, the ball in the example above moved 80 metres upwards. This is a change in displacement of +80 metres in 4 seconds, giving an average velocity of 80 4 = 20 metres per second. Average velocity thus equals the gradient of the secant OB on the displacement–time graph — but be careful, because there are different scales on the two axes. Hence the formula for average velocity is the familiar gradient formula. 23 Average velocity Suppose that a particle has displacement x = x1 at time t = t1 , and displacement x = x2 at time t = t2 . Then change in displacement x2 − x1 = . average velocity = change in time t2 − t 1 That is, on the displacement–time graph, average velocity = gradient of the secant. During its descent, the ball moved 80 metres downwards in 4 seconds, which is a change in displacement of 0 − 80 = −80 metres. The average velocity is therefore −20 metres per second, which is equal to the gradient of the secant BD. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 442 10K Chapter 10 Differentiation Example 29 Finding average velocities Consider again the example x = 5t(8 − t). Find the average velocities of the ball: a during the first second, b during the fifth second. Solution The first second stretches from t = 0 to t = 1, and the fifth second stretches from t = 4 to t = 5. The displacements at these times are given in the table to the right. a Average velocity during 1st second t 0 1 4 5 x 0 35 80 75 b Average velocity during 5th second x 2 − x1 t2 − t1 35 − 0 = 1−0 = 35 m/s. x2 − x1 t2 − t 1 75 − 80 = 5−4 = −5 m/s. = = Distance travelled The change in displacement can be positive, negative, or zero. Distance, however, is always positive or zero. In the previous example, the change in displacement during the 4 seconds from t = 4 to t = 8 is −80 metres, but the distance travelled is 80 metres. The distance travelled by a particle also takes into account any journey and return. Thus the total distance travelled by the ball is 80 + 80 = 160 metres — even though the ball’s change in displacement over the first 8 seconds is zero, because the ball is back at its original position on the ground. 24 Distance travelled • The distance travelled takes into account any journey and return. • Distance travelled, like all distances, can never be negative. Average speed The average speed is the distance travelled divided by the time taken. Average speed, unlike average velocity, can never be negative. 25 Average Speed distance travelled time taken Average speed, like all speeds, can never be negative. average speed = During the 8 seconds of its flight, the change in displacement of the ball is zero, but the distance travelled is 160 metres, so 160 0−0 average speed = average velocity = 8−0 8 = 0 m/s, = 20 m/s. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10K Average velocity and average speed Example 30 443 Contrasting average velocity and average speed Consider once again the original example x = 5t(8 − t). Find the average velocity and the average speed of the ball: b during the last six seconds. a during the eighth second, Solution The eighth second stretches from t = 7 to t = 8 and the last six seconds stretch from t = 2 to t = 8. The displacements at these times are given in the table to the right. t 0 2 7 8 x 0 60 35 0 a During the eighth second, the ball moves 35 metres down from x = 35 to x = 0. 0 − 35 8−7 = −35 m/s. Hence average velocity = Also distance travelled = 35 metres, so average speed = 35 m/s. b During the last six seconds, the ball rises 20 metres from x = 60 to x = 80, and then falls 80 metres from x = 80 to x = 0. 0 − 60 Hence average velocity = 8−2 = −10 m/s. Also so distance travelled = 20 + 80 = 100 metres, 100 average speed = 6 = 16 23 m/s. Exercise 10K 1 FOUNDATION A particle is moving with displacement function x = t2 + 2, where time t is in seconds and displacement x is in metres. a Find the position when t = 0. b Find the position when t = 4. c Find the average velocity during the first 4 seconds, using the definition: average velocity = 2 change in displacement change in time For each displacement function below, copy and complete the table of values to the right. Hence find the average velocity during the first 2 seconds. The units in each part are seconds and centimetres. a x = 12t − t2 b x = (t − 2)2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 c x = t3 − 4t + 3 t 0 2 x d x = 2t © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 444 10K Chapter 10 Differentiation 3 A particle moves according to the equation x = t3 − 4t, where x is the displacement in centimetres from the origin O at time t seconds after time zero. a Copy and complete the table of values to the right. t b Hence find the average velocity: x i during the first second (that is, from t = 0 to t = 1) 0 1 2 3 4 2 3 ii during the second second (that is, from t = 1 to t = 2) iii during the third second (that is, from t = 2 to t = 3) iv during the fourth second (that is, from t = 3 to t = 4). 4 A particle moves according to the equation x = t2 − 4, where x is the displacement in metres from the origin O at time t seconds after time zero. a Copy and complete the table of values to the right. t b Hence find the average velocity: x 0 1 i during the first second ii during the first two seconds iii during the first three seconds iv during the third second. c Use the table of values above to sketch the displacement–time graph. Then add the secants corresponding to the average velocities calculated in part b. 5 A piece of cardboard is shot 120 metres vertically into the air by an explosion and floats back to the ground, landing at the same place. The graph to the right gives its height x metres above the ground t seconds after the explosion. a When was the cardboard ascending, and when descending? x 120 72 b Copy and complete the following table of values. t 0 4 8 4 12 8 12 t x c What is the total distance travelled by the cardboard? d Find the average speed of the cardboard during its travels, using the formula distance travelled . time taken e Find the average velocity during: average speed = i the ascent ii the descent iii the full 12 seconds. f At roughly what other time was the height the same as the height when t = 8? g What are the domain and range of the equation of motion in this graph? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10K Average velocity and average speed 6 445 A particle moves according to the equation x = 4t − t2 , where distance is in metres and time is in seconds. a Copy and complete the table of values to the right. t b Hence sketch the displacement–time graph. x 0 1 2 3 4 c Find the total distance travelled during the first 4 seconds. Then find the average speed during this time. d Find the average velocity during the time: i from t = 0 to t = 2 ii from t = 2 to t = 4 iii from t = 0 to t = 4. e Add to your graph the secants corresponding to the average velocities in part d. DEVELOPMENT 7 Michael the mailman rides his bicycle 1 km up a hill at a constant speed of 10 km/hr. He then turns around and rides back down the hill at a constant speed of 30 km/hr. a How many minutes does he take to travel: i the first kilometre, when he is riding up the hill ii the second kilometre, when he is riding back down again? b Use these values to draw a displacement–time graph, with the time axis in minutes. c What is his average speed over the total 2 km journey? d What is the average of his speeds up and down the hill? 8 Eleni is practising reversing in her driveway. Starting 8 metres from the gate, she reverses to the gate, and pauses. Then she drives forward 20 metres, and pauses. Then she reverses to her starting point. The graph to the right shows her distance x in metres from the front gate after t seconds. a What is her average velocity: x i during the first 8 seconds 20 ii while she is driving forwards iii while she is reversing the second time? 8 b Find the total distance she travelled, and her average speed, over the 30 seconds. c Find her change in displacement, and her average velocity, over the 30 seconds. d What would her average speed have been had she not paused at the gate and garage? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 8 12 17 24 30 t © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 446 10K Chapter 10 Differentiation 9 A girl is leaning over a bridge 4 metres above the water, playing with a weight on the end of a spring. The graph shows the height x in metres of the weight above the water as a function of time t seconds after she first drops it. ii at x = 1 17 iii at x = − 12 ? 4 −1 b At what times is the weight: i at the water surface 4 2 a How many times is the weight: i at x = 3 x 8 14 t ii above the water surface? c How far above the water does it rise again after it first touches the water, and when does it reach this greatest height? d What is the weight’s greatest depth under the water and when does it occur? e What happens to the weight eventually? f What is its average velocity: i during the first 4 seconds ii from t = 4 to t = 8 iii from t = 8 to t = 17? g What distance does it travel: i over the first 4 seconds ii over the first 8 seconds iii over the first 17 seconds iv eventually? h What is its average speed over the first: i 4 seconds ii 8 seconds iii 17 seconds? i What are the domain and range of the equation of motion in this graph? ENRICHMENT 10 √ A particle moves according to the equation x = 2 t , for t ≥ 0, where distance x is in centimetres and time t is in seconds. a Copy and complete the table of values to the right, and sketch the curve. t b Hence find the average velocity as the particle moves: x 0 i from x = 0 to x = 2, ii from x = 2 to x = 4, iii from x = 4 to x = 6, iv from x = 0 to x = 6. 2 4 6 8 c What does the equality of the answers to parts b ii and b iv tell you about the corresponding secants in part a? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10L Instantaneous velocity and speed 447 10L Instantaneous velocity and speed Learning intentions • Interpret instantaneous velocity as the gradient of a tangent. • Use the derivative to solve problems about instantaneous velocity and speed. If I drive the 160 km from Sydney to Newcastle in 2 hours, my average velocity is 80 km per hour. But my instantaneous velocity during the journey, as displayed on the speedometer, may range from zero at traffic lights to 110 km per hour on expressways. On the displacement–time graph: • An average velocity corresponds to the gradient of a secant. • An instantaneous velocity corresponds to the gradient of a tangent. Instantaneous velocity as a derivative Thus, as we have seen many times before in this chapter, we can find the instantaneous velocity by differentiating the displacement–time function. The usual symbol for instantaneous velocity ia v, but it can also be written as dx , or as ẋ. dt 26 Instantaneous velocity • The instantaneous velocity v of a particle in motion is the gradient of the tangent on the displacement–time graph, dx which can also be written as v = ẋ . v= dt • The dot over any symbol means differentiation only with respect to time t. The notation ẋ, originally introduced by Newton, is yet another way of writing the derivative. The dot over the x, dx or over any symbol, stands for differentiation only with respect to time t. Thus the symbols v, and ẋ all mean dt the velocity. Example 31 Constructing the velocity–time graph Here again is the displacement–time graph of the ball moving with equation x = 5t(8 − t). x 80 B a Differentiate to find the equation for the velocity v, draw up a table of values at 2-second intervals. and sketch the velocity–time graph. b Measure the gradients of the tangents that have been drawn at A, B. and C on the displacement–time graph, and compare your answers with the table of values in part a. c With what velocity was the ball originally hit? d What is its impact speed when it hits the ground? 60 A 40 20 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 C 4 6 8 t © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 448 10L Chapter 10 Differentiation Solution x = 40t − 5t2 , a Expanding, the motion equation is and differentiating, v = 40 − 10t. The graph of velocity is a straight line, with v-intercept 40 and gradient −10. t 0 2 4 6 8 v 40 20 0 −20 −40 v 40 8 2 4 t 6 − 40 b These values agree with the measurements of the gradients of the tangents at A where t = 2, at B where t = 4, and at C where t = 6. (Be careful to take account of the different scales on the two axes.) c When t = 0, v = 40, so the ball was originally hit upwards at 40 m/s. d When t = 8, v = −40, so the ball hits the ground again at 40 m/s. Instantaneous speed The instantaneous speed is the absolute value |v| of the velocity. This means that instantaneous speed, like average speed, can never be negative. 27 Instantaneous speed • The instantaneous speed is the absolute value |v| of the instantaneous velocity. • From now on, the unqualified words velocity and speed, written on their own, will always mean instantaneous velocity and instantaneous speed. Vector and scalar quantities • Displacement and velocity are called vector quantities, meaning that they have a direction built into them. In the example above, a negative velocity means the ball is going downwards, and a negative displacement would mean it was below ground level. • Distance and speed, however, are called scalar quantities — distance is the magnitude of the displacement, and speed is the magnitude of the velocity — and neither can be negative. Example 32 Constructing the graph of instantaneous speed From the previous worked example, construct the graph of the instantaneous speed as a function of time. Solution We are graphing |v|, so this is the graph of speed = |40 − 10t|. Section 5E explained how to graph such a function. speed 40 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 4 6 8 t © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10L Instantaneous velocity and speed Example 33 449 Contrasting vector quantities with scalar quantities A particle moves with displacement x = t3 − 6t − 2, in units of metres and seconds. a Differentiate to find the velocity function. b Find the displacement, distance from the origin, velocity, and speed when: ii t = 3. i t=0 Solution x = t3 − 6t − 2, a The displacement equation is v = 3t2 − 6. and differentiating, b i When t = 0, x = −2 and v = −6. Thus when t = 0, the displacement is −2 metres, the particle is 2 metres from the origin, the velocity is −6 m/s, and the speed is 6 m/s. ii When t = 3, x = 27 − 18 − 2 = 7, and v = 27 − 6 = 21. Thus when t = 3, the displacement is 7 metres, the particle is 7 metres from the origin, the velocity is 21 m/s, and the speed is 21 m/s. Finding when a particle is stationary dx = 0. This is the origin of the term dt ‘stationary point’ that we have been using to describe a point on a graph where the derivative is zero. For example, the ball in the first example was stationary for an instant at the top of its flight when t = 4, because the velocity was zero at the instant when its motion changed from upwards to downwards. A particle is said to be stationary when its velocity v is zero, that is, when 28 Finding when a Particle is Stationary • A particle is stationary when its velocity is zero. • To find when a particle is stationary, put v = 0 and solve for t. Example 34 Finding when a particle is stationary A particle moves so that its distance in metres from the origin at time t seconds is given by x = 13 t3 − 6t2 + 27t − 18. a Find the times when the particle is stationary. b Find its distance from the origin at these times. Solution a The displacement function is and differentiating, x = 13 t3 − 6t2 + 27t − 18, v = t2 − 12t + 27 = (t − 3)(t − 9), so the particle is stationary after 3 seconds and after 9 seconds. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 Continued on next page. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 450 10L Chapter 10 Differentiation b When t = 3, x = 9 − 54 + 81 − 18 = 18, and when t = 9, x = 243 − 486 + 243 − 18 = −18. Thus the particle is 18 metres from the origin on both occasions. Exercise 10L FOUNDATION Note: Most questions in this exercise are long in order to illustrate how the physical situation of the particle’s motion is related to the mathematics and the graph. The calculations should now be well known, but the physical interpretations can be confusing. 1 A particle is moving with displacement function x = 20 − t2 , in units of metres and seconds. a Differentiate to find the velocity v as a function of time t. b Find the displacement and velocity when t = 3. c What are the distance from the origin and the speed when t = 3? 2 For each displacement function below, differentiate to find the velocity function v. Then find the displacement and velocity when t = 1. The units are metres and seconds. a x = 5t2 − 10t b x = 3t − 2t3 c x = t4 − t2 + 4 3 A particle’s displacement function is x = t2 − 10t, in units of centimetres and seconds. a Differentiate to find the velocity v = ẋ as a function of t. b What are the displacement, the distance from the origin, the velocity, and the speed after 3 seconds? c When is the particle stationary, and where is it then? 4 A particle moves on a horizontal line so that its displacement x cm to the right of the origin at time t seconds is x = t3 − 6t2 . a Differentiate to find ẋ as a function of t. b Where is the particle initially, and what is its speed then? c At time t = 3, is the particle to the left or to the right of the origin? d At time t = 3, is the particle travelling to the left or to the right? e Show that the particle is stationary when t = 4, and find where it is at this time. f Show that the particle is at the origin when t = 6, and find its velocity and speed at this time. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 10L Instantaneous velocity and speed 5 451 A cricket ball is thrown vertically upwards. Its height x in metres at time t seconds after it is thrown is given by x = 20t − 5t2 . a Find v as a function of t. Then sketch graphs of x, and of v, as functions of t. b Write down the domain and range of the displacement and velocity functions. c Find the speed at which the ball was thrown. d Find when it returns to the ground (that is, when x = 0), and show that its speed then is equal to the initial speed. e Find its maximum height above the ground, and the time taken to reach this height. DEVELOPMENT 6 A smooth piece of ice is projected up a smooth inclined surface, as shown to the right. Its distance x in metres up the surface at time t seconds is x = 6t − t2 . x a Find the velocity v as a function of t. b Sketch the graphs of displacement x and velocity v. c In which direction is the ice moving: ii when t = 4? i when t = 2? d When is the ice stationary, for how long is it stationary, and where is it then? e Show that the average velocity over the first 2 seconds is 4 m/s. Then find the time and place at which the instantaneous velocity equals this average velocity. f Show that the average speed during the first 3 seconds, the next 3 seconds, and the first 6 seconds are all the same. 7 A stone was thrown vertically upwards. The graph to the right shows its height x metres at time t seconds after it was thrown. a What was the stone’s maximum height, how long did it take to reach it, and what was its average speed during this time? b Draw tangents and measure their gradients to find the velocity of the stone at times t = 0, 1, 2, 3, 4, 5 and 6. c For what length of time was the stone stationary at the top of its flight? d i Estimate the speed at t = 5. x 45 40 25 0 1 2 3 4 5 6 t ii Estimate the average speed from t = 1.5 to t = 4.5. e The graph is concave down everywhere. How is this relevant to the motion? f Draw graphs of the velocity, and the speed, of the stone from t = 0 to t = 6. g What does velocity–time graph tell you about what happened to the velocity during these 6 seconds? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 452 10L Chapter 10 Differentiation 8 A particle is moving horizontally so that its displacement x metres to the right of the origin at time t seconds is given by the graph to the right. x 8 a In the first 10 seconds, what is its maximum distance from the origin and when 4 does it occur? b By examining the gradient, find when the particle is: i stationary ii moving to the right 3 iii moving to the left. 6 9 12 t c When does it return to the origin, and what is its velocity then? d At about what times are: i the displacement ii the velocity about the same as those at t = 2? e Sketch (roughly) the graphs of the velocity v and the speed. ENRICHMENT 9 A particle is moving vertically according to the graph shown to the right, where upwards has been taken as positive. x 5 a At what times is this particle: 4 i below the origin? ii moving downwards? b At what time is its speed greatest? c At about what times are: −3 −5 8 12 16 t i the distance from the origin ii the velocity about the same as those at t = 3? d How many times between t = 4 and t = 12 is the instantaneous velocity equal to the average velocity during this time? e How far will the particle eventually travel? f Draw an approximate sketch of the graph of v as a function of time. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 10 review 453 Chapter 10 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 10 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist and Skillsheet • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Along with the checklist, download the Skillsheet to target specific learning intentions and practice the core skills of this chapter. Chapter Review Exercise Use the definition of the derivative f (x) = limh→0 first principles. a f (x) = x2 + 5x 2 a y = x3 − 2x2 + 3x − 4 b y = x6 − 4x4 d y = (x + 3)(x − 2) e y = (2x − 1)(2 − 3x) c f (x) = 3x2 − 2x + 7 1 2 h y = 3x + 3x c y = 3x2 (x − 2x3 ) − 12 f y = 3x−2 − 2x−1 i y = x−2 (x2 − x + 1) Find the first and second derivatives of each function: a f (x) = x4 + x3 + x2 + x + 1 4 b f (x) = 6 − x2 Write down the derivative of each function. You will need to expand any brackets. g y = 4x3 − 4x−3 3 f (x + h) − f (x) to find the derivative of each function by h Review 1 b f (x) = 5x−2 Find the family of curves whose derivative is: dy dy = 3x2 + 4 = 7 − 12x − 12x2 a b dx dx c f (x) = 8x− 2 1 c dy = 20x4 − 12x2 + 4 dx 5 Rewrite each function in index notation and then differentiate it. Write each final answer without negative or fractional indices. √ 1 3 a y= b y= 2 c y=7 x x 6x √ √ 6 d y = 144x e y = −3x x f y= √ x 6 Divide each function through by the numerator and then differentiate it. Give each final answer without negative or fractional indices. 3x4 − 2x3 x3 − x2 + 7x 5x3 − 7 a y= b y = c y = 2x√ x√ x2 √ 2 x + 2x + 1 4x + 5 x 2x2 x + 3x x d y= e y= f y= √ x x2 x CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 454 Chapter 10 Differentiation Review 7 Use the formula d (ax + b)n = an(ax + b)n−1 to differentiate: dx a y = (3x + 7)3 d y= 8 1 (2 − 7x)2 b y = (5 − 2x)2 e y= √ 5x + 1 1 5x − 1 1 f y= √ 1−x c y= Use the chain rule to differentiate: a y = (7x2 − 1)3 d y= 1 (x2 − 1)3 b y = (1 + x3 )−5 c y = (1 + x − x2 )8 e y= f y= √ √ 9 − x2 1 9 − x2 9 Differentiate each function, using the product or quotient rule. Factorise each answer completely. x2 a y = x9 (x + 1)7 b y= c y = x2 (4x2 + 1)4 1−x 2x − 3 x2 + 5 d y= e y = (x + 1)5 (x − 1)4 f y= 2x + 3 x−2 10 Differentiate y = x2 + 3x + 2. Hence find the gradient and the angle of inclination, correct to the nearest minute when appropriate, of the tangents at the points where: a x=0 11 b x = −1 c x = −2 a Find the equations of the tangent and normal to y = x3 − 3x at the origin. b Find the equation of the tangent and normal to the curve at the point P(1, −2). c By solving f (x) = 0, find the points on the curve where the tangent is horizontal. d Find the points on the curve where the tangent has a gradient of 9. 12 a Find the equations of the tangent and normal to y = x2 − 5x at the point P(2, −6). b The tangent and normal at P meet the x-axis at A and B respectively. Find the coordinates of A and B. c Find the length of the line segment AB and the area of the triangle PAB. 13 Find the tangents to the curve y = x4 − 4x3 + 4x2 + x at the origin and at the point where x = 2, and show that these two lines are the same. 14 Find any points on each curve where the tangent has the given angle of inclination. a y = 13 x3 − 7, θ = 45◦ b y = x2 + 13 x3 , θ = 135◦ 15 a Find the points on y = x3 − 4x where the tangents are parallel to : x + y + 2 = 0. b Find the equations of these two tangents, and show that is one of them. 16 Find the x-coordinates of any points on y = (4x − 7)3 where the tangent is: a parallel to y = 108x + 7 b perpendicular to x + 12y + 6 = 0 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 10 review 17 a Find the values of t where Q = t3 − 15t2 + 27t has stationary points. A ball is thrown vertically upwards from the ground, and its height h metres after t seconds is given by h = −5t2 + 50t. Review dQ dQ to solve < 0. dt dt c Hence say where Q is increasing and where it is decreasing. b Draw up a table of test points of 18 455 a When does the ball hit the ground again? b Differentiate to find the velocity function v of the ball. c At what speed did the ball leave the thrower’s hand? d When was the ball at the top of its flight, and how high does it go? e How far in total did the ball travel, and what was its average speed? f What was its average speed during the first 3 seconds? g At what times was it travelling at a speed less that 20 m/s? h At what times was it higher than 80 m above the ground? 19 A huge fireball high in the air is exploding at the speed of sound, which is about 13 km/s. a Using the formula V = 43 πr3 for the volume of a sphere, show that the volume V of the fireball after 3 t seconds is V = 4π 81 t . b Differentiate to find the rate at which the volume is expanding. c Find, correct to two significant figures, the volume and the rate after 0.1 seconds. d Find, correct to two significant figures, when the volume is increasing by 1 km3 /s. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11 Polynomials Chapter introduction Once the variable x has been introduced into arithmetic, polynomial expressions such as 4x3 − 7x + 5 arise naturally. Indeed every expression that can be formed using just the three operations of addition, subtraction and multiplication can be written as a polynomial — simply expand brackets, and collect like terms. The course has already discussed linear and quadratic functions in detail, and this chapter begins the systematic study of polynomials of higher degree. Polynomials were mentioned briefly in Section 3G, and some readers will already have familiarity with their graphs, with long division of polynomials, and with the remainder and factor theorems. The later sections in this chapter develop the relationships between the coefficients and the zeroes. The problem of factoring a given polynomial is a constant theme, and the final Section 11G applies the methods of the chapter to geometric problems about polynomial curves, circles, and rectangular hyperbolas. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11A The language of polynomials 457 11A The language of polynomials Learning intentions • Define polynomial, leading term and coefficient, degree, monic, constant. • Classify zero, linear, quadratic, cubic and quartic polynomials. • Add, subtract, and multiply polynomials, and observe the resulting degrees. • Define identically equal polynomials, and use the relationship to the coefficients. Polynomials are expressions such as the quadratic x2 − 5x + 6 or the quartic 3x4 − 23 x3 + 4x + 7. They have occurred already in the course, but now our language and notation needs to be more precise. Polynomials and polynomial functions 1 Polynomials • A polynomial is an expression of the form P(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , where a0 , a1 , . . . , an are real numbers, and n is a whole number. • A polynomial function is a function that can be written as a polynomial. The term a0 is called the constant term. This is the value of the polynomial at x = 0, so a0 is the y-intercept of its graph. Leading term and degree The term of highest index with non-zero coefficient is called the leading term of the polynomial. Its coefficient is called the leading coefficient, and its index is called the degree. For example, the polynomial P(x) = −5x6 − 3x4 + 2x3 + x2 − x + 9 has leading term −5x6 and leading coefficient −5, and has degree 6, written as deg P(x) = 6. For convenience, the constant term a0 is regarded in this context as being a term of index 0. This is because a0 can be written as a0 x0 provided that x 0. But never actually write the term a0 as a0 x0 , because 00 is undefined. A monic polynomial is a polynomial whose leading coefficient is 1. For example, P(x) = x3 − 2x2 − 3x + 4 is monic. Every non-zero polynomial is a unique multiple of a monic polynomial because an 0, so an−1 n−1 a1 a0 n n−1 n . x + ··· + x+ an x + an−1 x + · · · + a1 x + a0 = an x + an an an Some names of polynomials The zero polynomial, and polynomials of low degree, have standard names. • The zero polynomial Z(x) = 0 is a special case. It has a constant term 0. But it has no term with a non-zero coefficient. Hence it has no leading term, no leading coefficient, and most importantly, no degree. Its graph is the x-axis — meaning that every real number is a zero of the zero polynomial. • A constant polynomial is a polynomial whose only term is the constant term: P(x) = 4, Q(x) = − 35 , CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 R(x) = π, Z(x) = 0. © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 458 11A Chapter 11 Polynomials Apart from the zero polynomial, all constant polynomials have degree 0, have no zeroes, and are equal to their leading term and to their leading coefficient. • A linear polynomial is a polynomial of degree 1: P(x) = x − 3, Q(x) = 4x + 7, R(x) = − 12 x. Warning: A constant function is a linear function. But a constant polynomial is NOT a linear polynomial, because it does not have degree 1 — it has degree 0 or is the zero polynomial. • A polynomial of degree 2 is called a quadratic polynomial: P(x) = 3x2 + 4x − 1, Q(x) = − 12 x − x2 , R(x) = 9 − x2 . Notice that the coefficient of x2 must be non-zero for the degree to be 2. • Polynomials of higher degree are called: cubics (degree 3), 2 quartics (degree 4), quintics (degree 5), ... . The degree of a polynomial P(x) • The leading term of P(x) is the term of highest index with non-zero coefficient. The degree of P(x) is the index of the leading term. The leading coefficient of P(x) is the coefficient of the leading term. A monic polynomial is a polynomial whose leading coefficient is 1. • The zero polynomial Z(x) = 0 has no leading term, and so has no degree. • A constant polynomial P(x) = a0 is a polynomial with only a constant term. If a0 0, it has degree 0. But if a0 = 0, it is the zero polynomial Z(x) = 0. • A linear polynomial P(x) = a1 x + a0 is a polynomial of degree 1, so a1 0. • Polynomials of higher degree are called quadratic, cubic, quartic, quintic, . . . Addition and subtraction When two polynomials are added or subtracted, the results are again polynomials: (5x3 − 4x + 3) + (3x2 − 3x − 2) = 5x3 + 3x2 − 7x + 1 (5x3 − 4x + 3) − (3x2 − 3x − 2) = 5x3 − 3x2 − x + 5 The zero polynomial Z(x) = 0 is the zero for addition, in the usual sense that P(x) + Z(x) = P(x), for all polynomials P(x). The opposite polynomial −P(x) of any polynomial P(x) is obtained by taking the opposite of every coefficient. Then the sum of P(x) and −P(x) is the zero polynomial. For example, (4x4 − 2x2 + 3x − 7) + (−4x4 + 2x2 − 3x + 7) = 0. The degree of the sum or difference is usually the maximum of the degrees of the two polynomials, as in the two examples above, where the polynomials have degrees 2 and 3 and their sum has degree 3. If, however, the two polynomials have the same degree, the leading terms may cancel out and disappear, for example, (x2 − 3x + 2) + (9 + 4x − x2 ) = x + 11, which has degree 1, or the two polynomials may be opposites, in which case everything cancels out so that their sum is zero and thus has no degree. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11A The language of polynomials 3 459 Degree of the sum and difference Let P(x) and Q(x) be non-zero polynomials of degree n and m respectively. • If n m, then deg P(x) + Q(x) = maximum of m and n. • If n = m, then deg P(x) + Q(x) ≤ n or P(x) + Q(x) = 0. Multiplication Any two polynomials can be multiplied, giving another polynomial: (3x3 + 2x + 1) × (x2 − 1) = (3x5 + 2x3 + x2 ) − (3x3 + 2x + 1) = 3x5 − x3 + x2 − 2x − 1 The constant polynomial I(x) = 1 is the identity for multiplication, in the sense that P(x) × I(x) = P(x), for all polynomials P(x). Multiplication by the zero polynomial, on the other hand, always gives the zero polynomial. If two polynomials are non-zero, then the degree of their product is the sum of their degrees, because the leading term of the product is always the product of the two leading terms. 4 Degree of the product If P(x) and Q(x) are non-zero polynomials, then deg P(x) × Q(x) = deg P(x) + deg Q(x). Identically equal polynomials We need to be clear what is meant by saying that two polynomials are the same. 5 Identically equal polynomials • Two polynomials P(x) and Q(x) are called identically equal if they are equal for all values of x: P(x) = Q(x), for all x, often written as P(x) ≡ Q(x). • If two polynomials are identically equal, then the corresponding coefficients of the two polynomials are equal. The second dotpoint needs proof. This is developed in the Enrichment section of Exercise 11B. In the meantime, here is an example that uses the result. Example 1 Finding coefficients in identically equal polynomials Find a, b, c, d, and e if ax4 + bx3 + cx2 + dx + e = (x2 − 3)2 for all x. Solution Expanding, (x2 − 3)2 = x4 − 6x2 + 9. Now comparing coefficients, a = 1, b = 0, c = −6, d = 0 and e = 9. Factoring polynomials The most significant problem of this chapter is the factoring of a given polynomial. For example, x(x + 2)2 (x − 2)2 (x2 + x + 1) = x7 + x6 − 7x5 − 8x4 + 8x3 + 16x2 + 16x is a routine expansion of a factored polynomial, but it is not at all clear how to move in the other direction from the expanded form back to the factored form. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 460 11A Chapter 11 Polynomials Polynomial equations If P(x) is a polynomial, then the equation formed by putting P(x) = 0 is a polynomial equation. For example, using the polynomial in the previous paragraph, we can form the polynomial equation x7 + x6 − 7x5 − 8x4 + 8x3 + 16x2 + 16x = 0. Solving polynomial equations and factoring polynomial functions are closely related. Using the factoring of the previous paragraph, x(x + 2)2 (x − 2)2 (x2 + x + 1) = 0, so the solutions are x = 0, x = 2 (double root) and x = −2 (double root), where the quadratic factor x2 + x + 1 has no zeroes, because Δ = −3. Thus factoring a polynomial into linear and irreducible quadratic factors solves the corresponding polynomial equation. The solutions of a polynomial equation are called roots, whereas the zeroes of a polynomial function are the values of x where the value of the polynomial is zero. The distinction between the two words is not always strictly observed. Exercise 11A 1 State whether or not each expression is a polynomial. 1 a 3x2 − 7x b 2 +x x √ √ 2 d 3x 3 − 5x + 11 e 3x2 + 5x 7x13 + 3x g (x + 1)3 h 4 4 3 2 3 x − ex + πx c √ x−2 f 2x − 1 i loge x x−2 x+1 k 5 l i the degree, ii the leading coefficient, iii the leading term, iv the constant term, v whether or not the polynomial is monic. j 2 FOUNDATION For each polynomial, state: Expand the polynomial first where necessary. a 4x3 + 7x2 − 11 d x 12 g 0 3 4 b 10 − 4x − 6x3 c 2 e x (x − 2) f (x2 − 3x)(1 − x3 ) h x(x3 − 5x + 1) − x2 (x2 − 2) i 6x7 − 4x6 − (2x5 + 1)(5 + 3x2 ) 2 If P(x) = 5x + 2 and Q(x) = x2 − 3x + 1, find: a P(x) + Q(x) b Q(x) + P(x) c P(x) − Q(x) d Q(x) − P(x) e P(x) × Q(x) f Q(x) × P(x) If P(x) = 5x + 2, Q(x) = x2 − 3x + 1 and R(x) = 2x2 − 3, show, by simplifying the LHS and RHS separately, that: a (P(x) + Q(x)) + R(x) = P(x) + (Q(x) + R(x)) b P(x) (Q(x) + R(x)) = P(x)Q(x) + P(x)R(x) c (P(x)Q(x)) R(x) = P(x) (Q(x)R(x)) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11A The language of polynomials 461 DEVELOPMENT 5 6 Factor each polynomial completely, and write down all its zeroes. a x3 − 8x2 − 20x b 2x4 − x3 − x2 c x4 − 81 d x4 − 5x2 − 36 For each polynomial, determine: i the degree, ii the leading coefficient, a (2x3 − 3)3 7 iii the constant term. b (2x2 + 1)(3x3 − 2)(4x4 + 3)(5x5 − 4) a The polynomials P(x) and Q(x) have degrees p and q respectively, and p q. What is the degree of: ii P(x) + Q(x)? i P(x)Q(x), b What differences would it make if P(x)and Q(x) both had the same degree p? c Give an example of two polynomials, both of degree 2, which have a sum of degree 0. 8 Write down the monic polynomial whose degree, leading coefficient, and constant term are all equal. 9 Find a, b and c, given that the following pairs of polynomials are identically equal. a ax2 + bx + c = 3x2 − 4x + 1, for all x. b (a − b)x2 + (2a + b)x = 7x − x2 , for all x. c a(x − 1)2 + b(x − 1) + c = x2 , for all x. d a(x + 2)2 + b(x + 3)2 + c(x + 4)2 = 2x2 + 8x + 6, for all x. 10 a Suppose that P(x) = ax4 + bx3 + cx2 + dx + e is even, so that P(−x) = P(x). Show that b = d = 0. b Suppose that Q(x) = ax5 + bx4 + cx3 + dx2 + ex + f is odd, so that Q(−x) = −Q(x). Show that b = d = f = 0. c Give a general statement of the situation in parts a and b. 11 Suppose that P(x), Q(x), R(x) and S (x) are polynomials. Indicate whether the following statements are true or false. Provide a counter-example for any false statements. a If P(x) is even, then P (x) is odd. b If Q (x) is even, then Q(x) is odd. c If R(x) is odd, then R (x) is even. d If S (x) is odd, then S (x) is even. ENRICHMENT 12 a Find a and b so that x4 + 1 = (x2 + ax + 1)(x2 + bx + 1). b Find a and b so that x4 + x2 + 1 = (x2 + ax + 1)(x2 + bx + 1). c Find a and b so that x4 − x2 + 1 = (x2 + ax + 1)(x2 + bx + 1). d Show that all the quadratic factors in parts a, b and c are irreducible. 13 a What is the coefficient of x in the polynomial B(x) = (1 + x)n ? b What is the coefficient of xn in the polynomial G(x) = (1 + x + x2 + · · · + xn )2 ? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 462 11B Chapter 11 Polynomials 11B Graphs of polynomial functions Learning intentions • Sketch polynomials, using behaviour as x → ∞ and as x → −∞. • Define multiple (or repeated) zeroes, and simple (sometimes called single) zeroes. • Take account of simple and multiple zeroes in sketches. We have assumed that the graph of a polynomial function is continuous for all values of x. We assume also that it is smooth, without sharp corners anywhere. This section will concentrate on two main concerns. • How does the graph behave for large positive and negative values of x? • Given the full factoring of the polynomial, how does the graph behave near its various x-intercepts? In Year 12, we will pursue further questions about stationary points and inflections that are not also zeroes of the polynomial. The graphs of polynomial functions It should be intuitively obvious that for large positive and negative values of x, the behaviour of the curve is governed entirely by the sign of its leading term. For example, the cubic graph sketched on the right below is y P(x) = x − 4x = x(x − 2)(x + 2). 3 For large positive values of x, the leading term x3 completely swamps the other term −4x. Hence P(x) → ∞ as x → ∞. −2 2 x Similarly, for large negative values of x, the term −4x is negligible compared with the far bigger negative values of the leading term x3 . Hence P(x) → −∞ as x → −∞. Every polynomial of odd degree has a graph that similarly disappears off diagonally opposite corners. The curve is continuous, so it must be zero somewhere at least once. Here is the general situation. 6 Behaviour of polynomials for large x Suppose that P(x) is a polynomial of degree at least 1 with leading term an xn . • As x → ∞, P(x) → ∞ if an is positive, and P(x) → −∞ if an is negative. • As x → −∞, P(x) behaves the same as when x → ∞ if the degree is even, and P(x) behaves in the opposite way if the degree is odd. It follows that every polynomial of odd degree has at least one zero. See the Enrichment section for a formal proof. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11B Graphs of polynomial functions 463 Zeroes and sign If the polynomial can be completely factored, then its zeroes can be quickly read off, and we can construct a table of test values to decide its sign. Here, for example, is the table of test values and the sketch of P(x) = (x + 2)3 x2 (x − 2). −2 x −3 −2 −1 0 1 2 3 y 0 −3 0 −27 0 1125 45 y 2 x The function changes sign around x = −2 and x = 2, where the associated factors (x + 2)3 and (x − 2) have odd degrees, but not around x = 0, where the factor x2 has even degree. Because the curve is smooth — without sharp corners — at x = 0, we know that the curve will be increasing on the left of x = 0, decreasing on the right of x = 0, and stationary at x = 0. This produces a turning point at the origin — the x-axis is a tangent there, and the curve turns over smoothly from increasing to stationary to decreasing without crossing the x-axis. At x = −2, our table of values tells us that the curve crosses the x-axis. We shall see in Year 12 that the curve is momentarily flat there, with a horizontal inflection on the x-axis at x = −2 — the x-axis is a tangent to the curve that actually crosses the curve there. This corresponds to the factor (x + 2)3 having odd degree greater than 1. Proving all this requires calculus, but the result is obvious by comparison with the known graph of the very simple polynomial function y = x3 that we first drew in Section 3G. Multiple zeroes (or repeated zeroes) Some language is needed here. Take the polynomial P(x) = (x + 2)3 x2 (x − 2). • The zero x = −2 is called a triple zero. • The zero x = 0 is called a double zero (as with quadratics in Chapter 3). • The zero x = 2 is called a simple zero. The triple zero x = −2 and the double zero x = 0 are called multiple or repeated zeroes. 7 Zeroes and their multipliicity • Suppose that x − α is a factor of a polynomial P(x), and P(x) = (x − α)m Q(x), where Q(x) is not divisible by x − α. Then x = α is called a zero of multiplicity m. • A zero of multiplicity 1 is called a simple zero. • A zero of multiplicity 2 or greater is called a multiple zero or a repeated zero. • For low orders, we use the terms double zero, triple zero, and quadruple zero, . . . . Note: The term single zero is sometimes used in place of simple zero, but that term is best avoided because of the constant verbal confusion that it causes. Behaviour at simple and multiple zeroes Here is the statement of how a polynomial graph behaves when it crosses the x-axis at a zero, as justified above as far as is possible in Year 11. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 464 11B Chapter 11 Polynomials 8 Multiple roots and the shape of the curve Suppose that x = α is a zero of a polynomial P(x). • If x = α is a simple zero (sometimes called a single zero), then the curve crosses the x-axis at x = α at an angle, and is not tangent to the x-axis there. • If x = α has even multiplicity, then the curve is tangent to the x-axis at x = α, and does not cross the x-axis there. The point (α, 0) is a turning point. • If x = α has odd multiplicity at least 3, then the curve is tangent to the x-axis at x = α, but crosses the x-axis. The point (α, 0) is a horizontal inflection. Note: When we were dealing only with quadratics in Chapter 3, we often used the terms ‘equal roots’ and ‘distinct roots’. Never use those terms in the context of polynomials, because of the verbal ambiguities that inevitably arise. Example 2 Sketching factored polynomials near simple and multiple zeroes In each part, name the zeroes and their multiplicity, and then sketch, showing the behaviour near any x-intercepts: a P(x) = (x − 1)2 (x − 2) b Q(x) = x3 (x + 2)4 (x2 + x + 1) c R(x) = −2(x − 2)2 (x + 1)5 (x − 1) Solution In part b, x2 + x + 1 has no zeroes, because Δ = 1 − 4 < 0. a b y 1 2 x c y y 8 −2 x −1 −2 1 2 x 1 is a double zero, −2 is a double zero, −1 is a quintuple zero, 2 is a simple zero. 0 is a triple zero. 1 is a simple zero, 2 is a double zero. Example 3 Simple and multiple zeroes Give an example of a polynomial factored into linear factors that: a has two multiple (or repeated) zeroes, and three simple zeroes (sometimes called ‘single zeroes’). b has degree 9, and has two triple zeroes and one double zero, and is non-monic. Solution a (x + 5)2 (x − 1)(x − 2)(x − 3)x7 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 b 12(x + 8)3 (x + 7)3 (x + 5)2 (x + 3) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11B Graphs of polynomial functions Exercise 11B 1 465 FOUNDATION Write down the zeroes of each polynomial, and classify each zero as a simple zero or a multiple zero. b P(x) = x3 (x + 1) a P(x) = (x − 3)(x − 6)2 c P(x) = 8(x + 5)4 (x − 7)2 (x + 10) Note: Readers should be aware that the term ‘repeated zero’ is often used for ‘multiple zero’, and that the term ‘single zero’ is sometimes used for ‘simple zero’. 2 Write down the zeroes of each polynomial, and state the multiplicity of each zero. b P(x) = x(x + 7)3 (x − 5) a P(x) = (x + 2)(x − 3)2 3 Write down the roots of each polynomial equation, and identify each root as a simple, double, triple or quadruple root. a (x − 2)2 (x + 3) = 0 4 b (4 − x)3 (9 + x)4 = 0 6 b P(x) = x c P(x) = x − 4 d P(x) = 3 − 2x Sketch the graphs of these quadratic polynomials, clearly indicating all intercepts with the axes. a P(x) = x2 b P(x) = (x − 1)(x + 3) c P(x) = (x − 2)2 d P(x) = 9 − x2 e P(x) = 2x2 + 5x − 3 f P(x) = 4 + 3x − x2 Sketch the graphs of these cubic polynomials, clearly indicating all intercepts with the axes. b y = x3 + 2 c y = (x − 4)3 d y = (x − 1)(x + 2)(x − 3) e y = x(2x + 1)(x − 5) f y = (1 − x)(1 + x)(2 + x) g y = (2x + 1) (x − 4) h y = x (1 − x) i y = (2 − x)2 (5 − x) a y = x3 2 7 c x4 (x − 6)2 (x + 8)3 = 0 Sketch the graphs of these linear polynomials, clearly indicating all intercepts with the axes. a P(x) = 2 5 c P(x) = (1 − x)3 (3 + x)5 2 Sketch the graphs of these quartic polynomials, clearly indicating all intercepts with the axes. Identify all double zeroes, triple zeroes, and quadruple zeroes, a F(x) = x4 b F(x) = (x + 2)4 c F(x) = x(3x + 2)(x − 3)(x + 2) d F(x) = (1 − x)(x + 5)(x − 7)(x + 3) e F(x) = x (x + 4)(x − 3) f F(x) = (x + 2)3 (x − 5) g F(x) = (2x − 3)2 (x + 1)2 h F(x) = (1 − x)3 (x − 3) 2 i F(x) = (2 − x)2 (1 − x2 ) DEVELOPMENT 8 These polynomials are not factored, but the positions of their zeroes can be found by trial and error. Copy and complete each table of values, then sketch the graph, and state how many zeroes there are, and between which integers they lie. a y = x2 − 3x + 1 x −1 0 1 b 2 3 y 9 4 y = 1 + 3x − x3 x −2 −1 0 1 2 3 y Sketch each polynomial function, clearly indicating all intercepts with the axes. a P(x) = x(x − 2)3 (x + 1)2 c P(x) = x(2x + 3) (1 − x) 3 b P(x) = (x + 2)2 (3 − x)3 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 d P(x) = (x + 1)(4 − x2 )(x2 − 3x − 10) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 466 11B Chapter 11 Polynomials 10 11 Use the graphs drawn in the previous question to solve these inequations. a x(x − 2)3 (x + 1)2 > 0 b (x + 2)2 (3 − x)3 ≥ 0 c x(2x + 3)3 (1 − x)4 ≥ 0 d (x + 1)(4 − x2 )(x2 − 3x − 10) < 0 Factor each polynomial, then sketch it, showing all intercepts with the axes. a P(x) = x4 − 13x2 + 36 b P(x) = 4x4 − 13x2 + 9 c P(x) = (x2 − 5x)2 − 2(x2 − 5x) − 24 d P(x) = (x2 − 3x + 1)2 − 4(x2 − 3x + 1) − 5 ENRICHMENT 12 Consider the polynomial P(x) = x4 − 5x2 + 4x + 13. a Show that P(x) can be expressed in the form (x2 − a)2 + (x − b)2 . b How many x-intercepts does the graph of P(x) have? Explain your answer. 13 At what points do the graphs of the polynomials f (x) = (x + 1)n and g(x) = (x + 1)m intersect, where m n? (Hint: Consider the cases where m and n are odd and even.) 14 To prove: Let P(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 be a polynomial of degree at least 1. Then the leading term dominates the behaviour of P(x) for large positive values of x, and for large negative values of x. P(x) a Write down n . x b Take the limit of this expression as x → ∞. c Draw a conclusion about the behaviour of P(x) as x → ∞. d Draw a conclusion about the behaviour of P(x) as x → −∞. e What conclusion, if any, can you draw about the zeroes of a polynomial of odd degree, and about the zeroes of a polynomial of even degree? 15 To prove: Let P(x) be a polynomial such that P(x) = 0 for all x. Then P(x) is the zero polynomial Z(x) = 0 that has no term with a non-zero coefficient. a Use the previous question to prove that P(x) cannot have degree ≥ 1, and hence is a constant polynomial P(x) = a0 . b Explain why the constant a0 cannot be non-zero. 16 To prove: If P(x) = Q(x) for all x, then P(x) and Q(x) have the same coefficients. Use the previous question, and the polynomial A(x) = P(x) − Q(x), to prove this result. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11C Division of polynomials 467 11C Division of polynomials Learning intentions • Apply the division algorithm for polynomials, and compare it with integer division. Section 11A had examples of adding, subtracting, and multiplying polynomials, operations that are quite straightforward. The division of one polynomial by another, however, is more elaborate. Division of polynomials It can happen that the quotient of two polynomials is again a polynomial: x2 + 4x − 5 6x3 + 4x2 − 9x and = 2x2 + 43 x − 3 = x − 1. 3x x+5 But usually, division results in rational functions, not polynomials: 9 x+4 1 x4 + 4x2 − 9 = x2 + 4 − 2 and =1+ . x+3 x+3 x2 x There is a close analogy here between the set Z of all integers and the set of all polynomials. In both cases, everything works nicely for addition, subtraction, and multiplication, but the results of division do not usually lie within the set. For example, although 20 ÷ 5 = 4 is an integer, the division of two integers usually results in a fraction rather than an integer, as in 23 ÷ 5 = 4 35 . In both cases, the best way to handle division is to use remainders. The division algorithm for integers On the right is an example of the well-known long division algorithm for integers, applied here to 197 ÷ 12. The number 12 is called the divisor, 197 is called the dividend, 16 is called the quotient, and 5 is called the remainder. 16 12 5 The result of the division can be written as 197 12 = 16 12 , but we can avoid fractions completely by writing the result as: 197 = 12 × 16 + 5, remainder 5 197 12 77 72 5 dividend = divisor × quotient + remainder. The remainder 5 must be less than 12, otherwise the division process could be continued. Thus the general result for division of integers can be expressed as: 9 Division of integers • Let p (the dividend) and d (the divisor) be integers, with d > 0. Then there are unique integers q (the quotient) and r (the remainder) such that p = dq + r and 0 ≤ r < d. • When the remainder r is zero, d is a divisor of p, and the integer p factors as p = d × q. The division algorithm for polynomials The method of dividing one polynomial by another is similar to the method of dividing integers. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 468 11C Chapter 11 Polynomials 10 The method of long division of polynomials • At each step, divide the leading term of the remainder by the leading term of the divisor. Continue the process for as long as possible. • Unless otherwise specified, express the final answer in the form dividend = divisor × quotient + remainder. Note on missing terms: In the example below, it is vital to notice that there is no term in x2 in the dividend. There are two approaches to deal with this: • Leave a gap for the x2 column. This is what has been done below. • If you prefer not to have any gaps, write in the missing term 0x2 . Why not try both methods? Both are very satisfactory. Example 4 Long division of polynomials Divide 3x4 − 4x3 + 4x − 8 by: b x2 − 2 a x−2 Give results first in the standard manner, then using rational functions. Solution In each part, the steps have been annotated to explain the method. 3x3 + 2x2 + 4x + 12 a x − 2 3x4 − 4x3 + 4x − 8 3x4 − 6x3 (divide x into 3x4 , giving the 3x3 above) (multiply x − 2 by 3x3 and then subtract) + 4x − 8 2x3 2x3 − 4x2 (divide x into 2x3 , giving the 2x2 above) (multiply x − 2 by 2x2 and then subtract) 4x2 + 4x − 8 (divide x into 4x2 , giving the 4x above) 4x2 − 8x (multiply x − 2 by 4x and then subtract) 12x − 8 (divide x into 12x, giving the 12 above) 12x − 24 (multiply x − 2 by 12 and then subtract) 16 (this is the final remainder) Hence 3x4 − 4x3 + 4x − 8 = (x − 2)(3x3 + 2x2 + 4x + 12) + 16, or, writing the result using rational functions, 16 3x4 − 4x3 + 4x − 8 = 3x3 + 2x2 + 4x + 12 + . x−2 x−2 3x2 − 4x + 6 b x2 − 2 3x4 − 4x3 + 4x − 8 − 6x2 3x4 (divide x2 into 3x4 , giving the 3x2 above) (multiply x2 − 2 by 3x2 and then subtract) − 4x3 + 6x2 + 4x − 8 (divide x2 into −4x3 , giving the −4x above) − 4x3 + 8x (multiply x2 − 2 by −4x and then subtract) 6x2 − 4x − 8 (divide x2 into 6x2 , giving the 6 above) − 12 (multiply x2 − 2 by 6 and then subtract) 6x2 − 4x + 4 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (this is the final remainder) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11C Division of polynomials Hence 3x4 − 4x3 + 4x − 8 = (x2 − 2)(3x2 − 4x + 6) + (−4x + 4), or 3x4 − 4x3 + 4x − 8 −4x + 4 = 3x2 − 4x + 6 + 2 . x2 − 2 x −2 469 The division theorem The division process illustrated above can be continued until the remainder is zero or has degree less than the degree of the divisor. Thus the general result for polynomial division is: 11 Division of polynomials • Suppose that P(x) (the dividend) and D(x) (the divisor) are polynomials with D(x) 0. Then there are unique polynomials Q(x) (the quotient) and R(x) (the remainder) such that: 1 P(x) = D(x)Q(x) + R(x), 2 either deg R(x) < deg D(x), or R(x) = 0. • When the remainder R(x) is zero, then D(x) is called a divisor of P(x), and the polynomial P(x) factors into the product P(x) = D(x) × Q(x). For example, in the two long divisions in Example 4: • In part a, the remainder after division by the degree 1 polynomial x − 2 is the polynomial 16 of degree 0. • In part b, the remainder after division by the degree 2 polynomial x2 − 2 is the linear polynomial −4x + 4 of degree 1. The uniqueness of the quotient and remainder in the first dotpoint above is proven in the Enrichment section of the following exercise. Exercise 11C 1 Perform each integer division, and write the result in the form p = dq + r, where 0 ≤ r < d. For example, 30 ÷ 7 = 4, remainder 2, so 30 = 4 × 7 + 2. a 63 ÷ 5 2 FOUNDATION b 125 ÷ 8 c 324 ÷ 11 d 1857 ÷ 23 Use long division to perform each division. Express each result in the form P(x) = D(x) Q(x) + R(x). a (x2 − 4x + 1) ÷ (x + 1) b (x2 − 6x + 5) ÷ (x − 5) c (x3 − x2 − 17x + 24) ÷ (x − 4) d (2x3 − 10x2 + 15x − 14) ÷ (x − 3) e (4x3 − 4x2 + 7x + 14) ÷ (2x + 1) f (x4 + x3 − x2 − 5x − 3) ÷ (x − 1) g (6x4 − 5x3 + 9x2 − 8x + 2) ÷ (2x − 1) h (10x4 − x3 + 3x2 − 3x − 2) ÷ (5x + 2) 3 Express the answers to parts a–d of the previous question in rational form, that is, as R(x) P(x) = Q(x) + . D(x) D(x) 4 Use long division to perform each division. Express each result in the standard form P(x) = D(x) Q(x) + R(x). a (x3 + x2 − 7x + 6) ÷ (x2 + 3x − 1) b (x3 − 4x2 − 2x + 3) ÷ (x2 − 5x + 3) c (x4 − 3x3 + x2 − 7x + 3) ÷ (x2 − 4x + 2) d (2x5 − 5x4 + 12x3 − 10x2 + 7x + 9) ÷ (x2 − x + 2) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 470 11C Chapter 11 Polynomials 5 a If the divisor of a polynomial has degree 3, what are the possible degrees of the remainder? b On division by D(x), a polynomial has remainder R(x) of degree 2. What are the possible degrees of D(x)? DEVELOPMENT 6 Use long division to perform each division. Take care to ensure that the columns line up correctly. Express each result in the form P(x) = D(x) Q(x) + R(x). a (x3 − 5x + 3) ÷ (x − 2) b (2x3 + x2 − 11) ÷ (x + 1) c (x3 − 3x2 + 5x − 4) ÷ (x2 + 2) d (2x4 − 5x2 + x − 2) ÷ (x2 + 3x − 1) e (2x3 − 3) ÷ (2x − 4) f (x5 + 3x4 − 2x2 − 3) ÷ (x2 + 1) Write the answers to parts c and f above in rational form, that is, in the form 7 R(x) P(x) = Q(x) + . D(x) D(x) a Use long division to show that P(x) = x3 + 2x2 − 11x − 12 is divisible by x − 3, and hence express P(x) as the product of three linear factors. b Find the values of x for which P(x) > 0. 8 a Use long division to show that F(x) = 2x4 + 3x3 − 12x2 − 7x + 6 is divisible by x2 − x − 2, and hence express F(x) as the product of four linear factors. b Find the values of x for which F(x) ≤ 0. 9 a Find the quotient and remainder when x4 − 2x3 + x2 − 5x + 7 is divided by x2 + x − 1. b Hence find a and b so that x4 − 2x3 + x2 + ax + b is exactly divisible by x2 + x − 1. 10 a Use long division to divide the polynomial f (x) = x4 − x3 + x2 − x + 1 by the polynomial d(x) = x2 + 4. Express your answer in the form f (x) = d(x)q(x) + r(x). b Hence find the values of c and d such that x4 − x3 + x2 + cx + d is divisible by x2 + 4. 11 If x4 − 2x3 − 20x2 + mx + n is exactly divisible by x2 − 5x + 2, find m and n. 12 Suppose that P(x) = x4 + x3 − 5x2 − 22x + 5 and D(x) = x2 + 3x + 5. a Find the polynomials Q(x) and R(x), where R(x) is of lower degree than D(x), so that P(x) = D(x) Q(x) + R(x). b Hence explain why P(x) and D(x) cannot have a common zero. ENRICHMENT 13 Consider the cubic equation x3 − kx + (k + 11) = 0. 12 . x−1 b Hence find all the integer values of k for which the equation has at least one positive integer solution for x. a Use long division to show that k = x2 + x + 1 + 14 To prove: Let P(x) and D(x) be polynomials with D(x) 0. Then there exists unique polynomials Q(x) and R(x) such that P(x) = D(x)Q(x) + R(x) and either R(x) = 0 or deg R(x) < deg D(x). Existence follows from the long division algorithm, two examples of which have been given. To prove uniqueness, let Q1 (x) and R1 (x) be polynomials such that P(x) = D(x)Q1 (x) + R1 (x), and either R1 (x) = 0 or deg R1 (x) < deg D(x). a Show that D(x) Q(x) − Q1 (x) = R1 (x) − R(x). b Use the possible degrees of LHS and RHS to prove the result. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11D The remainder and factor theorems 471 11D The remainder and factor theorems Learning intentions • Prove and use the remainder theorem to find remainders. • Prove and use the factor theorem to find zeroes. • Identify possible zeroes by factoring the constant term. • Combine the factor theorem and long division to factor a polynomial. Long division of polynomials is a cumbersome process. It is therefore useful to have two theorems, called the remainder theorem and the factor theorem, that provide information about the results of a division without the division actually being carried out. In particular, the factor theorem gives a simple test whether a particular linear polynomial is a factor. The remainder theorem The remainder theorem is a remarkable result which, in the case of linear divisors, allows the remainder to be found without the long division ever being performed. 12 The remainder theorem Suppose that P(x) is a polynomial and α is a constant. Then the remainder after division of P(x) by x − α is P(α). Proof Because x − α is a polynomial of degree 1, the division theorem tells us that there are unique polynomials Q(x) and R(x) such that P(x) = (x − α)Q(x) + R(x), where either R(x) = 0 deg R(x) = 0. or Hence R(x) is zero or a non-zero constant, which we can write more simply as r, so that P(x) = (x − α)Q(x) + r. Substituting x = α gives P(α) = (α − α)Q(α) + r r = P(α), and rearranging, Example 5 as required. Applying the remainder theorem Find the remainder when 3x4 − 4x3 + 4x − 8 is divided by x − 2: b by the remainder theorem. a by long division, Solution a In Example 4 of Section 11C, performing the division showed that 3x4 − 4x3 + 4x − 8 = (x − 2)(3x3 + 2x2 + 4x + 12) + 16, that is, the remainder is 16. b Alternatively, substituting x = 2 into P(x), remainder = P(2) (This is the remainder theorem.) = 48 − 32 + 8 − 8 = 16, as expected. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 472 11D Chapter 11 Polynomials Example 6 Finding coefficients using the remainder theorem The polynomial P(x) = x4 − 2x3 + ax + b has remainder 3 after division by x − 1, and has remainder −5 after division by x + 1. Find a and b. Solution Applying the remainder theorem for each divisor, P(1) = 3 1−2+a+b=3 a + b = 4. (1) P(−1) = −5 Also 1 + 2 − a + b = −5 −a + b = −8. Adding (1) and (2), 2b = −4, and subtracting them, 2a = 12. (2) Hence a = 6 and b = −2. The factor theorem The remainder theorem tells us that the number P(α) is the remainder after division by x − α. But x − α is a factor if and only if the remainder after division by x − α is zero, so: 13 The factor theorem Suppose that P(x) is a polynomial and α is a constant. Then x − α is a factor of P(x) if and only if P(α) = 0. This is a quick and easy way to test whether x − α is a factor of P(x). Example 7 Using the factor theorem and long division to factor a polynomial a Show that x − 3 is a factor of P(x) = x3 − 2x2 + x − 12, and x + 1 is not. b Then use long division to factor the polynomial completely. Solution a P(3) = 27 − 18 + 3 − 12 = 0, so x − 3 is a factor. P(−1) = −1 − 2 − 1 − 12 = −16 0, so x + 1 is not a factor. b Long division of P(x) = x3 − 2x2 + x − 12 by x − 3 (which we omit) gives P(x) = (x − 3)(x2 + x + 4), and because Δ = 1 − 16 = −15 < 0 for the quadratic, this factoring is complete. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11D The remainder and factor theorems 473 Factoring polynomials — the initial approach The factor theorem gives us the beginnings of an approach to factoring polynomials. This approach will be further refined in the next two sections. 14 Factoring polynomials — the initial approach • Use trial and error to find an integer zero x = α of P(x). • Then use long division to factor P(x) in the form P(x) = (x − α)Q(x). If the coefficients of P(x) are all integers, then all the integer zeroes of P(x) are divisors of the constant term. Proof We must prove the claim that if the coefficients of P(x) are integers, then every integer zero of P(x) is a divisor of the constant term. P(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 , Let where the coefficients an , an−1 , . . . a1 , a0 are all integers, and let x = α be an integer zero of P(x). Substituting into P(α) = 0 gives an αn + an−1 αn−1 + · · · + a1 α + a0 = 0 a0 = −an αn − an−1 αn−1 − · · · − a1 α = α(−an αn−1 − an−1 αn−2 − · · · − a1 ), so a0 is an integer multiple of α. Example 8 Factoring the constant term to find possible zeroes Factor P(x) = x4 + x3 − 9x2 + 11x − 4 completely. Solution Because all the coefficients are integers, any integer zero is a divisor of the constant term −4. Thus we test 1, 2, 4, −1, −2 and −4. P(1) = 1 + 1 − 9 + 11 − 4 = 0, so x − 1 is a factor. After long division (omitted), P(x) = (x − 1)(x3 + 2x2 − 7x + 4). Let Q(x) = x3 + 2x2 − 7x + 4, then Q(1) = 1 + 2 − 7 + 4 = 0, so x − 1 is a factor. Again after long division (omitted), P(x) = (x − 1)(x − 1)(x2 + 3x − 4). Factoring the quadratic, P(x) = (x − 1)3 (x + 4). Note: In the next two sections, and again in Year 12, we will develop methods that will often allow long division to be avoided. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 474 11D Chapter 11 Polynomials Exercise 11D 1 2 3 FOUNDATION Without division, find the remainder when P(x) = x3 − x2 + 2x + 1 is divided by: a x−1 b x−3 c x+2 d x+1 e x−5 f x+3 Without division, find which of the following are factors of F(x) = x3 + 4x2 + x − 6. a x−1 b x+1 c x−2 d x+2 e x−3 f x+3 a Find k, if x − 1 is a factor of P(x) = x3 − 3x2 + kx − 2. b Find m, if −2 is a zero of the function F(x) = x3 + mx2 − 3x + 4. c When the polynomial P(x) = 2x3 − x2 + px − 1 is divided by x − 3, the remainder is 2. Find p. d For what value of a is 3x4 + ax2 − 2 divisible by x + 1? 4 5 Factor each polynomial and sketch its graph, indicating all intercepts with the axes. Remember that any integer zeroes are divisors of the constant term. a P(x) = x3 + 2x2 − 5x − 6 b P(x) = x3 + 3x2 − 25x + 21 c P(x) = −x3 + x2 + 5x + 3 d P(x) = x4 − x3 − 19x2 − 11x + 30 Solve these equations by first factoring the LHS. a x3 + 3x2 − 6x − 8 = 0 b x3 − 4x2 − 3x + 18 = 0 c 6x3 − 5x2 − 12x − 4 = 0 d 2x4 + 11x3 + 19x2 + 8x − 4 = 0 DEVELOPMENT 6 a Show that P(x) = x3 − 8x2 + 9x + 18 is divisible by both x − 3 and x + 1. b By considering the leading term and constant term, express P(x) as a product of three linear factors. 7 a Show that P(x) = 2x3 − x2 − 13x − 6 is divisible by both x − 3 and 2x + 1. b By considering the leading term and constant term, express P(x) as a product of three linear factors. 8 a When the polynomial P(x) = 2x3 − x2 + ax + b is divided by x − 1 the remainder is 16, and when it is divided by x + 2 the remainder is −17. Find a and b. b The polynomial P(x) is given by P(x) = x3 + ax2 + bx − 18. Find a and b given that x + 2 is a factor of P(x), and −24 is the remainder when P(x) is divided by x − 1. 9 Without division, find the remainder when P(x) = x3 + 2x2 − 4x + 5 is divided by: a 2x − 1 10 b 2x + 3 c 3x − 2 a If P(x) = 2x3 + x2 − 13x + 6, evaluate P 12 . Hence use long division to express P(x) in fully factored form. b Given P(x) = 6x3 + x2 − 5x − 2, evaluate P − 23 then express P(x) in factored form. 11 Is either x + 1 or x − 1 a factor of xn + 1, where n is a positive integer? (Hint: Consider the cases where n is even or odd separately.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11D The remainder and factor theorems 12 475 a P(x) is an odd polynomial of degree 3. It has x + 4 as a factor, and when it is divided by x − 3 the remainder is 21. Find P(x). b Find p so that x − p is a factor of 4x3 − (10p − 1)x2 + (6p2 − 5)x + 6. 13 When the polynomial P(x) is divided by (x − 1)(x + 3), the quotient is Q(x) and the remainder is 2x + 5. a Write down a division identity based on this information. b Hence, by evaluating P(1), find the remainder when P(x) is divided by x − 1. c What is the remainder when P(x) is divided by x + 3? 14 The polynomial P(x) is divided by (x − 1)(x + 2). Suppose that the quotient is Q(x) and the remainder is R(x). a Explain why the general form of R(x) is ax + b, where a and b are constants. b If P(1) = 2 and P(−2) = 5, find a and b. (Hint: Use the division identity.) 15 The polynomial P(x) is divided by (x + 4)(x − 3). If P(−4) = 11 and P(3) = −3, use the same approach as the previous question to find the remainder. 16 a When a polynomial is divided by (2x + 1)(x − 3), the remainder is 3x − 1. What is the remainder when the polynomial is divided by 2x + 1? b When x5 + 3x3 + ax + b is divided by x2 − 1, the remainder is 2x − 7. Find a and b. c When a polynomial P(x) is divided by x2 − 5, the remainder is x + 4. Find the remainder when P(x) + P(−x) is divided by x2 − 5. (Hint: Write down the division identity.) ENRICHMENT 17 When the polynomial P(x) is divided by x2 − k2 , the remainder is ax + b. 1 1 a Show that a = P (k) − P (−k) and b = P (k) + P (−k) . 2k 2 b Given that P(x) = 8x5 − 4x4 + 6x3 − 11x2 − 2x + 3 and k = 12 , find a and b, and hence factor P(x) fully. 18 a Use the factor theorem to prove that a + b + c is a factor of a3 + b3 + c3 − 3abc. Then find the other factor. (Hint: Regard it as a polynomial in a.) b Factor ab3 − ac3 + bc3 − ba3 + ca3 − cb3 . 19 a If all the coefficients of a monic polynomial are integers, prove that all the rational zeroes are integers. (Hint: Look carefully at the proof under Box 14.) b If all the coefficients of a polynomial are integers, prove that the denominators of all the rational zeroes (in lowest terms) are divisors of the leading coefficient. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 476 11E Chapter 11 Polynomials 11E Consequences of the factor theorem Learning intentions • Develop consequences of the factor theorem to help factor a polynomial. • Develop bounds on the number of zeroes of a polynomial of a certain degree. • Develop a test for two polynomials to be identically equal. • Develop geometric consequences of the factor theorem. The factor theorem has a number of straightforward but very useful consequences. They are presented here as six successive theorems. A. Several distinct zeroes Suppose that several distinct zeroes of a polynomial have been found, probably using test substitutions into the polynomial. 15 Distinct zeroes Suppose that α1 , α2 , . . . α s are distinct zeroes of a polynomial P(x). Then (x − α1 )(x − α2 ) · · · (x − α s ) is a factor of P(x). Proof Because α1 is a zero, x − α1 is a factor, and P(x) = (x − α1 )P1 (x). Because P(α2 ) = 0 but α2 − α1 0, P1 (α2 ) must be zero. Hence x − α2 is a factor of P1 (x), and P1 (x) = (x − α2 )P2 (x), so P(x) = (x − α1 )(x − α1 )P2 (x). Continuing similarly for s steps, (x − α1 )(x − α2 ) · · · (x − α s ) is a factor of P(x). B. All distinct zeroes If n distinct zeroes of a polynomial of degree n can be found, then the factoring is complete, and the polynomial is the product of distinct linear factors. 16 All distinct zeroes If α1 , α2 , . . . , αn are n distinct zeroes of a polynomial P(x) of degree n, then P(x) = a(x − α1 )(x − α2 ) · · · (x − αn ), where a is the leading coefficient of P(x). Proof By the previous theorem, (x − α1 )(x − α2 ) · · · (x − αn ) is a factor of P(x), so P(x) = (x − α1 )(x − α2 ) · · · (x − αn )Q(x), for some polynomial Q(x). But P(x) and (x − α1 )(x − α2 ) · · · (x − αn ) both have degree n, so Q(x) is a constant. Equating coefficients of xn , the constant Q(x) is the leading coefficient. Factoring polynomials — finding several zeroes first If we can find several zeroes of a polynomial, then we have a quadratic or cubic factor, and the long divisions required can be reduced, or even avoided completely. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11E Consequences of the factor theorem 477 17 Factoring polynomials — finding several zeroes first • Use trial and error to find as many integer zeroes of P(x) as possible. • Using long division, divide P(x) by the product of the known factors. If the coefficients of P(x) are all integers, then any integer zero of P(x) must be one of the divisors of the constant term. When this procedure is applied to the polynomial factored in the previous section, one rather than two long divisions is required. Example 9 Finding several zeroes before long division Factor P(x) = x4 + x3 − 9x2 + 11x − 4 completely (done in Example 8). Solution As before, all the coefficients are integers, so any integer zero is a divisor of the constant term −4. That is, we test 1, 2, 4, −1, −2 and −4. P(1) = 1 + 1 − 9 + 11 − 4 = 0, so x − 1 is a factor. P(−4) = 256 − 64 − 144 − 44 − 4 = 0, so x + 4 is a factor. After long division by (x − 1)(x + 4) = x2 + 3x − 4 (omitted), P(x) = (x2 + 3x − 4)(x2 − 2x + 1). Factoring both quadratics, P(x) = (x − 1)(x + 4) × (x − 1)2 = (x − 1)3 (x + 4). Note: The methods of the next section will allow this particular factoring to be done with no long divisions. The next example involves a polynomial that factors into distinct linear factors — nothing more than the factor theorem is required to complete the task. Example 10 Finding all the zeroes by the factor theorem Factor P(x) = x4 − x3 − 7x2 + x + 6 completely. Solution The divisors of the constant term 6 are 1, 2, 3, 6, −1, −2, −3 and −6. P(1) = 1 − 1 − 7 + 1 + 6 = 0, so x − 1 is a factor. P(−1) = 1 + 1 − 7 − 1 + 6 = 0, so x + 1 is a factor. P(2) = 16 − 8 − 28 + 2 + 6 = −12 0, so x − 2 is not a factor. P(−2) = 16 + 8 − 28 − 2 + 6 = 0, so x + 2 is a factor. P(3) = 81 − 27 − 63 + 3 + 6 = 0, so x − 3 is a factor. We now have four distinct zeroes of a polynomial of degree 4. Hence P(x) = (x − 1)(x + 1)(x + 2)(x − 3). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 (Notice that P(x) is monic.) © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 478 11E Chapter 11 Polynomials C. The maximum number of zeroes If a polynomial of degree n had n + 1 zeroes, then by the first theorem (Box 15), it would be divisible by a polynomial of degree n + 1, which is impossible. 18 Maximum number of zeroes A polynomial of degree n has at most n zeroes. D. A vanishing condition The previous theorem translates easily into a condition for a polynomial to be the zero polynomial. 19 A vanishing condition • Suppose that P(x) is a polynomial that has no term of degree more than n, yet is zero for at least n + 1 distinct values of x. Then P(x) is the zero polynomial. • In particular, the only polynomial that is zero for all values of x is the zero polynomial (as was proven in Exercise 11B Enrichment). Proof Suppose that P(x) had a degree. This degree must be at most n because there is no term of degree more than n. But the degree must also be at least n + 1 because there are n + 1 distinct zeroes. This is a contradiction, so P(x) has no degree, and is therefore the zero polynomial. Note: This again highlights the fact that the zero polynomial Z(x) = 0 is quite different in nature from all other polynomials. It is the only polynomial with an infinite number of zeroes — in fact every real number is a zero of Z(x). Associated with this is the fact that x − α is a factor of Z(x) for all real values of α, because Z(x) = (x − α)Z(x) (which is trivially true, because both sides are zero for all x). No wonder then that the zero polynomial does not have a degree! E. A condition for two polynomials to be identically equal A most important consequence of this last theorem is a condition for two polynomials P(x) and Q(x) to be identically equal. 20 An identically equal condition • Suppose that P(x) and Q(x) are degree n polynomials that have the same values for at least n + 1 values of x. • Then the polynomials P(x) and Q(x) are identically equal (meaning that they are equal for all values of x), and their coefficients are equal. • In particular: A linear polynomial is determined by two values. A quadratic polynomial is determined by three values. A cubic polynomial is determined by four values. A quartic polynomial is determined by five values. Proof Let F(x) = P(x) − Q(x). Because F(x) is zero whenever P(x) and Q(x) have the same value, it follows that F(x) is zero for at least n + 1 values of x, so by the previous theorem, F(x) is the zero polynomial, so P(x) = Q(x) for all values of x. And we stated before in Box 5 of Section 11A — proven in Exercise 11B Enrichment — that it now follows that the coefficients are equal. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11E Consequences of the factor theorem Example 11 479 Using the condition that polynomials are identically equal Find a, b, c and d, if x3 − x = a(x − 2)3 + b(x − 2)2 + c(x − 2) + d for at least four values of x. Solution Because the cubics are equal for four values of x, they are identically equal. Substituting x = 2, 6 = d. 1 = a. Equating coefficients of x3 , Hence the identity is now Substituting x = 0, x − x = (x − 2)3 + b(x − 2)2 + c(x − 2) + 6. 3 0 = −8 + 4b − 2c + 6 2b − c = 1. Substituting x = 1, (1) 0 = −1 + b − c + 6 b − c = −5. (2) Solving (1) and (2) simultaneously, b = 6 and c = 11. F. Geometric implications of the factor theorem Here are some of the geometric versions of the factor theorem — they translate the consequences above into the language of coordinate geometry. You will already have seen them in operation when dealing with graphs of quadratics. 21 Geometric implications of the factor theorem 1 The graph of a polynomial function of degree n is completely determined by any n + 1 points on the curve. 2 The graphs of two distinct polynomial functions cannot intersect in more points than the maximum of the two degrees. 3 A line cannot intersect the graph of a polynomial of degree n in more than n points. Example 12 Examining intersections of curves using the factor theorem By factoring the difference F(x) = P(x) − Q(x), describe the intersections between the curves P(x) = x4 + 4x3 + 2 and Q(x) = x4 + 3x3 + 3x, and find where P(x) is above Q(x). Solution Subtracting, F(x) = x3 − 3x + 2. Substituting, F(1) = 1 − 3 + 2 = 0, so x − 1 is a factor. F(−2) = −8 + 6 + 2 = 0, so x + 2 is a factor. After long division by (x − 1)(x + 2) = x2 + x − 2, F(x) = (x − 1)2 (x + 2). Hence y = P(x) and y = Q(x) are tangent at x = 1, but do not cross there, and also intersect also at x = −2, where they cross at an angle — see Box 22 about this step. Because F(x) is positive for −2 < x < 1 or x > 1, and negative for x < −2, P(x) is above Q(x) for −2 < x < 1 or x > 1, and below it for x < −2. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 480 11E Chapter 11 Polynomials Simple and multiple roots when finding intersections of polynomials When two polynomial graphs cross, as in Example 12, solving them simultaneously results in a polynomial equation, which may have multiple roots. The behaviour at a multiple root is exactly analogous to the behaviour of a single polynomial at a multiple zero, as described in Box 8 in Section 11B: 22 The behaviour at an intersection of two polynomial graphs Suppose that x = å is a root of the polynomial equation obtained when two polynomial graphs y = P(x) and y = Q(x) are solved simultaneously. • If x = α is a simple root (sometimes called a single root), then the curves cross each other at an angle, and are not tangent to each other there. • If x = α has even multiplicity, then the curves are tangent to each other, and do not cross. • If x = α has odd multiplicity at least 3, then the curves are tangent to each other, and cross at the point of intersection. Again, the proof relies on Year 12 calculus. But this result does follow immediately from the earlier statement in Box 8, if one considers instead the difference F(x) = P(x) − Q(x), as was done in Example 12. A note for Extension 2 students The fundamental theorem of algebra cannot be proven in the Extension 2 course, but the theorem helps us to understand the importance of complex numbers for polynomials. It tells us that every polynomial equation of degree n ≥ 1 has exactly n roots, provided first that roots are counted according to their multiplicity, and secondly that complex roots are also counted. For example: • x3 = 0 has one root x = 0, but this root has multiplicity 3. • x3 − 1 = 0, which factors to (x − 1)(x2 + x + 1) = 0, has root x = 1, but also has the two complex roots of x2 + x + 1 = 0. This means that the graph of a polynomial of degree n ≥ 2 intersects every line in exactly n points, provided first that points where the line is a tangent are counted according to their multiplicity, and secondly that complex points of intersection are also counted. This theorem provides the fundamental link between the algebra of polynomials and the geometry of their graphs, and allows the degree of a polynomial to be defined algebraically as the highest index, or geometrically as the number of times every line crosses it. Exercise 11E 1 FOUNDATION Use the factor theorem to write down in factored form: a a monic cubic polynomial with zeroes −1, 3 and 4, b a monic quartic polynomial with zeroes 0, −2, 3 and 1, c a cubic polynomial with leading coefficient 6 and zeroes at 13 , − 12 and 1. 2 a Show that 2 and 5 are zeroes of P(x) = x4 − 3x3 − 15x2 + 19x + 30. b Hence explain why (x − 2)(x − 5) is a factor of P(x). c Divide P(x) by (x − 2)(x − 5) and hence express P(x) as the product of four linear factors. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11E Consequences of the factor theorem 3 481 Use trial and error to find as many integer zeroes of P(x) as possible. Use long division to divide P(x) by the product of the known factors and hence express P(x) in factored form. a P(x) = 2x4 − 5x3 − 5x2 + 5x + 3 b P(x) = 2x4 − 5x3 − 5x2 + 20x − 12 c P(x) = 6x4 − 25x3 + 17x2 + 28x − 20 d P(x) = 9x4 − 51x3 + 85x2 − 41x + 6 4 Refer to Box 19 to answer parts a and b. a The polynomial (a − 2)x2 + (1 − 3b)x + (5 − 2c) has three zeroes. What are the values of a, b and c? b The polynomial (a + 1)x3 + (b − 3)x2 + (2c − 1)x + (5 − 4d) has four zeroes. What are the values of a, b, c and d? DEVELOPMENT 5 a If 3x2 − 4x + 7 = a(x + 2)2 + b(x + 2) + c for all x, find a, b and c. b If 2x3 − 8x2 + 3x − 4 = a(x − 1)3 + b(x − 1)2 + c(x − 1) + d for all x, find a, b, c and d. c Use similar methods to express x3 + 2x2 − 3x + 1 as a polynomial in (x + 1). d If the polynomials 2x2 + 4x + 4 and a(x + 1)2 + b(x + 2)2 + c(x + 3)2 are equal for three values of x, find a, b and c. 6 a A polynomial of degree 3 has a double zero at 2. When x = 1 it takes the value 6 and when x = 3 it takes the value 8. Find the polynomial. b Two zeroes of a polynomial of degree 3 are 1 and −3. When x = 2 it takes the value −15 and when x = −1 it takes the value 36. Find the polynomial. 7 Show that x2 − 3x + 2 is a factor of P(x) = xn (2m − 1) + xm (1 − 2n ) + (2n − 2m ), where m and n are positive integers. 8 If two polynomials have degrees m and n, where m > n, what is the maximum number of intersection points of their graphs? 9 Explain why the graph of a cubic polynomial with three distinct zeroes must have two turning points. The line y = k meets the curve y = ax3 + bx2 + cx + d four times. Find the values of a, b, c and d in terms of k. √ √ 11 Find, in expanded form, the monic degree five polynomial whose zeroes are 0, −1, 1, 2 − 2 and 2 + 2. 10 12 By factoring the difference F(x) = P(x) − Q(x), describe the intersections between the curves P(x) and Q(x). a P(x) = 2x3 − 4x2 + 3x + 1, Q(x) = x3 + x2 − 8 b P(x) = x4 + x3 + 10x − 4, Q(x) = x4 + 7x2 − 6x + 8 c P(x) = −2x3 + 3x2 − 25, Q(x) = −3x3 − x2 + 11x + 5 d P(x) = x4 − 3x2 − 2, Q(x) = x3 − 5x e P(x) = x4 + 4x3 − x + 5, Q(x) = x3 − 3x2 − 2x + 5 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 482 11E Chapter 11 Polynomials 13 If a and b are non-zero, and a + b = 0, prove that the polynomials A(x) = x3 + ax2 − x + b and B(x) = x3 + bx2 − x + a have a common factor of degree 2 but are not identical polynomials. What is the common factor? ENRICHMENT 14 a Factor xn − 1. b Suppose that the roots of the equation xn − 1 = 0 are x = 1, α1 , α2 , . . . , αn−1 . Show that (1 − α1 ) (1 − α2 ) . . . (1 − αn−1 ) = n. 15 Suppose that P(x) = x5 + x2 + 1 and Q(x) = x2 − 2. If r1 , r2 , r3 , r4 and r5 are the five zeroes of P(x), find the value of Q(r1 ) × Q(r2 ) × Q(r3 ) × Q(r4 ) × Q(r5 ). 16 Suppose that P(x) is a polynomial of odd degree n. It is known that P(k) = k for k = 0, 1, 2, . . . , n. k+1 a Write down the zeroes of the polynomial (x + 1) P(x) − x. b Let A be the leading coefficient of the polynomial (x + 1) P(x) − x. Factor the polynomial, and hence show that A= 1 1 = . 1 × 2 × 3 × · · · × n × (n + 1) (n + 1)! c Find P(n + 1). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11F Sums and products of zeroes 483 11F Sums and products of zeroes Learning intentions • Develop and use the sum-and-product-of-roots formulae for degrees 2, 3, and 4. • Use these formulae to help factor polynomials, and for other purposes. When expanding a monic polynomial with only linear factors, such as P(x) = (x − 2)(x − 3)(x − 5)(x − 7) = x4 − 17x3 + 101x2 − 247x + 210, the four zeroes 2, 3, 5 and 7 of the polynomial clearly determine the five coefficients 1, −17, 101, −247 and 210. But what are the formulae for the coefficients? If we study the coefficient −17 of the term in x3 , and the constant term 210, it does not take long to realise that −17 = −(2 + 3 + 5 + 7) 210 = +2 × 3 × 5 × 7 = −(sum of the zeroes) = +(product of the zeroes). It is not at all clear, however, how the coefficients 101 of x2 , and −247 of x, are related to the zeroes. (We invite readers to try working it out before continuing.) This section answers these questions for quadratic, cubic, and quartic polynomials. Knowing these formulae is a helpful technique when trying to factor a polynomial, and has other more general applications. The zeroes of a quadratic Let P(x) = ax2 + bx + c be a quadratic with zeroes α and β and leading coefficient a. Then P(x) = a(x − α)(x − β) by the factor theorem. Expanding, P(x) = a(x − α)(x − β) = ax2 − a(α + β)x + aαβ. Hence the coefficient of x is −a times the sum of the zeroes, and the constant is +a times the product of the zeroes, b = −a(α + β) and c = +aαβ. This is just what happened with the quartic above, except for multiplication by the leading coefficient a. Solving for the sum and product of the zeroes: 23 Sum and product of zeroes of a quadratic Let P(x) = ax2 + bx + c have zeroes α and β. Then b c α+β=− and αβ = + . a a CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 484 11F Chapter 11 Polynomials Example 13 a Using the sum and product of zeroes of a quadratic i Show by substitution that x = 5 is a zero of P(x) = 3x2 − 30x + 75. ii Use sum-of-zeroes to find the other zero. iii Use product-of-zeroes to find the other zero. b Find the sum of the zeroes of Q(x) = x2 + 7x − 11, and hence find its axis of symmetry. Solution a i P(5) = 3 × 25 − 30 × 5 + 75 = 0. ii α + β = + 30 3 iii α + 5 = 10 α = 5, αβ = 75 3 α × 5 = 25 α = 5, so 5 is a double zero. so again, 5 is a double zero. b α + β = − 71 = −7. The axis of symmetry is midway between zeroes, so it is x = −3 12 b (This calculation is exactly the same as the formula x = − for the axis.) 2a The zeroes of a cubic Let P(x) = ax3 + bx2 + cx + d be a cubic with zeroes α, β and γ and leading coefficient a. Then P(x) = a(x − α)(x − β)(x − γ) by the factor theorem. Expanding, P(x) = a(x − α)(x − β)(x − γ) = ax3 − a(α + β + γ)x2 + a(αβ + βγ + γα)x − aαβγ, so as we saw before with the quartic and the quadratic, b = −a(α + β + γ) and d = −aαβγ, where the negative sign of the product comes from the cube of −1. But the new phenomenon here is the coefficient of x, c = a(αβ + βγ + γα) = +(sum of products of pairs of zeroes). Solving for the sums and products of the zeroes: 24 Sums and products of zeroes of a cubic Let P(x) = ax3 + bx2 + cx + d have zeroes α, β and γ. Then b α+β+γ=− (sum of the zeroes) a c (sum of products of pairs of zeroes) αβ + βγ + γα = + a d (product of the zeroes) αβγ = − a CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11F Sums and products of zeroes Example 14 485 Using the sum and product of zeroes of a cubic a Show that −6 is a zero of P(x) = x3 − 4x2 − 39x + 126. b Use sum and product of zeroes to find the other two zeroes. c Check the formula for the sum of products of pairs of zeroes. Solution a P(−6) = −216 − 144 + 234 + 126 = 0. −4 126 and αβ × (−6) = − 1 1 α + β = 10 αβ = 21. Hence by inspection, α and β are 3 and 7. (You have been using this ‘by inspection’ approach for years to solve quadratics.) c αβ + βγ + γα = 3 × 7 + 7 × (−6) + (−6) × 3 b α+β−6=− = −39, which is the coefficient of x. The zeroes of a quartic Suppose that the four zeroes of the quartic polynomial P(x) = ax4 + bx3 + cx2 + dx + e are α, β, γ and δ. By the factor theorem (see Box 16 in Section 11E), P(x) is a multiple of the product (x − α)(x − β)(x − γ)(x − δ): P(x) = a(x − α)(x − β)(x − γ)(x − δ) = ax4 − a(α + β + γ + δ)x3 + a(αβ + αγ + αδ + βγ + βδ + γδ)x2 − a(αβγ + βγδ + γδα + δαβ)x + aαβγδ. Equating coefficients of terms in x3 , x2 , x and constants now gives: 25 Zeroes and coefficients of a quartic b a c αβ + αγ + αδ + βγ + βδ + γδ = + a d αβγ + βγδ + γδα + δαβ = − a e αβγδ = + a α+β+γ+δ=− (sum of the zeroes) (sum of products of pairs of zeroes) (sum of products of triples of zeroes) (product of the zeroes) The second formula gives the sum of the products of pairs of zeroes, and the third formula gives the sum of the products of triples of zeroes. The general case The method is the same for all degrees. Notation is unfortunately a major difficulty here, and the results are better written in words. Suppose that α1 , α2 , . . . αn are the n zeroes of the degree n polynomial P(x) = an xn + an−1 xn−1 + · · · + a1 x + a0 . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 486 11F Chapter 11 Polynomials 26 Zeroes and coefficients of a polynomial an−1 an an−2 sum of products of pairs of zeroes = + an an−3 sum of products of triples of zeroes = − an ······ sum of the zeroes = α1 + α2 + · · · + αn = − product of the zeroes = α1 α2 · · · αn = (−1)n a0 an Notice the alternating signs of the successive results. It is unlikely that anything apart from the first and last formulae would be required. Example 15 Using the sum and product of zeroes of a cubic Let α, β and γ be the roots of the cubic equation x3 − 3x + 2 = 0. Use the formulae above to find: a α+β+γ d 1 1 1 + + α β γ b αβγ c αβ + βγ + γα e α2 + β 2 + γ 2 f α2 β + αβ2 + β2 γ + βγ2 + γ2 α + γα2 Check the result with the factoring x3 − 3x + 2 = (x − 1)2 (x + 2) obtained in the solution of Example 12 in Section 11E. Solution a α + β + γ = −0 1 =0 b αβγ = − 21 = −2 c αβ + βγ + γα = −3 1 = −3 d 1 1 1 βγ + γα + αβ + + = α β γ αβγ 3 = −3 −2 = 2 e (α + β + γ)2 = α2 + β2 + γ2 + 2αβ + 2βγ + 2γα, f α2 β + αβ2 + β2 γ + βγ2 + γ2 α + γα2 02 = α2 + β2 + γ2 + 2 × (−3) so = αβ(α + β + γ) + βγ(β + γ + α) α + β + γ = 6. 2 2 + γα(γ + α + β) − 3αβγ 2 = (αβ + βγ + γα)(α + β + γ) − 3αβγ = (−3) × 0 − 3 × (−2) =6 Because x3 − 3x + 2 = (x − 1)2 (x + 2), the actual roots are 1, 1 and −2, hence a α+β+γ=1+1−2=0 b αβγ = 1 × 1 × (−2) = −2 c αβ + βγ + γα = 1 − 2 − 2 = −3 d 1 1 1 + + = 1 + 1 − 12 = 1 12 α β γ f α2 β + αβ2 + β2 γ + βγ2 + γ2 α + γα2 e α2 + β 2 + γ 2 = 1 + 1 + 4 = 6 = 1 × 1 + 1 × 1 + 1 × (−2) + 1 × 4 + 4 × 1 + (−2) × 1 =6 all of which agree with the previous calculations. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11F Sums and products of zeroes 487 Factoring polynomials using the factor theorem and the sum and product of zeroes Long division can be avoided in many situations by applying the sum and product of zeroes formulae after one or more zeroes have been found. Here are the steps that we have developed so far for factoring polynomials: 27 Factoring polynomials — the steps so far • Use trial and error to find as many integer zeroes of P(x) as possible. • Use sum and product of zeroes to find the other zeroes. • Alternatively, use long division of P(x) by the product of the known factors. If the coefficients of P(x) are all integers, then any integer zero of P(x) must be one of the divisors of the constant term. In the next example, we factor a polynomial factored twice already, but this time there is no need for any long division at all. Example 16 Using the sum and product of zeroes of a quartic Factor F(x) = x4 + x3 − 9x2 + 11x − 4 completely. Solution As before, F(1) = 1 + 1 − 9 + 11 − 4 = 0, and F(−4) = 256 − 64 − 144 − 44 − 4 = 0. Let the zeroes be 1, −4, α and β. Then α + β + 1 − 4 = −1 α + β = 2. Also (1) αβ × 1 × (−4) = −4 αβ = 1. (2) From (1) and (2), α = β = 1, so F(x) = (x − 1)3 (x + 4). Example 17 Using the sum and product of zeroes of a cubic Factor completely the cubic G(x) = x3 − x2 − 4. Solution First, G(2) = 8 − 4 − 4 = 0. Let the zeroes be 2, α and β. Then 2+α+β=1 α + β = −1, and (1) 2×α×β=4 αβ = 2. (2) Substituting (1) into (2), α(−1 − α) = 2 α2 + α + 2 = 0 This is an irreducible quadratic, because Δ = −7, so the complete factoring is G(x) = (x − 2)(x2 + x + 2). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 488 11F Chapter 11 Polynomials Note: This procedure — developing the irreducible quadratic factor from the sum and product of zeroes — is really little easier than the long division it avoids. Forming identities with the coefficients If some information can be gained about the roots of a polynomial equation, it may be possible to form an identity with the coefficients of the polynomial. Example 18 Identities on the coefficients of a cubic If one zero of the cubic f (x) = ax3 + bx2 + cx + d is the opposite of another, prove that ad = bc. Solution We know that one of the zeroes is the opposite of another, so we can begin with the following sentence, which expresses this fact symbolically: Let the three zeroes of the cubic be α, −α and β. Now we can use the formula for the sum of the roots, b α + (−α) + β = − a aβ = −b. (1) Then using the formula for the product of the roots, d α × (−α) × β = − a 2 aα β = d. (2) There are three products of pairs of roots, namely −α2 and −αβ and βα, so using the sum of products of pairs of roots, c −α2 − αβ + βα = a aα2 = −c. (3) Now we must eliminate the roots α and β from equations (1), (2), and (3). Substituting (3) into (2), −cβ = d, b and dividing (1) by (4), −ac = − , d that is, ad = bc. (4) Exercise 11F 1 FOUNDATION If α and β are the roots of the quadratic equation x2 − 4x + 2 = 0, find: a α+β b αβ c α2 β + αβ2 1 1 + α β α β + g β α e (α + 2)(β + 2) f α2 + β 2 h αβ + α β 1 i α+ α d CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 3 3 1 β+ β © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11F Sums and products of zeroes 2 489 If α, β and γ are the roots of the equation x3 + 2x2 − 11x − 12 = 0, find: a α+β+γ b αβ + αγ + βγ c αβγ 1 1 1 d + + α β γ g (αβ)2 γ + (αγ)2 β + (βγ)2 α 1 1 1 e + + αβ αγ βγ h α2 + β 2 + γ 2 f (α + 1)(β + 1)(γ + 1) i (αβ)−2 + (αγ)−2 + (βγ)−2 Now find the roots of the equation x3 + 2x2 − 11x − 12 = 0 by factoring the LHS. Hence check your answers for the expressions in parts a–i. 3 If α, β, γ and δ are the roots of the equation x4 − 5x3 + 2x2 − 4x − 3 = 0, find: a α+β+γ+δ b αβ + αγ + αδ + βγ + βδ + γδ c αβγ + αβδ + αγδ + βγδ d αβγδ 1 1 1 1 + + + e α β γ δ 1 1 1 1 g + + + αβγ αβδ αγδ βγδ 4 f 1 1 1 1 1 1 + + + + + αβ αγ αδ βγ βδ γδ h α2 + β 2 + γ 2 + δ 2 If α and β are the roots of the equation 2x2 + 5x − 4 = 0, find: b αβ a α+β c α +β 2 d |α − β| (Hint: Square it.) 2 DEVELOPMENT 5 a The polynomial P(x) = 2x3 − 5x2 − 14x + 8 has zeroes at −2 and 4. Use the sum of the zeroes to find the other zero. b Suppose that x − 3 and x + 1 are factors of P(x) = x3 − 6x2 + 5x + 12. Use the product of the zeroes to find the other factor of P(x). 6 Consider the polynomial P(x) = x3 − x2 − x + 10. a Show that −2 is a zero of P(x). b Suppose that the zeroes of P(x) are −2, α and β. Show that α + β = 3 and αβ = 5. c By solving the two equations in part b simultaneously, show that α2 − 3α + 5 = 0. d Hence show that there are no such real numbers α and β. e Hence state how many times the graph of the cubic crosses the x-axis. 7 8 Show that x = 1 and x = −2 are zeroes of P(x), and use the sum and product of zeroes to find the other one or two zeroes. Note any multiple zeroes. a P(x) = x3 − 2x2 − 5x + 6 b P(x) = 2x3 + 3x2 − 3x − 2 c P(x) = x4 + 3x3 − 3x2 − 7x + 6 d P(x) = 3x4 − 5x3 − 10x2 + 20x − 8 a Find the values of a and b for which x3 + ax2 − 10x + b is exactly divisible by x2 + x − 12, and then factor the cubic. b Find the values of a and b for which x2 − x − 20 is a factor of x4 + ax3 − 23x2 + bx + 60, and then find all the zeroes. 9 a If one of the roots of the equation x2 + bx + c = 0 is twice the other root, show that 2b2 = 9c. (Hint: Let the roots be α and 2α.) b If one of the roots of the equation x2 + px + q = 0 is one more than the other root, show that p2 = 4q + 1. (Hint: Let the roots be α and α + 1.) CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 490 11F Chapter 11 Polynomials 10 a Find the roots of the equation 3x3 − x2 − 48x + 16 = 0 given that the sum of two of them is zero. (Hint: Let the roots be α, −α and β.) b Find the roots of the equation 2x3 − 5x2 − 46x + 24 = 0 given that the product of two of them is 3. 3 (Hint: Let the roots be α, and β.) α c Find the roots of the equation x3 − 27x − 54 = 0 given that two of them are equal. (Hint: Let the roots be α, α and β.) d Find the zeroes of the polynomial P(x) = 2x3 − 13x2 + 22x − 8 given that one of them is the product of the other two. (Hint: Let the zeroes be α, β and αβ.) 11 a Two of the roots of the equation x3 + 3x2 − 4x + a = 0 are opposites. Find the value of a and the three roots. b Two of the roots of the equation 4x3 + ax2 − 47x + 12 = 0 are reciprocals. Find the value of a and the three roots. 12 Find the zeroes of the polynomial x4 − 3x3 − 8x2 + 12x + 16 if they are α, 2α, β and 2β. 13 a If one root of the equation x3 − bx2 + cx − d = 0 is equal to the product of the other two, show that (c + d)2 = d(b + 1)2 . b The polynomial P(x) = x4 + ax3 + bx2 + cx + d has two zeroes that are opposites and another two zeroes that are reciprocals. Show that: ii c = ad i b=1+d 14 The cubic equation x3 − Ax2 + 3A = 0, where A > 0, has roots α, β and α + β. a Use the sum of the roots to show that α + β = 12 A. b Use the sum of the products of pairs of roots to show that αβ = − 14 A2 . √ c Show that A = 2 6. 15 The polynomial P(x) = x3 − Lx2 + Lx − M has zeroes α, a Show that: 1 and β. α 1 β ii 1 + αβ + = L +β= L α α b Show that either M = 1 or M = L − 1. i α+ 16 iii β = M Suppose that the polynomial P(x) = ax3 + bx2 + cx + d has zeroes α, β and γ. b2 − 2ac a Show that α2 + β2 + γ2 = . a2 b Hence explain why the polynomial 2x3 − 3x2 + 5x − 8 cannot have three real zeroes. ENRICHMENT 17 If α, β and γ are the roots of the equation x3 + 5x − 4 = 0, evaluate α3 + β3 + γ3 . (Hint: x = α, x = β and x = γ satisfy the given equation.) 18 Suppose that the equation x2 + bx + c = 0 has roots α and β. a Show that b and c are the roots of the equation x2 + (α + β − αβ)x − αβ(α + β) = 0. b Find the non-zero values of b and c for which the roots of the equation in part a are: i α and β CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii α2 and β2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11G Geometry using polynomial techniques 491 11G Geometry using polynomial techniques Learning intentions • Use the methods of polynomials to solve geometric problems. This final section adds the methods of the preceding sections, particularly the sum and product of zeroes, to the available techniques for studying the geometry of various curves. The standard technique is to examine the roots of the equation formed in the process of solving two curves simultaneously. This section is rather demanding, and could all be regarded as Enrichment. Midpoints and tangents When two curves intersect, we can form the equation whose solutions are the x- or y-coordinates of points of intersection of the two curves. The midpoint of two points of intersection can then be found using the average of the roots. Tangents can be identified as corresponding to double roots. The next example could also be done using quadratic equations, but it is a clear example of the use of sum and product of roots. Example 19 Dealing with tangents using double roots The line y = 2x meets the parabola y = x2 − 2x − 8 at the two points A(α, 2α) and B(β, 2β). a Show that α and β are roots of x2 − 4x − 8 = 0, and hence find the coordinates of the midpoint M of AB. b Use the identity (α − β)2 = (α + β)2 − 4αβ to find the horizontal distance |α − β| from A to B. Then use Pythagoras’ theorem and the gradient of the line to find the length of AB. c Find the value of b for which y = 2x + b is a tangent to the parabola, and find the point T of contact. Solution a Solving the line and the parabola simultaneously, y x2 − 2x − 8 = 2x A x − 4x − 8 = 0. 2 Hence and β α + β = 4, αβ = −8. Averaging the roots, M has x-coordinate x = 2, and substituting into the line, M = (2, 4). b We know that −2 B −8 M 4 α x T (α − β)2 = (α + β)2 − 4αβ = 16 + 32 = 48 √ so |α − β| = 4 3. √ Because the line has gradient 2, the vertical√distance is 8 3, √ so using Pythagoras, AB2 = (4 3 )2 + (8 3 )2 = 16 × 15 √ AB = 4 15. Continued on next page. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 492 11G Chapter 11 Polynomials c Solving y = x2 − 2x − 8 and y = 2x + b simultaneously, x2 − 4x − (8 + b) = 0 Because the line is a tangent, let the roots be θ and θ. Then using the sum of roots, θ + θ = 4, so θ = 2, Using the product of roots, θ2 = −8 − b and because θ = 2, b = −12. So the line y = 2x − 12 is a tangent at T (2, −8). Geometric problems using sum and product of roots The sum and product of roots can make some interesting geometric problems quite straightforward. Example 20 Tangents and the sum and product of roots A line with gradient m through the point P(−1, 0) on the cubic y = x3 − x crosses the curve at two further points A and B with x-coordinates α and β respectively. a Sketch the situation. b Show that α and β satisfy the cubic equation x3 − (m + 1)x − m = 0. c Show that the midpoint M of AB lies on the vertical line x = 12 . d Find the line through P tangent to the cubic at a point distinct from P. Solution y = x(x − 1)(x + 1), so the zeroes are x = −1, x = 0 and x = 1. b The line through P(−1, 0) has equation a The cubic factors as y A P −1 α y = m(x + 1). M B 1β x Solving the line simultaneously with the cubic, x3 − x = mx + m x3 − (m + 1)x − m = 0. c The cubic has roots α, β and 1. Using the sum of roots, α + β + (−1) = 0 (there is no term in x2 ) so α + β = 1. Hence the midpoint M of AB has x-coordinate 12 (α + β) = 12 , so M lies on the line x = 12 . d The line PM is a tangent at another point on the curve when the points A, M and B coincide, that is, when α = β, and because α + β = 1, this means that Using the product of roots, α = β = 12 . α × β × (−1) = m (the constant term is − m) 1 1 2 × 2 × (−1) = m m = − 14 , so the tangent through P is CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y = − 14 (x + 1). © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 11G Geometry using polynomial techniques Exercise 11G 493 FOUNDATION Note: Sketches should be drawn in all these questions to make the situation clear. 1 a Show that the x-coordinates of the points of intersection of the parabola y = x2 − 6x and the line y = 2x − 16 satisfy the equation x2 − 8x + 16 = 0. b Solve this equation, and hence show that the line is a tangent to the parabola. Find the point T of contact. 2 a Show that the x-coordinates of the points of intersection of the line y = −2x + b and the parabola y = x2 − 6x satisfy the quadratic equation x2 − 4x − b = 0. b Suppose that the line is a tangent to the parabola, so that the roots of the quadratic equation are equal. Let these roots be α and α. i Using the sum of roots, show that α = 2. ii Using the product of roots, show that α2 = −b, and hence find b. iii Find the equation of the tangent and its point T of contact. 3 The line y = x + 1 meets the parabola y = x2 − 3x at A and B. a Show that the x-coordinates α and β of A and B satisfy the equation x2 − 4x − 1 = 0. b Find α + β, and hence find the coordinates of the midpoint M of AB. DEVELOPMENT 4 a Show that the x-coordinates of the points of intersection of the line y = 3 − x and the cubic y = x3 − 5x2 + 6x satisfy the equation x3 − 5x2 + 7x − 3 = 0. b Show that x = 1 and x = 3 are roots of the equation, and use the sum of the roots to find the third root. c Explain why the line is a tangent to the cubic, then find the point of contact and the other point of intersection. 5 a Show that the x-coordinates of the points of intersection of the line y = mx and the cubic y = x3 − 5x2 + 6x satisfy the equation x3 − 5x2 + (6 − m)x = 0. b Suppose now that the line is a tangent to the cubic at a point other than the origin, so that the roots of the equation are 0, α and α. i Using the sum of roots, show that α = 52 . ii Using the product of pairs of roots, show that α2 = 6 − m, and hence find m. iii Find the equation of the tangent and its point T of contact. 6 The line y = x − 2 meets the cubic y = x3 − 5x2 + 6x at F(2, 0), and also at A and B. a Show that the x-coordinates α and β of A and B satisfy x3 − 5x2 + 5x + 2 = 0. b Find α + β, and hence find the coordinates of the midpoint M of AB. c Show that αβ = −1, then use the identity (α − β)2 = (α + β)2 − 4αβ to show that the horizontal distance √ |α − β| between A and B is 13. d Hence use Pythagoras’ theorem to find the length AB. 7 A line passes through the point A(−1, −7) on the curve y = x3 − 3x2 + 4x + 1. Suppose that the line has gradient m and is tangent to the curve at another point P on the curve whose x-coordinate is α. a Show that the line has equation y = mx + (m − 7). b Show that the cubic equation whose roots are the x-coordinates of the points of intersection of the line and the curve is x3 − 3x2 + (4 − m)x + (8 − m) = 0. c Explain why the roots of this equation are −1, α and α, and hence find the point P and the value of m. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 494 11G Chapter 11 Polynomials 8 The point P(p, p3 ) lies on the curve y = x3 . A line through P with gradient m intersects the curve again at A and B. a Find the equation of the line through P. b Show that the x-coordinates of A and B satisfy the equation x3 − mx + mp − p3 = 0. c Hence find the x-coordinate of the midpoint M of AB, and show that for fixed p, M always lies on a line that is parallel to the y-axis. 9 a The equation x3 − (m + 1)x + (6 − 2m) = 0 has a root at x = −2 and a double root at x = α. Find α and m. b Write down the equation of the line passing through the point P(−2, −3) with gradient m. c The diagram shows the curve y = x3 − x + 3 and the point P(−2, −3) on the curve. The line cuts the curve at P, and is tangent to the curve at another point A on the curve. Find the equation of the line . 10 y 3 x P(−2,−3) a Use the factor theorem to factor the polynomial y = x4 − 4x3 − 9x2 + 16x + 20, given that there are four distinct zeroes, then sketch the curve. b The line : y = mx + b touches the quartic y = x4 − 4x3 − 9x2 + 16x + 20 at two distinct points A and B. Explain why the x-coordinates α and β of A and B are double roots of the equation x4 − 4x3 − 9x2 + (16 − m)x + (20 − b) = 0. c Use the theory of the sum and product of roots to write down four equations involving α, β, m and b. d Hence find m and b, and write down the equation of . Note: If two curves touch each other at P, then they are tangent to each other at P. 11 a Find k and the points of contact if the parabola y = x2 − k touches the quartic y = x4 at two points. b Find k and the point T of contact if the parabola y = x2 − k touches the cubic y = x3 . c Find k and the points of contact if the parabola y = x2 − k touches the circle x2 + y2 = 1 at two points. ENRICHMENT 12 A circle passing through the origin O is tangent to the hyperbola xy = 1 at A, and intersects the hyperbola again at two distinct points B and C. Prove that OA ⊥ BC. 13 The diagram to the right shows the circle x2 + y2 = 1 and the parabola y = (λx − 1)(x − 1), where λ is a constant. The circle and parabola meet in the four points P(1, 0), Q(0, 1), A(α, ϕ), 1 B(β, ψ). The point M is the midpoint of the chord AB. a Show that the x-coordinates of the points of intersection of the two curves satisfy the equation y β α 1 φ M A x ψ λ2 x4 − 2λ(1 + λ)x3 + (λ2 + 4λ + 2)x2 − 2(1 + λ)x = 0. −1 B b Use the formula for the sum of the roots to show that the x-coordinate of M is λ+2 . 2λ c Use a similar method to find the y-coordinate of M, and hence show that M lies on the line through the origin O parallel to PQ. d For what values of λ is the parabola tangent to the circle in the fourth quadrant? e For what values of λ are the four points P, Q, A and B distinct, with real numbers as coordinates? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. Chapter 11 review 495 Chapter 11 Review Review activity • Create your own summary of this chapter on paper or in a digital document. Chapter 11 Multiple-choice quiz • This automatically-marked quiz is accessed in the Interactive Textbook. A printable PDF Worksheet version is also available there. Skills Checklist • Available in the Interactive Textbook, use the checklist to track your understanding of the learning intentions. Checklist Printable PDF and word document versions are also available there. Chapter Review Exercise 2 Consider the polynomial P(x) = 2x3 − 5x2 − 6x − 11. State: a the degree of P(x), b the leading coefficient of P(x), c the leading term of P(x), d the constant term of P(x). Review 1 The polynomial P(x) has degree 3. Write down the degree of the polynomial: b (P(x))3 a 3 P(x) 3 Find the coefficient of x2 in the polynomial P(x) = (x2 − 3x − 7)(2x2 + 4x − 9). 4 a Sketch the graph of the polynomial function y = (x + 2)2 (x − 1)(x − 3), showing all intercepts with the coordinate axes. b Hence find the values of x for which (x + 2)2 (x − 1)(x − 3) < 0. 5 Sketch the graph of the polynomial P(x) = x3 − x5 . 6 Suppose that the polynomial P(x) = 2x3 + 7x2 − 4x + 5 is divided by D(x) = x − 3. a Find the quotient Q(x) and the remainder R(x). b Write down a division identity using the information above. 7 Without long division, find the remainder when P(x) = x3 − 5x2 + 1 is divided by: a x−3 8 b x+2 a Use the factor theorem to show that x − 2 is a factor of P(x) = x3 − 19x + 30. b Hence factor P(x) fully. 9 Find the value of k given that x + 3 is a factor of P(x) = x3 + 4x2 + kx − 12. 10 Find the values of b and c given that x + 1 is a factor of P(x) = x3 + bx2 + cx − 7, and the remainder is −12 when P(x) is divided by x − 5. 11 Find the values of h and k given that x + 2 is a factor of Q(x) = (x + h)2 + k, and the remainder is 16 when Q(x) is divided by x. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 496 Chapter 11 Polynomials Review 12 The polynomial P(x) is divided by (x + 1)(x − 2). Suppose that the quotient is Q(x) and the remainder is R(x). a Explain why the general form of R(x) is ax + b, where a and b are constants. b When P(x) is divided by x + 1 the remainder is 10, and when P(x) is divided by x − 2 the remainder is −8. Find a and b. (Hint: Use the division identity.) 13 Suppose that the polynomial Q(x) = x2 − 6x − 4 has zeroes α and β. Without finding the zeroes, find the value of: a α+β d 14 15 1 1 + α β b αβ c α2 β + β 2 α e (α − 3)(β − 3) f α2 + β 2 If α, β and γ are the roots of the equation x3 + 10x2 + 5x − 20 = 0, find: a α+β+γ b αβ + αγ + βγ c αβγ 1 1 1 + + d α β γ 1 1 1 + + e αβ αγ βγ f (α + 2)(β + 2)(γ + 2) g α2 β2 γ + α2 γ2 β + β2 γ2 α h α2 + β 2 + γ 2 i 1 1 1 + + α2 β2 α2 γ2 β2 γ2 The equation x3 + 5x2 + cx + d = 0 has roots −3, 7 and α. a Use the sum of the roots to find α. b Use the product of the roots to find d. c Use the sum of the roots in pairs to find c. 16 The equation 6x3 − 17x2 − 5x + 6 = 0 has roots α, β and γ, where αβ = −2. a Use the product of the roots to find γ. b Use the sum of the roots to find the other two roots. 17 One root of the equation ax2 + 2bx + c = 0 is the reciprocal of the square of the other root. Show that a3 + c3 + 2abc = 0. 18 Solve the equation 9x3 − 27x2 + 11x + 7 = 0 given that the roots are α − β, α and α + β. 19 Find the zeroes of the polynomial P(x) = 8x3 − 14x2 + 7x − 1 given that they are α , α and αβ. β 20 The line y = 9x + 5 is the tangent to the curve y = x3 − 3x2 at the point A(−1, −4). The line intersects the curve at another point B. Suppose that the x-coordinate of B is α. a Write down the cubic equation whose roots are the x-coordinates of A and B. b Explain why the roots of this equation are −1, −1 and α. c Hence find the point B. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12 Euler’s number Chapter introduction Chapter 9 dealt with exponential functions such as y = 2 x and y = 10 x , where the variable is in the exponent (index). We saw there how vitally important these functions are when modelling many common natural phenomena — where something is dying away, such as radioactive decay or the noise of a plucked guitar string, or where something is growing, such as populations or inflation. But we cannot yet differentiate these functions, so we cannot work with the rates associated with these phenomena. Thus some of the most important functions in science are still beyond our reach, because so far, calculus has only been developed for algebraic functions such as √ 1 f (x) = x3 + 8x and f (x) = x and f (x) = x2 − 2 x that involve powers and the four operations of arithmetic. This short chapter explains only the very first steps in extending calculus to exponential functions. Next year we will develop all this much further, applying the standard methods of calculus. Later, we shall apply calculus also to the trigonometric functions so that we can model wavelike phenomana. Differentiating exponential functions requires replacing our usual bases such as 2 and 10 by a new base e 2.7183. This new number e is called Euler’s number, because it was introduced by the great 18th century Swiss mathematician Leonhard Euler. The number e is just as important in mathematics as π, but only makes its first appearance in this chapter, in the exponential function y = e x . Both numbers π and e are real numbers. But neither number is a fraction, although the proofs of this are difficult. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 498 12A Chapter 12 Euler’s number 12A The exponential function base e Learning intentions • Define Euler’s number using tangents to exponential functions. d • Understand that (e x ) = e x — that e x is its own derivative.. dx • Graph y = e x and identify its characteristic tangent property.. As indicated in the Introduction, differentiating exponential functions requires a new base e 2.7183. The fundamental result of this section is surprisingly simple — the function y = e x is its own derivative: d x e = ex . dx An investigation of the graph of y = 2 x We begin by examining the tangents to the familiar exponential function y = 2 x . Its curve is drawn below, and we will need our assumption in Section 10B that, there being no obvious problem, the curve is differentiable for every value of x. Before reading the following argument, carry out the investigation in Question 1 of Exercise 12A. The investigation looks at the tangent to y = 2 x at each of several points on the curve, and relates its gradient to the height of the curve at the point. The procedures demonstrate graphically what the argument proves. Differentiating y = 2 x Below is a sketch of y = 2 x , with the tangent drawn at its y-intercept A(0, 1). Differentiating y = 2 x requires first-principles differentiation, because the theory so far hasn’t provided any rule for differentiating 2 x . The formula for first-principles differentiation is f (x + h) − f (x) . f (x) = lim h→0 h Applying this formula to the function f (x) = 2 x , 2 x+h − 2 x h→0 h 2 x × 2h − 2 x , because 2 x × 2h = 2 x+h , = lim h→0 h and taking out the common factor 2 x in the numerator, f (x) = lim y 2 1 A 1 x 2h − 1 h→0 h f (x) = 2 x × lim f (x) = 2 x × m, where m = limh→0 2h − 1 . h We assumed in Section 10B that all our functions are differentiable at every point, unless there is an obvious problem, so we are assuming y = 2 x is differentiable at x. 2h − 1 exists, but it cannot be found algebraically. h Substituting x = 0, however, gives a very simple geometric interpretation of the limit: Hence this limit m = limh→0 f (0) = 20 × m = m, because 20 = 1, so m is just the gradient of the tangent to y = 2 x at its y-intercept (0, 1). CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12A The exponential function base e 499 The conclusion of all this is that d x 2 = 2 x × m, where m is the gradient of y = 2 x at its y-intercept. dx The investigation in Question 1 of Exercise 12A shows that m 0.7 This argument applied to the base 2. But exactly the same argument can be applied to any exponential function y = a x , whatever the base a, simply by replacing 2 by a in the argument above: 1 Differentiating y = a x For all real numbers a > 0, d x a = a x × m, where m is the gradient of y = a x at its y-intercept. dx That is, the derivative of an exponential function is a multiple of itself. The gradient m of y = a x at its y-intercept changes as the base a changes. • When a = 1, the function is the constant function y = 1, so m = 0, • As the base a increases, the curve y = a x tilts up more and more steeply at its y-intercept (0, 1) — see the investigations in the following exercise. The definition of Euler’s number e It now makes sense to choose the base that will make the gradient exactly 1 at the y-intercept, because the value of m will then be exactly 1. This base is called Euler’s number — it is given the symbol e, and has the value e 2.7183. 2 The definition of e • Define Euler’s number e to be the number such that the exponential function y = e x has gradient exactly 1 at its y-intercept. Then e 2.7183. • The function y = e x is called the exponential function to distinguish it from all other exponential functions y = a x . Investigations to approximate e Question 5 of Exercise 12A uses a simple argument to prove that e is between 2 and 4. The investigation in Question 6 uses technology to find ever closer approximations to e. Both questions are based on the definition of e given above. The derivative of e x The fundamental result of this section then follows immediately by combining the two boxed results developed so far in this chapter. By Box 1, d x e = e x × m, where m is the gradient of y = e x at its y-intercept, dx = e x × 1, by the definition of e, in Box 2 = ex . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 500 12A Chapter 12 Euler’s number 3 The exponential function y = e x is its own derivative d x (e ) = e x . dx We will not develop the calculus of e x any further until next year, when y = e x will be the foundation function for a large number of very common rate situations. The graph of e x The graph of y = e x has been drawn below on graph paper. The tangent has been drawn at the y-intercept (0, 1) — it has gradient exactly 1. This graph of y = e x is one of the most important graphs in the whole course, and its shape and properties need to be memorised. y 3 e y=x +1 2 1 y=e x 1 e −2 −1 0 1 x Some vitally important properties of the graph of y = e x • The domain is all real x, or we can write domain = (−∞, ∞). The range is y > 0, or we can write range = (0, ∞). • There are no zeroes. • • • • The curve is always above the x-axis. The x-axis y = 0 is a horizontal asymptote to the curve on the left. dy → 0. That is, as x → −∞, y → 0 and dx On the right-hand side, the curve rises steeply at an increasing rate. dy → ∞. That is, as x → ∞, y → ∞ and dx The curve has gradient 1 at its y-intercept (0, 1). The curve is always increasing — increasing at an increasing rate — and is always concave up. CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12A The exponential function base e 501 Gradient equals height The fact that the derivative of the exponential function e x is the same function has a striking geometrical dy = e x , which means that for this function interpretation in terms of its graph. If y = e x , then dx dy = y, that is, gradient = height. dx dy of the curve is equal to the height y of the curve above Thus at each point on the curve y = e x , the gradient dx the x-axis. We have already seen this happening at the y-intercept (0, 1), where the gradient is 1 and the height is also 1. 4 Gradient equals height At each point on the graph of the exponential function y = e x , dy = y. dx That is, the gradient of the curve is always equal to its height above the x-axis. We shall see next year why this property of the exponential function is the reason why the function is so important in calculus. An investigation of the graph of y = e x The investigation in Question 2 of Exercise 12A confirms these properties of the graph of y = e x , particularly the fact that the gradient and the height are equal at every point on the curve. Exercise 12A INVESTIGATION Technology: The first two investigations are written as graph-paper exercises, so that they are independent of any device. The instructions are identical if graphing software is used. When using the graph-paper method, the diagrams should first be photocopied. 1 [Graph paper, but easily transferred to graphing software] The graph is the function y = 2 x . y a Photocopy the graph of y = 2 x , and on it draw the tangent at the point (0, 1). Extend the tangent across the diagram. b Use ‘rise over run’ to measure the gradient dy of this tangent. The run should be dx chosen as 10 or 20 little divisions, then count how many little divisions the rise is. c Similarly, draw tangents at four more points on the curve where x = −2, −1, 1 and 2. dy Measure the gradient of each tangent. dx 3 2 1 −1 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 0 1 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 502 12A Chapter 12 Euler’s number d Copy and complete the table of values to the right (the values 2 −2 x will only be rough). e What do you notice about the ratios of gradient to height? f Hence copy and complete: dy = ···y dx −1 0 1 2 height y dy gradient dx gradient height [Graph paper, but easily transferred to graphing software] y 3 e 2 1 1 e −2 −1 0 1 x These questions refer to the graph of y = e x drawn above. a Photocopy the graph of y = e x above and on it draw the tangent at the point (0, 1) where the height is 1. Extend the tangent across the diagram. b Measure the gradient of this tangent and confirm that it is equal to the height of the exponential graph at the point of contact. c Copy and complete the table of values to the right by measuring the dy of the tangent at the points where the height y is 12 , 1, 2 gradient dx and 3. d What do you notice about the ratios of gradient to height? e What does this tell you about the derivative of y = e x ? 3 height y 1 2 1 2 3 dy gradient dx gradient height a Photocopy again the graph of y = e x in Question 2 (or use graphing software). b Draw the tangents at the points where x = −2, −1, 0 and 1, extending each tangent down to the x-axis. c Measure the gradient of each tangent and confirm that it is equal to the height of the graph at the point. d What do you notice about the x-intercepts of all the tangents? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12A The exponential function base e 503 DEVELOPMENT 4 This question will confirm the result in Q3(d). Let P(a, ea ) lie on the curve y = e x . a Use the fact that y = e x to find the gradient of the tangent at P. b Find the equation of the tangent at P. c Show that the x-intercept of the tangent is 1 unit to the left of P. 5 Approximating e to many significant figures is difficult. The argument below, however, at least shows very quickly that the number e lies between 2 and 4. Here are tables of values and sketches of y = 2 x and y = 4 x . On each graph, the tangent at the yintercept A(0, 1) has been drawn. For y = 2 x y 2 y 2 For y = 4 x B x 0 1 x − 12 0 1 2 2x 1 2 4x 1 2 1 2 1 A C − 12 x 1 a 1 A 1 2 1 2 x i In the first diagram, find the gradient of the chord AB. ii Hence explain why the tangent at A has gradient less than 1. b i In the second diagram, find the gradient of the chord CA. ii Hence explain why the tangent at A has gradient greater than 1. c Use these two results to explain why 2 < e < 4. 6 [Technology] This question requires graphing software. a Use graphing software to graph y = 2 x and y = x + 1 on the same number plane, and hence observe that the gradient of y = 2 x at (0, 1) is less than 1. b Similarly, graph y = 3 x and y = x + 1 on the same number plane and hence observe that the gradient of y = 3 x at (0, 1) is greater than 1. c Choose a sequence of bases between 2 and 3 that make the line y = x + 1 more and more like a tangent to the curve at (0, 1) — that is, the gradient at (0, 1) becomes closer and closer to 1. In this way a reasonable approximation for e can be obtained. 7 x x Consider the five functions y = 13 , y = 12 , y = 1 x , y = 2 x , and y = 3 x . x x a Explain why 13 = 3−x , and 12 = 2−x . b Sketch these graphs, or locate their graphs in this section or in Section 8F. c From the graphs, identify where each of these is increasing, and decreasing. d What is happening to these rates of increase, or decrease, as x increases? e What is happening to the five rates of increase as the base is increasing from 13 to 3? CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 504 12A Chapter 12 Euler’s number 8 [Technology] At the start of this section, it was shown by first-principles differentiation that the gradient of 2h − 1 y = 2 x at its y-intercept is given by limh→0 . The investigation in Question 1 showed that the limit is h 2h − 1 , correct to five decimal places, at least for the about 0.7. Use a calculator or a spreadsheet to evaluate h following values of h: a 1 b 0.1 c 0.01 d 0.001 e 0.0001 f 0.000 01 ENRICHMENT 9 This is an alternative way to define e. We proved above that d x 2 = 2 x × m, where m is the gradient of y = 2 x at its y-intercept (0, 1). dx a Explain why a horizontal dilation with factor m creates a graph with y-intercept (0, 1) whose gradient at its y-intercept is 1. b Write down the equation of the dilated function. c Use the chain rule to prove that the dilated function is its own derivative. d Hence define e so that the dilated function is y = e x . CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12B Transformations of exponential functions 505 12B Transformations of exponential functions Learning intentions • Use the calculator to evaluate expressions involving e. • Apply transformations to the graph of y = e x . Transformations of exponential functions with bases other than e were discussed in Chapter 9. This short exercise first practises using the calculator to evaluate powers of e, and is then concerned with transformations of y = e x . Using the calculator to approximate powers of e A calculator will provide approximate values of e x correct to about ten significant figures. Most calculators label the function ex and locate it above the button labelled ln . Pressing shift first may be required. Example 1 Approximating powers of e Use your calculator to find, correct to four significant figures: b e−4 a e2 1 c e3 Solution Using the function labelled ex on the calculator: b e−4 0.01832 a e2 7.389 1 c e 3 1.396 The next three examples must first be put into index form before using the calculator. In particular, e itself must be written as e1 , so that an approximation for the number e can be found using the function ex with input x = 1. Example 2 Approximating powers of e Write each as a power of e, then approximate it correct to five decimal places. √ 1 a e b c e e Solution a e = e1 b 2.71828 1 = e−1 e 0.36788 c √ 1 e = e2 1.64872 The graphs of e x and e−x The graphs of y = e x and y = e−x are fundamental to this whole course. Because x has been replaced by −x in the second equation, the two graphs y = e x and y = e−x are reflections of each other in the y-axis. For y = e x : For y = e−x : x −2 −1 0 1 e2 1 e 1 x −2 −1 0 1 2 1 1 e 1 e2 y y e2 e 1 y = e−x 2 e2 e CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 y = ex 1 The two curves cross at (0, 1). By the definition of e, the gradient of y = e x at its y-intercept (0, 1) is 1. y e −1 1 x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 506 12B Chapter 12 Euler’s number Hence by reflection, the gradient of y = e−x at its y-intercept (0, 1) is −1. Thus the curves are perpendicular at their point of intersection. Note: The function y = e−x is as important as y = e x in applications, or even more important. It describes a great many physical situations where a quantity ‘dies away exponentially’, such as the dying away of the sound of a plucked string. Transformations of y = e x The usual methods of shifting and reflecting graphs can be applied to y = e x . When the graph is shifted vertically, the horizontal asymptote at y = 0 will be shifted also. A small table of approximate values can be a very useful check. All our standard transformations of y = e x have domain all real x, but the range may change. Example 3 Transformations of y = e x Use transformations of the graph of y = e x , confirmed by a table of values, to generate a sketch of each function. Show and state the y-intercept and the horizontal asymptote, and state the range, in both notations. a y = ex + 3 b y = e x−2 c y = e2x Solution a Graph y = e x + 3 by shifting y = e x up 3 units. x −1 0 1 y e−1 + 3 4 e+3 approximation 3.37 4 5.72 y e+3 4 3 y-intercept: (0, 4) asymptote: y=3 range: y > 3, b Graph y = e x−2 by shifting y = e x to the right by 2 units. x 0 1 2 3 y −2 −1 1 e 0.37 1 2.72 e approximation 0.14 e −2 x 1 or we can write (3, ∞) y 1 e−2 y-intercept: (0, e ) 2 asymptote: y=0 (the x-axis) range: y > 0, or we can write (0, ∞) c Graph y = e2x by dilating y = e x horizontally with factor 12 . x − 12 0 y e 1 e approximation 0.37 1 2.72 1 2 −1 y e 1 y-intercept: (0, 1) asymptote: y=0 (the x-axis) range: y > 0, or we can write (0, ∞) 1 2 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 x x © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 12B Transformations of exponential functions Exercise 12B 507 FOUNDATION Technology: The transformations of the graph of the exponential function y = e x throughout this exercise can be confirmed and further developed using graphing software, after which experimentation with further graphs can easily be done. 1 Use the function ex on your calculator to approximate correct to four decimal places. a e2 b e10 f e−2 g e2 c e0 h e− 2 1 1 d e1 e e−1 i e−0.001 j e−6 2 Write each expression as a power of e. Then approximate correct to four significant figures. √ 1 1 1 1 a b 4 c 3e d √ e 20 f e30 e e e e 3 Write each expression as a power of e. Then approximate correct to four significant figures. √ √ 4 5 1 6 a 5e2 b 64 e c 7 e d 35 e e f e 7e4 4 Sketch the graph of y = e x , then use your knowledge of transformations to graph these functions, showing the horizontal asymptote. For each function, state the transformation, give the equation of the asymptote, and state the range in both notations. 5 a y = ex + 1 b y = ex + 2 c y = ex − 1 d y = ex − 2 e y = 2e x f y = e3 x 1 a Copy and complete the following tables of values for the functions y = e x and y = e−x , giving your answers correct to two decimal places. x e −2 −1 0 1 2 x x e −2 −1 0 1 2 −x b Sketch both graphs on one number plane, and draw the tangents at each y-intercept. c What transformation exchanges these graphs? d We saw that the tangent to y = e x at its y-intercept has a gradient of 1. What is the gradient of y = e−x at its y-intercept? Explain why the two tangents are perpendicular. e Hence use point–gradient form to find the equations of the tangents to y = e x and y = e−x at the point A(0, 1) where they intersect. 6 a Sketch the graph of y = e−x . State the horizontal asymptote and the range. b Then use your knowledge of transformations to graph each function below. State the transformation of y = e−x , give the equation of the asymptote, and state the range. i y = e−x + 1 ii y = e−x + 2 iii y = e−x − 1 iv y = e−x − 2 DEVELOPMENT Technology: The graph transformations in this exercise can be confirmed and further developed using graphing software. 7 a What transformation maps y = e x to y = e x−1 ? b Sketch y = e x−1 . c Similarly sketch these functions: i y = e x−3 CambridgeMATHS NSW Stage 6 – Mathematics Extension 1 Year 11 ii y = e x+1 iii y = e x+2 © Pender et al. 2025 Cambridge University Press & Assessment ISBN 978-1-009-65480-7 Photocopying is restricted under law and this material must not be transferred to another party. 508 12B Chapter 12 Euler’s number 8 Let A, B and C be the points on y = e x with x-coordinates 0, 1 and 2. a Find the y-coordinates of A, B and C, and sketch the situation. b Find the gradient of the chord AB, find its equation, then show that its x-intercept is − 1 . e−1 1 . e−1 d Let P and Q be the points on y = e x with x-coordinates a and a + 1. Show similarly that the x-intercept of 1 . the chord PQ is a − e−1 c Similarly show that the x-intercept of the chord BC is 1 − 9 Use the graph of y = e x and your knowledge of transformations to graph these functions. Show the horizontal asymptote and state the range in each case, using both notations. a y = −e x b y = − e−x c y = 1 − ex e y = 1 − e−x f y = e−3x − 2 g y = 2 − 12 e− 2 x d y = 3 − ex 1 h y = e−|x| ENRICHMENT 10 Technology: A spreadsheet will allow this investigation to be extended. The function e x may be approximated by adding up a few terms of the infinite power series x x2 x3 x4 + + + + ··· . 1 1×2 1×2×3 1×2×3×4 Use this power series to approximate each power of e, correct to two decimal places. Then compare your answers with those given by your calculator. ex = 1 + a e b e−1 c e2 d e−2 e e0 When x is large, many more terms are needed to approximate e x correct to say two decimal places. Why is this so? Use your spreadsheet to investigate this. Note: Th

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