www.ck12.org
MHF4U Advanced Functions
ii
www.ck12.org
Patricia Garcia
Andrew Gloag
Eve Rawley
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
Brenda Meery
Kaitlyn Spong
Anne Gloag
CK-12
Mark Spong
Katie Sinclear
Melissa Kramer
Sue Boskat Murray
Jen Kershaw
Lori Jordan
Kate Dirga
Bill Zahner
Bradley Hughes
Larry Ottman
Andrea Hayes
Ck12 Science
Jean Brainard, Ph.D.
Dana Desonie, Ph.D.
Dan Greenberg
Victor Cifarelli
Jim Sconyers
James H Dann, Ph.D.
iii
Say Thanks to the Authors
Click http://www.ck12.org/saythanks
(No sign in required)
www.ck12.org
To access a customizable version of this book, as well as other
interactive content, visit www.ck12.org
EDITORS
Kaitlyn Spong
Larame Spence
SOURCE
Anne Gloag
CK-12 Foundation is a non-profit organization with a mission to
reduce the cost of textbook materials for the K-12 market both in
the U.S. and worldwide. Using an open-source, collaborative, and
web-based compilation model, CK-12 pioneers and promotes the
creation and distribution of high-quality, adaptive online textbooks
that can be mixed, modified and printed (i.e., the FlexBook®
textbooks).
Copyright © 2019 CK-12 Foundation, www.ck12.org
The names “CK-12” and “CK12” and associated logos and the
terms “FlexBook®” and “FlexBook Platform®” (collectively
“CK-12 Marks”) are trademarks and service marks of CK-12
Foundation and are protected by federal, state, and international
laws.
Any form of reproduction of this book in any format or medium,
in whole or in sections must include the referral attribution link
http://www.ck12.org/saythanks (placed in a visible location) in
addition to the following terms.
Except as otherwise noted, all CK-12 Content (including CK-12
Curriculum Material) is made available to Users in accordance
with the Creative Commons Attribution-Non-Commercial 3.0
Unported (CC BY-NC 3.0) License (http://creativecommons.org/
licenses/by-nc/3.0/), as amended and updated by Creative Commons from time to time (the “CC License”), which is incorporated
herein by this reference.
Complete terms can be found at http://www.ck12.org/about/
terms-of-use.
Printed: December 15, 2019
CONTRIBUTOR
Chris Addiego
AUTHORS
Patricia Garcia
Andrew Gloag
Eve Rawley
Raja Almukahhal
Larame Spence
Mara Landers
Nick Fiori
Art Fortgang
Melissa Vigil
Brenda Meery
Kaitlyn Spong
Anne Gloag
CK-12
Mark Spong
Katie Sinclear
Melissa Kramer
Sue Boskat Murray
Jen Kershaw
Lori Jordan
Kate Dirga
Bill Zahner
Bradley Hughes
Larry Ottman
Andrea Hayes
Ck12 Science
Jean Brainard, Ph.D.
Dana Desonie, Ph.D.
Dan Greenberg
Victor Cifarelli
Jim Sconyers
James H Dann, Ph.D.
Sanjeev Narayanaswamy
v
Contents
www.ck12.org
Contents
1
vi
Unit 1
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20
1.21
1.22
1.23
1.24
1.25
1.26
1.27
1.28
1.29
1.30
1.31
1.32
1.33
1.34
1.35
1.36
1.37
1.38
1.39
1.40
Algebraic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Domain and Range . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Function Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Intervals and Interval Notation . . . . . . . . . . . . . . . . . . . . . . . . .
Determining the Equation of a Line . . . . . . . . . . . . . . . . . . . . . . .
Transformations of functions . . . . . . . . . . . . . . . . . . . . . . . . . .
Transformations of Quadratic Functions . . . . . . . . . . . . . . . . . . . .
Vertex Form of a Quadratic Function . . . . . . . . . . . . . . . . . . . . . .
Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Maximums and Minimums . . . . . . . . . . . . . . . . . . . . . . . . . . .
Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Zeroes of a Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Graphs of Polynomials Using Zeros . . . . . . . . . . . . . . . . . . . . . .
Graphs of Polynomials Using Transformations . . . . . . . . . . . . . . . . .
Increasing and Decreasing . . . . . . . . . . . . . . . . . . . . . . . . . . .
Average Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Instantaneous Rate of Change . . . . . . . . . . . . . . . . . . . . . . . . . .
Long Division . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Multiplication of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . .
Factorization using Difference of Squares . . . . . . . . . . . . . . . . . . .
Factorization of Quadratic Expressions . . . . . . . . . . . . . . . . . . . . .
Quadratic Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polynomial Long Division and Synthetic Division . . . . . . . . . . . . . . .
Real Zeros of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . .
Relationship between Zeros and Coefficients of a Polynomial . . . . . . . . .
Factorization of Special Cubics . . . . . . . . . . . . . . . . . . . . . . . . .
Zero Product Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Polynomial and Rational Inequalities . . . . . . . . . . . . . . . . . . . . . .
Domain and Range of a Function . . . . . . . . . . . . . . . . . . . . . . . .
Slope of a Line Using Two Points . . . . . . . . . . . . . . . . . . . . . . . .
Methods for Solving Quadratic Functions . . . . . . . . . . . . . . . . . . .
Quadratic Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Graphing Rational Functions in Standard Form . . . . . . . . . . . . . . . . .
Graphs of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
Holes in Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . .
Zeroes of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . .
Graphing when the Degrees of the Numerator and Denominator are Different
Graphing when the Degrees of the Numerator and Denominator are the Same
Analysis of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . .
Solving Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
1
2
7
13
21
27
33
45
52
59
73
79
87
95
103
114
120
125
131
134
139
143
148
152
157
165
167
170
175
183
187
197
202
209
217
224
229
232
238
245
251
www.ck12.org
1.41
1.42
1.43
2
3
Contents
Clearing Denominators in Rational Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
Unit 1 Test Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271
Unit 2
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
2.15
2.16
2.17
2.18
2.19
2.20
2.21
2.22
2.23
2.24
2.25
2.26
2.27
2.28
2.29
2.30
2.31
2.32
2.33
2.34
2.35
2.36
2.37
2.38
Trigonometric Ratios with a Calculator . . . . . . . . . . . . . . . . .
Introduction to Trig Identities . . . . . . . . . . . . . . . . . . . . . .
Multiplication of Polynomials by Binomials . . . . . . . . . . . . . .
Distance Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Reference Angles and Angles in the Unit Circle . . . . . . . . . . . .
Conversion between Degrees and Radians . . . . . . . . . . . . . . .
Radian Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Arc Length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Applications of Radian Measure . . . . . . . . . . . . . . . . . . . .
The Unit Circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Trigonometric Ratios on the Unit Circle . . . . . . . . . . . . . . . .
Reciprocal Trigonometric Functions . . . . . . . . . . . . . . . . . .
Cofunction Identities and Reflection . . . . . . . . . . . . . . . . . .
Finding Exact Trig Values using Sum and Difference Formulas . . . .
Sine Sum and Difference Formulas . . . . . . . . . . . . . . . . . . .
Cosine Sum and Difference Formulas . . . . . . . . . . . . . . . . . .
Tangent Sum and Difference Formulas . . . . . . . . . . . . . . . . .
Solving Trig Equations using Sum and Difference Formulas . . . . . .
Finding Exact Trig Values using Double and Half Angle Formulas . .
Sum and Difference Identities . . . . . . . . . . . . . . . . . . . . . .
Double Angle Identities . . . . . . . . . . . . . . . . . . . . . . . . .
Simplifying Trig Expressions using Double and Half Angle Formulas .
Double, Half, and Power Reducing Identities . . . . . . . . . . . . . .
Six Trigonometric Functions and Radians . . . . . . . . . . . . . . .
Graphing Sine and Cosine . . . . . . . . . . . . . . . . . . . . . . . .
Amplitude, Period, and Frequency . . . . . . . . . . . . . . . . . . .
Horizontal Translations or Phase Shifts . . . . . . . . . . . . . . . . .
Graphing Tangent . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Vertical Translations . . . . . . . . . . . . . . . . . . . . . . . . . . .
Sine and Cosecant Graphs . . . . . . . . . . . . . . . . . . . . . . . .
Cosine and Secant Graphs . . . . . . . . . . . . . . . . . . . . . . . .
Tangent and Cotangent Graphs . . . . . . . . . . . . . . . . . . . . .
Putting it all Together . . . . . . . . . . . . . . . . . . . . . . . . . .
Trigonometric Equations Using Factoring . . . . . . . . . . . . . . .
Trigonometric Equations Using the Quadratic Formula . . . . . . . .
Trig. Word Problems and IROC . . . . . . . . . . . . . . . . . . . . .
Unit 2 Test Review . . . . . . . . . . . . . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
275
276
282
286
293
300
308
311
317
326
336
343
349
352
354
357
361
367
371
375
378
383
389
392
396
400
408
414
421
426
432
444
454
469
475
480
485
489
498
Unit 3
3.1
3.2
3.3
3.4
3.5
3.6
Graphs of Exponential Functions . . . . . . . . . . . .
Laws of Exponents for Real Numbers . . . . . . . . .
Zero and Negative Exponents . . . . . . . . . . . . . .
Inverse Functions . . . . . . . . . . . . . . . . . . . .
Graphs of Logs as the Inverse of Exponential Functions
Defining Logarithms . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
500
501
511
512
520
528
534
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
vii
Contents
4
5
viii
www.ck12.org
3.7
3.8
3.9
3.10
3.11
3.12
3.13
3.14
3.15
3.16
3.17
3.18
3.19
3.20
3.21
3.22
3.23
3.24
Graphs of Logarithmic Functions . . . . . . . .
Graphing Logarithmic Functions . . . . . . . .
Change of Base . . . . . . . . . . . . . . . . .
pH . . . . . . . . . . . . . . . . . . . . . . . .
Intensity and Loudness of Sound . . . . . . . .
Earthquake Magnitude Scales . . . . . . . . . .
Applications . . . . . . . . . . . . . . . . . . .
Simplifying Radicals . . . . . . . . . . . . . .
Solving Exponential Equations . . . . . . . . .
Exponential Equations . . . . . . . . . . . . . .
Half Life . . . . . . . . . . . . . . . . . . . . .
Product and Quotient Properties of Logarithms .
Properties of Logs . . . . . . . . . . . . . . . .
Power Property of Logarithms . . . . . . . . .
Solving Logarithmic Equations . . . . . . . . .
Logarithmic Models . . . . . . . . . . . . . . .
Unit 3 Test Review . . . . . . . . . . . . . . .
References . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
538
547
552
555
558
562
565
570
573
577
584
588
591
595
599
603
608
616
Unit 4
4.1
4.2
4.3
4.4
4.5
4.6
4.7
CK-12 Calculus Study Guide - Arithmetic Operations on Functions
Operations on Functions . . . . . . . . . . . . . . . . . . . . . . .
CK-12 Calculus Study Guide - Compositions of Functions . . . .
Function Composition . . . . . . . . . . . . . . . . . . . . . . . .
Composition of Functions . . . . . . . . . . . . . . . . . . . . . .
Composition of Functions and Invertible Functions . . . . . . . .
Function Operations . . . . . . . . . . . . . . . . . . . . . . . . .
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
617
618
627
634
645
650
654
666
Exam Review
667
5.1
Exam Review Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 668
5.2
Exam Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 674
www.ck12.org
Chapter 1. Unit 1
C HAPTER
1
Unit 1
Chapter Outline
1.1
A LGEBRAIC F UNCTIONS
1.2
D OMAIN AND R ANGE
1.3
F UNCTION N OTATION
1.4
I NTERVALS AND I NTERVAL N OTATION
1.5
D ETERMINING THE E QUATION OF A L INE
1.6
T RANSFORMATIONS OF FUNCTIONS
1.7
T RANSFORMATIONS OF Q UADRATIC F UNCTIONS
1.8
V ERTEX F ORM OF A Q UADRATIC F UNCTION
1.9
P OLYNOMIAL F UNCTIONS
1.10
M AXIMUMS AND M INIMUMS
1.11
E VEN AND O DD F UNCTIONS
1.12
Z EROES OF A P OLYNOMIAL
1.13
G RAPHS OF P OLYNOMIALS U SING Z EROS
1.14
G RAPHS OF P OLYNOMIALS U SING T RANSFORMATIONS
1.15
I NCREASING AND D ECREASING
1.16
AVERAGE R ATE OF C HANGE
1.17
I NSTANTANEOUS R ATE OF C HANGE
1.18
L ONG D IVISION
1.19
M ULTIPLICATION OF P OLYNOMIALS
1.20
FACTORIZATION USING D IFFERENCE OF S QUARES
1.21
FACTORIZATION OF Q UADRATIC E XPRESSIONS
1.22
Q UADRATIC F ORMULA
1.23
P OLYNOMIAL L ONG D IVISION AND S YNTHETIC D IVISION
1.24
R EAL Z EROS OF P OLYNOMIALS
1.25
R ELATIONSHIP BETWEEN Z EROS AND C OEFFICIENTS OF A P OLYNOMIAL
1.26
FACTORIZATION OF S PECIAL C UBICS
1.27
Z ERO P RODUCT P RINCIPLE
1.28
P OLYNOMIAL AND R ATIONAL I NEQUALITIES
1.29
D OMAIN AND R ANGE OF A F UNCTION
1.30
S LOPE OF A L INE U SING T WO P OINTS
1.31
M ETHODS FOR S OLVING Q UADRATIC F UNCTIONS
1.32
Q UADRATIC I NEQUALITIES
1.33
G RAPHING R ATIONAL F UNCTIONS IN S TANDARD F ORM
1
www.ck12.org
2
1.34
G RAPHS OF R ATIONAL F UNCTIONS
1.35
H OLES IN R ATIONAL F UNCTIONS
1.36
Z EROES OF R ATIONAL F UNCTIONS
1.37
G RAPHING WHEN THE D EGREES OF THE N UMERATOR AND D ENOMINATOR ARE
D IFFERENT
1.38
G RAPHING WHEN THE D EGREES OF THE N UMERATOR AND D ENOMINATOR ARE
THE S AME
1.39
A NALYSIS OF R ATIONAL F UNCTIONS
1.40
S OLVING R ATIONAL E QUATIONS
1.41
C LEARING D ENOMINATORS IN R ATIONAL E QUATIONS
1.42
U NIT 1 T EST R EVIEW
1.43
R EFERENCES
www.ck12.org
Chapter 1. Unit 1
1.1 Algebraic Functions
Learning Objectives
Here you’ll learn how to determine whether a relation is a function given its domain and range or its graph.
Algebraic Functions
A function is a special kind of relation. In a function, for each input there is exactly one output; in a relation, there
can be more than one output for a given input.
Looking at a Relation
Consider the relation that shows the heights of all students in a class. The domain is the set of people in the class and
the range is the set of heights. This relation is a function because each person has exactly one height. If any person
had more than one height, the relation would not be a function.
Notice that even though the same person can’t have more than one height, it’s okay for more than one person to have
the same height. In a function, more than one input can have the same output, as long as more than one output never
comes from the same input.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133265
3
1.1. Algebraic Functions
www.ck12.org
Determining if a Relation is a Function
Determine if the relation is a function.
The easiest way to figure out if a relation is a function is to look at all the x-values in the list or the table. If a value
of x appears more than once, and it’s paired up with different y-values, then the relation is not a function.
1. (1, 3), (-1, -2), (3, 5), (2, 5), (3, 4)
You can see that in this relation there are two different y-values paired with the x-value of 3. This means that this
relation is not a function.
2. (-3, 20), (-5, 25), (-1, 5), (7, 12), (9, 2)
Each value of x has exactly one y-value. The relation is a function.
3. x 2 1
y 12 10 8
0 1
6 4
2
In this relation there are two different y-values paired with the x-value of 2 and two y-values paired with the x-value
of 1. The relation is not a function.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133264
Determining if a Graph is a Function
When a relation is represented graphically, we can determine if it is a function by using the vertical line test. If you
can draw a vertical line that crosses the graph in more than one place, then the relation is not a function.
For the following graphs, determine whether they are functions.
Not a function. It fails the vertical line test.
4
www.ck12.org
Chapter 1. Unit 1
A function. No vertical line will cross more than one point on the graph.
Watch this video for help with the Examples above.
Examples
For the following graphs, determine whether they are functions.
Example 1
A function. No vertical line will cross more than one point on the graph.
Example 2
5
1.1. Algebraic Functions
www.ck12.org
Not a function. It fails the vertical line test.
Review
In 1-8, determine whether each relation is a function:
1. (1, 7), (2, 7), (3, 8), (4, 8), (5, 9)
2. (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3)
3.
x -4 -3 -2 -1 0
y 16 9 4 1 0
4. (2, -6), (1, -3), (0, 0), (1, 3), (2, 6)
5. (-2, -8), (-1, -1), (0, 0), (1, 1), (2, 8)
6. (-5, 10), (-1, 5), (0, 10), (1, 5), (5, 10)
7.
x 0 1 10 100 1000
y 2 -2
2
-2
2
8. Age2025253035
Number of jobs 3 4 7 4 2
(in the past 5 years)
In 9-10, use the vertical line test to determine whether each relation is a function.
9.
10.
6
www.ck12.org
Chapter 1. Unit 1
Review (Answers)
To view the Review answers, open this PDF file and look for section 1.15.
7
1.2. Domain and Range
www.ck12.org
1.2 Domain and Range
Learning Objectives
Here you will learn to identify the independent variable (the domain) and the dependent variable (the range) of
a function.
You and your 3 friends are debating which theater to choose for a trip to the movies on Friday. There are 4 theaters
in town, all with different prices for tickets. The most expensive theater charges $12.50 per person, but has stadium
seating and a screen 5 stories tall. The cheapest is only $5.50 per ticket, but only plays older movies and is much
less comfortable. The other two are between the extremes at $8.50 and $9.75 each.
If you were to graph the total costs for the 4 of you based on which theater you choose, which values would represent
the domain and which the range?
Finding Domain and Range
Independent Variable, Domain
The domain of a function is defined as the set of all x values for which the function is defined. For example, the
domain of the function y = 3x is the set of all real numbers, often written as R. This means that√x can be any real
number. Other functions have restricted domains. For example, the domain of the function y = x is the set of all
real numbers greater than or equal to zero. The domain of this function is restricted in this way because the square
root of a negative number is not a real number. Therefore, the domain is restricted to non-negative values of x so that
the function values will be defined.
The variable x is often referred to as the independent variable, while the variable y is referred to as the dependent
variable. We talk about x and y this way because the y values of a function depend on what the x values are. That
is why we also say that “y is a function of x.” For example, the value of y in the function y = 3x depends on what x
value we are considering. If x = 4, we can easily determine that y = 3(4) = 12.
Dependent Variable, Range
The range of a function is defined as the set of all y values for which a function is defined. Just as we did with domain,
we can examine a function and determine its range. Again, it is often helpful to think about what restrictions there
might be, and what the graph of the function looks like. Consider for example the function y = x2 . The domain of
this function is R, all real numbers, but what about the range?
The range of the function is the set of all real numbers greater than or equal to zero. This is the case because every
y value is the square of an x value. If we square positive and negative numbers, the result will always be positive. If
x = 0, then y = 0. We can also see the range if we look at a graph of y = x2 . Notice below that the y values are all
greater than or equal to zero.
8
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187052
Examples
Example 1
Earlier, you were given a problem about you and 3 friends, who are going to see a movie Friday.
If you were to graph the total costs for the 4 of you based on which theater you choose, which values would represent
the domain and which the range?
One function that describes this situation would be: P(t) = 4t, where P(t) would be total Price based on per-ticket
price, and it is equal to 4 x ticket price.
9
1.2. Domain and Range
www.ck12.org
In this case, your domain would be: {$5.50, $8.50, $9.75, $12.50} since these are the prices you would input in
place of the independent variable, x to get the total price for each theater. The range would be: {$22.00, $34.00,
$38.00, $50.00} since these are the output values given by the dependent variable, y.
Example 2
State the domain of each function.
a. y = x2
The domain of this function is the set of all real numbers. There are no restrictions.
b.
1
x
The domain of this function is the set of all real numbers except x = 0. The domain is restricted this way because a
fraction with denominator zero is undefined.
c. (2, 4), (3, 9), (5, 11)
The domain of this function is the set of x values: {2, 3, 5}.
Example 3
State the domain and range of the function y = 2x .
For this function, we can choose any x value except x = 0, since we cannot divide a number by zero. Therefore the
domain of the function is the set of all real numbers except x = 0.
The range is also restricted to the non-zero real numbers, but for a different reason. Because the numerator of the
fraction is 2, the numerator can never equal zero, so the fraction can never equal zero.
Example 4
Determine the domain of the function y =
√
x, shown in the graph below.
√
First consider what restrictions there might be and then look at the graph. We can see that y = x has a domain of
all real numbers greater than or equal to zero because the graph only exists for x values that are positive, since there
are no real square roots of negative numbers.
10
www.ck12.org
Chapter 1. Unit 1
Example 5
Give an example of a function for which the domain and range are equivalent to each other.
Answers will vary. y = x is an example.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187057
Review
Determine the domain and the range of the relations.
1. Relation: {(0,4), (3, 20), (90, 33)}
2. Relation: {(3, -4), (6, 37) , (10, -10), (-31, 2)}
3. Tina’s car travels about 30 miles on one gallon of gas. She has between 10 and 12 gallons of gas in the tank.
Find the domain and range of the function to calculate how far she can drive.
4. Joe and three of his friends plan on going bowling and plan on bowling one or two games each. Each game
costs $2.75. Find the domain and range of the function calculating the cost of the trip.
5. Bob had a summer job that paid $10.00 per hour and he worked between 20-25 hours every week. His weekly
salary can be modeled by the equation S = 10h, where S is his weekly salary and h is the number of hours he
worked per week. What is the independent variable for this problem? Describe the domain and range for this
problem.
6. What does each value in the ordered pair (20, 200) mean in context of the previous problem?
7. Which group of students represents the domain, and which the range, in these ordered pairs? (Jim, Kitty),
(Joe, Betty), (Brian, Alice), (Jesus, Anissa), (Ken, Kelli)
State the domain and range:
8.
9.
10.
11.
y = |x|
x = |y|
y = x2
x = y2
11
1.2. Domain and Range
12.
13.
14.
12
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
15.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.2.
13
1.3. Function Notation
www.ck12.org
1.3 Function Notation
Learning Objectives
Here you’ll learn how to use function notation when working with functions.
Suppose the value V of a digital camera t years after it was bought is represented by the function V (t) = 875 − 50t.
• Can you determine the value of V (4) and explain what the solution means in the context of this problem?
• Can you determine the value of t when V (t) = 525 and explain what this situation represents?
• What was the original cost of the digital camera?
Function Notation
A function is like a machine that shows how an input is changed to create an output. For example, a function
that triples the input and subtracts one from it to create the output, would convert x into 3(x) − 1. If that function
were named f , and 3 is fed into the machine, 3(3) − 1 = 8 comes out. In other words, function f of 3 = 3(3) − 1, or,
in standard function notation, f (3) = 3(3) − 1 = 8.
14
www.ck12.org
Chapter 1. Unit 1
When naming a function the symbol f (x) is often used. The symbol f (x) is pronounced as “ f of x.” This means that
the equation is a function that is written in terms of the variable x. An example of such a function is f (x) = 3x + 4.
Functions can also be written using a letter other than f and a variable other than x. For example, v(t) = 2t 2 − 5 and
d(h) = 4h − 3. In addition to representing a function as an equation, you can also represent a function:
•
•
•
•
As a graph
As ordered pairs
As a table of values
As an arrow or mapping diagram
When a function is represented as an equation, an ordered pair can be determined by evaluating various values of
the assigned variable. Suppose f (x) = 3x − 4. To calculate f (4), substitute:
f (4) = 3(4) − 4
f (4) = 12 − 4
f (4) = 8
Graphically, if f (4) = 8, this means that the point (4, 8) is a point on the graph of the line.
Let’s practice function notation by substituting the following values of x into the equation f (x) =
x2 2x5 :
1. f (2)
2. f (−7)
3. f (1.4)
To determine the value of the function for the assigned values of the variable, substitute the values into the
function.
15
1.3. Function Notation
www.ck12.org
f (x) = x2 + 2x + 5
f (x) = x2 + 2x + 5
↓
↓
↓
&
2
↓
&
2
f (x) = x2 + 2x + 5
↓
↓
&
2
f (2) = (2) + 2(2) + 5
f (−7) = (−7) + 2(−7) + 5
f (1.4) = (1.4) + 2(1.4) + 5
f (2) = 4 + 4 + 5
f (−7) = 49 − 14 + 5
f (1.4) = 1.96 + 2.8 + 5
f (−7) = 40
f (1.4) = 9.76
f (2) = 13
Functions can also be represented as mapping rules.
Now, if g : x → 5 − 2x , let’s find the following in simplest form:
1. g(y)
g(y) = 5 − 2y
2. g(y − 3)
g(y − 3) = 5 − 2(y − 3) = 5 − 2y + 6 = 11 − 2y
3. g(2y)
g(2y) = 5 − 2(2y) = 5 − 4y
Examples
Example 1
Earlier, you were told that the value V of a digital camera t years after it was bought is represented by the function
V (t) = 875 − 50t. We want to:
• Determine the value of V (4) and explain what the solution means to this problem.
• Determine the value of t when V (t) = 525 and explain what this situation represents.
• What was the original cost of the digital camera?
The camera is valued at $675, 4 years after it was purchased.
V (t) = 875 − 50t
V (4) = 875 − 50(4)
V (4) = 875 − 200
V (4) = $675
The digital camera has a value of $525, 7 years after it was purchased.
16
www.ck12.org
Chapter 1. Unit 1
V (t) = 875 − 50t
Let V (t) = 525
525 = 875 − 50t
Solve the equation
525 − 875 = 875 − 875 − 50t
− 350 = −50t
−350 −50t
=
−50
−50
7=t
The original cost of the camera was $875.
V (t) = 875 − 50t
Let t = 0.
V (0) = 875 − 50(0)
V (0) = 875 − 0
V (0) = $875
Example 2
If f (x) = 3x2 − 4x + 6 find:
• f (−3)
• f (a − 2)
f (x) = 3x2 − 4x + 6
Substitute (−3) for x in the function.
f (−3) = 3(−3)2 − 4(−3) + 6
Perform the indicated operations.
f (−3) = 3(9) + 12 + 6
Simplify
f (−3) = 27 + 12 + 6
f (−3) = 45
f (−3) = 45
f (x) = 3x2 − 4x + 6
f (a − 2) = 3(a − 2)2 − 4(a − 2) + 6
Write (a − 2)2 in expanded form.
f (a − 2) = 3(a − 2)(a − 2) − 4(a − 2) + 6
Perform the indicated operations.
f (a − 2) = (3a − 6)(a − 2) − 4(a − 2) + 6
f (a − 2) = 3a2 − 6a − 6a + 12 − 4a + 8 + 6
Simplify
2
f (a − 2) = 3a − 16a + 26
f (a − 2) = 3a2 − 16a + 26
17
1.3. Function Notation
www.ck12.org
Example 3
If f (m) =
m+3
2m−5
12
13
find ‘m’ if f (m) =
m+3
2m − 5
12
m+3
=
13 2m − 5
12
m+3
(13)(2m − 5) = (13)(2m − 5)
13
2m − 5
12
m+3
(13)(2m
− 5) = (13)
(2m
− 5)
13
2m−
5
(2m − 5)12 = (13)m + 3
f (m) =
Solve the equation for m.
24m − 60 = 13m + 39
24m − 60+60 = 13m + 39+60
24m = 13m + 99
24m−13m = 13m−13m + 99
11m = 99
11m 99
=
11
11
9
11m
99
=
11
11
m=9
Example 4
The emergency brake cable in a truck parked on a steep hill breaks and the truck rolls down the hill. The distance in
feet, d, that the truck rolls is represented by the function d = f (t) = 0.5t 2 .
How far will the truck roll after 9 seconds? How long will it take the truck to hit a tree which is at the bottom of the
hill 600 feet away? Round your answer to the nearest second.
d = f (t) = 0.52
d = f (t) = 0.52
2
f (9) = 0.5(9)
f (9) = 0.5(81)
f (9) = 40.5 f eet
18
Substitute 9 for t.
Perform the indicated operations.
www.ck12.org
Chapter 1. Unit 1
After 9 seconds, the truck will roll 40.5 feet.
d = f (t) = 0.5t 2
600 = 0.5t
Substitute 600 for d.
2
Solve for t.
600 0.5t 2
=
0.5
0.5
1200
2
600
0.5t
=
0.5
0.5
1200 = t 2
√
√
1200 = t 2
34.64 seconds ≈ t
The truck will hit the tree in approximately 35 seconds.
Example 5
Let P(a) =
2a−3
a+2 .
First, evaluate:
• P(0)
• P(1) • P − 12
Then find a value of ‘a’ where P(a) does not exist, find P(a − 2) in simplest form, and find ‘a’ if P(a) = −5
Evaluating for 0, 1, and − 21 :
2a − 3
a+2
2(0) − 3
P(0) =
(0) + 2
P(a) =
P(0) =
−3
2
2a − 3
a+2
2(1) − 3
P(1) =
(1) + 2
P(a) =
P(1) =
2−3
1+2
P(1) =
−1
3
2a − 3
P(a) =
a+2
2 − 12 − 3
1
P −
=
2
− 12 + 2
1 2 − 1 − 3
1
2
P −
=
2
− 12 + 42
1
−1 − 3
P −
=
3
2
2
1
3
P −
= −4 ÷
2
2
1
2
P −
= −4
2
3
1
−8
P −
=
2
3
Finding where the function is undefined:
The function will not exist if the denominator equals zero because division by zero is undefined.
19
1.3. Function Notation
www.ck12.org
a+2 = 0
a+2−2 = 0−2
a = −2
Therefore, if a = −2, then P(a) =
2a−3
a+2
does not exist.
Finding P(a − 2):
2a − 3
a+2
2(a − 2) − 3
P(a − 2) =
(a − 2) + 2
2a − 4 − 3
P(a − 2) =
a−2+2
2a − 7
P(a − 2) =
a
2a 7
−
P(a − 2) =
a
a
7
P(a − 2) = 2 −
a
P(a) =
Substitue a − 2 for a
Remove parentheses
Combine like terms
Express the fraction as two separate fractions and reduce.
Finding ‘a’ if P(a) = −5:
2a − 3
a+2
2a − 3
−5 =
a + 2 2a − 3
− 5(a + 2) =
(a + 2)
a+2
2a − 3
(
− 5a − 10 =
a+
2)
a+
2
− 5a − 10 = 2a − 3
P(a) =
− 5a − 10 − 2a = 2a − 2a − 3
− 7a − 10 = −3
− 7a − 10 + 10 = −3 + 10
− 7a = 7
7
−7a
=
−7
−7
a = −1
Review
If g(x) = 4x2 − 3x + 2, find expressions for the following:
1. g(a)
20
Let P(a) = −5
Multiply both sides by (a + 2)
Simplify
Solve the linear equation
Move 2a to the left by subtracting
Simplify
Move 10 to the right side by addition
Simplify
Divide both sides by -7 to solve for a.
www.ck12.org
2.
3.
4.
5.
Chapter 1. Unit 1
g(a − 1)
g(a + 2)
g(2a)
g(−a)
If f (y) = 5y − 3, determine the value of ‘y’ when:
6.
7.
8.
9.
10.
f (y) = 7
f (y) = −1
f (y) = −3
f (y) = 6
f (y) = −8
The value of a Bobby Orr rookie card n years after its purchase is V (n) = 520 + 28n.
11. Determine the value of V (6) and explain what the solution means.
12. Determine the value of n when V (n) = 744 and explain what this situation represents.
13. Determine the original price of the card.
Let f (x) =
3x
x+2 .
14. When is f (x) undefined?
15. For what value of x does f (x) = 2.4?
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.3.
21
1.4. Intervals and Interval Notation
www.ck12.org
1.4 Intervals and Interval Notation
Learning Objectives
Learn to identify real functions, and recognize closed and open intervals. Learn to interpret and express intervals in
interval notation.
Suppose you and 2 of your friends were out for lunch and decide to buy tacos. Together you have $15 to spend on
lunch, and tacos are $1.25 each. It is clear that the total cost could be graphed as a function of the number of tacos
purchased, but how would you specify that the graph should not include values greater than $15 or less than $3.75
(one taco each)?
Intervals and Interval Notation
Real Values and Intervals
A function is defined as a real function if both the domain and the range are sets of real numbers. Many of the
functions you have likely encountered before are real functions, and many of these functions have Domain = R.
Consider, for example, the function y = 3x. A section of the graph of this function is shown below.
You may already be familiar with the graphs of lines. In particular, you may already be in the habit of placing
arrows at the ends. We do this in order to indicate that the line will continue forever in both the positive and negative
directions, both in terms of the domain and the range. The line above, however, only shows the function y = 3x on
the interval [-3, 3]. The square brackets indicate that the graph includes the endpoints of the interval, where x = -3
and x = 3. We call this a closed interval. A closed interval contains its endpoints. In contrast, an open interval does
not contain its endpoints. We indicate an open interval with parentheses. For example, (-3, 3) indicates the set of
numbers between -3 and 3, not including -3 and 3. You may have noticed that the open interval notation looks like
22
www.ck12.org
Chapter 1. Unit 1
the notation for a point (x, y) in the plane. It is important to read an example or a homework problem carefully to
avoid confusing a point with an interval! The difference is generally quite clear from the context.
The table below summarizes the kinds of intervals you may need to consider while studying functions and their
domains:
TABLE 1.1:
Interval notation
[a, b]
Inequality notation
a≤x≤b
(a, b)
a<x<b
[a, b)
a≤x<b
(a, b]
a<x≤b
(a, ∞)
x>a
[a, ∞)
x≥a
(−∞, a)
(−∞, a]
x
x≤a
Description
The value of x is between a and b,
including a and b, where a, b are
real numbers.
The value of x is between a and b,
not including a and b.
The value of x is between a and b,
including a, but not including b.
The value of x is between a and b,
including b, but not including a.
The value of x is strictly greater than
a.
The value of x is greater than or
equal to a
The value of x is strictly less than a
The value of x is less than or equal
to a.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187083
Examples
Example 1
Earlier, you were given a problem about buying tacos for lunch.
Together, you have $15 to spend on lunch, and tacos are $1.25 each. It is clear that the total cost could be graphed
as a function of the number of tacos purchased, but how would you specify that the graph should not include values
greater than $15 or less than $3.75 (one taco each)?
To specify that the graph of the cost of lunch only includes values between $3.75 and $15.00, specify the interval of
the domain as: [3.75, 15].
23
1.4. Intervals and Interval Notation
www.ck12.org
Example 2
Identify the sets described:
a. (−3, 9]
The set of numbers between -3 and 9, “not including” the actual value of -3, but “including” 9.
b. [−23, 12]
The set of numbers between -23 and 12, “including” the values -23 and 12.
c. (−∞, 0)
All numbers less than 0, not including 0 itself.
Example 3
Sketch the graph of the function f (x) = 12 x − 6 on the interval [-4, 12).
The figure below shows a graph of f (x) = 21 x − 6 on the given interval:
Example 4
Describe the specified intervals, use interval notation:
a. All positive numbers
(0, +∞)
Zero is neither positive nor negative, so the “(“ is used to specify that zero is “not” included. Since there is no
maximum positive number, we specify that infinity is the upper value, and use “)” since it cannot be reached.
b. The numbers between negative eight and two hundred forty two, including both.
[−8, 242]
The “[“ is used on both ends, since both values are included.
24
www.ck12.org
Chapter 1. Unit 1
c. All negative numbers, zero, and the positive numbers up to and including nine.
(−∞, 9]
The “(“ denotes that negative infinity cannot be reached, and “]” on the other end specifies that 9 is included in the
set.
Example 5
Describe the domain and range in the sets in the images below using interval notation.
TABLE 1.2:
a.
b.
a. The domain is the set of x values starting with the included -6 and ending at 4, which is not included: [-6, 4).
The range is the set of y values from -3 (not included) to 4 (included): (-3, 4].
b. Domain: [-6, 7) (from above)
Range: [-1, 6) (from above)
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187062
Review
Write the following in interval notation.
1.
2.
3.
4.
−3 ≤ x < 1
0<x<2
x > −3
x≤2
Solve and put your answer in interval notation.
25
1.4. Intervals and Interval Notation
5. −2x + 3 < 1
6. 7x + 4 ≤ 2x − 6
For each number line, write the given set of numbers in interval notation.
7.
8.
9.
10.
Name the domain and range for each relation using interval notation.
11.
26
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
12.
Express the following sets using interval notation, then sketch them on a number line.
13.
14.
15.
16.
{x : 1 ≤ x ≤ 3}
{x : 2 ≤ x < 1}
A is the set of all numbers bigger than 2 but less than or equal to 5.
{x : 3 < x < ∞}
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.3.
27
1.5. Determining the Equation of a Line
www.ck12.org
1.5 Determining the Equation of a Line
Here you’ll learn how to write the equations of lines given their slope and y-intercept or two of their points.
Determining the Equation of a Line
You are probably aware that many real-world situations can be described with linear graphs and equations. In this
lesson, we’ll see how to find equations in a variety of situations.
Write an Equation Given Slope and y−Intercept
Recall that you may write an equation in slope-intercept form with a few simple steps: start with the general equation
for the slope-intercept form of a line, y = mx + b, and then substitute the given values of m and b into the equation.
For example, a line with a slope of 4 and a y−intercept of -3 would have the equation y = 4x − 3.
If you are given just the graph of a line, you can identify the slope and y−intercept from the graph and write the
equation from there. For example, on the graph below you can see that the line rises by 1 unit as it moves 2 units to
the right, so its slope is 21 . Also, you can see that the y−intercept is -2, so the equation of the line is y = 12 x − 2.
Write an Equation Given the Slope and a Point
Often, we don’t know the value of the y−intercept, but we know the value of y for a non-zero value of x. In this case,
it’s often easier to write an equation of the line in point-slope form. An equation in point-slope form is written as
y − y0 = m(x − x0 ), where m is the slope and (x0 , y0 ) is a point on the line.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133178
28
www.ck12.org
Chapter 1. Unit 1
Writing the Equation of a Line in Point-Slope Form
A line has a slope of 35 , and the point (2, 6) is on the line. Write the equation of the line in point-slope form.
Start with the formula y − y0 = m(x − x0 ).
Plug in
3
5
for m, 2 for x0 and 6 for y0 .
The equation in point-slope form is y − 6 = 35 (x − 2).
Notice that the equation in point-slope form is not solved for y. If we did solve it for y, we’d have it in y−intercept
form. To do that, we would just need to distribute the 53 and add 6 to both sides. That means that the equation of this
line in slope-intercept form is y = 53 x − 65 + 6, or simply y = 35 x + 24
5.
Write an Equation Given Two Points
Point-slope form also comes in useful when we need to find an equation given just two points on a line.
For example, suppose we are told that the line passes through the points (-2, 3) and (5, 2). To find the equation of
the line, we can start by finding the slope.
−y1
2−3
, we plug in the x− and y−values of the two points to get m = 5−(−2)
= −1
Starting with the slope formula, m = xy22 −x
7 .
1
We can plug that value of m into the point-slope formula to get y − y0 = − 17 (x − x0 ).
Now we just need to pick one of the two points to plug into the formula. Let’s use (5, 2); that gives us y − 2 =
− 17 (x − 5).
What if we’d picked the other point instead? Then we’d have ended up with the equation y − 3 = − 71 (x + 2), which
doesn’t look the same. That’s because there’s more than one way to write an equation for a given line in point-slope
form. But let’s see what happens if we solve each of those equations for y.
Starting with y − 2 = − 71 (x − 5), we distribute the − 71 and add 2 to both sides. That gives us y = − 17 x + 75 + 2, or
y = − 17 x + 19
7.
On the other hand, if we start with y − 3 = − 17 (x + 2), we need to distribute the − 17 and add 3 to both sides. That
gives us y = − 71 x − 27 + 3, which also simplifies to y = − 17 x + 19
7.
So whichever point we choose to get an equation in point-slope form, the equation is still mathematically the same,
and we can see this when we convert it to y−intercept form.
Writing an Equation in y−intercept form
A line contains the points (3, 2) and (-2, 4). Write an equation for the line in point-slope form; then write an equation
in y−intercept form.
Find the slope of the line: m =
y2 −y1
x2 −x1
=
4−2
−2−3
= − 25
Plug in the value of the slope: y − y0 = − 52 (x − x0 ).
Plug point (3, 2) into the equation: y − 2 = − 52 (x − 3).
The equation in point-slope form is y − 2 = − 52 (x − 3).
To convert to y−intercept form, simply solve for y:
29
1.5. Determining the Equation of a Line
www.ck12.org
2
2
6
y − 2 = − (x − 3) → y − 2 = − x +
5
5
5
2
6
→ y = − x+ +2
5
5
1
2
→ y = − x+3 .
5
5
The equation in y−intercept form is y = − 25 x + 3 15 .
Graph an Equation in Point-Slope Form
Another useful thing about point-slope form is that you can use it to graph an equation without having to convert
it to slope-intercept form. From the equation y − y0 = m(x − x0 ), you can just read off the slope m and the point
(x0 , y0 ). To draw the graph, all you have to do is plot the point, and then use the slope to figure out how many units
up and over you should move to find another point on the line.
Graphing an Equation of a Line
Make a graph of the line given by the equation y + 2 = 32 (x − 2).
To read off the right values, we need to rewrite the equation slightly: y − (−2) = 23 (x − 2). Now we see that point
(2, -2) is on the line and that the slope is 23 .
First plot point (2, -2) on the graph:
A slope of
point:
30
2
3
tells you that from that point you should move 2 units up and 3 units to the right and draw another
www.ck12.org
Chapter 1. Unit 1
Now draw a line through the two points and extend it in both directions:
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133179
Examples
A line contains the points (1, -2) and (0, 0). Write an equation for the line in point-slope form; then write an equation
in y−intercept form.
Find the slope of the line: m =
y2 −y1
x2 −x1
=
−2−0
1−0
=
−2
1
= −2
31
1.5. Determining the Equation of a Line
www.ck12.org
Plug in the value of the slope: y − y0 = −2(x − x0 ).
Plug point (1, -2) into the equation: y − (−2) = −2(x − 1).
The equation in point-slope form is y + 2 = −2(x − 1).
To convert to y−intercept form, simply solve for y:
y + 2 = −2(x − 1) → y + 2 = −2x + 2
→ y = −2x + 2 − 2
→ y = −2x.
The equation in y−intercept form is y = −2x.
Review
Find the equation of each line in slope-intercept form.
1.
2.
3.
4.
5.
The line has a slope of 7 and a y−intercept of -2.
The line has a slope of -5 and a y−intercept of 6.
The line has a slope of − 14 and contains the point (4, -1).
The line contains points (3, 5) and (-3, 0).
The line contains points (10, 15) and (12, 20).
Write the equation of each line in slope-intercept form.
6.
32
www.ck12.org
Chapter 1. Unit 1
7.
Find the equation of each linear function in slope-intercept form.
8.
9.
10.
11.
12.
13.
m = 5, f (0) = −3
m = −7, f (2) = −1
m = 13 , f (−1) = 32
m =4.2, f (−3) = 7.1
f 14 = 43 , f (0) = 54
f (1.5) = −3, f (−1) = 2
Write the equation of each line in point-slope form.
14.
15.
16.
17.
18.
19.
20.
21.
1
and goes through the point (10, 2).
The line has slope − 10
The line has slope -75 and goes through the point (0, 125).
The line has slope 10 and goes through the point (8, -2).
The line goes through the points (-2, 3) and (-1, -2).
The line contains the points (10, 12) and (5, 25).
The line goes through the points (2, 3) and (0, 3).
The line has a slope of 35 and a y−intercept of -3.
The line has a slope of -6 and a y−intercept of 0.5.
Write the equation of each linear function in point-slope form.
22.
23.
24.
25.
26.
27.
m = − 15 and f (0) = 7
m = −12 and f (−2) = 5
f (−7) = 5 and f (3) = −4
f (6) = 0 and f (0) = 6
m = 3 and f (2) = −9
m = − 59 and f (0) = 32
Review (Answers)
To view the Review answers, open this PDF file and look for section 5.1.
33
1.6. Transformations of functions
www.ck12.org
1.6 Transformations of functions
Learning objectives
•
•
•
•
Graph functions by shifting basic functions.
Graph functions by stretching or compressing basic functions.
Graph functions that are reflections of basic functions.
Combine transformations in order to graph more complicated functions.
Introduction
In the previous lesson we examined the key characteristics of different families of functions. In this lesson we will
focus on the relationships among functions within a given family. In particular, we will analyze how a function
is related to the parent function in the family (e.g., y = x2 , y = x3 ) so that we can efficiently identify the key
characteristics of a particular function and sketch its graph. We will consider several kinds of relationships between
functions and their parents: shifts, stretches or compressions, and reflections. We will begin with shifts.
Vertical shifts
Consider the graphs of three functions shown below: y = x2 , y = x2 - 3, and y = x2 + 4
At first glance, it may seem that the graphs have different widths. For example, it might look like y = x2 + 4,
the uppermost of the three parabolas, is thinner than the other two parabolas. However, this is not the case. The
34
www.ck12.org
Chapter 1. Unit 1
parabolas are congruent. That is, if we shifted the graph of y = x2 up four units, we would have the exact same
graph as y = x2 + 4. Similarly, if we shifted y = x2 down three units, we would have the graph of y = x2 - 3.
Numerically, we can determine that for any value of x, the value of y = x2 + 4 is 4 more than the value of y = x2 , and
the value of y = x2 - 3 is 3 less than the value of y = x2 . The table below shows several function values:
TABLE 1.3:
x
1
0
1
2
5
10
y = x2
5
0
1
4
25
100
y = x2 + 4
y = x2 - 3
4
5
8
29
104
1
22
97
We can formally express this relationship using function notation: for any function f (x), the function g(x) = f (x) + c
has a graph that is the same as f (x), shifted c units vertically. If c is positive, the graph is shifted up. If c is negative,
the graph is shifted down. We can then use this relationship to graph functions by hand.
Example 1. Use the graph of y = x2 to graph the function y = x2 - 5.
Solution:
The graph of y = x2 is a parabola with vertex at (0, 0). The graph of y = x2 - 5 is therefore a parabola with vertex (0,
-5). To quickly sketch y = x2 - 5, you can sketch several points on y = x2 , and then shift them down 5 units.
In sum, when we add or subtract a constant from the equation of a function, the graph of the new function is a
vertical shift from the original function. We can use similar reasoning to graph functions that are horizontal shifts of
parent graphs.
35
1.6. Transformations of functions
www.ck12.org
Horizontal shifts
Consider the functions f (x) = |x| and g(x) = |x - 3|. From the examples of vertical shifts above, you might think that
the graph of g(x) is the graph of f(x), shifted 3 units to the left. However, this is not the case. The graph of g(x) is
the graph of f (x), shifted 3 units to the right. The direction of the shift makes sense if we look at specific function
values.
TABLE 1.4:
x
0
1
2
3
4
5
6
g(x) = abs(x - 3)
3
2
1
0
1
2
3
From the table we can see that the vertex of the graph is the point (3, 0). The function values on either side of x = 3
are symmetric, and greater than 0.
We can formalize horizontal shifts in the same way we formalized vertical shifts: given a function f (x), and a constant
a > 0, the function g(x) = f (x - a) represents a horizontal shift a units to the right from f (x). The function h(x) = f (x
+ a) represents a horizontal shift a units to the left.
36
www.ck12.org
Chapter 1. Unit 1
Notice that in the case of a horizontal shift, the constant is part of the “input” of the function. For example, y =
(x + 7)2 represents a horizontal shift of the original function y=x2 , while y = x2 + 7 represents a vertical shift. It is
important to keep this distinction in mind as you interpret equations.
Example 2: What is the relationship between f (x) = x2 and g(x) = (x - 2)2 - 6?
Solution: The graph of g(x) is the graph of f (x), shifted 2 units to the right, and down 6 units.
In sum, adding or subtracting constants within the equation of a function creates a vertical or horizontal shift in the
graph. When the constant is “inside” the parentheses, the shift will be horizontal. Now we will consider the results
of multiplying a function or the input of a function (x) by a coefficient.
Stretching and compressing graphs
If we multiply a function by a coefficient, the graph of the function will be stretched or compressed. Consider the
graph of three parabolas below.
The widest parabola is the function y = (1/5) x2 . The middle parabola is the function y = x2 . The thinnest parabola is
the function y = 10x2 . From these graphs, we can see that multiplying by different coefficients has different effects
on the graph of y = x2 . The graph of y = 10x2 represents a vertical stretch of the graph of y = x2 . Every function
value of y = 10x2 is 10 times the equivalent function value of y = x2 . For example, y = x2 contains the point (2, 4),
while y = 10x2 contains the point (2, 40).
The coefficient of 1/5 has an opposite effect. We describe this as a vertical compression. Every function value of y
= (1/5) x2 is 1/5 the value of y = x2 . For example, y = x2 contains the point (10, 100), while y = (1/2) x2 contains the
point (10, 20).
37
1.6. Transformations of functions
www.ck12.org
We can formalize vertical stretching and compressing as follows:
• A function g(x) = cf (x) represents a vertical stretch if c > 1.
• A function h(x) = cf (x) represents a vertical compression if 0 < c < 1.
Notice that the graph of y = (1/5) x2 is wider than the other two parabolas. While 1/5 has the effect of compressing the
graph of y = x2 vertically, we can also describe the graph as having a horizontal stretch. Formally, we characterize
a transformation as a horizontal stretch or compression if the function is of the form g(x) = f (cx). That is, instead
of multiplying the function by a coefficient, the variable x is multiplied by c. However, if we analyze a quadratic
equation, we can see why a vertical compression looks like a horizontal stretch. (The parabola gets shorter/wider, or
taller/thinner.)
Consider for example the function y = (3x)2 . We have multiplied x by 3. This should affect the graph horizontally.
However, if we simplify the equation, we get y = 9x2 . Therefore the graph if this parabola will be taller/thinner than y
= x2 . Multiplying x by a number greater than 1 creates a horizontal compression, which looks like a vertical stretch.
Now consider the function y = ((1/2)x)2 . If we simplify this equation, we get y = (1/4) x2 . Therefore multiplying
x by a number between 0 and 1 creates a horizontal stretch, which looks like a vertical compression. That is, the
parabola will be shorter/wider.
We can see similar behavior in other functions. For example, consider the functions , , and , shown in the graph
below. If we simplify the second equation, we get . Therefore the second and third equations are actually the same
function, and we can describe the function as a vertical stretch or as a horizontal compression.
Given a function f (x), we can formalize compressing and stretching the graph of f (x) as follows:
38
www.ck12.org
Chapter 1. Unit 1
• A function g(x) represents a vertical stretch of f (x) if g(x) = cf (x) and c > 1.
• A function g(x) represents a vertical compression of f (x) if g(x) = cf (x) and 0 < c < 1.
• A function h(x) represents a horizontal compression of f (x) if h(x) = f (cx) and c > 1.
• A function h(x) represents a horizontal stretch of f (x) if h(x) = f (cx) 0 < c < 1.
Notice that a vertical compression or a horizontal stretch occurs when the coefficient is a number between 0 and 1.
Now we will consider the effects of a coefficient less than 0.
Reflecting graphs over the
Consider the graphs of the functions y = x2 and y = -x2 , shown below.
The graph of y = -x2 represents a reflection of y = x2 , over the x-axis. That is, every function value of y = -x2 is the
negative of a function value of y = x2 . In general, g(x) = -f (x) has a graph that is the graph of f (x), reflected over the
x-axis.
Example 3: Sketch a graph of y = x3 and y = -x3 on the same axes.
Solution:
39
1.6. Transformations of functions
www.ck12.org
At first the two functions might look like two parabolas. If you graph by hand, or if you set your calculator to
sequential mode (and not simultaneous), you can see that the graph of y = -x3 is in fact a reflection of y = x3 over the
x-axis.
However, if you look at the graph, you can see that it is a reflection over the y-axis as well. This is the case because
in order to obtain a reflection over the y-axis, we negate x. That is, h(x) = f (-x) is a reflection of f (x) over the y-axis.
For the function y = x3 , h(x) = (-x)3 = (-x) (-x) (-x) = -x3 . This is the same function as the one we have already
graphed.
It is important to note that this is a special case. The graph of y = x2 is also a special case. If we want to reflect y =
x2 over the y-axis, we will just get the same graph! This can be explained algebraically: y = (-x)2 = (-x) (-x) = x2 .
Now let’s consider a different function.
Example 4: Graph the functions and .
Solution:
40
www.ck12.org
Chapter 1. Unit 1
The equation might look confusing because of the -x under the square root. It is important to keep in mind that x means the opposite of x. Therefore the domain of this function is restricted to values ≤ 0. For example, if x = 4, . It is this domain, which includes all real numbers not in the domain of plus zero, that gives us a graph that is a
reflection over the y-axis.
In sum, a graph represents a reflection over the x-axis if the function has been negated (i.e. the y has been negated
if we think of y = f (x)). The graph represents a reflection over the y-axis if the variable x has been negated. Now
that we have considered shifts, stretches/compressions, and reflections, we will look at graphs that combine these
transformations.
Combining transformations
Consider the equation y = 2(x - 3)2 + 1. We can compare the graph of this function to the graph of the parent y = x2 :
the graph represents a vertical stretch by a factor of 2, a horizontal shift 3 units to the right, and a vertical shift of 1
unit.
We can use this relationship to graph the function y = 2(x - 3)2 + 1. You can start by sketching y = x2 or y = 2x2 .
Then you can shift the graph 3 units to the right, and up 1 unit.
41
1.6. Transformations of functions
www.ck12.org
It is slightly more complicated to graph functions that represent reflections of the parent graphs. Next we will
examine reflections combined with other transformations.
Example 5. Graph each function using your knowledge of the function y = |x| and your knowledge of transformations.
a. f (x) = -|x| + 3
b. g(x) = |-x + 3|
Solution:
a. f (x) = -|x| + 3
The parent graph of this function is the graph of y = |x|, reflected over the x-axis, and shifted up 3 units. The question
is: which transformation do you perform first?
We can answer this question if we consider a few key function values. The table below shows several function values
for f (x) = -|x| + 3:
TABLE 1.5:
x
-3
-2
-1
0
1
3
f (x) = - abs(x) + 3
-abs(-3) + 3 = -(+3) + 3 = -3 + 3 = 0
-abs(-2) + 3 = -(+2) + 3 = -2 + 3 = 1
2
3
2
0
From the function values in the table we can see that the function increases until a vertex at (0, 3), and then it
decreases again. This tells us that we can obtain the graph if we first reflect y = |x| over the x-axis (turn the “v” upside
down), and then shift the graph up 3 units.
42
www.ck12.org
Chapter 1. Unit 1
We can also justify this ordering of the transformations of we think about the order of operations. To find any
function value we take an x value, find its absolute value, find the negative of that number, and then add 3. This is
the same as the order of the transformation: reflection comes before shifting up.
b. g(x) = |-x + 3|
This function represents a horizontal shift of y = |x|, and a reflection over the x-axis. Before graphing, consider a few
function values:
TABLE 1.6:
x
abs(-(-3) + 3) = abs(3 + 3) = abs(6) = 6
abs(-(-2) + 3 ) = (2 + 3) = abs(5) = 5
4
0
1
2
3
4
5
6
g(x) =
3
2
1
0
1
2
3
From the values in the table, we can see that the vertex of the graph is at (3, 0). The graph is shown below.
43
1.6. Transformations of functions
www.ck12.org
The graph looks the same as the graph of y = |x - 3|. This is the case because y = |-x + 3| = |-(x - 3)|, and because
|- a| = |a| for all values of a, then |-(x - 3)| = |x - 3|. So the original function is equal to |x - 3|. We can still think of
this graph as a reflection: if we reflect y = |x| over the x-axis, the graph remains the same, as it is symmetric over
the x-axis. Then we shift the graph 3 units to the right. What is important to note here is that in order to “read” the
equation as a horizontal shift, the entire expression inside the function (in this case, inside the absolute value) must
be negated.
Lesson Summary
In this lesson we have explored transformations of functions. We have related different functions to their parent
graphs, in terms of vertical and horizontal shifts, vertical and horizontal stretches and compressions, and reflections
over the y-axis and the x-axis. By analyzing the functions, we can relate the graphs of functions to the graphs of their
parent functions, and we can use these relationships to sketch graphs. The issue of “inside” and “outside”, raised
throughout the lesson, and seen specifically in example 5b, will be examined more closely in the next lesson.
Points to Consider
1. How can you analyze an equation and determine if the graph represents a vertical or horizontal shift?
2. How can you analyze an equation and determine if the graph represents a stretch or compression?
3. How is it that different equations might yield the same graphs?
44
www.ck12.org
Chapter 1. Unit 1
Review Questions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
How is the graph of f (x) = x3 + 5 related to the graph of g(x) = x3 ?
How is the graph of f (x) = (x - 3)2 related to the graph of f (x) = (x + 3)2 ? Sketch these functions together.
Write an equation that represents a vertical stretch of the graph of by a factor of 5.
Describe the function f (x) = (5x)2 as both a stretch and a compression of y = x2 .
Graph the function and on the same grid. What is the equation of ?
Describe the relationship between the graphs of f (x) = 4(x + 8)3 - 3 and g(x) = x3 .
Graph the functions and on the same graph. State the asymptotes of both functions.
Graph the functions f (x) = 3x + 1 and f (-x) on the same grid. What is the equation of f (-x)?
Explain how a function can be its own reflection over an axis.
Give an example of two equations that can be represented by the same graph.
45
1.7. Transformations of Quadratic Functions
www.ck12.org
1.7 Transformations of Quadratic Functions
Here you’ll learn how to transform the basic quadratic functions (y = x2 and y = −x2 ) to make new quadratic
functions.
Look at the parabola below. How is this parabola different from y = −x2 ? What do you think the equation of this
parabola is?
Watch This
Khan Academy Graphing a Quadratic Function
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/58468
Guidance
This is the graph of y = x2 :
46
www.ck12.org
Chapter 1. Unit 1
This is the graph of y = x2 that has undergone transformations:
The vertex of the red parabola is (3, 1). The sides of the parabola open upward but they appear steeper and longer
than those on the blue parabola.
As shown above, you can apply changes to the graph of y = x2 to create a new parabola (still a ’U’ shape) that no
longer has its vertex at (0, 0) and no longer has y-values of 1, 4 and 9. These changes are known as transformations.
The vertex of (0, 0) will change if the parabola undergoes either a horizontal translation and/or a vertical translation.
These transformations cause the parabola to slide left or right and up or down.
If the parabola undergoes a vertical stretch, the y-values of 1, 4 and 9 can increase if the stretch is a whole number.
This will produce a parabola that will appear to be narrower than the original base graph. If the vertical stretch is a
fraction, the values of 1, 4 and 9 will decrease. This will produce a parabola that will appear to be wider than the
original base graph.
Finally, a parabola can undergo a vertical reflection that will cause it to open downwards as opposed to upwards. For
example, y = −x2 is a vertical reflection of y = x2 .
Example A
Look at the two parabolas below. Describe the transformation from the blue parabola to the red parabola. What is
the coordinate of the vertex of the red parabola?
47
1.7. Transformations of Quadratic Functions
www.ck12.org
Solution:
The blue parabola is the graph of y = x2 . Its vertex is (0, 0). The red graph is the graph of y = x2 that has been moved
four units to the right. When the graph undergoes a slide of four units to the right, it has undergone a horizontal
translation of +4. The vertex of the red graph is (4, 0). A horizontal translation changes the x−coordinate of the
vertex of the graph of y = x2 .
Example B
Look at the two parabolas below. Describe the transformation from the blue parabola to the red parabola. What is
the coordinate of the vertex of the red parabola?
48
www.ck12.org
Chapter 1. Unit 1
Solution:
The blue parabola is the graph of y = x2 . Its vertex is (0, 0). The red graph is the graph of y = x2 that has been moved
four units to the right and three units upward. When the graph undergoes a slide of four units to the right, it has
undergone a horizontal translation of +4. When the graph undergoes a slide of three units upward, it has undergone
a vertical translation of +3.The vertex of the red graph is (4, 3). A horizontal translation changes the x−coordinate
of the vertex of the graph of y = x2 while a vertical translation changes the y−coordinate of the vertex.
Example C
Look at the parabola below. How is this parabola different from y = x2 ? What do you think the equation of this
parabola is?
Solution:
This is the graph of y = 12 x2 . The points are plotted from the vertex as right and left one and up one-half, right and
left 2 and up two, right and left three and up four and one-half. The original y-values of 1, 4 and 9 have been divided
by two or multiplied by one-half. When the y-values are multiplied, the y-values either increase or decrease. This
transformation is known as a vertical stretch.
49
1.7. Transformations of Quadratic Functions
www.ck12.org
Concept Problem Revisited
This is the graph of y = − 21 x2 . The points are plotted from the vertex as right and left one and down one-half, right
and left 2 and down two, right and left three and down four and one-half. The original y-values of 1, 4 and 9 have
been multiplied by one-half and then were made negative because the graph was opening downward. When the
y-values become negative, the direction of the opening is changed from upward to downward. This transformation
is known as a vertical reflection. The graph is reflected across the x-axis.
Vocabulary
Horizontal translation
The horizontal translation is the change in the base graph y = x2 that shifts the graph right or left. It changes
the x−coordinate of the vertex.
Transformation
A transformation is any change in the base graph y = x2 . The transformations that apply to the parabola are a
horizontal translation (HT ), a vertical translation (V T ), a vertical stretch (V S) and a vertical reflection (V R).
Vertical Reflection
The vertical reflection is the reflection of the image graph in the x-axis. The graph opens downward and the
y-values are negative values.
Vertical Stretch
The vertical stretch is the change made to the base function y = x2 by stretching (or compressing) the graph
vertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the original
base graph of y = x2 .
Vertical Translation
The vertical translation is the change in the base graph y = x2 that shifts the graph up or down. It changes the
y−coordinate of the vertex.
50
www.ck12.org
Chapter 1. Unit 1
Guided Practice
1. Use the following tables of values and identify the transformations of the base graph y = x2 .
X
−3
−2
−1
Y
9
4
1
X
−4
−3
Y
15
5
0
1
2
3
0
1
4
9
−2
−1
0
1
2
−1
−3
−1
5
15
2. Identify the transformations of the base graph y = x2 .
3. Draw the image graph of y = x2 that has undergone a vertical reflection, a vertical stretch by a factor of 12 , a
vertical translation up 2 units, and a horizontal translation left 3 units.
Answers:
1. To identify the transformations from the tables of values, determine how the table of values for y = x2 compare to
the table of values for the new image graph.
• The x-values have moved one place to the left. This means that the graph has undergone a horizontal translation
of -1.
51
1.7. Transformations of Quadratic Functions
www.ck12.org
• The y−coordinate of the vertex is -3. This means that the graph has undergone a vertical translation of -3. The
vertex is easy to pick out from the tables since it is the point around which the corresponding points appear.
X
−4
−3
−2
−1
0
1
2
Y
15
5
−1
−3
−1
5
15
• The points from the vertex are plotted left and right one and up two, left and right two and up eight. This
means that the base graph has undergone a vertical stretch of 2.
• The y-values move upward so the parabola will open upward. Therefore the image is not a vertical reflection.
2. The vertex is (1, 6). The base graph has undergone a horizontal translation of +1 and a vertical translation of
+6. The parabola opens downward, so the graph is a vertical reflection. The points have been plotted such that
the y-values of 1 and 4 are now 2 and 8. It is not unusual for a parabola to be plotted with five points rather than
seven. The reason for this is the vertical stretch often multiplies the y-values such that they are difficult to graph on
a Cartesian grid. If all the points are to be plotted, a different scale must be used for the y-axis.
3. The vertex given by the horizontal and vertical translations and is (-3, 2). The y-values of 1, 4 and 9 must be
multiplied by 12 to create values of 12 , 2 and 4 12 . The graph is a vertical reflection which means the graph opens
downward and the y-values become negative.
52
www.ck12.org
Chapter 1. Unit 1
1.8 Vertex Form of a Quadratic Function
Learning Objectives
Here you will learn to write the equation for a parabola that has undergone transformations.
Given the equation y = 3(x + 4)2 + 2, list the transformations of y = x2 .
Watch This
James Sousa: Find the Equation of a Quadratic Function from a Graph
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/58471
Guidance
The equation for a basic parabola with a vertex at (0, 0) is y = x2 . You can apply transformations to the graph of
y = x2 to create a new graph with a corresponding new equation. This new equation can be written in vertex form.
The vertex form of a quadratic function is y = a(x − h)2 + k where:
• |a| is the vertical stretch factor. If a is negative, there is a vertical reflection and the parabola will open
downwards.
• k is the vertical translation.
• h is the horizontal translation.
Given the equation of a parabola in vertex form, you should be able to sketch its graph by performing transformations
on the basic parabola. This process is shown in the examples.
Example A
Given the following function in vertex form, identify the transformations of y = x2 .
y = − 21 (x − 2)2 − 1
Solution:
• a - Is a negative? YES. The parabola will open downwards.
• a - Is there a number in front of the squared portion of the equation? YES. The vertical stretch factor is the
absolute value of this number. Therefore, the vertical stretch of this function is 12 .
53
1.8. Vertex Form of a Quadratic Function
www.ck12.org
• k - Is there a number after the squared portion of the equation? YES. The value of this number is the vertical
translation. The vertical translation is -1.
• h - Is there a number after the variable ‘x’? YES. The value of this number is the opposite of the sign that
appears in the equation. The horizontal translation is +2.
Example B
Given the following transformations, determine the equation of the image of y = x2 in vertex form.
• Vertical stretch by a factor of 3
• Vertical translation up 5 units
• Horizontal translation left 4 units
Solution:
• a - The image is not reflected in the x-axis. A negative sign is not required.
• a - The vertical stretch is 3, so a = 3.
• k - The vertical translation is 5 units up, so k = 5.
• h - The horizontal translation is 4 units left so h = −4.
The equation of the image of y = x2 is y = 3(x + 4)2 + 5.
Example C
Using y = x2 as the base function, identify the transformations that have occurred to produce the following image
graph. Use these transformations to write the equation in vertex form.
54
www.ck12.org
Chapter 1. Unit 1
Solution:
a - The parabola does not open downward so a will be positive.
a - The y-values of 1 and 4 are now up 3 and up 12. a = 3.
k - The y−coordinate of the vertex is -5 so k = −5.
h - The x−coordinate of the vertex is +3 so h = 3.
The equation is y = 3(x − 3)2 + 5.
Example D
When the equation of the basic quadratic function is written in vertex form, the function can also be expressed in
mapping notation form. This form describes how to obtain the image of a given graph by using the changes in the
ordered pairs.
The standard base table of values for the base quadratic function y = x2 is given by:
X
−3
Y
9
−2
4
−1
0
1
1
0
2
1
3
4
9
When these ordered pairs are plotted, we get the base parabola. The mapping rule used to generate the image of a
quadratic function is (x, y) → (x0 , y0 ) where (x0 , y0 ) are the coordinates of the image graph. The resulting mapping
rule from the equation y = a(x − h)2 + k is (x, y) → (x + h, ay + k). A mapping rule details the transformations that
were applied to the coordinates of the base function y = x2 .
Given the following quadratic equation, y = 2(x + 3)2 + 5 write the mapping rule and create a table of values for the
mapping rule.
55
1.8. Vertex Form of a Quadratic Function
www.ck12.org
Solution:
The mapping rule for this function will tell exactly what changes were applied to the coordinates of the base quadratic
function.
y = 2(x + 3)2 + 5 :
(x, y) → (x − 3, 2y + 5)
These new coordinates of the image graph can be plotted to generate the graph.
Concept Problem Revisited
Given the equation y = 3(x + 4)2 + 2, list the transformations of y = x2 .
a = 3 so the vertical stretch is 3. k = 2 so the vertical translation is up 2. h = −4 so the horizontal translation is left
4.
Vocabulary
Horizontal translation
The horizontal translation is the change in the base graph y = x2 that shifts the graph right or left. It changes
the x−coordinate of the vertex.
Mapping Rule
The mapping rule is another form used to express a quadratic function. The mapping rule defines the
transformations that have occurred to the base quadratic function y = x2 . The mapping rule is (x, y) → (x0 , y0 )
where (x0 , y0 ) are the coordinates of the image graph.
Transformation
A transformation is any change in the base graph y = x2 . The transformations that apply to the parabola are
a horizontal translation, a vertical translation, a vertical stretch and a vertical reflection.
Vertex form of y = x2
The vertex form of y = x2 is the form of the quadratic base function y = x2 that shows the transformations of
the image graph. The vertex form of the equation is y = a(x − h)2 + k.
56
www.ck12.org
Chapter 1. Unit 1
Vertical Reflection
The vertical reflection is the reflection of the image graph in the x-axis. The graph opens downward and the
y-values are negative values.
Vertical Stretch
The vertical stretch is the change made to the base function y = x2 by stretching (or compressing) the graph
vertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the original
base graph of y = x2 .
Vertical Translation
The vertical translation is the change in the base graph y = x2 that shifts the graph up or down. It changes the
y−coordinate of the vertex.
Guided Practice
1. Identify the transformations of y = x2 for the quadratic function −2(y + 3) = (x − 4)2
2. List the transformations of y = x2 and graph the function = −(x + 5)2 + 4
3. Graph the function y = 2(x − 2)2 + 3 using the mapping rule method.
Answers:
1. a - a is negative so the parabola opens downwards.
a - The vertical stretch of this function is 12 .
k - The vertical translation is -3.
h - The horizontal translation is +4.
2.
57
1.8. Vertex Form of a Quadratic Function
www.ck12.org
a → negative
a→1
k → +4
h → −5
3. Mapping Rule (x, y) → (x + 2, 2y + 3)
Make a table of values:
TABLE 1.7:
x → x+2
−3
-2
-1
0
1
2
3
Draw the Graph
58
-1
0
1
2
3
4
5
y → 2y + 3
9
4
1
0
1
4
9
21
11
5
3
5
11
21
www.ck12.org
Chapter 1. Unit 1
59
1.9. Polynomial Functions
www.ck12.org
1.9 Polynomial Functions
Learning Objectives
• Understand and apply methods to find the zeros of a factored polynomial
• Use the zeros of polynomial functions to sketch a graph of the function
Standard Form of Polynomial Function
You have already studied many different kinds of functions, for example linear functions, constant functions, and
quadratic functions. All three these functions belong to a larger group of functions called the polynomial functions.
Polynomial Functions
If P(x) is a polynomial function, then it is given by
P(x) = an xn + an−1 xn−1 + an−2 xn−2 + · · · + a1 x + a0
where the coefficients an , an−1 , · · · , a1 , a0 are real numbers and the exponents are positive integers.
We call the first nonzero coefficient an the leading coefficient. The term an xn is called the leading term. The degree
of the polynomial is n. For example, the quadratic equation f (x) = −2x2 + 3x − 5 has a leading coefficient of -2, a
leading term of −2x2 and of degree n = 2. On the other hand, the polynomial f (x) = 1 is a polynomial of with a
leading coefficient of 1, and the leading term is 1x0 = 1, so the degree is n = 0.
A very interesting property of polynomials is that they are all continuous, that is, they have no holes, breaks.
Informally, we say you can draw the graph of a continuous function without lifting your pencil from the paper.
In calculus you will study function continuity in more depth and you will use the continuity of polynomials to
simplify the analysis of other, more complicated functions.
In addition to their continuity, the domain of all polynomial functions is the set of all real numbers. That is, the
domain is x ∈ (−∞, ∞).
Sometimes it is helpful to consider a counterexample. There are many functions that we use in mathematics which
are not polynomials. For example the floor function, y = int(x), is a function that gives the greatest integer less than
or equal to x, so int(5.67) = 5. This function is NOT a polynomial and it is NOT continuous.
Power Function (even, odd)
The most simple polynomial is called a power function. A power function is a polynomial of the form f (x) = axn
where a is a real number and n is an integer with n ≥ 1.
If n is even, then the power function is also called “even,” and if n is odd, then the power function is “odd.” The
graphs of the first five power functions are shown below.
60
www.ck12.org
Chapter 1. Unit 1
Notice that each power function has only one x− and y−intercept, the origin (0, 0).
The end behavior of a function describes the y−values as x gets very large (x → ∞ in symbols) or as x gets very
small (x → −∞).
• For even powers n, the power function f (x) = axn is U-shaped (like a parabola) and as x → ∞, f (x) → ∞.
Likewise as x → −∞, f (x) → ∞.
• For odd powers n, the power function goes from the third quadrant to the first quadrant (like the line y = x).
As x → ∞, f (x) → ∞, and as x → −∞, f (x) → −∞.
As with quadratics and polynomials, the leading coefficient a changes the vertical “stretching” of power functions.
Graph Polynomial Functions Using Transformations
Just like quadratics, polynomial functions can be graphed using transformations of a known graph. The basic
transformations are vertical and horizontal shifts and reflections about the x− and y−axis.
Given a polynomial p(x) and constant real numbers c and a
•
•
•
•
•
•
p(x) + c is a vertical shift of the graph of p(x) by c units up (so the function shifts down if c < 0).
p(x − c) is a horizontal shift of the graph of p(x) by c units to the right. (So the function shifts left if c < 0).
−p(x) is a reflection of the graph of p(x) about the x−axis.
p(−x) is a reflection of the graph of p(x) about the y−axis.
ap(x) is a vertical stretch by a multiple of a.
p(ax) is a horizontal compression by a multiple of a.
Example 1
The graph of f (x) is shown below. Use the graph of f (x) to graph each of the following: 1) f (x) + 3, 2) f (x + 4),
and 3) f (−x) + 3
61
1.9. Polynomial Functions
Solution
1) This is a vertical shift of f (x) up by 3 units.
2) This is a horizontal shift of f (x) to the left by 4 units.
62
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
3) This is a reflection of f (x) about the y−axis and a vertical shift up by 3 units.
Graph Polynomial Functions Given Zeros (
Recall that the zeros or the roots of a polynomial are the x−intercepts or the solution set of the polynomial function.
For example, the polynomial
h(x) = x3 + 2x2 − 5x − 6
63
1.9. Polynomial Functions
www.ck12.org
can be factored into
y = h(x) = x3 + 2x2 − 5x − 6 = (x + 1)(x − 2)(x + 3)
To find the zeros, we set
h(x) = y = 0
and solve for x.
(x + 1)(x − 2)(x + 3) = 0
This gives
x+1 = 0
x−2 = 0
x+3 = 0
or
x = −1
x=2
x = −3
So we say that the solution set is {−3, −1, 2}. They are the zeros of the function h(x). The zeros of h(x) are the
x−intercepts of the graph y = h(x) below.
64
www.ck12.org
Chapter 1. Unit 1
The previous example illustrates a strategy for graphing a polynomial if you know the zeros of the function. If you
know the zeros of a polynomial you can use “test values” between the zeros to see whether the function is above or
below the x−aixs in the interval between the zeros. While this cannot tell you the height (y−values) of the function
between the zeros, you can use the zeros to sketch an approximation of the graph. The following examples illustrate
first finding the zeros, and then a method for graphing a polynomial once you know the zeros.
Example 2
Find the zeros of g(x) = −(x − 2)(x − 2)(x + 1)(x + 5)(x + 5)(x + 5).
Solution
The polynomial can be written as
g(x) = −(x − 2)2 (x + 1)(x + 5)3
To solve the equation, we simply set it equal to zero
−(x − 2)2 (x + 1)(x + 5)3 = 0
this gives
x−2 = 0
x+1 = 0
x+5 = 0
65
1.9. Polynomial Functions
www.ck12.org
or
x=2
x = −1
x = −5
Notice the occurrence of the zeros in the function. The factor (x − 2) occurred twice (because it was squared), the
factor (x + 1) occurred once and the factor (x + 5) occurred three times. We say that the zero we obtain from the
factor (x − 2) has a multiplicity k = 2 and the factor (x + 5) has a multiplicity k = 3.
To graph g(x), use the zeros to create a table of intervals and see whether the function is above or below the x−axis
in each interval:
TABLE 1.8:
Interval
Test value x
g(x)
Sign of g(x)
(−∞, −5)
x = −5
(-5, -1)
x = −1
(-1, 2)
x=2
(2, ∞)
-6
-5
-2
-1
0
2
3
320
0
144
0
-100
+
NA
+
NA
NA
-
-256
Finally, use this information and the test points to sketch a graph of g(x).
Example 3
Find the zeros and sketch a graph of the polynomial
f (x) = x4 − x2 − 56
66
Location of graph
relative to x−axis
Above
Above
Below
Below
www.ck12.org
Chapter 1. Unit 1
Solution
This is a factorable equation,
f (x) = x4 − x2 − 56
= (x2 − 8)(x2 + 7)
Setting f (x) = 0,
(x2 − 8)(x2 + 7) = 0
the first term gives
x2 − 8 = 0
x2 = 8
√
x=± 8
√
= ±2 2
and the second term gives
x2 + 7 = 0
x2 = −7
√
x = ± −7
√
= ±i 7
√
√
So the solutions are ±2 2 and ±i 7, a total of four zeros of f (x).
Keep in mind that only the real √
zeros of a function correspond to the x−intercept of its√graph. For√our case in this
example, only the two zeros ±2 2 correspond to actual x−intercepts (Figure 9) but +i 7 and −i 7 do not, since
they are complex. These are given more attention later in the book ((Add cross-referencing link?))
67
1.9. Polynomial Functions
www.ck12.org
Summarize Analysis and Graphing of Polynomial Functions
It is helpful to consider the following facts when graphing any polynomial function.
Two Basic Facts About the Graphs of Polynomial Functions
If f (x) is a polynomial function with degree n ≥ 1, then
• The maximum number of real zeros (x−intercepts) is n.
• The maximum number of turning points is n − 1.
It is also helpful to learn about the behavior of the function as x becomes very large (x → +∞) or very small
(x → −∞).
The Leading-Term Test
If an xn is the leading term of a polynomial. Then the behavior of the graph as x → ∞ or x → −∞ can be known by
one the four following behaviors:
1. If an > 0 and n even
2. If an < 0 and n even
3. If an > 0 and n odd
68
www.ck12.org
Chapter 1. Unit 1
4. If an < 0 and n odd
The Leading Term Test implies that if you “zoom out” far enough, then all polynomials look like the power function
made by the leading term of the polynomial. That is, if the exponent of the leading term is even, then the polynomial
is U-shaped, and if the exponent of the leading term is odd, then the polynomial is shaped like y = x3 .
Zeros of a Polynomial Function
• Every polynomial function with degree n ≥ 1 has at least one zero and at most n zeros (counting imaginary or
complex zeros).
or
• Every polynomial function with degree n ≥ 1 has exactly n zeros, if and only if the multiplicities are taken
into account (again, counting imaginary zeros).
Example 4
Graph the polynomial function f (x) = −3x4 + 2x3 .
Solution
Since the leading term here is −3x4 then an = −3 < 0, and n = 4 even. Thus the end behavior of the graph as x → ∞
and x → −∞ is that of Box #2, item 2.
69
1.9. Polynomial Functions
www.ck12.org
We can find the zeros of the function by simply setting f (x) = 0 and then solving for x.
−3x4 + 2x3 = 0
−x3 (3x − 2) = 0
This gives
x=0
or x =
2
3
So we have two x−intercepts, at x = 0 and at x = 23 , with multiplicity k = 3 for x = 0 and multiplicity k = 1 for
x = 23 .
To find the y−intercept, we find f (0), which gives
f (0) = 0
So the graph passes the y−axis at y = 0.
Since the x−intercepts are 0 and 32 , they divide the x−axis into three intervals: (−∞, 0), 0, 32 , and 32 , ∞ . Now we
are interested in determining at which intervals the function f (x) is negative and at which intervals it is positive. To
do so, we construct a table and choose a test value for x from each interval and find the corresponding f (x) at that
value.
TABLE 1.9:
Interval
Test Value x
f (x)
Sign of f (x)
(−∞,0)
0, 23 2
3,∞
-1
-5
1
2
1
16
1
-1
+
-
Location of points
on the graph
below the x−axis
above the x−axis
below the x−axis
1
Those test points give us three additional points to plot: (−1, −5), 12 , 16
, and (1, -1). Now we are ready to plot our
graph. We have a total of three intercept points, in addition to the three test points. We also know how the graph is
behaving as x → −∞ and x → +∞. This information is usually enough to make a rough sketch of the graph. If we
need additional points, we can simply select more points to complete the graph (Figure 10).
70
www.ck12.org
Chapter 1. Unit 1
To summarize, the following procedure can be followed when graphing a polynomial function.
Graphing a Polynomial Function
1.
2.
3.
4.
Use the leading-term test to determine the end behavior of the graph.
Find the x−intercept(s) of f (x) by setting f (x) = 0 and then solving for x.
Find the y−intercept of f (x) by setting y = f (0) and finding y.
Use the x−intercept(s) to divide the x−axis into intervals and then choose test points to determine the sign of
f (x) on each interval.
5. Plot the test points.
6. If necessary, find additional points to determine the general shape of the graph.
Example 5
Graph the polynomial function
g(x) = 2x3 + 3x2 − 50x − 75
Solution
Notice that the leading term is 2x3 , where n = 3 odd and an = 2 > 0. This tells us that the end behavior will take the
shape of item 3 in Box 2. Next we find the x− and y−intercepts. Setting
g(x) = 0
3
2
2x + 3x − 50x − 75 = 0
71
1.9. Polynomial Functions
www.ck12.org
We can factor this polynomial by grouping,
(2x + 3)(x2 ) − (2x + 3)(25) = 0
(2x + 3)(x2 − 25) = 0
(x + 5)(x − 5)(2x + 3) = 0
The zeros are -5, 5, and
−3
2 .
And they divide the x−axis into four intervals:
(−∞, −5)
−3
−5,
2
−3
,5
2
(5, ∞)
The y−intercept is when y = g(0). Thus the graph intercepts the y−axis at y = −75.
Now we choose test points from each interval and find g(x).
TABLE 1.10:
Interval
Test value x
Value of g(x) and its sign
(−∞, −5)
−5, −3
2
−3
2 ,5
(5, ∞)
-6
-2
0
6
-99
21
-75
165
Location of points on
graph
below the x−axis
above the x−axis
below the x−axis
above the x−axis
From the information obtained, we can roughly sketch the graph (below).
Applications, Technological Tools
You can use the same tools on a graphing calculator to analyze polynomials that you used to analyze quadratics. In
particular, you can use the Y = menu to graph a polynomial, use WINDOW and ZOOM to set up the view, and
72
www.ck12.org
Chapter 1. Unit 1
then use the CALC menu to find MINIMUM, MAXIMUM, and ZEROs of the function.
Two additional tools that are helpful for analyzing graphs with TI-83/84 calculators are the TRACE and TABLE
functions.
If you have graphed a function using Y = and set the WINDOW settings properly to allow you to see the graph,
then you can use TRACE to “read” pairs of (x, y) coordinates on the graph.
((Insert screen shot of graph with the TRACE function turned on))
While this is useful, a drawback of the TRACE function is that when you use the arrow keys ( < and > ) to “trace”
along the function values you cannot easily control the amount by which the x−coordinate changes each time you
press the arrow key. One solution to find a particular value of the function is to press TRACE and then enter a
number on the keypad. For example the sequence TRACE 5 will show the coordinates of the selected function
when x = 5. (Note: this assumes that x is between XMIN and XMAX. If this is not the case, you will get an error).
((Insert screen shot of function with TRACE 5. It should show X = 5 in the lower left corner, and the corresponding
y−value in the lower right))
A third way to analyze a function is to use the table. If you enter 2nd GRAPH then you see a table of function
values.
((insert screen shot of table))
If you go to TABLSET (2ND WINDOW), you can set the interval on the table, or change the independent variable
to ASK and then type in x−values directly in the table.
Exercises
1. State the degree, leading coefficient, domain and range for each.
a. f (x) = −3x4
b. h(x) = 21 x5
c. q(x) = 4x8
2. Using a graphing calculator, determine the end behaviour of the following functions: a) g(x) = 23 x5 − 4x3 +
7x − π
b) k(x) = −3x4 + 18x2 − 5x + 3
3. Without graphing the function, what can you say about the symmetry of g(x) = − 13 x6 ? What happens as
x → ∞? What happens as x → −∞?
1) a) Degree: 4, leading coefficient: - 3, {x ∈ R}, {y ∈ R, y ≤ 0}
b) Degree: 5, leading coefficient: 1/2, {x ∈ R}, {y ∈ R}
c) Degree: 8, leading coefficient: 4, {x ∈ R}, {y ∈ R, y ≥ 0}
2. a) Extends from quadrant 3 to quadrant 1. x → ∞ , y → ∞. x → −∞ , y → −∞.
b) Extends from quadrant 3 to quadrant 4. x → ∞ , y→−∞ . x → −∞ , y → −∞.
3. Line symmetry along the y-axis. x → ∞ , y→−∞ . x → −∞ , y → −∞. y ≥ 0}
73
1.10. Maximums and Minimums
www.ck12.org
1.10 Maximums and Minimums
Here you will learn to identify the maximums and minimums in various graphs and be able to differentiate between
global and relative extreme values.
When riding a roller coaster there is always one point that is the absolute highest off the ground. There are usually
many other places that reach fairly high, just not as high as the first. There are also places where the roller coaster
dips with one being the absolute lowest the roller coaster can go. How do you identify and distinguish between these
different peaks and valleys in a precise way?
Finding Maximums and Minimums
A global maximum refers to the point with the largest y value possible on a function. A global minimum refers
to the point with the smallest y value possible. Together these two values are referred to as global extrema. There
can only be one global maximum and only one global minimum. Global refers to entire space where the function is
defined. Global extrema are also called absolute extrema.
In addition to global maximums and global minimums, there are also local extrema or relative maximums and
relative minimums. The word relative is used because in relation to some neighborhood, these values stand out as
being the highest or the lowest.
Calculus uses advanced analytic tools to compute extreme values, but for the purposes of PreCalculus it is sufficient
to be able to identify and categorize extreme values graphically or through the use of technology. For example, the
TI-84 has a maximum finder when you select <2nd > then <trace>.
74
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/57938
Examples
Example 1
Earlier, you were asked to identify and distinguish between different peaks and valleys on a graph. Maximums and
minimums should be intuitive because they simply identify the highest points and the lowest points, or the peaks and
the valleys, in a graph. There is a formal distinction about whether a maximum is the highest on some local open
interval (does not matter how small), or whether it is simply the highest overall.
Example 2
Identify and categorize all extrema.
There are no global or local maximums or minimums. The function flattens, but does not actually reach a peak or a
valley.
Example 3
Identify and categorize all the extrema in the graph below:
75
1.10. Maximums and Minimums
www.ck12.org
Since the function appears from the arrows to increase and decrease beyond the display, there are no global extrema.
There is a local maximum at approximately (0, 3) and a local minimum at approximately (2.8, -7).
Example 4
Identify and categorize all the extrema in the graph below:
Since the function seems to abruptly end at the end points and does not go beyond the display, the endpoints are
important.
There is a global minimum at (0, 0). There is a local maximum at (-1, 1) and a global maximum at (5, 5).
Example 5
Identify and categorize all the extrema in the graph below:
76
www.ck12.org
Chapter 1. Unit 1
Since this function appears to increase to the right as indicated by the arrow there is no global maximum. There are
not any other high points either, so there are no local maximums. There is only the end point at (0, 0) which is a
global minimum.
Review
Use the graph below for 1-2.
1. Identify any global extrema.
2. Identify any local extrema.
Use the graph below for 3-4.
3. Identify any global extrema.
4. Identify any local extrema.
Use the graph below for 5-6.
77
1.10. Maximums and Minimums
www.ck12.org
5. Identify any global extrema.
6. Identify any local extrema.
Use the graph below for 7-8.
7. Identify any global extrema.
8. Identify any local extrema.
Use the graph below for 9-10.
9. Identify any global extrema.
10. Identify any local extrema.
11. Explain the difference between a global maximum and a local maximum.
12. Draw an example of a graph with a global minimum and a local maximum, but no global maximum.
13. Draw an example of a graph with local maximums and minimums, but no global extrema.
78
www.ck12.org
Chapter 1. Unit 1
14. Use your graphing calculator to identify and categorize the extrema of:
f (x) = 12 x4 + 2x3 − 6.5x2 − 20x + 24.
15. Use your graphing calculator to identify and categorize the extrema of:
g(x) = −x4 + 2x3 + 4x2 − 2x − 3.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.5.
79
1.11. Even and Odd Functions
www.ck12.org
1.11 Even and Odd Functions
While looking at the equation of a function that needed to be graphed for homework, a student recognized that the
task would go quickly because the sign of the values in the domain did not matter in calculating the function values,
and only half the values in the domain needed to be used. What kind of function was the student evaluating?
Types of Functions: Even, Odd, or Neither
A function f (x) is an even function if: f (x) = f (−x), i.e., the value of the function acting on an argument x is the
same as the value of the function when acting on the argument −x. So, for example, if f (x) is an even function,
then f (2) has the same answer as f (−2); f (5) has the same answer as f (−5), and so on.
Even functions are symmetric about the y-axis.
To show that a function is even, show that the definition f (x) = f (−x) holds. Take the function y = f (x) = x2 .
f (−x) = (−x)2 = x2 = f (x). So the function is an even function. The function graph below shows that the function
is symmetric with respect to the y-axis.
In contrast to an even function, a function f (x) is an odd function if: − f (x) = f (−x), i.e., the function is odd
when the negative of the function’s answer for a given argument is the same as the function acting on the negative
argument. If a function f (x) were odd, then f (−2) = − f (2); f (−5) = − f (5), and so on.
Odd functions are not symmetric about the y-axis, but they are symmetric about the origin.
To show that a function is odd, show that the definition − f (x) = f (−x) holds. Take the function y = x3
f (−x) = (−x)3 = −x3 = − f (x). So the function is an odd function.
80
www.ck12.org
Chapter 1. Unit 1
It is important to remember that a function does not have to be even or odd. Most functions are neither even nor
odd.
To determine whether the function y = 3(x + 2)2 + 4 is even or odd, apply the test for both types.
Apply the test for an even function: f (−x) = 3(−x + 2)2 + 4 = 3(x − 2)2 + 4 6= f (x). The function is not an even
function.
Apply the test for an odd function: f (−x) = 3(x − 2)2 + 4 6= − f (x). The function is not an odd function.
The above function is neither even nor odd. As can be seen in the graph below, the function is neither symmetric
about the y-axis nor symmetric about the origin.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/189633
81
1.11. Even and Odd Functions
www.ck12.org
Examples
Example 1
Earlier, you were told about a problem where the student only needed half of the values in the domain to graph his
function because the signs did not matter. Either all negative values or all positive values could be used to graph the
function. To be able to graph a function without having to worry about the sign of values in the domain means that
the function must be an even function, f (x) = f (−x). This means that the graph is is symmetric about the y − axis .
Example 2
Here are eight basic functions that are often encountered. Use their function graphs to determine whether they are
even, odd, or neither.
• Linear : f (x) = x
Domain = All reals
Range = All reals
This linear function is symmetric about the origin and is an odd function: f (x) = f (−x).
• Square(Quadratic) f (x) = x2
82
Domain = All reals
Range = {y ≥ 0}
www.ck12.org
Chapter 1. Unit 1
As shown earlier in the concept, this quadratic function is symmetric about the y-axis and is an even function:
f (x) = f (−x).
• Cube(Polynomial) f (x) = x3
Domain = All reals
Range = All reals
As shown earlier in the concept, this cubic function is symmetric about the origin and is an odd function:
f (−x) = − f (x).
• Square Root f (x) =
√
x
Domain = {x ≥ 0}
Range = {y ≥ 0}
This square root function is neither symmetric about the origin or the y-axis, and is neither odd nor even.
• Absolute Value f (x) = |x| Domain = All reals
Range = {y ≥ 0}
83
1.11. Even and Odd Functions
www.ck12.org
This absolute value function is symmetric about the y-axis and is an even function: f (x) = f (−x).
• Rational f (x) =
1
x
Domain = {x 6= 0}
Range = {y 6= 0}
This rational value function is symmetric about the origin and is an odd function: f (−x) = − f (x).
• Sine f (x) = sin x
84
Domain = All reals
Range = {−1 ≤ y ≤ 1}
www.ck12.org
Chapter 1. Unit 1
This sine function is symmetric about the origin and is an odd function: f (−x) = − f (x).
• Cosine f (x) = cos x
Domain = All reals
Range = {−1 ≤ y ≤ 1}
This sine function is symmetric about the y-axis and is an even function: f (x) = f (−x).
Review
For #1-4, is the function even, odd, or neither?
85
1.11. Even and Odd Functions
1.
2.
3.
86
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
4.
For #5-15, determine algebraically whether the function is even, odd or neither.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
y = −4x
f (x) = 2|x|+3
y = (x + 3)2 − 1
3 −2x
h(x) = xx4 +3x
2
g(x) = 1 −p
x5 + x23
k(x) = 5x x4 − 7x2
y = 3 sin2 (x) − 4
h(x) = sinx x
p
f (x) = p|x|
f (x) = 4 |x − 1|
k(x) = (x + 3)(x − 4)
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.4.
87
1.12. Zeroes of a Polynomial
www.ck12.org
1.12 Zeroes of a Polynomial
To find the Zeroes of a Linear and Quadratic Polynomial
What if you had a polynomial equation like 3x2 + 4x − 4 = 0? How could you factor the polynomial to solve the
equation? After completing this Concept, you’ll be able to solve polynomial equations by factoring and by using the
zero-product property.
Watch This
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133003
Watch the video at: https://www.youtube.com/watch?v=seHcVM2SU40
Guidance
The most useful thing about factoring is that we can use it to help solve polynomial equations.
Example A
Consider an equation like 2x2 + 5x − 42 = 0. How do you solve for x?
Solution:
There’s no good way to isolate x in this equation, so we can’t solve it using any of the techniques we’ve already
learned. But the left-hand side of the equation can be factored, making the equation (x + 6)(2x − 7) = 0.
How is this helpful? The answer lies in a useful property of multiplication: if two numbers multiply to zero, then at
least one of those numbers must be zero. This is called the Zero-Product Property.
What does this mean for our polynomial equation? Since the product equals 0, then at least one of the factors on
the left-hand side must equal zero. So we can find the two solutions by setting each factor equal to zero and solving
each equation separately.
Setting the factors equal to zero gives us:
(x + 6) = 0
88
OR
(2x − 7) = 0
www.ck12.org
Chapter 1. Unit 1
Solving both of those equations gives us:
2x − 7 = 0
x+6 = 0
x = −6
2x = 7
OR
x=
7
2
Notice that the solution is x = −6 OR x = 27 . The OR means that either of these values of x would make the product
of the two factors equal to zero. Let’s plug the solutions back into the equation and check that this is correct.
7
2
(x + 6)(2x − 7) =
7
7
+6
2· −7 =
2
2
19
(7 − 7) =
2
19
(0) = 0
2
Check : x = −6;
Check : x =
(x + 6)(2x − 7) =
(−6 + 6)(2(−6) − 7) =
(0)(−19) = 0
Both solutions check out.
Factoring a polynomial is very useful because the Zero-Product Property allows us to break up the problem into
simpler separate steps. When we can’t factor a polynomial, the problem becomes harder and we must use other
methods that you will learn later.
As a last note in this section, keep in mind that the Zero-Product Property only works when a product equals zero.
For example, if you multiplied two numbers and the answer was nine, that wouldn’t mean that one or both of the
numbers must be nine. In order to use the property, the factored polynomial must be equal to zero.
Example B
Solve each equation:
a) (x − 9)(3x + 4) = 0
b) x(5x − 4) = 0
c) 4x(x + 6)(4x − 9) = 0
Solution:
Since all the polynomials are in factored form, we can just set each factor equal to zero and solve the simpler
equations separately
a) (x − 9)(3x + 4) = 0 can be split up into two linear equations:
x−9 = 0
x=9
3x + 4 = 0
or
3x = −4
4
x=−
3
89
1.12. Zeroes of a Polynomial
www.ck12.org
b) x(5x − 4) = 0 can be split up into two linear equations:
5x − 4 = 0
x=0
5x = 4
4
x=
5
or
c) 4x(x + 6)(4x − 9) = 0 can be split up into three linear equations:
4x = 0
0
x=
4
4x − 9 = 0
x+6 = 0
or
x = −6
or
x=0
4x = 9
x=
9
4
Solve Simple Polynomial Equations by Factoring
Now that we know the basics of factoring, we can solve some simple polynomial equations. We already saw how we
can use the Zero-Product Property to solve polynomials in factored form—now we can use that knowledge to solve
polynomials by factoring them first. Here are the steps:
a) If necessary, rewrite the equation in standard form so that the right-hand side equals zero.
b) Factor the polynomial completely.
c) Use the zero-product rule to set each factor equal to zero.
d) Solve each equation from step 3.
e) Check your answers by substituting your solutions into the original equation
Example C
Solve the following polynomial equations.
a) x2 − 2x = 0
b) 2x2 = 5x
c) 9x2 y − 6xy = 0
Solution:
a) x2 − 2x = 0
Rewrite: this is not necessary since the equation is in the correct form.
Factor: The common factor is x, so this factors as x(x − 2) = 0.
Set each factor equal to zero:
x=0
90
or
x−2 = 0
www.ck12.org
Chapter 1. Unit 1
Solve:
x=0
x=2
or
Check: Substitute each solution back into the original equation.
x = 0 ⇒ (0)2 − 2(0) = 0
works out
x = 2 ⇒ (2)2 − 2(2) = 4 − 4 = 0
works out
Answer: x = 0, x = 2
b) 2x2 = 5x
Rewrite: 2x2 = 5x ⇒ 2x2 − 5x = 0
Factor: The common factor is x, so this factors as x(2x − 5) = 0.
Set each factor equal to zero:
x=0
2x − 5 = 0
or
Solve:
x=0
2x = 5
5
x=
2
or
Check: Substitute each solution back into the original equation.
x = 0 ⇒ 2(0)2 = 5(0) ⇒ 0 = 0
2
5
5
5
25 25
25 25
x= ⇒2
= 5· ⇒ 2·
=
⇒
=
2
2
2
4
2
2
2
Answer: x = 0, x =
works out
works out
5
2
c) 9x2 y − 6xy = 0
Rewrite: not necessary
Factor: The common factor is 3xy, so this factors as 3xy(3x − 2) = 0.
Set each factor equal to zero:
3 = 0 is never true, so this part does not give a solution. The factors we have left give us:
x=0
or
y=0
or
3x − 2 = 0
91
1.12. Zeroes of a Polynomial
www.ck12.org
Solve:
x=0
y=0
2
x=
3
or
or
3x = 2
Check: Substitute each solution back into the original equation.
x = 0 ⇒ 9(0)y − 6(0)y = 0 − 0 = 0
works out
y = 0 ⇒ 9x2 (0) − 6x(0) = 0 − 0 = 0
2
2
2
4
2
y − 6 · y = 9 · y − 4y = 4y − 4y = 0
x = ⇒ 9·
3
3
3
9
works out
Answer: x = 0, y = 0, x =
works out
2
3
Watch this video for help with the Examples above.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133004
Watch the video at: https://www.youtube.com/watch?v=wiRrlXoL0R8
Vocabulary
• Polynomials can be written in expanded form or in factored form. Expanded form means that you have
sums and differences of different terms:
• The factored form of a polynomial means it is written as a product of its factors.
• Zero Product Property: The only way a product is zero is if one or more of the terms are equal to zero:
a · b = 0 ⇒ a = 0 or b = 0.
Guided Practice
Solve the following polynomial equation.
9x2 − 3x = 0
Solution: 9x2 − 3x = 0
Rewrite: This is not necessary since the equation is in the correct form.
Factor: The common factor is 3x, so this factors as: 3x(3x − 1) = 0.
92
www.ck12.org
Chapter 1. Unit 1
Set each factor equal to zero.
3x = 0
or
x−2 = 0
Solve:
x=0
x=2
or
Check: Substitute each solution back into the original equation.
x=0
(0)2 − 2(0) = 0
x=2
(2)2 − 2(2) = 0
Answer: x = 0, x = 2
Real World Example - Jump Height
Topic
What’s your vertical jump height? How can we represent this as a quadratic function?
Student Exploration
One of the things that basketball players love to show off and brag about is their vertical jump height. If you walk
past a basketball hoop, there will always be someone walking past or before playing basketball to jump and try to
reach the rim. Some make it, and some don’t.
93
1.12. Zeroes of a Polynomial
www.ck12.org
Vertical jump height is discussed quite frequently among NBA players and fans. Vertical jump height refers to the
distance between the highest point a person can reach after a big jump and the standing reach height.
David Noel, from the NBA Draft of 2006, had a vertical jump height of 34 inches. The quadratic equation that can
best describe his vertical jump height over time is h = 162t − 192t 2 , with h representing the vertical jump height
and t represents time in seconds. We can use this equation to find how long it took for him to land on the ground.
Applying our understanding of factoring, we can factor out 6t from both terms on the right side of the equation.
Once we do that, we have h = 6t(27 − 32t). We can use the Zero Product Principle to set each factor equal to zero
and solve for t when h = 0.
0 = 6t(27 − 32t) We factored 6t from both terms because 6t is the greatest common factor of both terms.
6t = 0 and 27 − 32t = 0 We used the zero product principle to solve for t.
t = 0 and t =
27
32
= 0.84375, rounded to 0.84
What do these numbers mean?
This means that at 0 and 0.84 seconds Noel’s vertical jump height is 0 ft. Why are there two numbers?
The two numbers represent the time that Noel started at 0 seconds and the time that it took for him to reach the
ground after jumping.
For the sake of this activity, we’re going to look at the quadratic equation if someone were to find their vertical jump
height if the person were jumping off a bench. How do you think the quadratic function would change?
Let’s say Noel jumped off a bench 1ft off the ground and had the same jump height. How would the quadratic
equation that represents his jump height change?
From reading the “vertical shifts of quadratic functions” concept, we know that jumping from a bench would mean
that this is a vertical shift, assuming that his vertical jump height is still the same. The new equation would add
a “c” value. In this case, we would add 12 (since 1 f t = 12 inches) to the equation. Our new equation is h =
162t − 192t 2 + 12.
Try graphing the two quadratic equations and see what the similarities and differences are.
Extension Investigation
Advanced: Try finding your own vertical jump height and determine with your friends when you reach the highest
height and what time you land. Using this information, try to find the quadratic equation that best represents your
vertical jump height over time.
Practice:
Solve the following polynomial equations. Verify your answers using graphing software.
1. x(x + 12) = 0
2. (2x + 1)(2x − 1) = 0
3. (x − 5)(2x + 7)(3x − 4) = 0
4. 2x(x + 9)(7x − 20) = 0
5. x(3 + y) = 0
6. x(x − 2y) = 0
7. 18y − 3y2 = 0
8. 9x2 = 27x
94
www.ck12.org
Chapter 1. Unit 1
9. 4a2 + a = 0
10. b2 − 35 b = 0
11. 4x2 = 36
12. x3 − 5x2 = 0
Resources Cited
• http://www.topendsports.com/testing/results/vertical-jump.htm
95
1.13. Graphs of Polynomials Using Zeros
www.ck12.org
1.13 Graphs of Polynomials Using Zeros
Learning Objectives
Here you will explore the graphs of polynomial functions with powers greater than 2. The graphs of these functions
often have rather strange shapes and may seem daunting at first, but by understanding how to locate the places where
the function values cross the x and y axes of a graph, you will learn that sketching the approximate shape is much
less difficult than it appears.
How is finding and using the zeroes of a higher-degree polynomial related to the same process you have used in the
past on quadratic functions?
Graphing Polynomials Using Zeros
The following procedure can be followed when graphing a polynomial function.
•
•
•
•
Use the leading-term test to determine the end behavior of the graph.
Find the x−intercept(s) of f (x) by setting f (x) = 0 and then solving for x.
Find the y−intercept of f (x) by setting y = f (0) and finding y.
Use the x−intercept(s) to divide the x−axis into intervals and then choose test points to determine the sign of
f (x) on each interval.
• Plot the test points.
• If necessary, find additional points to determine the general shape of the graph.
The Leading-Term Test
If an xn is the leading term of a polynomial. Then the behavior of the graph as x → ∞ or x → −∞ can be known by
one the four following behaviors:
TABLE 1.11:
1. If an > 0 and n even:
96
2. If an < 0 and n even:
www.ck12.org
Chapter 1. Unit 1
TABLE 1.12:
3. If an > 0 and n odd:
4. If an < 0 and n odd:
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187261
Examples
Example 1
Earlier, you were asked to identify some similarities in graphing using zeroes between quadratic functions and
higher-degree polynomials.
Despite the more complex nature of the graphs of higher-degree polynomials, the general process of graphing using
zeroes is actually very similar. In both cases, your goal is to locate the points where the graph crosses the x or y
axis. In both cases, this is done by setting the y value equal to zero and solving for x to find the x axis intercepts, and
setting the x value equal to zero and solving for y to find the y axis intercepts.
Example 2
Find the roots (zeroes) of the polynomial:
h(x) = x3 + 2x2 − 5x − 6
Start by factoring:
h(x) = x3 + 2x2 − 5x − 6 = (x + 1)(x − 2)(x + 3)
To find the zeros, set h(x)=0 and solve for x.
(x + 1)(x − 2)(x + 3) = 0
97
1.13. Graphs of Polynomials Using Zeros
www.ck12.org
This gives
x+1 = 0
x−2 = 0
x+3 = 0
or
x = −1
x=2
x = −3
So we say that the solution set is {−3, −1, 2}. They are the zeros of the function h(x). The zeros of h(x) are the
x−intercepts of the graph y = h(x) below.
98
www.ck12.org
Chapter 1. Unit 1
Example 3
Find the zeros of g(x) = −(x − 2)(x − 2)(x + 1)(x + 5)(x + 5)(x + 5).
The polynomial can be written as
g(x) = −(x − 2)2 (x + 1)(x + 5)3
To solve the equation, we simply set it equal to zero
−(x − 2)2 (x + 1)(x + 5)3 = 0
this gives
x−2 = 0
x+1 = 0
x+5 = 0
or
x=2
x = −1
x = −5
Notice the occurrence of the zeros in the function. The factor (x − 2) occurred twice (because it was squared), the
factor (x + 1) occurred once and the factor (x + 5) occurred three times. We say that the zero we obtain from the
factor (x − 2) has a multiplicity k = 2 and the factor (x + 5) has a multiplicity k = 3.
Example 4
Graph the polynomial function f (x) = −3x4 + 2x3 .
Since the leading term here is −3x4 then an = −3 < 0, and n = 4 even. Thus the end behavior of the graph as x → ∞
and x → −∞ is that of Box #2, item 2.
We can find the zeros of the function by simply setting f (x) = 0 and then solving for x.
−3x4 + 2x3 = 0
−x3 (3x − 2) = 0
This gives
x=0
or x =
2
3
99
1.13. Graphs of Polynomials Using Zeros
www.ck12.org
So we have two x−intercepts, at x = 0 and at x = 23 , with multiplicity k = 3 for x = 0 and multiplicity k = 1 for
x = 23 .
To find the y−intercept, we find f (0), which gives
f (0) = 0
So the graph passes the y−axis at y = 0.
Since the x−intercepts are 0 and 32 , they divide the x−axis into three intervals: (−∞, 0), 0, 32 , and 32 , ∞ . Now we
are interested in determining at which intervals the function f (x) is negative and at which intervals it is positive. To
do so, we construct a table and choose a test value for x from each interval and find the corresponding f (x) at that
value.
TABLE 1.13:
Interval
Test Value x
f (x)
Sign of f (x)
(−∞,0)
0, 23 2
3,∞
-1
-5
1
2
1
16
1
-1
+
-
Location of points
on the graph
below the x−axis
above the x−axis
below the x−axis
1
Those test points give us three additional points to plot: (−1, −5), 12 , 16
, and (1, -1). Now we are ready to plot our
graph. We have a total of three intercept points, in addition to the three test points. We also know how the graph is
behaving as x → −∞ and x → +∞. This information is usually enough to make a rough sketch of the graph. If we
need additional points, we can simply select more points to complete the graph.
Example 5
Find the zeros and sketch a graph of the polynomial
f (x) = x4 − x2 − 56
100
www.ck12.org
Chapter 1. Unit 1
This is a factorable equation,
f (x) = x4 − x2 − 56
= (x2 − 8)(x2 + 7)
Setting f (x) = 0,
(x2 − 8)(x2 + 7) = 0
the first term gives
x2 − 8 = 0
x2 = 8
√
x=± 8
√
= ±2 2
and the second term gives
x2 + 7 = 0
x2 = −7
√
x = ± −7
√
= ±i 7
√
√
So the solutions are ±2 2 and ±i 7, a total of four zeros of f (x). Keep in mind that only the real zeros of a
function correspond to the x−intercept of its graph.
101
1.13. Graphs of Polynomials Using Zeros
www.ck12.org
Example 6
Graph g(x) = −(x − 2)2 (x + 1)(x + 5)3 .
Use the zeros to create a table of intervals and see whether the function is above or below the x−axis in each interval:
TABLE 1.14:
Interval
Test value x
g(x)
Sign of g(x)
(−∞, −5)
x = −5
(-5, -1)
x = −1
(-1, 2)
x=2
(2, ∞)
-6
-5
-2
-1
0
2
3
320
0
144
0
-100
0
-256
+
NA
+
NA
NA
-
Location of graph
relative to x−axis
Above
Above
Below
Below
Finally, use this information and the test points to sketch a graph of g(x).
Review
1. If c is a zero of f, then c is a/an _________________________ of the graph of f.
2. If c is a zero of f, then (x - c) is a factor of ___________________?
3. Find the zeros of the polynomial: P(x) = x3 − 5x2 + 6x
Consider the function: f (x) = −3(x − 3)4 (5x − 2)(2x − 1)3 (4 − x)2 .
4. How many zeros (x-intercepts) are there?
5. What is the leading term?
Find the zeros and graph the polynomial. Be sure to label the x-intercepts, y-intercept (if possible) and have correct
end behavior. You may use technology for questions 9-12.
6.
7.
8.
9.
10.
102
P(x) = −2(x + 1)2 (x − 3)
P(x) = x3 + 3x2 − 4x − 12
f (x) = −2x3 + 6x2 + 9x + 6
f (x) = −4x2 − 7x + 3
f (x) = 2x5 + 4x3 + 8x2 + 6x
www.ck12.org
Chapter 1. Unit 1
11. f (x) = x4 − 3x2
12. g(x) = x2 − |x|
13. Given: P(x) = (3x + 2)(x − 7)2 (9x + 2)3
State:
a. The leading term:
b. The degree of the polynomial:
c. The leading coefficient:
Determine the equation of the polynomial based on the graph:
14.
15.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.4.
103
1.14. Graphs of Polynomials Using Transformations
www.ck12.org
1.14 Graphs of Polynomials Using
Transformations
Learning Objectives
Here you will explore the graphs of polynomial functions with powers greater than 2. The graphs of these functions
often have rather strange shapes and may seem daunting at first, but the rules for transformations are no more
complicated than for the more common functions you have dealt with before.
Penny has been commissioned to paint a large mural on the side of the humane society. The society has a mascot, an
African lion that the society supports by donating money to the local zoo for his care. Penny is expected to sketch
the lion, submit the sketch to the board of the humane society for approval, and then scale up the drawing to 22
square feet so it will cover the side of the building.
Shortly after Penny actually begins the job of painting the mural, she is visited by the chairman of the board of the
humane society. He tells Penny that the Fire Chief has just notified the society that the additional activity the mural is
expected to attract means that the building will need another entrance/exit for fire safety. Unfortunately, that means
that the mural will need to be moved up and to the right about 5 feet.
What kinds of transformations of complex lines (like the ones in this lesson) will Penny have used by the time she
completes the job?
Graphing Polynomials Using Transformations
Polynomial Functions
You have already studied many different kinds of functions, for example linear functions, constant functions, and
quadratic functions. All three these functions belong to a larger group of functions called the polynomial functions.
The most simple polynomial is called a power function. A power function is a polynomial of the form f (x) = axn
where a is a real number and n is an integer with n ≥ 1.
If n is even, then the power function is also called “even,” and if n is odd, then the power function is “odd.” The
graphs of the first five power functions are shown below.
104
www.ck12.org
Chapter 1. Unit 1
Notice that each power function has only one x− and y−intercept, the origin (0, 0).
The end behavior of a function describes the y−values as x gets very large (x → ∞ in symbols) or as x gets very
small (x → −∞).
• For even powers n, the power function f (x) = axn is U-shaped (like a parabola) and as x → ∞, f (x) → ∞.
Likewise as x → −∞, f (x) → ∞.
• For odd powers n, the power function goes from the third quadrant to the first quadrant (like the line y = x).
As x → ∞, f (x) → ∞, and as x → −∞, f (x) → −∞.
As with quadratics and polynomials, the leading coefficient a changes the vertical “stretching” of power functions.
Graph Polynomial Functions Using Transformations
Just like quadratics, polynomial functions can be graphed using transformations of a known graph. The basic
transformations are vertical and horizontal shifts and reflections about the x− and y−axis.
Given a polynomial p(x) and constant real numbers c and a
•
•
•
•
•
•
p(x) + c is a vertical shift of the graph of p(x) by c units up (so the function shifts down if c < 0).
p(x − c) is a horizontal shift of the graph of p(x) by c units to the right. (So the function shifts left if c < 0).
−p(x) is a reflection of the graph of p(x) about the x−axis.
p(−x) is a reflection of the graph of p(x) about the y−axis.
ap(x) is a vertical stretch by a multiple of a.
p(ax) is a horizontal compression by a multiple of a.
105
1.14. Graphs of Polynomials Using Transformations
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187260
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187148
Examples
Example 1
Earlier, you were asked a question about the kinds of transformations of complex lines that Penny will have used in
her sketch by the time she completes it.
When Penny first sketches the lion, she will be (probably unconsciously) applying a vertical and horizontal compression to the sketch, unless she is using a very large piece of paper!
Once the sketch is approved, Penny will then need to greatly stretch the image vertically and horizontally to make
the image big enough to cover the side of the building.
After the Fire Chief visit, Penny was required to apply a horizontal and vertical shift to move the image out of the
way of the new door.
Example 2
The graph of f (x) is shown below. Use the graph of f (x) to graph f (x + 4).
106
www.ck12.org
Chapter 1. Unit 1
This is a horizontal shift of f (x) to the left by 4 units.
Example 3
The graph of f (x) is shown below. Use the graph of f (x) to graph f (−x) + 3.
107
1.14. Graphs of Polynomials Using Transformations
www.ck12.org
This is a reflection of f (x) about the y−axis and a vertical shift up by 3 units.
Example 4
Describe the transformations to a graph of the function y = x2 necessary to make a reasonable approximation of
f (x) = −x4 .
Since f (x) = −x4 is an even function, the function f (x) = −x2 is a reasonable approximation, requiring only a
reflection of y = x2 across the x-axis.
108
www.ck12.org
Chapter 1. Unit 1
Example 5
Describe the transformations necessary to make a graph of the reference function f (x) = x3 look like the graph of
y = −2x3 + 2.
The transformations required to replicate the function y = −2x3 + 2 are reflect f (x) = x3 across the x-axis, stretch
f (x) = −x3 by 2, and shift f (x) = −2x3 upward by 2.
Example 6
Describe the end behavior of f (x) = −7x3 + 6x2 − 3x using the leading coefficient test.
According to the leading coefficient test, given f (x) = axn where a is the leading coefficient and n is the degree, if n
is odd and a is negative, the graph goes up on the left, and down on the right.
∴ f (x) = −7x3 + 6x2 − 3x grows without bound toward ∞ in Quadrant II and grows negatively without bound in
Quadrant IV.
Review
1. Given: P(x) = 7x4 − 5x3 + x2 − 7x + 6
State:
a. The leading term:
b. The degree of the polynomial:
c. The leading coefficient:
Describe the transformation given in each question below:
2.
3.
4.
5.
Original Function:
Original Function:
Original Function:
Original Function:
g(x) = 3x3 Transformed Function: f (x) = 3x3 + 3
g(x) = 2x3 + 3 Transformed Function: f (x) = 2x3 + 7
g(x) = x4 + 2 Transformed Function: f (x) = 3(x4 + 2)
g(x) = 5x3 Transformed Function: f (x) = 21 (x3 )
Graph the following by using transformations of parent functions:
6. f (x) = 2x5 − 4
7. f (x) = (x − 4)3 + 6
8. The graph of f (x) = −2x4 + x2 is shown below.
109
1.14. Graphs of Polynomials Using Transformations
Describe each transformation based on the images below:
9. From:
To:
110
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
10. From:
To:
111
1.14. Graphs of Polynomials Using Transformations
11. From:
To:
112
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
12. From:
To:
113
1.14. Graphs of Polynomials Using Transformations
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.3.
114
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
1.15 Increasing and Decreasing
Here you will apply interval notation to identify when functions are increasing and decreasing.
It is important to be able to distinguish between when functions are increasing and when they are decreasing. In
business this could mean the difference between making money and losing money. In physics it could mean the
difference between speeding up and slowing down.
How do you decide when a function is increasing or decreasing?
Increasing and Decreasing Functions
Increasing means places on the graph where the slope is positive.
The formal definition of an increasing interval is: an open interval on the x axis of (a, d) where every b, c ∈ (a, d)
with b < c has f (b) ≤ f (c).
A interval is said to be strictly increasing if f (b) < f (c) is substituted into the definition.
Decreasing means places on the graph where the slope is negative. The formal definition of decreasing and strictly
decreasing are identical to the definition of increasing with the inequality sign reversed.
115
1.15. Increasing and Decreasing
www.ck12.org
A function is called monotonic if the function only goes in one direction and never switches between increasing and
decreasing.
Out of the basic functions, the monotonically increasing functions are:
√
f (x) = x, f (x) = x3 , f (x) = x, f (x) = ex , f (x) = ln x, f (x) = 1+e1 −x
The only basic functions that are not monotonically increasing are:
f (x) = x2 , f (x) = |x|, f (x) = 1x , f (x) = sin x
Identifying analytically where functions are increasing and decreasing often requires Calculus. For PreCalculus, it
will be sufficient to be able to identify intervals graphically and through your knowledge of what the parent functions
look like.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/57943
Examples
Example 1
Earlier, you were asked how to determine if a function is increasing or decreasing. Increasing is where the function
has a positive slope and decreasing is where the function has a negative slope. A common misconception is to look
at the squaring function and see two curves that symmetrically increase away from zero. Instead, you should always
read functions from left to right and draw slope lines and decide if they are positive or negative.
Example 2
Estimate where the following function is increasing and decreasing.
116
www.ck12.org
Chapter 1. Unit 1
Increasing: x ∈ (−∞, −1.5) ∪ (1.5, ∞).
Decreasing: x ∈ (−1.5, 1.5)
Example 3
Estimate where the following function is increasing and decreasing.
Increasing x ∈ (−∞, −4) ∪ (−4, −2.7) ∪ (−1, 2) ∪ (2, ∞).
Decreasing x ∈ (−2.7, −1)
Example 4
Estimate the intervals where the function is increasing and decreasing.
Increasing: x ∈ (−∞, −4) ∪ (−2, 1.5)
Decreasing: x ∈ (−4, −2) ∪ (1.5, ∞)
117
1.15. Increasing and Decreasing
www.ck12.org
Notice that open intervals are used because at x = −4, −2, 1.5 the slope of the function is zero. This is where the
slope transitions from being positive to negative. The reason why open parentheses are used is because the function
is not actually increasing or decreasing at those specific points.
Example 5
A continuous function has a global maximum at the point (3, 2), a global minimum at (5, -12) and has no relative
extrema or other places with a slope of zero. What are the increasing and decreasing intervals for this function?
Increasing x ∈ (−∞, 3) ∪ (5, ∞).
Decreasing x ∈ (3, 5)
Note: The y coordinates are not used in the intervals. A common mistake is to want to use the y coordinates.
Review
Use the graph below for 1-2.
1. Identify the intervals (if any) where the function is increasing.
2. Identify the intervals (if any) where the function is decreasing.
Use the graph below for 3-4.
3. Identify the intervals (if any) where the function is increasing.
4. Identify the intervals (if any) where the function is decreasing.
Use the graph below for 5-6.
118
www.ck12.org
Chapter 1. Unit 1
5. Identify the intervals (if any) where the function is increasing.
6. Identify the intervals (if any) where the function is decreasing.
Use the graph below for 7-8.
7. Identify the intervals (if any) where the function is increasing.
8. Identify the intervals (if any) where the function is decreasing.
Use the graph below for 9-10.
119
1.15. Increasing and Decreasing
www.ck12.org
9. Identify the intervals (if any) where the function is increasing.
10. Identify the intervals (if any) where the function is decreasing.
11. Give an example of a monotonically increasing function.
12. Give an example of a monotonically decreasing function.
13. A continuous function has a global maximum at the point (1, 4), a global minimum at (3, -6) and has no relative
extrema or other places with a slope of zero. What are the increasing and decreasing intervals for this function?
14. A continuous function has a global maximum at the point (1, 1) and has no other extrema or places with a slope
of zero. What are the increasing and decreasing intervals for this function?
15. A continuous function has a global minimum at the point (5, -15) and has no other extrema or places with a
slope of zero. What are the increasing and decreasing intervals for this function?
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.7.
120
www.ck12.org
Chapter 1. Unit 1
1.16 Average Rate of Change
Learning Objectives
Here you will learn about identifying the average slope for a portion of a graph, called an interval.
Suppose you and your family are planning to drive to Disney World. You know that the trip includes the following
approximate sections:
200 Miles East
100 Miles South
350 Miles East
50 Miles North
150 Miles East
450 Miles South
How could you generalize the description of the journey?
Average Rate of Change
Consider the following situation: you are on a week long road trip with your friend. When you begin to drive on
the second day, you have already driven a total of 200 miles. After 6 hours of driving on the second day, you have
driven a total of 500 miles. On average, how many miles did you drive per hour on the second day of the trip?
The graph below shows this situation, with the x axis representing the number of hours driving (on the second day),
and the y axis representing the number of miles driven. The first point on the graph, (0, 200), says that at the
beginning of the second day you have already driven 200 miles. The second point on the graph, (6, 500), says that
after 6 hours of driving on the second day you have driven 500 miles total.
Notice that in total, during your 6 hours of driving, you have driven 300 miles. The rate at which you drove is 300
miles in 6 hours, or 50 miles per hour. We refer to this rate as the average rate of change because it is an average
across the 6 hours. That is, you did not necessarily drive 50 miles every hour. There could have been one hour where
you drove 70 miles and another hour where you drove only 30 miles.
121
1.16. Average Rate of Change
www.ck12.org
We can represent the average rate of change on the graph by indicating how much each quantity has changed: The y
values increased by 300, and the x values increased by 6. The average rate of change is the ratio of these changes in
each variable. This is how we can define average rate of change in general:
change in y
Average rate of change = change in x
We can examine the average rate of change of a function, whether it is represented as data, as in the previous
example, or by an equation.
Notice that the average rate of change of the function f (x) = 4x is the slope of the line, 4. While a linear function
has a constant slope, other functions, such as f (x) = x2 , will not. You will explore this idea in greater detail in your
study of calculus.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187072
Examples
Example 1
Earlier, you were given a problem about your trip to Disney World.
One way to generalize the information about your trip is to identify your average rate of change in position from
home to Disney World.
Average Rate of Change = (Change in y) / (Change in x)
The trip segments were:
200 Miles East
100 Miles South
350 Miles East
50 Miles North
150 Miles East
450 Miles South
If we assume that travelling South is -y, and traveling East is +x:
The total change in y = -100 +50 -450 = -500
The total change in x = +200 +350 +150 = 700
Your trip includes a total of approximately 1200 miles of travel, and you will average 5 miles South for every 7 miles
East. Your average rate of change is -5/7.
122
www.ck12.org
Chapter 1. Unit 1
Example 2
Find the average rate of change on the given interval:
f (x) = x2 on [0, 2]
The endpoints of the interval are (0, 0) and (2, 4). Therefore the change in y is 4 and the change in x is 2. The
average rate of change is 4/2 = 2.
Example 3
Find the average rate of change on the given interval:
f (x) = 4x on [1, 7]
The endpoints of the interval are (1, 4) and (7, 28). Therefore the change in y is 28 - 4 = 24 and the change in x is 7
- 1 = 6. The average rate of change is 24/6 = 4.
Example 4
What is the average rate of change of the function f (x) = 3x2 on the interval [2, 5]?
The two points are (2, 12) and (5, 75). The average rate of change is 63/3 = 21.
Example 5
If Kelli spends $5 at 1pm, $7 at 2:30pm, $12 at 4pm, and $2 at 4:30pm, what is her average rate of spending?
Kelli spent a total of $26 over a total of 3 1/2 hours:
$26
7/2hr
$52
7
≈
Multiply top and bottom by 2 to remove the fraction in denominator.
$7.43
hr
Example 6
What is the average rate of change shown in the image below?
123
1.16. Average Rate of Change
www.ck12.org
The two points in the image are:(−6, −1) : (7, 6).
The average rate of change is the same as the slope of the line.
Recall that m =
∆y
∆x
∴
y2 −y1
x2 −x1
(6)−(−1)
(7)−(−6)
7
13
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187074
Review
For y = x3 , find the average rate of change as:
1.
2.
3.
4.
124
x increases from 1 to 3
x increases from -4 to -1
Suppose f (1) = 2 and the average rate of change of f between 1 and 5 is 3. Find f (5)
Jamie went on a bicycle trip and stopped regularly at half-hour intervals. At each break he recorded his total
distance since leaving home. What was his average speed in km/h, during the first half of the trip? During the
last half? Jamie hoped to average at least 11.5km/h over the course of the trip. Did he? Explain.
www.ck12.org
Chapter 1. Unit 1
TABLE 1.15:
Stops
1
2
3
4
5
6
Time (h)
0.5
1.0
1.5
2.0
2.5
3.0
Distance (km)
7
15
21
24
28
36
5. On Monday, the price of a gallon of gas was $3.74, and on Saturday, the price had risen to $4.09. What is the
average rate of change of the price of a gallon of gas?
6. According to census figures, the population of Clovis was 31,194 in 1980 and 32,511 in 2001. What was the
average rate of change of the population over that time interval?
7. Griego and Sons will deliver 12 yd3 of gravel for $240 and 30 yd3 for $575. What is the average rate of change
of the cost as the number of cubic yards varies from 12 to 30?
8. When a load of 5 pounds is placed on a spring, its length is 6 inches, and when a load of 9 pounds is placed
on the spring, its length is 8 inches. What is the average rate of change of the length of the spring as the load
varies from 5 pounds to 9 pounds?
9. Driving at a speed of 75 mph, a Mini Cooper’s fuel efficiency is 29 miles per gallon. If the driver slows to a
speed of 60 mph, he will have a fuel efficiency of 34 miles per gallon. What is the average rate of change of
the fuel efficiency as the speed drops from 75 mph to 60 mph?
10. Let y = f (x) = x2 + x + 2 a) Find the average rate of change of y with respect to x between x = −1 and x = 2
b) Draw the graph of f and the graph of the secant line through (−1,−2) and (2, 4)
11. Amy takes a trip from Chicago to Milwaukee. Due to road construction, she drives the first 10 miles at a
constant speed of 20 mph. For the next 30 miles she maintains a constant speed of 60 mph and then stops at
McDonald’s for 10 minutes for a snack. She drives the next 45 miles at a constant speed of 45 mph. a) What
was Amy’s average driving speed for the trip? b) What is her average speed for the entire trip (including the
stop at McDonald’s)?
t2
12. The weight w(t) (in grams) of a tumor t weeks after it forms is given by w(t) = 15
. Find the average rate at
which the tumor is growing during the fifth week after it was formed.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.4.
125
1.17. Instantaneous Rate of Change
www.ck12.org
1.17 Instantaneous Rate of Change
Learning Objectives
Here you will learn about instantaneous rate of change and the concept of a derivative.
When you first learned about slope you learned the mnemonic device “rise over run” to help you remember that to
calculate the slope between two points you use the following formula:
m=
y2 − y1
x2 − x1
In Calculus, you will learn that for curved functions, it makes more sense to discuss the slope at one precise point
rather than between two points. The slope at one point is called the slope of the tangent line and the slope between
two separate points is called a secant line.
Consider a car driving down the highway and think about its speed. You are probably thinking about speed in terms
of going a given distance in a given amount of time. The units could be miles per hour or feet per second, but the
units always have time in the denominator. What happens when you consider the instantaneous speed of the car at
one instant of time? Wouldn’t the denominator be zero?
Instantaneous Rate of Change
The slope at a point P represents the instantaneous rate of change at that point. The slope at a point P(otherwise
known as the slop of the tangent line) can be approximated by the slope of secant lines as the “run” of each secant
line approaches zero. A secant line is a line that passes through two distinct points on a function. A tangent line to
a function at a given point is the straight line that just touches the curve at that point.
Because you are interested in the slope as the “run” approaches zero, this is a limit question. One of the main reasons
that you study limits in calculus is so that you can determine the slope of a curve at a point or the slope of the tangent
line.
For a graph of lines, it is easy to estimate the slopes of the tangent lines since the slope of the tangent is the same as
the slope of the line. Estimate the slope of the following function at -3, -2, -1, 0, 1, 2, 3. Organize the slopes in a
table.
126
www.ck12.org
Chapter 1. Unit 1
By mentally drawing a tangent line at the following x values you can estimate the following slopes.
TABLE 1.16:
slope
0
0
-1
-1
2
0
0
x
-3
-2
-1
0
1
2
3
If you graph these points you will produce a graph of what’s known as the derivative of the original function. A
derivative is a function of the slopes of the original function.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/63172
Examples
Example 1
Earlier, you were asked about the instantaneous speed of a car. If you write the ratio of distance to time and use limit
notation to allow time to go to zero you do seem to get a zero in the denominator.
lim distance
time
time→0
The great thing about limits is that you have learned techniques for finding a limit even when the denominator goes to
zero. Instantaneous speed for a car essentially means the number that the speedometer reads at that precise moment
in time. You are no longer restricted to finding slope from two separate points.
127
1.17. Instantaneous Rate of Change
www.ck12.org
Example 2
√
Estimate the slope of the function f (x) = x at the point (1, 1) by calculating 4 successively close secant lines. If
you had to guess what the slope was at the point (1, 1) what would you guess the slope to be?
Calculate the slope between (1, 1) and 4 other points on the curve:
√
√ 5−1 ≈ 0.309
• The slope of the line between 5, 5 and (1, 1) is: m1 = 5−1
2−1
• The slope of the line between (4, 2) and (1, 1) is: m2 = 4−1
≈ 0.333
√
√ 3−1 ≈ 0.366
• The slope of the line between 3, 3 and (1, 1) is: m3 = 3−1
√
√ 2−1 ≈ 0.414
• The slope of the line between 2, 2 and (1, 1) is: m4 = 2−1
√
1−1 . Unfortunately, this cannot be computed
Notice that the pattern in the previous example is leading up to 1−1
directly because there is a zero in the denominator. Luckily, you know how to evaluate using limits.
√ x−1
To determine what the slope would be at the point (1, 1), you would evaluate the limit lim
x−1
x→1
√
√
( x − 1) ( x + 1)
√
m = lim
·
x→1
(x − 1) ( x + 1)
(x − 1)
√
= lim
x→1 (x − 1) ( x + 1)
1
√
= lim
x→1 ( x + 1)
1
= √
1+1
1
=
2
= 0.5
The slope of the function f (x) =
√
x at the point (1, 1) is exactly m = 12 .
Example 3
Sketch a complete cycle of a sine graph. Estimate the slopes at 0, π2 , π, 3π
2 , 2π.
128
www.ck12.org
Chapter 1. Unit 1
TABLE 1.17:
x
0
π
2
π
3π
2
2π
Slope
1
0
-1
0
1
You should notice that these are the exact values of cosine evaluated at those points.
Example 4
Logan travels by bike at 20 mph for 3 hours. Then she gets in a car and drives 60 mph for 2 hours. Sketch both the
distance vs. time graph and the rate vs. time graph.
Distance vs. Time:
Rate vs. Time: (this is the graph of the derivative of the original function shown above)
129
1.17. Instantaneous Rate of Change
www.ck12.org
Example 5
Approximate the slope of y = x3 at (1, 1) by using secant lines from the left. Will the actual slope be greater or less
than the estimates?
0.73 −1
0.7−1 ≈ 2.19
0.83 −1
3
(0.8, 0.8 ) and (1, 1) is: m2 = 0.8−1 ≈ 2.44
3 −1
(0.9, 0.93 ) and (1, 1) is: m3 = 0.9
0.9−1 ≈ 2.71
3 −1
(0.95, 0.953 ) and (1, 1) is: m1 = 0.95
0.95−1 ≈ 2.8525
3 −1
(0.975, 0.9753 ) and (1, 1) is: m1 = 0.975
0.975−1 ≈ 2.925625
• The slope of the line between (0.7, 0.73 ) and (1, 1) is: m1 =
• The slope of the line between
• The slope of the line between
• The slope of the line between
• The slope of the line between
The slope at (1, 1) will be slightly greater than the estimates because of the way the slope curves. The slope at (1, 1)
appears to be about 3.
Review
1. Approximate the slope of y = x2 at (1, 1) by using secant lines from the left. Will the actual slope be greater or
less than the estimates?
2. Evaluate the following limit and explain how it confirms your answer to #1.
130
www.ck12.org
lim
x→1
x2 −1
x−1
Chapter 1. Unit 1
3. Approximate the slope of y = 3x2 + 1 at (1, 4) by using secant lines from the left. Will the actual slope be greater
or less than the estimates?
4. Evaluate the following limit and explain how it confirms your answer to #3.
2
+1−4
lim 3x x−1
x→1
5. Approximate the slope of y = x3 − 2 at (1, −1) by using secant lines from the left. Will the actual slope be greater
or less than the estimates?
6. Evaluate the following limit and explain how it confirms your answer to #5.
3
lim x −2−(−1)
x−1
x→1
7. Approximate the slope of y = 2x3 − 1 at (1, 1) by using secant lines from the left. Will the actual slope be greater
or less than the estimates?
8. What limit could you evaluate to confirm your answer to #7?
9. Sketch a complete cycle of a cosine graph. Estimate the slopes at 0, π2 , π, 3π
2 , 2π.
10. How do the slopes found in the previous question relate to the sine function? What function do you think is the
derivative of the cosine function?
11. Sketch the line y = 2x + 1. What is the slope at each point on this line? What is the derivative of this function?
12. Logan travels by bike at 30 mph for 2 hours. Then she gets in a car and drives 65 mph for 3 hours. Sketch both
the distance vs. time graph and the rate vs. time graph.
13. Explain what a tangent line is and how it relates to derivatives.
14. Why is finding the slope of a tangent line for a point on a function the same as the instantaneous rate of change
at that point?
15. What do limits have to do with finding the slopes of tangent lines?
Review (Answers)
To see the Review answers, open this PDF file and look for section 14.8.
131
1.18. Long Division
www.ck12.org
1.18 Long Division
Learning Objectives
• Divide multi-digit numbers using the standard algorithm.
Practice Long Division
Use interactive below to practice dividing with the standard method.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/255748
Practice More Difficult Long Division
Use these interactives to practice some more challenging and advanced long division problems! Can you answer the
most difficult ones?
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/255750
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/255749
132
www.ck12.org
Chapter 1. Unit 1
What percentage of the year is each season?
Does each season have the same number of days? Are the seasons in the Northern hemisphere the same length as
the seasons in the Southern hemisphere. Use long division in the interactive below to find the percent of the year
for each season.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/254058
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/254070
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/255371
Can you decode the puzzle?
Here is a challenging word puzzle based on long division. Can you decode the hidden message?
Hint: The questions below the puzzle may help!
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/254059
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/255772
133
1.18. Long Division
www.ck12.org
Ready for a new challenge?
Once you’ve decoded the puzzle in the interactive, try creating your own long division puzzles by setting up two
words as a divisor and a dividend and then assigning a specific number to each letter. Using the numbers in place
of the letters, you can work out the long division problem in full. You should expect to have a remainder in your
division problems. Afterwards, translate each number that you originally assigned to a letter back into the letter you
chose.There will probably be a few numbers in the worked-out solution that do not have letters associated with them,
and that’s a good thing! If there aren’t any, pick a couple of letters to decode at the start to help solve the puzzle.
Challenge one of your classmates to solve your problem. You can also share it in the cK-12 cafe!
134
www.ck12.org
Chapter 1. Unit 1
1.19 Multiplication of Polynomials
Learning Objectives
Here you will learn how to multiply polynomials using the distributive property.
Jack was asked to frame a picture. He was told that the width of the frame was to be 5 inches longer than the glass
width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass and finds
the height to width ratio is 4:3. Write the expression to determine the area of the picture frame.
Multiplying Polynomials
To multiply polynomials you will need to use the distributive property. Recall that the distributive property says that
if you start with an expression like 3(5x + 2), you can simplify it by multiplying both terms inside the parentheses
by 3 to get a final answer of 15x + 6.
When multiplying polynomials, you will need to use the distributive property more than once for each problem.
Let’s find the product of the following polynomials:
1. (x + 6)(x + 5)
To answer this question you will use the distributive property. The distributive property would tell you to
multiply x in the first set of parentheses by everything inside the second set of parentheses , then multiply 6 in
the first set of parentheses by everything in the second set of parentheses . Here is what that looks like:
135
1.19. Multiplication of Polynomials
www.ck12.org
2. (2x + 5)(x − 3)
Again, use the distributive property. The distributive property tells you to multiply 2x in the first set of
parentheses by everything inside the second set of parentheses , then multiply 5 in the first set of parentheses
by everything in the second set of parentheses . Here is what that looks like:
3. (4x + 3)(2x2 + 3x − 5)
Even though at first this question may seem different, you can still use the distributive property to find the
product. The distributive property tells you to multiply 4x in the first set of parentheses by everything inside
the second set of parentheses, then multiply 3 in the first set of parentheses by everything in the second set of
parentheses. Here is what that looks like:
136
www.ck12.org
Chapter 1. Unit 1
Examples
Example 1
Earlier, you were told that Jack had to frame a picture and that the width of the frame was to be 5 inches longer than
the glass width and the height of the frame was to be 7 inches longer than the glass height. Jack measures the glass
and finds the height to width ratio is 4:3. Write the expression to determine the area of the picture frame.
What is known?
• The width is 5 inches longer than the glass
• The height is 7 inches longer than the glass
• The glass has a height to width ratio of 4:3
The equations:
• The height of the picture frame is 4x + 7
• The width of the picture frame is 3x + 5
The formula:
Area = w × h
Area = (3x + 5)(4x + 7)
Area = 12x2 + 21x + 20x + 35
Area = 12x2 + 41x + 35
Example 2
Find the product: (x + 3)(x − 6)
(x + 3)(x − 6)
137
1.19. Multiplication of Polynomials
www.ck12.org
Example 3
Find the product: (2x + 5)(3x2 − 2x − 7)
(2x + 5)(3x2 − 2x − 7)
Example 4
An average football field has the dimensions of 160 ft by 360 ft. If the expressions to find these dimensions were
(3x + 7) and (7x + 3), what value of x would give the dimensions of the football field?
138
www.ck12.org
Chapter 1. Unit 1
Area = l × w
Area = 360 × 160
(7x + 3) = 360
7x = 360 − 3
7x = 357
x = 51
(3x + 7) = 160
3x = 160 − 7
3x = 153
x = 51
The value of x that satisfies these expressions is 51.
Review
Use the distributive property to find the product of each of the following polynomials:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
(x + 4)(x + 6)
(x + 3)(x + 5)
(x + 7)(x − 8)
(x − 9)(x − 5)
(x − 4)(x − 7)
(x + 3)(x2 + x + 5)
(x + 7)(x2 − 3x + 6)
(2x + 5)(x2 − 8x + 3)
(2x − 3)(3x2 + 7x + 6)
(5x − 4)(4x2 − 8x + 5)
9a2 (6a3 + 3a + 7)
−4s2 (3s3 + 7s2 + 11)
(x + 5)(5x3 + 2x2 + 3x + 9)
(t − 3)(6t 3 + 11t 2 + 22)
(2g − 5)(3g3 + 9g2 + 7g + 12)
Review (Answers)
To see the Review answers, open this PDF file and look for section 7.2.
139
1.20. Factorization using Difference of Squares
www.ck12.org
1.20 Factorization using Difference of Squares
Learning Objectives
Here you’ll learn how to factor polynomials that are the difference of two squares.
Factorization Using Difference of Squares
When you learned how to multiply binomials we talked about two special products.
The sum and difference formula: (a + b)(a − b) = a2 − b2
The square of a binomial formulas:
(a + b)2 = a2 + 2ab + b2
(a − b)2 = a2 − 2ab + b2
In this section we’ll learn how to recognize and factor these special products.
Factor the Difference of Two Squares
We use the sum and difference formula to factor a difference of two squares. A difference of two squares is any
quadratic polynomial in the form a2 − b2 , where a and b can be variables, constants, or just about anything else. The
factors of a2 − b2 are always (a + b)(a − b); the key is figuring out what the a and b terms are.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133021
Factoring the Difference of Squares
1. Factor the difference of squares:
a) x2 − 9
Rewrite x2 − 9 as x2 − 32 . Now it is obvious that it is a difference of squares.
The difference of squares formula is:
a2 − b2 = (a + b)(a − b)
140
www.ck12.org
Chapter 1. Unit 1
Let’s see how our problem matches with the formula:
x2 − 32 = (x + 3)(x − 3)
The answer is:
x2 − 9 = (x + 3)(x − 3)
We can check to see if this is correct by multiplying (x + 3)(x − 3):
x+3
x−3
− 3x − 9
x2 + 3x
x2 + 0x − 9
The answer checks out.
Note: We could factor this polynomial without recognizing it as a difference of squares. With the methods we
learned in the last section we know that a quadratic polynomial factors into the product of two binomials:
(x )(x )
We need to find two numbers that multiply to -9 and add to 0 (since there is no x−term, that’s the same as if the
x−term had a coefficient of 0). We can write -9 as the following products:
− 9 = −1 · 9
and
−1+9 = 8
− 9 = 1 · (−9)
and
1 + (−9) = −8
− 9 = 3 · (−3)
and
3 + (−3) = 0
T hese are the correct numbers.
We can factor x2 − 9 as (x + 3)(x − 3), which is the same answer as before. You can always factor using the methods
you learned in the previous section, but recognizing special products helps you factor them faster.
b) x2 − 100
Rewrite x2 − 100 as x2 − 102 . This factors as (x + 10)(x − 10).
c) x2 − 1
Rewrite x2 − 1 as x2 − 12 . This factors as (x + 1)(x − 1).
2. Factor the difference of squares:
a) 16x2 − 25
Rewrite 16x2 − 25 as (4x)2 − 52 . This factors as (4x + 5)(4x − 5).
b) 4x2 − 81
Rewrite 4x2 − 81 as (2x)2 − 92 . This factors as (2x + 9)(2x − 9).
c) 49x2 − 64
Rewrite 49x2 − 64 as (7x)2 − 82 . This factors as (7x + 8)(7x − 8).
141
1.20. Factorization using Difference of Squares
www.ck12.org
3. Factor the difference of squares:
a) x2 − y2
x2 − y2 factors as (x + y)(x − y).
b) 9x2 − 4y2
Rewrite 9x2 − 4y2 as (3x)2 − (2y)2 . This factors as (3x + 2y)(3x − 2y).
c) x2 y2 − 1
Rewrite x2 y2 − 1 as (xy)2 − 12 . This factors as (xy + 1)(xy − 1).
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133022
Examples
Factor the difference of squares:
Example 1
x4 − 25
Rewrite x4 − 25 as (x2 )2 − 52 . This factors as (x2 + 5)(x2 − 5).
Example 2
16x4 − y2
Rewrite 16x4 − y2 as (4x2 )2 − y2 . This factors as (4x2 + y)(4x2 − y).
Example 3
x2 y8 − 64z2
Rewrite x2 y4 − 64z2 as (xy2 )2 − (8z)2 . This factors as (xy2 + 8z)(xy2 − 8z).
Review
Factor the following differences of squares.
142
www.ck12.org
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Chapter 1. Unit 1
x2 − 4
x2 − 36
−x2 + 100
x2 − 400
9x2 − 4
25x2 − 49
9a2 − 25b2
−36x2 + 25
4x2 − y2
16x2 − 81y2
Review (Answers)
To view the Review answers, open this PDF file and look for section 9.10.
143
1.21. Factorization of Quadratic Expressions
www.ck12.org
1.21 Factorization of Quadratic Expressions
Learning Objectives
Here you’ll learn to determine what the factors of a quadratic expression are.
Suppose your height above sea level in feet when traveling over a hill can be represented by the expression −x2 +
18x + 63, where x is the horizontal distance traveled. If you wanted to factor this expression, could you do it? What
would be the steps that you would follow?
Factoring Quadratic Expressions
Previously, we factored common monomials, so you already know how to factor quadratic polynomials where c = 0.
Now, we will learn how to factor quadratic polynomials for different values of a, b, and c.
Quadratic polynomials are polynomials of degree 2. The standard form of a quadratic polynomial is ax2 + bx + c,
where a, b, and c are real numbers.
A quadratic of the form x2 + bx + c factors to the product of two binomials: (x + m)(x + n) where m + n = b and mn =
c.
There are some important rules to know when factoring quadratics:
• If b and c are positive then both m and n are positive.
– Example: x2 + 8x + 12 factors as (x + 6)(x + 2).
• If b is negative and c is positive then both m and n are negative.
– Example: x2 − 6x + 8 factors as (x − 2)(x − 4).
• If c is negative then either m is positive and n is negative or vice-versa.
– Example: x2 + 2x − 15 factors as (x + 5)(x − 3).
– Example: x2 + 34x − 35 factors as (x + 35)(x − 1).
• If a = −1, factor a common factor of -1 from each term in the trinomial and then factor as usual. The answer
will have the form −(x + m)(x + n).
– Example: −x2 + x + 6 factors as −(x − 3)(x + 2).
Let’s factor the following polynomials:
1. x2 + 5x + 6
We are looking for an answer that is the product of the two binomials in parentheses: (x +
)(x +
).
To fill in the blanks, we want two numbers m and n that multiply to 6 and add to 5. A good strategy is to list
the possible ways we can multiply two numbers to give us 6 and then see which of these pairs of numbers add
to 5.
Since both b and c are positive, then both spaces will be filled by positive numbers.
The number six can be written as the product of 1 and 6 or 2 and 3. Testing the product and sums we get:
144
www.ck12.org
Chapter 1. Unit 1
6 = 1×6
and
1+6 = 7
6 = 2×3
and
2+3 = 5
So the answer is (x + 2)(x + 3) because 2 and 3 multiply to 6 and sum to 5.
We can check to see if this is correct by multiplying (x + 2)(x + 3).
x is multiplied by x and 3 to get x2 + 3x.
2 is multiplied by x and 3 to get 2x + 6.
Combine the like terms: x2 + 5x + 6.
2. x2 − 6x + 8
We are looking for an answer that is the product of the two binomials in parentheses: (x +
)(x +
).
Since b is negative and c is positive, both of the blanks are going to be negative. We only want the negative
factors of 8.
The number 8 can be written as the product of -1 and -8 or -2 and -4. Testing the product and sums, we get:
−x2 + 16x + 63 = −(x2 − 16x − 63)
-2 and -4 are the factors that multiply to 8 and add to -6. This is the correct choice.
The answer is (x − 2)(x − 4).
3. x2 + 2x − 15
We are looking for an answer that is the product of the two binomials in parentheses: (x ±
)(x ±
).
Since b is positive and c is negative, one of the blanks will be negative and one will be positive. The number
-15 can be written as the product of -1 and 15, 1 and -15, 3 and -5, or -3 and 5. Testing the product and sums,
we get:
−15 = −1 · 15
and
−1 + 15 = 14
−15 = 1 · −15
and
1 + (−15) = −14
−15 = −3 · 5
and
(−3) + 5 = 2
−15 = 3 · −5
and
3 + −5 = −2
-3 and 5 are the factors that multiply to -15 and add to 2. The answer is (x − 3)(x + 5).
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180124
145
1.21. Factorization of Quadratic Expressions
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180126
Examples
Example 1
Earlier, you were told that your height above sea level when traveling over a hill can be represented by the expression −x2 + 18x + 63, where x is the horizontal distance traveled. What is the factored form of this expression?
First, factor the common factor of -1 from each term in the trinomial. Factoring -1 changes the signs of each term in
the expression.
−x2 + 18x + 63 = −(x2 − 18x − 63)
We are looking for an answer that is the product of the two binomials in parentheses: (x ±
)(x ±
).
Now our job is to factor x2 − 18x − 63. Since b is negative and c is negative, one of the blanks is positive and one is
negative.
The number -63 can be written as the product of -1 and 63, 1 and -63, -3 and 21, 3 and -21, -7 and 9, or 7 and -9.
Testing the products and sums, we get:
− 63 = (−1) × 63
and
(−1) + 63 = 62
− 63 = 1 × (−63)
and
1 + (−63) = −62
− 63 = (−3) × 21
and
(−3) + 21 = 18
− 63 = 3 × (−21)
and
3 + (−21) = −18
− 63 = (−7) × 9
and
(−7) + 9 = 2
− 63 = 7 × (−9)
and
7 + (−9) = −2
3 and -21 are the factors that multiply to -63 and sum to -18.The answer is −(x − 21)(x + 3).
Example 2
Factor −x2 + x + 6.
Like in Example 1, first factor the common factor of -1 from each term in the trinomial.
−x2 + x + 6 = −(x2 − x − 6)
146
www.ck12.org
Chapter 1. Unit 1
We are looking for an answer that is the product of the two binomials in parentheses: (x ±
)(x ±
).
Now our job is to factor x2 − x − 6.
The number -6 can be written as the product of the following numbers.
− 6 = (−1) × 6
and
(−1) + 6 = 5
− 6 = 1 × (−6)
and
1 + (−6) = −5
− 6 = (−2) × 3
and
(−2) + 3 = 1
− 6 = 2 × (−3)
and
2 + (−3) = −1
-3 and 2 are the factors that multiply to -6 and add to -1. The answer is −(x − 3)(x + 2).
Review
Factor the following quadratic polynomials.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
x2 + 10x + 9
x2 + 15x + 50
x2 + 10x + 21
x2 + 16x + 48
x2 − 11x + 24
x2 − 13x + 42
x2 − 14x + 33
x2 − 9x + 20
x2 + 5x − 14
x2 + 6x − 27
x2 + 7x − 78
x2 + 4x − 32
x2 − 12x − 45
x2 − 5x − 50
x2 − 3x − 40
x2 − x − 56
−x2 − 2x − 1
−x2 − 5x + 24
−x2 + 18x − 72
−x2 + 25x − 150
x2 + 21x + 108
−x2 + 11x − 30
x2 + 12x − 64
x2 − 17x − 60
Mixed Review
25.
26.
27.
28.
Evaluate f (2) when f (x) = 21 x2 − 6x + 4.
√
Simplify 405.
√
√
3
Graph the following on a number line: −π, 2, 35 , − 10
, 16.
What is the multiplicative inverse of 49 ?
147
1.21. Factorization of Quadratic Expressions
www.ck12.org
Quick Quiz
1.
2.
3.
4.
5.
6.
Name the following polynomial. State its degree and leading coefficient: 6x2 y4 z + 6x6 − 2y5 + 11xyz4 .
Simplify (a2 b2 c + 11abc5 ) + (4abc5 − 3a2 b2 c + 9abc).
A rectangular solid has dimensions (a + 2) by (a + 4) by (3a). Find its volume.
Simplify −3h jk3 (h2 j4 k + 6hk2 ).
Find the solutions to (x − 3)(x + 4)(2x − 1) = 0.
Multiply (a − 9b)(a + 9b).
Review (Answers)
To see the Review answers, open this PDF file and look for section 9.8.
148
www.ck12.org
Chapter 1. Unit 1
1.22 Quadratic Formula
Learning Objectives
Here you’ll learn how to solve quadratic equations by using the quadratic formula.
The quadratic equation x2 + x − 5 = 0 describes the shape of the cut-away of a ditch. You can graph the function to
get a general idea of the shape. However you would want more precise data in order to learn the length of bridge
you will need to cross it. The roots of the function would represent the borders of the ditch. How can you accurately
identify the roots of this quadratic function that does not easily factor? The Quadratic Formula is a useful tool for
identifying accurate roots of quadratic functions that aren’t easy to factor.
Quadratic Formula
There are a number of methods to solve a quadratic equation:
• Graphing to find the zeros
• Using square roots
• Completing the square
Here you will learn a fourth way to solve a quadratic equation: using the quadratic formula.
History of the Quadratic Formula
As early as 1200 BC, people were interested in solving quadratic equations. The Babylonians solved simultaneous
equations involving quadratics. In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit formula
to solve a quadratic equation. The quadratic formula was written as it is today by the Arabic mathematician AlKhwarizmi. It is his name upon which the word “Algebra” is based.
The solution to any quadratic equation in standard form, 0 = ax2 + bx + c, is:
x=
−b ±
√
b2 − 4ac
2a
This formula is called the quadratic formula.
Let’s solve the following equations using the quadratic formula:
1. x2 + 10x + 9 = 0
Applying the quadratic formula and a = 1, b = 10, and c = 9, we get:
149
1.22. Quadratic Formula
www.ck12.org
p
(10)2 − 4(1)(9)
x=
2(1)
√
−10 ± 100 − 36
x=
√2
−10 ± 64
x=
2
−10 ± 8
x=
2
−10 + 8
−10 − 8
x=
or x =
2
2
x = −1 or x = −9
−10 ±
2. −4x2 + x + 1 = 0
Quadratic formula
Substitute a = −4, b = 1, c = 1
Simplify
Separate the two options
Solve
√
b2 − 4ac
x=
2a
p
−1 ± (1)2 − 4(−4)(1)
x=
2(−4)
√
√
−1 ± 1 + 16 −1 ± 17
x=
=
−8
−8
√
√
−1 + 17
−1 − 17
x=
and x =
−8
−8
x ≈ −.39 and x ≈ .64
−b ±
3. 8t 2 + 10t + 3 = 0
Quadratic formula
x=
Plug in the values a = 8, b = 10, c = 3
x=
Simplify
x=
Separate the two options
x=
x=
Solve
150
x=
√
b2 − 4ac
2a
p
−10 ± (10)2 − 4(8)(3)
2(8)
√
√
−10 ± 100 + 96 −10 ± 196
=
16
16 √
√
−10 + 196
−10 − 196
and x =
16
16
−10 + 14
−10 − 14
and x =
16
16
1
3
and x = −
4
2
−b ±
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180240
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180242
Examples
Example 1
Earlier, you were told that the quadratic equation x2 + x − 5 = 0 describes the shape of the cut-away of a ditch. To
find the borders of the ditch, you need to know the roots of the function. How can you accurately identify the roots
of this quadratic function?
The quadratic formula is a useful tool for identifying accurate roots of quadratic equations that aren’t easy to factor.
Quadratic formula
Plug in the values a = 1, b = 1, c = −5
Simplify
Separate the two options
Solve
√
b2 − 4ac
x=
2a
p
−1 ± (1)2 − 4(1)(−5)
x=
2(1)
√
√
−1 ± 1 + 20 −1 ± 21
=
x=
2
2 √
√
−1 + 21
−1 − 21
x=
and x =
2√
√2
1
21
1
21
x=− +
and − −
2
2
2
2
−b ±
Notice that 21 is not a perfect square so we will leave the square root in the results to keep the answer more exact. If
necessary, the approximate solutions are -2.79 and 1.79. The borders of the ditch are at (-2.79, 0) and (1.79, 0).
Example 2
Solve 3k2 + 11k = 4 using the quadratic formula.
151
1.22. Quadratic Formula
www.ck12.org
First, we must make it so one side is equal to zero:
3k2 + 11k = 4 ⇒ 3k2 + 11k − 4 = 0
Now it is in the correct form for using the quadratic formula.
Quadratic formula
x=
Plug in the values a = 3, b = 11, c = −4
x=
Simplify
x=
Separate the two options
x=
x=
Solve
x=
√
b2 − 4ac
2a
p
−11 ± (11)2 − 4(3)(−4)
2(3)
√
√
−11 ± 121 + 48 −11 ± 169
=
6 √
√6
−11 + 169
−11 − 169
and x =
6
6
−11 + 13
−11 − 13
and x =
6
6
1
and x = −4
3
−b ±
Review
1. What is the quadratic formula? When is the most appropriate situation to use this formula?
2. When was the first known solution of a quadratic equation recorded?
Solve the following quadratic equations using the quadratic formula.
3.
4.
5.
6.
7.
8.
9.
10.
x2 + 4x − 21 = 0
x2 − 6x = 12
3x2 − 21 x = 38
2x2 + x − 3 = 0
−x2 − 7x + 12 = 0
−3x2 + 5x = 0
4x2 = 0
x2 + 2x + 6 = 0
Review (Answers)
To see the Review answers, open this PDF file and look for section 10.7.
152
www.ck12.org
Chapter 1. Unit 1
1.23 Polynomial Long Division and Synthetic
Division
Here you will learn how to perform long division with polynomials. You will see how synthetic division abbreviates
this process. In addition to mastering this procedure, you will see the how the remainder root theorem and the
rational root theorem operate. While you may be experienced in factoring, there will always be polynomials that do
not readily factor using basic or advanced techniques. How can you identify the roots of these polynomials?
Rational Roots and Dividing Polynomials
There are numerous theorems that point out relationships between polynomials and their factors. For example
there is a theorem that a polynomial of degree n must have exactly n solutions/factors that may or may not be real
numbers. The Rational Root Theorem and the Remainder Theorem are two theorems that are particularly useful
starting places when manipulating polynomials.
The Rational Root Theorem
The
Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction
Rational
p
x = q , where p is an integer factor of the constant term and q is an integer factor of the leading coefficient.
Let’s identify all the possible rational solutions of the following polynomial using the Rational Root Theorem.
12x18 − 91x17 + x16 + · · · + 2x2 − 14x + 5 = 0
The integer factors of 5 are 1, 5. The integer factors of 12 are 1, 2, 3, 4, 6 and 12. Since pairs of factors could both
be negative, remember to include ±.
1 5 5 5 5 5 5
± qp = 11 , 12 , 13 , 14 , 16 , 12
, 1 , 2 , 3 , 4 , 6 , 12
The possible solutions can be found from these 24 possible rational answers. If this question required you to find a
solution, then the Rational Root Theorem would give you a great starting place. Once you have one root, you can
use either polynomial long division or synthetic to divide the factor out and to keep reducing the expression. We will
use the Rational Root Theorem in Example 3.
Polynomial Long Division and the Remainder Theorem
Polynomial long division is identical to regular long division except the dividend and divisor are both polynomials
instead of numbers.
The Remainder Theorem states that the remainder of a polynomial f (x) divided by a linear divisor (x − a) is equal
to f (a). The Remainder Theorem is only useful after you have performed polynomial long division because you
are usually never given the divisor and the remainder to start. The main purpose of the Remainder Theorem in this
setting is a means of double checking your application of polynomial long division.
Let’s put this knowledge to use and use polynomial long division to divide:
x3 +2x2 −5x+7
x−3
First note that it is clear that 3 is not a root of the polynomial because of the Rational Root Theorem and so there
will definitely be a remainder. Start a polynomial long division question by writing the problem like a long division
problem with regular numbers:
x − 3)x3 + 2x2 − 5x + 7
153
1.23. Polynomial Long Division and Synthetic Division
www.ck12.org
Just like with regular numbers ask yourself “how many times does x go into x3 ?” which in this case is x2 .
x2
3
x − 3)x + 2x2 − 5x + 7
Now multiply the x2 by x − 3 and copy below. Remember to subtract the entire quantity.
x2
3
x − 3)x + 2x2 − 5x + 7
−(x3 − 3x2 )
Combine the rows, bring down the next number and repeat.
x2 +5x+10
3
x − 3)x + 2x2 − 5x + 7
−(x3 − 3x2 )
5x2 − 5x
−(5x2 − 15x)
10x + 7
−(10x − 30)
37
The number 37 is the remainder. There are two things to think about at this point. First, interpret in an equation:
x3 +2x2 −5x+7
x−3
37
= (x2 + 5x + 10) + x−3
Second, check your result with the Remainder Theorem which states that the original function evaluated at 3 must
be 37. Notice the notation indicating to substitute 3 in for x.
(x3 + 2x2 − 5x + 7)|x=3 = 33 + 2 · 32 − 5 · 3 + 7 = 27 + 18 − 15 + 7 = 37
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/60732
Synthetic Division
Synthetic division is an abbreviated version of polynomial long division where only the coefficients are used. Synthetic division is mostly used when the leading coefficients of the numerator and denominator are equal to 1 and the
divisor is a first degree binomial. Let’s use synthetic division to divide the same expression that we divided above
with polynomial long division:
x3 +2x2 −5x+7
x−3
Instead of continually writing and rewriting the x symbols, synthetic division relies on an ordered spacing.
154
www.ck12.org
Chapter 1. Unit 1
+3| 1 2 -5 7
Notice how only the coefficients for the denominator are used and the divisor includes a positive three rather than
a negative three. The first coefficient is brought down and then multiplied by the three to produce the value which
goes beneath the 2.
+3| 1 2 -5 7
↓3
1
Next the new column is added. 2 + 3 = 5, which goes beneath the 2nd column. Now, multiply 5 · +3 = 15, which
goes underneath the -5 in the 3rd column. And the process repeats. . .
+3| 1 2 -5 7
↓ 3 15 30
1 5 10 37
The last number, 37, is the remainder. When writing out the resulting expression, you will put this remainder over
the divisor. The three other numbers represent the quadratic that is identical to the solution to the result from dividing
the expression using polynomial long division. Note that when writing out the expression, you decrease the exponent
of the leading coefficient of the original by 1.
37
(1x2 + 5x + 10) + x−3
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/60734
Examples
Example 1
Earlier, you were asked about identifying roots of polynomials that do not readily factor using the techniques you
have learned so far. Identifying roots of polynomials by hand can be tricky business. The best way to identify
roots is to use the rational root theorem to quickly identify likely candidates for solutions and then use synthetic or
polynomial long division to quickly and effectively test them to see if their remainders are truly zero.
Example 2
Divide the following polynomials.
x3 +2x2 −4x+8
x−2
155
1.23. Polynomial Long Division and Synthetic Division
www.ck12.org
Since the leading coefficients of the numerator and denominator are both 1 and the denominator is a binomial,
synthetic division is a good method to use here.
x3 +2x2 −4x+8
x−2
16
= x2 + 4x + 4 + x−2
Example 3
Completely factor the following polynomial.
x4 + 6x3 + 3x2 − 26x − 24
Notice that possible roots are ±1, 2, 3, 4, 6, 8, 24. Of these 14 possibilities, four will yield a remainder of zero. When
you find one, use long division or synthetic division to factor out the root that you found. Then find another zero and
repeat the process.
x4 + 6x3 + 3x2 − 26x − 24
= (x + 1)(x3 + 5x2 − 2x − 4)
= (x + 1)(x − 2)(x2 + 7x + 12)
= (x + 1)(x − 2)(x + 3)(x + 4)
The first zero found was -1. It was divided out of the original expression to find the remaining non-factored portion
of the expression. The second zero found was 2 from the remaining piece and was divided out. Once you get down
to a quadratic expression, you can use the other factoring techniques you know to factor the rest of the expression.
Example 4
Divide the following polynomials.
3x5 −2x2 +10x−5
x−1
Since the first coefficient of the numerator is not 1, polynomial long division is a good method to use here.
3x5 −2x2 +10x−5
x−1
6
= 3x4 + 3x3 + 3x2 + x + 11 + x−1
Review
Identify all possible rational solutions of the following polynomials using the Rational Root Theorem.
1. 15x14 − 12x13 + x12 + · · · + 2x2 − 5x + 5 = 0
2. 18x11 + 42x10 + x9 + · · · + x2 − 3x + 7 = 0
3. 12x16 + 11x15 + 3x14 + · · · + 6x2 − 2x + 11 = 0
4. 14x7 − 7x6 + x5 + · · · + x2 + 6x + 3 = 0
5. 9x9 − 10x8 + 3x7 + · · · + 4x2 − 2x + 2 = 0
Completely factor the following polynomials.
6. 2x4 − x3 − 21x2 − 26x − 8
7. x4 + 7x3 + 5x2 − 31x − 30
8. x4 + 3x3 − 8x2 − 12x + 16
9. 4x4 + 19x3 − 48x2 − 117x − 54
156
www.ck12.org
Chapter 1. Unit 1
10. 2x4 + 17x3 − 8x2 − 173x + 210
Divide the following polynomials.
11.
x4 +7x3 +5x2 −31x−30
x+4
12.
x4 +7x3 +5x2 −31x−30
x+2
13.
x4 +3x3 −8x2 −12x+16
x+3
14.
2x4 −x3 −21x2 −26x−8
x3 −x2 −10x−8
15.
x4 +8x3 +3x2 −32x−28
x3 +10x2 +23x+14
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.5.
157
1.24. Real Zeros of Polynomials
www.ck12.org
1.24 Real Zeros of Polynomials
Learning Objectives
Here you will learn about using polynomial long division and synthetic division to find roots, and how to apply
synthetic division to find the rational zeros of an unfactored polynomial (of degree >2).
In the real world, problems do not always easily fit into quadratic or even cubic equations. Financial models,
population models, fluid activity, etc., all often require many degrees of the input variable in order to approximate
the overall behavior. While it can be challenging to model some of these more complex interactions, the effort can
be well worth it. Mathematical models of stocks are used constantly as a way to "look into the future" of finance and
make the kinds of educated guesses that are behind some of the largest fortunes in the world.
What benefits can you think of to modeling the behavior of large populations? Can you think of other useful
applications not mentioned here?
Finding Real Zeros of Polynomials
There are three theorems and a rule that we will be referring to during this lesson in order to help make the discovery
of the roots of polynomial functions easier. You should review them and be prepared to refer to them often during
the practice problems.
The Remainder Theorem
If a polynomial f (x) of degree n > 0 is divided by x − c, then the remainder R is a constant and it is equal to the
value of the polynomial when c is substituted for x. That is
f (c) = R
The Factor Theorem
If f (x) is a polynomial of degree n > 0 and f (c) = 0, then x − c is a factor of the polynomial f (x). Further, if x − c
is a factor, then c is a zero of f .
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187406
158
www.ck12.org
Chapter 1. Unit 1
The Rational Zero Theorem
Given the polynomial
f (x) = an xn + an−1 xn−1 + · · · + a1 x + a0
an 6= 0 and n is a positive integer. If the coefficients are integers and
divisor of a0 and q is a divisor of an .
p
q
is a rational zero in lowest terms, then p is a
Descartes’ Rule of Signs
Given any polynomial, p(x),
1.
2.
3.
4.
5.
Write it with the terms in descending order, i.e. from the highest degree term to the lowest degree term.
Count the number of sign changes of the terms in p(x). Call the number of sign changes n.
Then the number of positive roots of p(x) is less than or equal to n.
Further, the possible number of positive roots is n, n − 2, n − 4, . . .
To find the number of negative roots of p(x), write p(−x) in descending order as above (i.e. change the sign
of all terms in p(x) with odd powers), and repeat the process above. Then the maximum number of negative
roots is n.
Examples
Example 1
Earlier, you were asked if you could identify some valuable real-world uses for modeling higher-degree polynomials.
Here are a few possibilities:
• Identifying what times of the day people are most likely to want coffee (market research like this is key to
running your own business)
• Predicting population growth in a particular neighborhood or area of town (useful for identifying a good
location to start a small business)
• Identifying which stocks are likely to rise or fall based on weather or season
• Forecasting the weather
• Calculating the right head start to give a slower car to make a drag race exciting
There are many, many more.
Example 2
Use synthetic division and the remainder and factor theorems to find the quotient Q(x) and the remainder R if
f (x) = 2x3 − 3x2 + 6 is divided by x − 5.
5 )2 − 3 0
6
↓ 10 35 175
2 7 35 181
159
1.24. Real Zeros of Polynomials
www.ck12.org
Hence
2x3 − 3x2 + 6 = (2x2 + 7x + 35)(x − 5) + 181
Notice that the remainder is 181 and it can also be obtained if we simply substituted x = 5 into f (x),
f (5) = 2(5)3 − 3(5)2 + 6
= 250 − 75 + 6
= 181
Example 3
Use the rational zero theorem and synthetic division to find all the possible rational zeros of the polynomial
f (x) = x3 − 2x2 − x + 2
From the rational zero theorem, qp is a rational zero of the polynomial f . So p is a divisor of 2 and q is a divisor of
1. Hence, p can take the following values: -1, 1, -2, 2 and q can be either -1 or 1. Therefore, the possible values of
p
q are
p
: −1, 1, −2, 2
q
So there are four possible zeros. Of these four, not more than three can be zeros of f because f is a polynomial with
degree 3. To test which of the four possible candidates are zeros of f , we use the synthetic division. Recall from the
remainder theorem if f (c) = 0, then c is a zero of f . We have
2 )1 − 2 − 1
↓ 2
1
2
0 −2
0 −1
0
Hence, 2 is a zero of f . Further, by the division algorithm,
f (x) = (x − c)Q(x) + R(x)
= (x − 2)(x2 − 1) + 0
The remaining zeros of f are simply the zeros of Q(x) = x2 − 1 which is easier to manipulate,
Q(x) = x2 − 1
= (x − 1)(x + 1)
and thus the remaining zeros are -1 and 1. Thus the rational zeros of f are -1, 1, and 2.
160
www.ck12.org
Chapter 1. Unit 1
Example 4
Graph the polynomial function h(x) = 2x3 − 9x2 + 12x − 5.
Notice that the leading term is 2x3 , where n = 3 odd and an = 2 > 0. This tells us that the end behavior will take the
shape of a power function with an odd exponent.
Here, as you can see, there is no straight-forward way to find the zeros of h(x). However, with the use of the factor
theorem and the synthetic division, we can find the rational roots of h(x).
First, we use the rational zero theorem and find that the possible rational zeros are
p
5 5
: −1, 1, −2, 2, −5, 5, − ,
q
2 2
testing all these numbers by the synthetic division,
− 1 )2 − 9 12 − 5
↓ −2 11 − 23
2 − 11 23 − 28
-1 is not a root. Now let’s test x = 1.
1 )2 − 9 12 − 5
↓ −2 − 7
5
2 − 7 5 − 28
we find that 1 is a zero of h and so we can re-write h(x),
h(x) = (2x2 − 7x + 5)(x − 1)
Looking at quadratic part,
2x2 − 7x + 5 = (2x − 5)(x − 1)
and so
h(x) = (2x − 5)(x − 1)2
Thus 1 and
5
2
are the x−intercepts of h(x). The y−intercept is
h(0) = −5
Further, the synthetic division can be also used to form a table of values for the graph of h(x):
161
1.24. Real Zeros of Polynomials
www.ck12.org
x
−1
0
1
2
h(x)
− 28
−5
0
−1
5
2
0
3
4
We choose test points from each interval and find g(x).
TABLE 1.18:
Interval
Test Value x
h(x)
Sign of h(x)
(−∞,1)
1, 25 5
2,∞
-1
-28
3
2
−11
4
3
4
+
Location of points
on the graph
below the x−axis
below the x−axis
above the x−axis
From this information, the graph of h(x) is shown in the two graphs below. Notice that the second graph is a
magnification of h(x) in the vicinity of the x−axis.
162
www.ck12.org
Chapter 1. Unit 1
Example 5
Use the ’rational zero’ theorem and synthetic division to find all the possible rational zeros of the polynomial
f (x) = x3 − 2x2 − 5x + 6
Assume qp is a rational zero of f . By the rational zero theorem, p is a divisor of 6 and q is a divisor of 1. Thus p and
q can assume the following respective values
p : 1, −1, 2, −2, 3, −3, 6, −6
and
q : −1, 1
Therefore, the possible rational zeros will be
p
: −1, 1, −2, 2, −3, 3, −6, 6
q
Notice that with these choices for p and q there could be 8 · 2 = 16 rational zeros. But, eight of them are duplicates.
1
For example −1
= −1
1 = −1. The next step is to test all these values by the synthetic division (we’ll let you do this
on your own for practice) and we finally find that
1, −2, and 3
are zeros of f . That is
f (x) = x3 − 2x2 − 5x + 6
= (x − 1)(x + 2)(x − 3)
163
1.24. Real Zeros of Polynomials
www.ck12.org
Example 6
Use Descartes’ Rule of Signs to identify the possible number of positive and negative roots of
f (x) = −2x3 + x2 − 3x5 + 5x − 1.
First, re-write f (x) in descending order
f (x) = −3x5 − 2x3 + x2 + 5x − 1.
The number of sign changes of f (x) is 2, so the number of positive roots is either 2 or 0.
For the negative roots, write
f (−x) = −3x5 + 2x3 + x2 − 5x − 1
The number of sign changes of f (−x) is 2, so the maximum number of negative roots is 2.
The graph of f (x) below shows that there is one negative root and two positive roots.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187376
164
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187407
Review
Questions 1 - 3: Use a) long division and b) synthetic division to perform the divisions. Express each result in the
"corresponding statement" form: f (x) = D(x) · Q(x) + R.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
5x5 − 3x4 + 2x3 + x2 − 7x + 3 by x − 2
−4x6 − 5x3 + 3x2 + x + 7 by x − 1
2x3 − 5x2 + 5x + 11 by x − 12
Use synthetic division to find Q(x) and f (c) so that f (x) = (x − c)Q(x) + f (c) if f (x) = −3x4 − 3x3 + 3x2 +
2x − 4 and c = −2
If f (x) = x3 + 2x2 − 10x + 10, determine the following: a) f (−1) b) f (−3) c) f (0) d) f (4) e) What are the
factors of f (x)?
Find k so that x − 2 is a factor of f (x) = 3x3 + 4x2 + kx − 20
Determine all the zeros of the polynomials: a) f (x) = 3x3 − 7x2 + 8x − 2 b) g(x) = 4x4 − 4x3 − 7x2 + 4x + 3
Graph the polynomial function f (x) = x3 − 2x2 − 5x + 6 by using synthetic division to find the x−intercepts
and locate the y−intercept.
Graph the polynomial function h(x) = x3 − 3x2 + 4 by to find the x−intercepts and locate the y−intercepts.
Write a 3rd degree equation of a polynomial function with the zeroes: 0, 2, and -5.
Write a 7th degree equation of a polynomial function with the zeroes: 0 (order 2), 2 (order 3), and -5 (order
2).
Write a quadratic equation which has 4 (order 2) as the zero and opens downward.
Write a 3rd degree polynomial function with the zeroes: -2, 2, and 6, passing through the point (3, 4)
Let f (x) = 2x3 − 5x2 − 4x + 3 and find the solutions: a) f (x) = 0 b) f (2x) = 0
Graph and find the solution set of the inequality x3 − 2x2 − 5x + 6 ≤ 0.
Use the graph of f (x) = x(x − 1)(x + 2) to find the solution set of the inequality x(x − 1)(x + 2) > 0.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.12.
165
1.25. Relationship between Zeros and Coefficients of a Polynomial
www.ck12.org
1.25 Relationship between Zeros and
Coefficients of a Polynomial
Roots to Determine a Quadratic Function
Example A
Find a quadratic function with the roots: 3 ±
√
5.
Solution:
The solutions to the quadratic equation are:
√
√
x = 3 + 5 and x = 3 − 5
√
√
The factors are (x − (3 + 5)) and (x − (3 − 5)).
One possible function in factored form is:
√ √ y = x−3− 5 x−3+ 5
Multiply and simplify:
√
√
√
√
√
y = x2 − 3x + √5x − 3x + 9 − 3 √5 − √5x + 3 √5 − 25
y = x2 − 3x + 5x − 3x + 9 − 3 5 − 5x + 3 5 − 5
y = x2 − 3x − 3x + 9 − 5
y = x2 − 6x + 4
Keep in mind that any multiple of the right side of the above function would also have the given roots.
Relationship between Zeros of Polynomial and Coefficients
a) For Quadratic Polynomial
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/165519
Watch the video at: https://www.youtube.com/watch?v=dr100S63SmI
b) For Cubic Polynomial
166
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/165521
Watch the video at: https://www.youtube.com/watch?v=MtYhFXo3tU8
Videos in context with problems based on the relationship between zeros and coefficients of cubic polynomials.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/165512
Watch the video at: https://www.youtube.com/watch?v=S15ITrMWTvE
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/165514
Watch the video at: https://www.youtube.com/watch?v=XLr_DbaOohY
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/165517
Watch the video at: https://www.youtube.com/watch?v=6nNjY3XF8Jk
167
1.26. Factorization of Special Cubics
www.ck12.org
1.26 Factorization of Special Cubics
Learning Objectives
Here you’ll learn to factor the sum and difference of perfect cubes.
Factor the following cubic polynomial: 375x3 + 648.
Factorization of Special Cubics
While many cubics cannot easily be factored, there are two special cases that can be factored quickly. These special
cases are the sum of perfect cubes and the difference of perfect cubes.
• Factoring the sum of two cubes follows this pattern:
x3 + y3 = (x + y)(x2 − xy + y2 )
• Factoring the difference of two cubes follows this pattern:
x3 − y3 = (x − y)(x2 + xy + y2 )
The acronym SOAP can be used to help you remember the positive and negative signs when factoring the sum and
difference of cubes. SOAP stands for "Same", "Opposite", "Always Positive". "Same" refers to the first sign in the
factored form of the cubic being the same as the sign in the original cubic. "Opposite" refers to the second sign in
the factored cubic being the opposite of the sign in the original cubic. "Always Positive" refers to the last sign in the
factored form of the cubic being always positive. See below:
Let’s factor the following cubics:
1. x3 + 27
This is the sum of two cubes and uses the factoring pattern: x3 + y3 = (x + y)(x2 − xy + y2 ).
x3 + 33 = (x + 3)(x2 − 3x + 9).
2. x3 − 343
168
www.ck12.org
Chapter 1. Unit 1
This is the difference of two cubes and uses the factoring pattern: x3 − y3 = (x − y)(x2 + xy + y2 ).
x3 − 73 = (x − 7)(x2 + 7x + 49).
3. 64x3 − 1
This is the difference of two cubes and uses the factoring pattern: x3 − y3 = (x − y)(x2 + xy + y2 ).
(4x)3 − 13 = (4x − 1)(16x2 + 4x + 1).
Examples
Example 1
Earlier, you were asked to factor the following cubic polynomial: 375x3 + 648.
First you need to recognize that there is a common factor of 3. 375x3 + 648 = 3(125x3 + 216)
Notice that the result is the sum of two cubes. Therefore, the factoring pattern is x3 + y3 = (x + y)(x2 − xy + y2 ).
375x3 + 648 = 3(5x + 6)(25x2 − 30x + 36)
Example 2
Factor the following cubic.
x3 + 512
x3 + 83 = (x + 8)(x2 − 8x + 64).
Example 3
Factor the following cubic.
8x3 + 125
(2x)3 + 53 = (2x + 5)(4x2 − 10x + 25).
Example 4
x3 − 216
x3 − 63 = (x − 6)(x2 + 6x + 36).
Review
Factor each of the following cubics.
1.
2.
3.
4.
5.
6.
x3 + h3
a3 + 125
8x3 + 64
x3 + 1728
2x3 + 6750
h3 − 64
169
1.26. Factorization of Special Cubics
7.
8.
9.
10.
11.
12.
13.
14.
15.
s3 − 216
p3 − 512
4e3 − 32
2w3 − 250
x3 + 8
y3 − 1
125e3 − 8
64a3 + 2197
54z3 + 3456
Review (Answers)
To see the Review answers, open this PDF file and look for section 7.11.
170
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
1.27 Zero Product Principle
Learning Objectives
Here you’ll find the solutions to polynomial equations by using the Zero Product Principle.
Suppose that you know that the area of a chalkboard is 48 square feet, and you also know that the area in square feet
can be represented by the expression x2 + 2x. How can you find the value of x if you wrote an equation, moved all
the terms to one side, and factored?
Zero Product Principle
Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and
differences of different terms:
6x4 + 7x3 − 26x2 − 17x + 30
Notice that the degree of the polynomial is four.
The factored form of a polynomial means it is written as a product of its factors.
The factors are also polynomials, usually of lower degree. Here is the same polynomial in factored form.
(x − 1)(x + 2)(2x − 3)(3x + 5)
Suppose we want to know where the polynomial 6x4 + 7x3 − 26x2 − 17x + 30 equals zero. It is quite difficult to solve
this using the methods we already know. However, we can use the Zero Product Property to help.
The Zero Product Property states that the only way a product is zero is if one or both of the terms are zero.
By setting the factored form of the polynomial equal to zero and using this property, we can easily solve the original
polynomial.
Let’s solve for x in the following equations:
1. 6x4 + 7x3 − 26x2 − 17x + 30
As shown earlier, the factored version of this equation is:
(x − 1)(x + 2)(2x − 3)(3x + 5) = 0
According to the property, for the original polynomial to equal zero, we have to set each factor equal to zero
and solve.
171
1.27. Zero Product Principle
www.ck12.org
(x − 1) = 0 → x = 1
(x + 2) = 0 → x = −2
3
(2x − 3) = 0 → x =
2
5
(3x + 5) = 0 → x = −
3
The solutions to 6x4 + 7x3 − 26x2 − 17x + 30 = 0 are the following: x = −2, − 53 , 1, 32 .
2. (x − 9)(3x + 4) = 0
Separate the factors using the Zero Product Property: (x − 9)(3x + 4) = 0.
x−9 = 0
or
3x + 4 = 0
3x = −4
−4
x=
3
x=9
Solving Simple Polynomial Equations by Factoring
We already saw how we can use the Zero Product Property to solve polynomials in factored form. Here you will
learn how to solve polynomials in expanded form. These are the steps for this process.
Step 1: Rewrite the equation in standard form such that: Polynomial expression = 0.
Step 2: Factor the polynomial completely.
Step 3: Use the zero-product rule to set each factor equal to zero.
Step 4: Solve each equation from step 3.
Step 5: Check your answers by substituting your solutions into the original equation.
Let’s solve the following polynomial equation:
10x3 − 5x2 = 0.
First, factor by removing the greatest common factor. Since both terms have at least 5x2 as a factor, we will remove
that.
10x3 − 5x2 = 0
5x2 (2x − 1) = 0
Separate each factor and set equal to zero:
5x2 = 0
x2 = 0
x=0
172
or
2x − 1 = 0
2x = 1
1
x=
2
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180120
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180122
Examples
Example 1
Earlier, you were told that the area of a chalkboard is 48 square feet and that the area in square feet could be
represented by the expression x2 + 2x. What is the value of x?
The starting equation is:
x2 + 2x = 48
To find the value of x, we need to follow the steps for solving an equation in expanded form.
Rewrite:
x2 + 2x = 48
x2 + 2x − 48 = 0
Factor:
x2 + 2x − 48 = 0
(x + 8)(x − 6) = 0
Note that factoring this expression requires a technique that has not yet been covered. Similar to the first problem in
this Concept, the factored form is given to you.
Set each factor equal to zero and solve:
173
1.27. Zero Product Principle
www.ck12.org
(x + 8) = 0 → x = −8
(x − 6) = 0 → x = 6
Check:
x = −8
(−8)2 + 2(−8) = 64 − 16 = 48
x=6
(6)2 + 2(6) = 36 + 12 = 48
The check shows us that the answers are correct. x is either -8 or 6.
Example 2
Solve the following polynomial equation.
x2 − 2x = 0
x2 − 2x = 0
Rewrite: This is not necessary since the equation is in the correct form.
Factor: The common factor is x, so this factors as: x(x − 2) = 0.
Set each factor equal to zero.
x=0
or
x−2 = 0
Solve:
x=0
x=2
or
Check: Substitute each solution back into the original equation.
x=0
(0)2 − 2(0) = 0
x=2
(2)2 − 2(2) = 0
The answer is x = 0, x = 2.
Review
1. What is the Zero Product Property? How does this simplify solving complex polynomials?
In 2-6, explain why the Zero Product Property can’t be used to solve the polynomial.
174
www.ck12.org
2.
3.
4.
5.
6.
Chapter 1. Unit 1
(x − 2)(x) = 2
(x + 6) + (3x − 1) = 0
(x−3 )(x + 7) = 0
(x + 9) − (6x − 1) = 4
(x4 )(x2 − 1) = 0
In 7-16, solve the polynomial equations.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
x(x + 12) = 0
(2x + 3)(5x − 4) = 0
(2x + 1)(2x − 1) = 0
24x2 − 4x = 0
60m = 45m2
(x − 5)(2x + 7)(3x − 4) = 0
2x(x + 9)(7x − 20) = 0
18y − 3y2 = 0
9x2 = 27x
4a2 + a = 0
Review (Answers)
To see the Review answers, open this PDF file and look for section 9.7.
175
1.28. Polynomial and Rational Inequalities
www.ck12.org
1.28 Polynomial and Rational Inequalities
Learning Objectives
Here you will learn about the properties of polynomial and rational inequalities, and will apply your understanding
of polynomial and rational inequalities to draw graphs.
Often it is easier to use and remember new terms when you have a ’hook’ or comparison to a term you know already.
"Polynomial inequality" is a term generally used to refer to inequalities where the x variable has a degree of 3 or
greater.
The prefix "poly" means ’multiple’ or ’many’, and the root word "nomial" means ’name’ or ’term’.
Therefore a "polynomial" is literally: "many terms".
The prefix "in" means ’not’, and the root word "equal" of course means ’the same’.
Therefore and "inequality" refers to things which are "not same" or "not equal".
Can you use this logic to identify the origin of some of the other terms used in this lesson?
Polynomial and Rational Inequalities
Polynomial Inequalities
Solving polynomial inequalities is very similar to solving quadratic inequalities. The basic steps are the same:
1.
2.
3.
4.
Set up the inequality in the form p(x) > 0 (or p(x) < 0, p(x) ≤ 0, p(x) ≥ 0)
Find the solutions to the equation p(x) = 0.
Divide the number line into intervals based on the solutions to p(x) = 0.
Use test points to find solution sets to the equation.
Rational Inequalities
There is one step added to the process of solving rational inequalities because a rational function can also change
signs at its vertical asymptotes or at a break in the graph. For instance, look at the graph of the function r(x) = x2x−9
below.
176
www.ck12.org
Chapter 1. Unit 1
If we want to solve the inequality x2x−9 > 0, then we need to use the following critical points: x = 0, x = 3, and
x = −3. x = 0 is the solution of setting the numerator equal to 0, and this gives us the only root of the function.
x = ±3 are the vertical asymptotes, the x−coordinates that make the function undefined because putting in 3 or -3
for x will cause a division by zero.
Testing the intervals between each critical point to see if the values in that interval satisfy the function gives us:
TABLE 1.19:
Interval
(−∞, −3)
(-3, 0)
(0, 3)
(3, +∞)
Thus, the solutions to
Test Point
-4
-2
2
4
x
x2 −9
Positive/Negative?
+
+
Part of Solution Set?
no
yes
no
yes
> 0 are x ∈ (−3, 0) ∪ (3, +∞).
Examples
Example 1
Earlier, you were given a question about identifying the origins of other terms in this lesson.
How many of the terms we have used recently were you able to track down the origins of? A few are listed below,
did you find others?
Bi-nomial: "two-named" or "two-terms" - from "bi", meaning ’two’ and "nomial", meaning ’name’ or ’term’. Quadratic: "related to a square" - from "quadratus", meaning ’square’. Ratio-nal: "related to a ratio" - from "ratio",
meaning ’reason’ (as in "to reason" or "calculate") and "-al", meaning "related to".
177
1.28. Polynomial and Rational Inequalities
www.ck12.org
Example 2
Solve x3 − 3x2 + 2x ≥ 0.
The polynomial is already in the correct form p(x) ≥ 0 so we solve the equation
x3 − 3x2 + 2x = 0
x(x2 − 3x + 2) = 0
x(x − 2)(x − 1) = 0
The zeros are at x = 0, x = 1, and x = 2.
TABLE 1.20:
Interval
(−∞, 0)
(0, 1)
(1, 2)
(2, +∞)
Test Point
-5
Positive/Negative?
+
+
1
2
3
2
3
Part of Solution Set?
no
yes
no
yes
Notice that this inequality is greater than or equal to zero, so we include the zeros in the solution set. Therefore the
solutions are x ∈ [0, 1] ∪ [2, +∞].
Example 3
Solve 6x4 + 5x2 < 25.
First we will change the inequality to 6x4 + 5x2 − 25 < 0. Now, solve the equation 6x4 + 5x2 − 25 = 0.
6x4 + 5x2 − 25 = 0
(3x2 − 5)(2x2 + 5) = 0
r
The first term yields the solutions x = ±
5
and there are no real solutions for the second term.
3
TABLE 1.21:
Interval
r !
5
−∞, −
3
r r !
5
5
−
,
3
3
!
r
5
, +∞
3
Test Point
Positive/Negative?
Part of Solution Set?
-3
+
no
0
-
yes
3
+
no
r
Finally, the solution set is x ∈
178
−
5
,
3
r !
5
.
3
www.ck12.org
Chapter 1. Unit 1
Example 4
Find the solution set of the inequality
4x − 12
<0
3x − 2
From the numerator we solve 4x − 12 = 0 or x = 3. In the denominator, solve 3x − 2 = 0 and we find the critical
point x = 23 .
Making the table
TABLE 1.22:
Interval
−∞, 23
2
3,3
(3, +∞)
Test Point
0
1
5
Positive/Negative?
+
+
Therefore,
the solution set includes the numbers in the interval
2
x| 3 < x < 3 .
Part of Solution Set?
no
yes
no
2
3,3
. Or in set-builder notation, the solution is
Example 5
(using technology) The McNeil Surf Company makes wetsuits. For a given number of wetsuits x, McNeil’s profit,
in dollars, is given by the function P(x) = −0.01x2 + 25x − 3000.
a. If the manager of McNeil wants the profit to stay above $9,000, what is the minimum and maximum number
of wetsuits they can manufacture to maintain that level of profit?
Set up the inequality
−0.01x2 + 25x − 3000 > 9000
−0.01x2 + 25x − 12000 > 0
With a calculator you can graph the function Y1 = −0.01x2 + 25x − 12000.
On a TI-83: use the window [−1000, 4000] × [−5000, 15000].
The settings are:
Xmin = −1000, Xmax = 4000, Xscl = 500
Y min = −5000,Y max = 5000,Y scl = 1000 xres = 1
On a software grapher, the image looks like this with the window x : 0 → 2400 and y : 0 → 3500
179
1.28. Polynomial and Rational Inequalities
www.ck12.org
Using the CALC menu (2ND TRACE), and selecting the option ZEROS, we can see that the zeros of Y1 =
−0.01x2 + 25x − 12000 are at x = 647.920 and x = 1852.080.
By inspecting the graph, we can see that the solution set to the inequality −0.01x2 + 25x − 12000 > 0 is x ∈
(648, 1852).
Visually that looks like:
b. What is the maximum profit McNeil can make?
180
www.ck12.org
Chapter 1. Unit 1
Keeping the same graph open, use CALC MAXIMUM to solve for the maximum profit. The maximum is at (1250,
3625), indicating that the maximum profit is $3625 above the minimum we set: of $9000.
So the actual maximum profit is $12,625 when 1250 wetsuits are produced.
c. Can you explain why this shape might make sense for a profit function?
One possible reason the profit function might take this shape is labor costs. If McNeil wants to make a very large
number of wetsuits in a short period of time, then that may require paying overtime for workers, and this could
reduce the profit margin.
Example 6
For the following rational function, determine limitations on the domain and the asymptotes, and then sketch the
graph.
f (x) ≥
2x+5
x−1
To identify the graph of the inequality f (x) ≥
For f (x) ≥
2x+5
x−1
2x+5
x−1 ,
first treat it as if it were the equality f (x) ≥
2x+5
x−1
:
To find the critical points, identify the value(s) which make the denominator = 0: x 6= 1
That gives us a vertical asymptote of x = 1
The horizontal asymptote becomes apparent as x becomes truly huge and the "+5" and "-1" no longer matter. At that
point, we have f (x) = 2x
x → f (x) = 2 So the horizontal asymptote is y = 2
Now that you know the shape of the graph, simply shade the area above the lines, since the original function was
f (x) is greater-than function, and leave the lines solid since it was a greater-than or equal to.
The final graph should look like:
181
1.28. Polynomial and Rational Inequalities
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187409
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187410
Review
Find the solution set of the following inequalities without using a calculator. Display the solution set on a number
line.
1. x2 + 2x − 3 ≤ 0
2. 3x2 − 7x + 2 > 0
3. −6x2 − 13x + 5 ≥ 0
4. 5x−1
x−2 > 0
5. 1−x
x <1
6. Solve: 4x3 − 4x2 − 3x > 0
4
7. Solve: x4 − x2 < 0
8. Solve: 4x3 − 8x2 − x + 2 ≥ 0
3 −2n2 −n+2
9. nn3 +3n
2 +4n+12 < 0
10.
182
n3 +3n2 −4n−12
n3 −5n2 +4n−20
≤0
www.ck12.org
11.
12.
Chapter 1. Unit 1
2n3 +5n2 −18n−45
≥0
3n3 −n2 +27n−9
12n3 +16n2 −3n−4
>0
8n3 +12n2 +10n+15
Use a calculator to solve the following inequalities. Round your answer to three places after the decimal.
13. −9.8t 2 + 357.6t ≥ 0
14. x3 − 5x + 7 ≤ −4x2 + 18
2 −2x
15. xx−5
> x2 − 25
2
−4
16. Solve and graph: f (x) > 9x
3x+2
R2
17. The total resistance of two electronics components wired in parallel is given by RR11+R
where R1 and R2 are
2
the individual resistances (in Ohms) of the two components. a) If the resistance of R1 is 20 Ohms, what is
the maximum resistance of R2 if the total resistance must be less than 15 Ohms? b) What is the maximum
theoretical resistance of this circuit? How do you know?
18. A rectangular lot of land has a length that is 7 meters more than twice its width. If the area of the lot is greater
than 60 square meters, what are the possible values of the widths of the lot?
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.10.
183
1.29. Domain and Range of a Function
www.ck12.org
1.29 Domain and Range of a Function
Learning Objectives
Here you’ll build on your previous knowledge of functions by learning how to create a table of values for a function
and how to identify its domain and range.
Suppose you have a function that allows you to input the number of years you have until retirement and which
outputs the amount of money you should have saved. How would you go about determining the domain of such a
function? How would you decide on the range?
The Domain and Range of a Function
Using Tables to Represent Functions
A function really is an equation. Therefore, a table of values can be created by choosing values to represent the
independent variable. The answers to each substitution represent f (x).
Recall from the previous concept that Joseph and his friends decided to go to a theme park where each ride costs
$2.00. Joseph can represent the cost of r rides at a theme park by the function J(r) = 2r.
Let’s use Joseph’s function to generate a table of values. Because the variable represents the number of rides Joseph
will pay for, negative values do not make sense and are not included in the list of values of the independent variable.
TABLE 1.23:
R
0
1
2
3
184
J(r) = 2r
2(0) = 0
2(1) = 2
2(2) = 4
2(3) = 6
www.ck12.org
Chapter 1. Unit 1
TABLE 1.23: (continued)
J(r) = 2r
2(4) = 8
2(5) = 10
2(6) = 12
R
4
5
6
As you can see, the list cannot include every possibility. A table allows for precise organization of data. It also
provides an easy reference for looking up data and offers a set of coordinate points that can be plotted to create a
graphical representation of the function. A table does have limitations; namely it cannot represent infinite amounts
of data and it does not always show the possibility of fractional values for the independent variable.
Domain and Range
The set of all possible input values for the independent variable is called the domain. The domain can be expressed
in words, as a set, or as an inequality. The values resulting from the substitution of the domain represent the range
of a function.
Let’s once again take a look at Joseph’s function about the cost of the rides at the theme park.
The domain of the function representing Joseph’s situation will not include negative numbers because it does not
make sense to ride negative rides. He also cannot ride a fraction of a ride, so decimals and fractional values do not
make sense as input values. Therefore, the values of the independent variable r will be whole numbers beginning at
zero.
Domain: All whole numbers
The values resulting from the substitution of whole numbers are whole numbers times two. Therefore, the range of
the function representing Joseph’s situation is still whole numbers, just twice as large.
Range: All even whole numbers
Let’s find the domain and range of the following situations:
1. A tennis ball is bounced from a height and bounces back to 75% of its previous height.
The function representing this situation is h(b) = 0.75b, where b represents the previous bounce height.
Domain: The previous bounce height can be any positive number, so b ≥ 0.
Range: The new height is 75% of the previous height, and therefore will also be any positive number (decimal
or whole number), so the range is all positive real numbers.
2. f (x) = 2x − 3 when the domain is 0, 1, 2, 3.
Domain: The domain is given to be 0, 1, 2, 3
Since the range is the output, we plug in the values in the domain to see what values we will get out.
f (0) = 2(0) − 3 = −3
f (1) = 2(1) − 3 = −1
f (2) = 2(2) − 3 = 1
f (3) = 2(3) − 3 = 3
Range: −3, −1, 1, 3.
185
1.29. Domain and Range of a Function
www.ck12.org
Notice that we used function notation to keep track of which input value gave us which output value. This
will be useful later.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/56125
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79599
Examples
Example 1
Earlier, you were asked to pretend that you had a function that allows you to input the number of years you have until
retirement and which outputs the amount of money you should have saved. How would you go about determining
the domain of such a function? How would you decide on the range?
Since input is the number of years until retirement, the domain should not include negative numbers and the numbers
should be whole numbers. Theoretically, the domain is all positive whole numbers but due to the lifespan of humans,
the domain is more likely to be all positive whole numbers below 100. The range will all positive numbers since the
amount you have saved should be positive and there can be decimals when talking about money. Depending on how
much you make, the amount you have saved can be large.
For Examples 2-4, use the following information: Eli makes $20 an hour tutoring math.
Example 2
Write a function expressing the amount of money Eli earns.
Let M(h) represent money earned for h hours. Then the function is M(h) = 20h.
Example 3
What are the domain and range of the function from Example 2?
Since hours worked can only be zero or positive, h ≥ 0 is the domain. If Eli works for zero hours, she will earn zero
dollars. She could also earn any positive amount of money, so the range is also all non-negative real numbers. That
is, M ≥ 0.
186
www.ck12.org
Chapter 1. Unit 1
Example 4
Suppose Eli will only work for either 1, 1.5 or 2 hours. Express this domain and the corresponding range in a table.
First we plug the domain into our function:
M(1) = 20(1) = 20
M(1.5) = 20(1.5) = 30
M(2) = 20(2) = 40.
Putting this into a table, we get:
TABLE 1.24:
h
1
1.5
2
M(h)
20
30
40
Review
1. Define domain.
2. True or false? Range is the set of all possible inputs for the independent variable.
3. Generate a table from −5 ≤ x ≤ 5 for f (x) = −(x)2 − 2.
In 4-8, identify the domain and range of the function.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Dustin charges $10 per hour for mowing lawns.
Maria charges $25 per hour for math tutoring, with a minimum charge of $15.
f (x) = 15x − 12
f (x) = 2x2 + 5
f (x) = 1x
Make up a situation in which the domain is all real numbers but the range is all whole numbers.
What is the range of the function y = x2 − 5 when the domain is −2, −1, 0, 1, 2?
What is the range of the function y = 2x − 43 when the domain is −2.5, 1.5, 5?
Angie makes $6.50 per hour working as a cashier at the grocery store. Make a table of values that shows her
earnings for the input values 5, 10, 15, 20, 25, 30.
13. The area of a triangle is given by: A = 12 bh. If the base of the triangle is 8 centimeters, make a table of values
that shows the area of the triangle for heights 1,√2, 3, 4, 5, and 6 centimeters.
14. Make a table of values for the function f (x) = 2x + 3 for the input values −1, 0, 1, 2, 3, 4, 5.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.11.
187
1.30. Slope of a Line Using Two Points
www.ck12.org
1.30 Slope of a Line Using Two Points
Learning Objectives
In this concept, you will learn to find the slope of a line.
Hadiya has a very active dog that she would like to train for agility. Hadiya’s dad is helping her make a ramp off
their back deck so Hadiya and Max can practice. Measuring from the ground at the base of the deck, the ramp will
be placed out 4 feet, the horizontal distance, and reach the 7-foot-high deck, the vertical distance. Using the base of
the deck as point (0,0), solve for the slope of the agility ramp.
In this concept, you will learn to solve for the slope of a line using two points on a line, or coordinates.
Solving for Slope of a Line Using Coordinates
Coordinates are values on a line that show a point’s exact position. They are indicated by parentheses, (x,y). The
point (7,-4) means that the x value of a particular point is 7 units to the right of point (0,0) and the y value of the
same point is 4 units below point (0,0).
If the graph is given, the slope of a line can be found by choosing two points on the line and counting the units
∆y
.
vertically and horizontally to find the change in y over the change in x, ∆x
∆y
∆x
=
y2 −y1
x2 −x1
y2 − y1 represents the vertical change, or difference, between two points on a line.
x2 − x1 represents the horizontal change, or difference, between two points on a line.
Since slope is a ratio, improper fractions are not to be converted to mixed numbers.
188
www.ck12.org
Chapter 1. Unit 1
Examples
Example 1
Earlier, you were given a problem about Hadiya and her agile dog, Max.
Hadiya’s dad is making an agility ramp that rises 7 feet and stretches out 4 feet from a point (0,0) at the base of the
deck in their backyard. What is the slope of the ramp?
First, write the equation.
slope =
∆y
∆x
Next, substitute in the given information.
The change in y, ∆y , goes from 0 to 7 feet, a change of +7.
The change in x, ∆x, goes from 0 to 4 feet. Assume to the right and a change of +4.
Then, slope = 74 .
The answer is
7
4
.
Example 2
Find the slope of line CD below.
First, write the formula for the slope.
∆y
∆x
=
y2 −y1
x2 −x1
189
1.30. Slope of a Line Using Two Points
www.ck12.org
Next, fill in the values that are known from the coordinates given, using C as Point 1 and D as Point 2.
y−y1
x2 −x1
= 9−3
9−7
Then, simplify.
6
2
=3
The answer is that the slope of line CD equals 3. The slope is positive.
Example 3
Find the slope of line FG below.
First, write the formula.
∆y
∆x
=
y2 −y1
x2 −x1
Next, fill in the values that are known from the coordinates given using F as Point 1 and G as Point 2.
7−3 y−y1
3−6 x2 −x1
=
Then, simplify. Remember that slopes are not given as mixed numbers.
4
−3
= − 43
− 43 The answer is that the slope of line FG equals
The slope is negative
190
www.ck12.org
Chapter 1. Unit 1
Example 4
Draw a line that goes through a point at (-4, -1) and has a slope of 37 . What are the coordinates of the second point?
First, on a coordinate plane, plot the given point at (-4, -1)
Next, remember the formula for the slope of a line.
slope =
∆y
∆x
Then substitute the values that are given.
3
7
=
∆y
∆x
This means that the change in y, ∆y, is equal to +3.
The change in x, ∆x, is equal to +7.
Next, starting at the given point (-4, -1), move up +3 units on the y-axis and right +7 units on the x-axis.
Then, mark the point, read its location, and draw a line through both points.
The answer is (3, 2).
Example 5
Draw a line that goes passes through (-5, 4) and has a slope of − 23 . Give the coordinates of a second point.
First, on a coordinate plane, plot the given point at (-5, 4).
191
1.30. Slope of a Line Using Two Points
www.ck12.org
Next, remember the formula for the slope of a line.
slope =
∆y
∆x
Then substitute the values that are given.
− 32 =
∆y
∆x
Since the slope is negative, the minus sign can be assigned to either value.
The change in y, ∆y, is equal to -2.
The change in x, ∆x, is equal to +3.
Next, starting at the given point (-5, 4), move down 2 units on the y-axis and right 3 units on the x-axis.Then, mark
the point, read its location, and draw a line through both points.
The answer is (-2, 2).
There are an infinite number of points along this line, one of which could be identified by solving the above with the
negative assigned to the change in x rather than the change in y.
Review
Find the slope of each line shown.
192
www.ck12.org
Chapter 1. Unit 1
1.
2.
3.
193
1.30. Slope of a Line Using Two Points
4.
5.
6.
194
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
7.
8.
9.
195
1.30. Slope of a Line Using Two Points
10.
11.
12. On the coordinate grid below, draw a line that passes through (-3, 2) and has a slope of 21 .
196
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
13. Is this slope positive or negative?
14. On the coordinate grid below, draw a line that passes through (-2, 5) and has a slope of -4.
15. Is this slope positive or negative?
Review (Answers)
To see the Review answers, open this PDF file and look for section 5.8.
Resources
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/183389
197
1.31. Methods for Solving Quadratic Functions
www.ck12.org
1.31 Methods for Solving Quadratic Functions
Learning Objectives
Here you will review multiple methods of solving quadratic functions for the x intercepts. This is a far-reaching
lesson with concepts such as: applying the quadratic formula, completing the square, and solving by factoring, that
you will be returning to time and again as you progress further into your mathematics studies.
Can you measure height with a stopwatch? How could that be possible?
This lesson is all about quadratic functions, like the one used in Physics and described later in the lesson, which
allows a good approximation of the height of a tall object to be calculated with time!
Methods for Solving Quadratic Functions
A quadratic function can be described as:
Formally: A function f defined by f (x) = ax2 + bx + c, where a, b, and c are real numbers and a 6= 0.
Informally: The defining characteristic of a quadratic function is that it is a polynomial whose highest exponent is 2.
There are several ways to write quadratic functions:
• standard form, the form of the quadratic function above: f (x) = ax2 + bx + c
• vertex form, commonly used for quick sketching: f (x) = a(x − h)2 + k
• factored form, excellent for finding x-intercepts: f (x) = a(x − r1 )(x − r2 )
You can convert between forms of quadratic functions using algebra and you will see that there are uses for each of
these forms when working with quadratic functions.
When graphing, the y−intercept of a quadratic function in standard form is (0, c) and it is found by substituting 0
for x in f (x) = ax2 + bx + c.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187255
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187256
198
www.ck12.org
Chapter 1. Unit 1
Examples
Example 1
Earlier, you were asked if you could measure height with a stopwatch.
What formula could be used to calculate the height of an object based on the time required for an object to fall from
the top?
One formula is h = 12 gt 2
This is not completely accurate, as it ignores the effects of air resistance of an object, but offers a very close
approximate for most common objects.
The value g is gravity. In Metric units this is 9.81m/s/s
The value t is the time taken for the object to fall in seconds
The result h is the distance or the height in meters.
Example 2
Solve x2 − 5x + 6 = 0 using the "Factor to Solve" method.
The factor method is based on writing the quadratic equation in factored form. That is, as a product of two linear
expressions. So our equation maybe solved by the following way:
x2 − 5x + 6 = 0
(x − 3)(x − 2) = 0
Recall, that
a · b = 0 if and only if a = 0 and b = 0
This tells us that
x−3 = 0
or
x−2 = 0
which give the roots (or zeros)
x=3
and
199
1.31. Methods for Solving Quadratic Functions
www.ck12.org
x=2
In other words, the solution set is {2, 3}.
Example 3
Solve x2 − 5x + 6 = 0 by the "Completing the Square" method.
To solve the above equation by completing the square, first move the “c” term to the other side of the equation,
x2 − 5x = −6
Next, make the left-hand side a “perfect square” by adding the appropriate number. To do so, take one-half of the
coefficient of x (the b coefficient) and square it and then add the result to both sides of the equation:
b = −5
b −5
=
2
2
2
b
25
=
2
4
Adding to both sides of the equation,
25
25
= −6 +
4
4
25 1
2
x − 5x +
=
4
4
2
5
1
x−
=
2
4
x2 − 5x +
this last equation can be easily solved by taking the square root of both sides,
r
5
1
1
x−
=
=±
2
4
2
Hence
1 5
x=± +
2 2
and the solutions are
x=3
200
www.ck12.org
Chapter 1. Unit 1
and
x=2
which are identical to answers of the factor method.
Example 4
Solve x2 − 5x + 6 = 0 using the Quadratic Formula.
The quadratic formula works for finding the x-intercepts in all quadratic equations, therefore it is highly encouraged
that you memorize the formula. The other methods can be much faster though, so it is well worth understanding
each of the different methods.
Recall, if ax2 + bx + c = 0 where a, b, and c are real numbers and a 6= 0, then the roots of the equation can be
determined by the quadratic formula.
x=
−b ±
√
b2 − 4ac
.
2a
Here the coefficients are a = 1, b = −5, and c = 6.
Substituting into the quadratic formula, we get:
p
(−5)2 − 4(1)(6)
x=
2(1)
√
5 ± 25 − 24
=
2
5±1
=
2
= 3 or 2
−(−5) ±
which is, again, identical to our two solutions above.
Example 5
Solve the following equation by factoring: 4x2 + 7x = 2.
4x2 + 7x − 2 = 0
0 = (4x − 1)(x + 2) ∴ x = 41 , −2
Example 6
Solve the following equation by completing the square: 23 x2 − x + 13 = 0.
x2 − 32 x + 21 = 0
9
x2 − 32 x + 16
=
9
16
(x − 34 )(x − 43 ) =
− 21
1
16
x = 21 , 1
201
1.31. Methods for Solving Quadratic Functions
www.ck12.org
Example 7
Solve the following equation by using the quadratic formula: (z + 6)2 + 2z = 0.
√
−14± 142 − 4 · 1 · 36
2
z + 14z + 36 = 0 ∴ x =
2·1
√
x = −7 ± 13
Review
1. List three ways to write a quadratic function.
Solve with the quadratic formula:
2.
3.
4.
5.
6.
7.
m2 − 5m − 14 = 0
b2 − 4b + 4 = 0
4b2 + 8b + 7 = 0
2m2 − 7m − 13 = −10
5r2 = 80
k2 − 31 − 2k = −6 − 3k2 − 2k
8. If you have an equation with a power of 4, explain how you could solve it using the quadratic formula.
Solve by completing the square:
9. 2x2 − 12x + 26 = 10
10. x2 − 12x + 29 = −3
11. 7x2 − 14x − 64 = −8
Solve by the most efficient method:
12. x4 + 13x2 + 36 = 0
13. x4 + 16x2 − 225 = 0
14. 41 x2 − 13 x + 1 = 0
3
15. 72 c2 − 21 c − 14
=0
In the quadratic formula b2 − 4ac is called the discriminant. The values of the discriminant tell us the nature of the
solutions or roots of a quadratic equation, ax2 + bx + c = 0
16. What value(s) of the discriminant result in two unique real solutions?
17. What value(s) of the discriminant result in one unique real solutions?
18. What value(s) of the discriminant result in two unique imaginary solutions?
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.1.
202
www.ck12.org
Chapter 1. Unit 1
1.32 Quadratic Inequalities
Learning Objectives
Here you will learn about graphing quadratic inequalities.
How would you express the following as a function?
You are supposed to mow your square-shaped lawn for your parents, but the mower only has part of a tank of gas. If
you can mow 2500 sf per gallon, and the mower has approximately 2.5 gallons in it, what is the maximum length of
one side of the lawn you can mow? If your lawn is 75 feet long, will you need more gas?
Quadratic Inequalities
Quadratic inequalities are inequalities that have one of the following forms
ax2 + bx + c > 0
and
ax2 + bx + c < 0
We can solve these inequalities by using the techniques that we have learned about solving quadratic equations. For
example, consider the graph of the equation:
y = f (x) = x2 + x − 6
203
1.32. Quadratic Inequalities
www.ck12.org
Notice that the curve intersects the x−axis at -3 and 2. From graph, we notice the followings
• If x < −3 then f (x) > 0
• If −3 < x < 2, then f (x) < 0
• If x > 2, then f (x) > 0
Therefore, x2 + x − 6 > 0 whenever x or x > 2, and x2 + x − 6 < 0 when −3 < x < 2.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187345
Examples
Example 1
Earlier, you were given a question about mowing a lawn.
204
www.ck12.org
Chapter 1. Unit 1
If you can mow 2500 sf of grass per gallon of gas, and the mower has 2.5 gallons in it, what is the maximum length
of one side of the lawn you can mow?
If your lawn is 75 ft long, will you need more gas?
We know that the function S2 < 6250 describes the possible side lengths of square shapes you could mow before
running out of gas (if you’re having trouble understanding why, see Example 4 for a similar process).
Solving for S gives:
√
√
s2 = 6250
s = 79
With 2.5 gal of gas, you could mow a square up to apx 79 ft on each side.
You should not need more gas if the lawn is only 75 ft long on each side.
Example 2
What is the solution set of the inequality 2x2 + 7x − 4 < 0?
It is best to graph the function f (x) = 2x2 + 7x − 4 and look for the values of x such that the inequality f (x) < 0 is
true.
Thus from graph, 2x2 + 7x − 4 < 0 only if
−4 < x <
1
2
205
1.32. Quadratic Inequalities
www.ck12.org
So the solution set is x ∈ −4, 12 or in set builder notation, x|−4 < x <
1
2
Although the method of graphing to find the solution set of an inequality is easy to follow, another algebraic method
can be used. The algebraic method involves finding the x−intercepts of the graph and then dividing the x−axis into
intervals separated by the x−intercepts. The examples below illustrate the method.
Example 3
Find the solution set of the quadratic inequality x2 + 2x − 8 > 0 without graphing.
To find the solution set without graphing, first factor:
x2 + 2x − 8 = 0
(x + 4)(x − 2) = 0
Recalling the zero product rule, we can see that the two solutions to this quadratic equation are x = −4 and x = 2,
thus, the x−intercepts of the function f (x) = x2 + 2x − 8 are -4 and 2.
These points divide the x−axis into three intervals: (−∞, −4)|(−4, 2)|(2, ∞). We can choose a test point from each
interval, substitute it into f (x) and see if the function is negative or positive with that value as x. This procedure can
be simplified by making a table as shown below:
TABLE 1.25:
Interval
Test Point
(−∞, −4)
(−4, 2)
(2, +∞)
−5
1
3
Is x2 + 2x − 8 positive or
negative?
+
−
+
Part of Solution set?
yes
no
yes
From the table, we conclude that since x2 + 2x − 8 > 0 if and only if x < −4 and x > 2. The solution set can also be
written as:
x ∈ (−∞, −4) ∪ (2, +∞)
Some problems in science involve quadratic inequalities. The example below illustrates one such application.
Example 4
A rectangle has a length 10 meters more than twice the width. Find all of the possible widths that result in the area
of the rectangle not exceeding 100 squared meters.
Let w be the width of the rectangle and l its length. Given the information in the question, we can say:
l = 10 + 2w
Then we can use the formula for the area of a rectangle:
area = l × w
Substituting 10 + 2w in for l gives:
area = w × (10 + 2w)
= 2w2 + 10w
206
www.ck12.org
Chapter 1. Unit 1
The area cannot exceed 100 m2 , so
10w + 2w2 < 100
or
2w2 + 10w − 100 < 0
Simplify by dividing both sides by 2:
w2 + 5w − 50 < 0
Factor the trinomial:
w2 + 5w − 50 = (w + 10)(w − 5)
So the partition points are 5 and -10, which means we have three intervals. Since width cannot be negative, we can
safely ignore -10. That means the maximum area is 100 m2
∴ w < 5.
The width must be less than 5 meters.
Example 5
Find the solution set of the inequality x2 ≤ 16.
Set the function equal to zero:
x2 − 16 = 0
Factor to find the critical values (points where the graph crosses the x axis, thereby changing signs):
(x − 4)(x + 4) = 0
By the zero product rule: x = 4 or x = −4
That gives us three sections on the graph:
x ≤ −4
−4 ≤ x ≤ 4
4≤x
Test one sample value from each division to identify possible solution sets.
TABLE 1.26:
Set
Test value
f (x) true with value?
x ≤ −4
−5
4 · (−5)2 = 100 16 ∴ No
−4 ≤ x ≤ 4
0
4 · 02 = 0 ≤ 16 ∴ Yes
4≤x
5
4 · 52 = 100 16 ∴ No
207
1.32. Quadratic Inequalities
www.ck12.org
Therefore the solution set is −4 ≤ x ≤ 4.
Example 6
Graph the solution set: (x − 3)(x + 4) ≥ 0.
The solutions to (x − 3)(x + 4) ≥ 0 can be identified with the rules for multiplying negative numbers:
Recall from Pre-Algebra that an even number of negatives yields a positive answer, and an odd number of negatives
yields a negative answer.
Since (x − 3)(x + 4) ≥ 0 we know we need a positive answer or zero.
Therefore either:
Case #1: (x − 3) ≥ 0 → x ≥ 3 and (x + 4) ≥ 0 → x ≥ −4
or
Case #2: (x − 3) ≤ 0 → x ≤ 3 and (x + 4) ≤ 0 → x ≤ −4
Since any number greater than 3 is already greater than -4, from Case #1 we get: x ≥ 3
Since any number less than -4 is already less than 3, from Case #2 we get x ≤ −4
Therefore our answer is x ≤ −4 or x ≥ 3
In set notation: x ∈ (−∞, −4] ∪ [3, +∞)
To graph this information, we draw a line graph, and mark the values that x can be, with solid dots on the end
numbers to indicate that those values are included.
Visually that is:
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187409
Review
Graph the solutions sets below on a number line:
1. x < 3 or x > 4
2. x ≥ −5 and x ≥ 3
3. x < 6 and x ≥ −2
208
www.ck12.org
Chapter 1. Unit 1
4. x > 7 or x ≥ −4
5. x ≤ −8 and x > 3
Identify critical points, solve, and graph:
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
x2 + 9x > −14
x2 − 5x ≤ 50
x2 + 2x ≤ 48
x − 20
x − 8 < 0 (hint: multiply both sides by x first)
x + 10 ≥ − 21
x
(x + 6)(x − 3) > 0
(x − 8)(x + 1) > 0
x2 − x ≥ 90
3x2 − 23x ≤ 8
x2 + x − 6 > 0
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.9.
209
1.33. Graphing Rational Functions in Standard Form
www.ck12.org
1.33 Graphing Rational Functions in Standard
Form
Learning Objectives
Here you’ll learn how to graph basic rational functions.
The standard form of a rational function is given by the equation f (x) =
equation?
a
x−h
+ k. What are the asymptotes of this
Rational Function
A rational function is in the form
rational functions is y =
1
x,
p(x)
q(x)
where p(x) and q(x) are polynomials and q(x) 6= 0. The parent graph for
and the shape is called a hyperbola.
TABLE 1.27:
x
−4
−2
−1
− 21
1
2
1
2
4
210
y
− 14
− 12
−1
−2
2
1
1
2
1
4
www.ck12.org
Chapter 1. Unit 1
Notice the following properties of this hyperbola: the x-axis is a horizontal asymptote, the y-axis is a vertical
asymptote, and the domain and range are all real numbers except where the asymptotes are. Recall that the vertical
asymptote is the value that makes the denominator zero because we cannot divide by zero. For the horizontal
asymptote, it is the value where the range is not defined.
The two parts of the graph are called branches. In the case with a hyperbola, the branches are always symmetrical
about the point where the asymptotes intersect. In this example, they are symmetrical about the origin.
In this lesson, all the rational functions will have the form f (x) =
Let’s graph f (x) =
−2
x
a
x−h
+ k.
and find any asymptotes, the domain, range, and any zeros.
Let’s make a table of values.
TABLE 1.28:
x
1
2
4
y
−2
−1
− 12
Notice that these branches are in the second and fourth quadrants. This is because of the negative sign in front of the
2, or a. The horizontal and vertical asymptotes are still the x and y-axes. There are no zeros, or x-intercepts, because
the x-axis is an asymptote. The domain and range are all non-zero real numbers (all real numbers except zero).
Now, let’s graph y =
1
x−5
+ 2 and find all asymptotes, zeros, the domain and range.
1
+ 2, the vertical asymptote is x = 5 because that would make the denominator zero and we cannot divide
For y = x−5
by zero. When x = 5, the value of the function would be y = 10 + 2, making the range undefined at y = 2. The shape
and location of the branches are the same as the parent graph, just shifted to the right 5 units and up 2 units.
211
1.33. Graphing Rational Functions in Standard Form
www.ck12.org
Therefore, for the general form of a rational function, y =
horizontal asymptote.
a
x−h
+ k, x = h is the vertical asymptote and y = k is the
The domain is all real numbers; x 6= 5 and the range is all real numbers; y 6= 2. To find the zero, set the function
equal to zero and solve for x.
1
+2
x−5
1
−2 =
x−5
−2x + 10 = 1
0=
−2x = −9
9
x = = 4.5
2
To find the y-intercept, set x = 0, and solve for y. y =
1
0−5
+ 2 = − 15 + 2 = 1 45 .
Finally, let’s find the equation of the hyperbola below.
We know that the numerator will be negative because the branches of this hyperbola are in the second and fourth
a
quadrants. The asymptotes are x = −3 and y = −4. So far, we know y = x+3
− 4. In order to determine a, we can
use the given x-intercept.
212
www.ck12.org
Chapter 1. Unit 1
a
−4
−3.75 + 3
a
4=
−0.75
−3 = a
0=
The equation is y =
−3
−4
x+3
Examples
Example 1
Earlier, you were asked to find the asymptotes of the equation f (x) =
a
x−h
+ k.
From the previous problems, we’ve seen that the vertical asymptote occurs when the denominator of the equation
equals zero and the horizontal asymptote occurs when the range is undefined.
When x = h, the denominator of f (x) =
a
x−h
+ k is zero, so x = h is the vertical asymptote.
The value of the function at x = h would be y = 0a + k, making the range undefined at y = k. Therefore, y = k is the
horizontal asymptote.
Example 2
What are the asymptotes for f (x) =
−1
x+6
+ 9? Is (−5, −8) on the graph?
The asymptotes are x = −6 and y = 9. To see if the point (−5, −8) is on the graph, substitute it in for x and y.
−1
+9
−5 + 6
−8 = −1 + 9
−8 =
− 8 6= 8, therefore, the point (−5, −8) is not on the graph.
For Examples 3 & 4, graph the rational functions. Find the zero, y-intercept, asymptotes, domain and range.
213
1.33. Graphing Rational Functions in Standard Form
www.ck12.org
Example 3
y = 4x − 2
There is no y-intercept because the y-axis is an asymptote. The other asymptote is y = −2. The domain is all real
numbers; x 6= 0. The range is all real numbers; y 6= −2. The zero is:
4
−2
x
4
2=
x
2x = 4
0=
x=2
Example 4
y=
2
x−1
+3
The asymptotes are x = 1 and y = 3. Therefore, the domain is all real numbers except 1 and the range is all real
2
numbers except 3. The y-intercept is y = 0−1
+ 3 = −2 + 3 = 1 and the zero is:
214
www.ck12.org
Chapter 1. Unit 1
2
+3
x−1
2
−3 =
x−1
−3x + 3 = 2
0=
−3x = −1 → x =
1
3
Example 5
Determine the equation of the hyperbola.
215
1.33. Graphing Rational Functions in Standard Form
The asymptotes are x = −1, y = 3, making the equation y =
a
+3
0+1
1=a
4=
www.ck12.org
a
x+1
+ 3. Taking the y-intercept, we can solve for a.
The equation is y =
1
+ 3.
x+1
Review
1.
2.
3.
4.
2
What are the asymptotes for y = x+8
− 3?
Is (−6, −2) a point on the graph from #1?
1
What are the asymptotes for y = 6 − x−4
?
Is (5, 4) a point on the graph from #3?
For problems 5-13, graph each rational function, state the equations of the asymptotes, the domain and range and
the intercepts.
5.
6.
7.
8.
9.
10.
11.
12.
13.
y = 3x
y = 1x + 6
y = − 1x
1
y = − x+3
1
y = x+5
1
y = x−3
−4
2
y = x+4 − 3
y = 5x + 2
1
y = 3 − x+2
Write the equations of the hyperbolas. You may assume that a = 1.
14.
216
www.ck12.org
Chapter 1. Unit 1
15.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 9.4.
217
1.34. Graphs of Rational Functions
www.ck12.org
1.34 Graphs of Rational Functions
Learning Objectives
Here you’ll investigate the graphs of rational functions using a graphing calculator.
Can you use your graphing calculator to help you sketch the graph of the rational function? Does this function have
any zeros or asymptotes?
y=
4
x2 +1
Graphing Rational Functions
A rational function is the quotient of two polynomial functions. In general,
f (x) =
A(x)
B(x)
where A and B are polynomials and B 6= 0. You can use your graphing calculator to graph a rational function and
x
look for important features. Consider the function y = x−3
.
Notice that the function has two pieces. In between those two pieces are the asymptotes.
• A vertical asymptote occurs at the x value(s) that cause the denominator of the function to be equal to zero
(which is undefined). This function has a vertical asymptote at x = 3.
• A horizontal asymptote occurs at the y value(s) that cause the denominator of the function to be zero if the
function is rewritten and solved for x instead of y. This is what it looks like to solve for x:
218
www.ck12.org
Chapter 1. Unit 1
y=
x
x−3
x
(x − 3)(y) = (x − 3)
x−3
x
xy − 3y = (x
− 3)
x−
3
xy − 3y = x
xy − x = 3y
x(y − 1) = 3y
x=
3y
y−1
Thus, y 6= 1 and this function has a horizontal asymptote at y = 1.
The image below shows the graph with the asymptotes drawn in and labelled. For rational functions, the asymptotes
represent the lines that the function will approach but never touch.
It is also important to note the x-intercepts (zeros) of the function. The zeros of the function will be the values for x
that cause the numerator, but not also the denominator, to be equal to zero.
Let’s graph the following rational functions:
1. y =
1
x2 −9
=
1
(x+3)(x−3)
Here is the sketch from the calculator:
219
1.34. Graphs of Rational Functions
www.ck12.org
It can sometimes be hard to interpret what you see on the graphing calculator screen. Use algebra to find the
asymptotes and zeros and sketch those first.
There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.
The zeros of the denominator are 3 and -3. This means that there are two vertical asymptotes. One vertical
asymptote is the line x = 3 and the other is the line x = −3.
Another way to determine a horizontal asymptote besides solving the equation for x is to look at the degrees
of the numerators and denominators. The degree is the highest exponent. The degree of the numerator is 0
and the degree of the denominator is 2. In general, if the degree of the numerator is less than the degree of the
denominator, there will be a horizontal asymptote at y = 0.
Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide
what the rest of the graph looks like. Here is the graph with the asymptotes labelled.
2. y = − 1x
Here is the sketch from the calculator:
Use algebra to find the asymptotes and zeros and sketch those first.
There are no zeros for the numerator. Therefore, there are no x-intercepts for the function.
The zero of the denominator is 0. This means that there is one vertical asymptote, the line x = 0.
The horizontal asymptote is y = 0. You can use algebra to solve the equation for x and look for the values of
y that will cause the denominator to be equal to zero:
220
www.ck12.org
Chapter 1. Unit 1
1
x
xy = −1
xy −1
=
y
y
y=−
x=−
1
y
Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide
what the rest of the graph looks like. Here is the graph with the asymptotes labelled.
3. y =
x+2
x−3
Here is the sketch from the calculator:
Use algebra to find the asymptotes and zeros and sketch those first.
The zero of the numerator is -2 so the zero (x-intercept) of the function is (-2, 0).
The zero of the denominator is 3. This means that there is one vertical asymptote, the line x = 3.
The horizontal asymptote is y = 1. You can use algebra to solve the equation for x and look for the values of
y that will cause the denominator to be equal to zero:
221
1.34. Graphs of Rational Functions
www.ck12.org
y=
x+2
x−3
x+2
(x − 3)(y) =
(x − 3)
x−3
xy − 3y = x + 2
xy − x = 2 + 3y
x(y − 1) = 2 + 3y
x=
3y + 2
y−1
Once you have sketched the asymptotes, use the table and/or graph from the graphing calculator to decide
what the rest of the graph looks like. Here is the graph with the asymptotes labelled.
Examples
Example 1
Earlier, you were asked if the following function has any zeros or asymptotes:
y=
4
x2 +1
Here is a sketch of the function:
• There are no zeros for this function since there are no zeros for the numerator. The graph does not cross the
x-axis.
• There are no zeros for the denominator. Therefore, there are no vertical asymptotes.
222
www.ck12.org
Chapter 1. Unit 1
• Because the degree of the numerator is less than the degree of the denominator, there is a horizontal asymptote
at y = 0.
Example 2
Sketch the graph of the rational function: y =
x+1
x
The zero of the denominator is 0. This means that there will be a vertical asymptote at x = 0 (the y-axis). Because
1
the degree of the denominator and numerator are the same, you can solve the equation for x and get x = y−1
. The
zero of the denominator is now 1. This means there is a horizontal asymptote at y = 1. The numerator has a zero at
-1, so there is an x-intercept at -1.
Example 3
Sketch the graph of the rational function: y =
x2
x−3
The zero of the denominator is 3. This means that there will be a vertical asymptote at x = 3. Because the degree
of the numerator is greater than the degree of the denominator, there are no horizontal asymptotes. The zero of the
numerator is 0, so there is an x-intercept at 0.
Example 4
Find the asymptotes of the function: y =
1
x2 −16
The zeros of the denominator are 4 and -4. This means that there will be two vertical asymptotes. One vertical
asymptote will be the line x = 4 and the other will be the line x = −4. Because the degree of the numerator is less
than the degree of the denominator, there is a horizontal asymptote at y = 0. Though you did not need to sketch the
graph, here is the graph of the function:
223
1.34. Graphs of Rational Functions
www.ck12.org
Review
Sketch the graph of each of the following rational functions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
y=
y=
y=
y=
y=
y=
y=
y=
y=
y=
2
x+3
x
x−1
1
x2 −4
x+2
x
1
x2 +2
x
x+2
1
x2 −x−12
x−1
x+3
x−1
x+4
5
x2 +1
Without graphing the following rational functions, state what you know about their asymptotes and zeros.
11.
12.
13.
14.
15.
1
y = x2 −x−2
2
y = − x−4
y = − x22+1
y = x26+1
x−1
y = x+3
Review (Answers)
To see the Review answers, open this PDF file and look for section 8.4.
224
www.ck12.org
Chapter 1. Unit 1
1.35 Holes in Rational Functions
Here you will start factoring rational expressions that have holes known as removable discontinuities. In a function
, you should note that the factor (x − 1) clearly cancels leaving only 3x − 1. This appears to
like f (x) = (3x+1)(x−1)
(x−1)
be a regular line. What happens to this line at x = 1?
Holes and Rational Functions
A hole on a graph looks like a hollow circle. It represents the fact that the function approaches the point, but is not
actually defined on that precise x value.
Take a look at the graph of the following equation:
f (x) = (2x + 2) ·
(x+ 21 )
(x+ 21 )
The reasonwhy this function is not defined at − 21 is because − 12 is not in the domain of the function. As you can
see, f − 12 is undefined because it makes the denominator of the rational part of the function zero which makes the
whole function undefined. Also notice that once the factors are canceled/removed then you are left with a normal
function which in this case is 2x + 2. The hole in this situation is at (− 12 , 1) because after removing the factors that
cancel, f (− 12 ) = 1.
This is the essence of dealing with holes in rational functions. You should cancel what you can and graph the
function like normal making sure to note what x values make the function undefined. Once the function is graphed
without holes go back and insert the hollow circles indicating what x values are removed from the domain. This is
why holes are called removable discontinuities.
Watch the first part of this video and focus on holes in rational equations.
225
1.35. Holes in Rational Functions
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/205738
Examples
Example 1
Earlier, you were asked what happens to the equation f (x) = (3x+1)(x−1)
at x = 1. Since this function that is not
(x−1)
defined at x = 1 there is a removable discontinuity that is represented as a hollow circle on the graph. Otherwise the
function behaves precisely as 3x + 1.
Example 2
Graph the following rational function and identify any removable discontinuities.
f (x) =
−x3 +3x2 +2x−4
x−1
This function requires some algebra to change it so that the removable factors become obvious. You should suspect
that (x − 1) is a factor of the numerator and try polynomial or synthetic division to factor. When you do, the function
becomes:
f (x) =
(−x2 +2x+4)(x−1)
(x−1)
The removable discontinuity occurs at (1, 5).
Example 3
Graph the following rational function and identify any removable discontinuities.
f (x) =
x6 −6x5 +5x4 +27x3 −48x2 −9x+54
x3 −7x−6
This is probably one of the most challenging rational expressions with only holes that people ever try to graph
by hand. There are multiple ways to start, but a good habit to get into is to factor everything you possibly can
226
www.ck12.org
Chapter 1. Unit 1
initially. The denominator seems less complicated with possible factors (x ± 1), (x ± 2), (x ± 3), (x ± 6). Using
polynomial division, you will find the denominator becomes:
f (x) =
x6 −6x5 +5x4 +27x3 −48x2 −9x+54
(x+1)(x+2)(x−3)
The factors of the denominator are strong hints as to the factors of the numerator so use polynomial division and try
each. When you fully factor the numerator you will have:
f (x) =
(x3 −6x2 +12x−9)(x+1)(x+2)(x−3)
(x+1)(x+2)(x−3)
Note the factors that cancel (x = −1, −2, 3) and then work with the cubic function that remains.
f (x) = x3 − 6x2 + 12x − 9
At this point it is probably reasonable to make a table and plot points to get a sense of where this cubic function
lives. You also could notice that the coefficients are almost of the pattern 1 3 3 1 which is the binomial expansion. By
separating the -9 into -8 -1 you can factor the first four terms.
f (x) = x3 − 6x2 + 12x − 8 − 1 = (x − 2)3 − 1
This is a cubic function that has been shifted right by two units and down one unit.
Note that there are two holes that do not fit in the graph window. When this happens you still need to note where
they would appear given a properly sized window. To do this, substitute the invalid x values: x = −1, −2, 3 into the
factored cubic that remained after canceling.
f (x) = (x − 2)3 − 1
Holes: (3, 0); (-1, -28); (-2, -65)
Example 4
Without graphing, identify the location of the holes of the following function.
f (x) =
x3 +4x2 +x−6
x2 +5x+6
First factor everything. Then, identify the x values that make the denominator zero and use those values to find the
exact location of the holes.
f (x) =
(x+2)(x+3)(x−1)
(x+3)(x+2)
Holes: (-3, -4); (-2, -3)
227
1.35. Holes in Rational Functions
www.ck12.org
Example 5
What is a possible equation for the following rational function?
The function seems to be a line with a removable discontinuity at (1, -1). The line is has slope 1 and y-intercept of
-2 and so has the equation:
f (x) = x − 2
The removable discontinuity must not allow the x to be 1 which implies that it is of the form
function is:
f (x) =
(x−2)(x−1)
x−1
Review
1. How do you find the holes of a rational function?
2. What’s the difference between a hole and a removable discontinuity?
3. If you see a hollow circle on a graph, what does that mean?
Without graphing, identify the location of the holes of the following functions.
4. f (x) =
x2 +3x−4
x−1
5. g(x) =
x2 +8x+15
x+3
6. h(x) =
x3 +6x2 +2x−8
x2 +x−2
7. k(x) =
x3 +6x2 +2x−8
x2 −3x+2
8. j(x) =
x3 +4x2 −17x−60
x2 −9
9. f (x) =
x3 +4x2 −17x−60
x2 −5x+4
10. g(x) =
x3 −4x2 −19x−14
x2 −8x+7
11. What is a possible equation for the following rational function?
228
x−1
x−1 .
Therefore, the
www.ck12.org
Chapter 1. Unit 1
12. What is a possible equation for the following rational function?
Sketch the following rational functions.
13. f (x) =
x3 +4x2 −17x−60
x2 −x−12
14. g(x) =
x3 +4x2 −17x−60
x2 +8x+15
15. h(x) =
x3 −4x2 −19x−14
x2 −6x−7
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.7.
229
1.36. Zeroes of Rational Functions
www.ck12.org
1.36 Zeroes of Rational Functions
Learning Objectives
Here you will revisit how to find zeroes of functions by focusing on rational expressions and what to do in special
cases where the zeroes and holes seem to overlap.
The zeroes of a function are the collection of x values where the height of the function is zero. How do you find
these values for a rational function and what happens if the zero turns out to be a hole?
Finding Zeroes of Rational Functions
Zeroes are also known as x-intercepts, solutions or roots of functions. They are the x values where the height of the
function is zero. For rational functions, you need to set the numerator of the function equal to zero and solve for the
possible x values. If a hole occurs on the x value, then it is not considered a zero because the function is not truly
defined at that point.
Take the following rational function:
f (x) =
(x−1)(x+3)(x+3)
x+3
Notice how one of the x + 3 factors seems to cancel and indicate a removable discontinuity. Even though there are
two x + 3 factors, the only zero occurs at x = 1 and the hole occurs at (-3, 0).
Watch the video below and focus on the portion of this video discussing holes and x-intercepts.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/60830
230
www.ck12.org
Chapter 1. Unit 1
Examples
Example 1
Earlier, you were asked how to find the zeroes of a rational function and what happens if the zero is a hole. To find
the zeroes of a rational function, set the numerator equal to zero and solve for the x values. When a hole and a zero
occur at the same point, the hole wins and there is no zero at that point.
Example 2
Create a function with zeroes at x = 1, 2, 3 and holes at x = 0, 4.
There are an infinite number of possible functions that fit this description because the function can be multiplied by
any constant. One possible function could be:
f (x) =
(x−1)(x−2)(x−3)x(x−4)
x(x−4)
Note that 0 and 4 are holes because they cancel out.
Example 3
Identify the zeroes, holes and y intercepts of the following rational function without graphing.
f (x) =
x(x−2)(x−1)(x+1)(x+1)(x+2)
(x−1)(x+1)
The holes occur at x = −1, 1. To get the exact points, these values must be substituted into the function with the
factors canceled.
f (x) = x(x − 2)(x + 1)(x + 2)
f (−1) = 0, f (1) = −6
The holes are (-1, 0); (1, 6). The zeroes occur at x = 0, 2, −2. The zero that is supposed to occur at x = −1 has
already been demonstrated to be a hole instead.
Example 4
Identify the y intercepts, holes, and zeroes of the following rational function.
f (x) =
6x3 −7x2 −x+2
x−1
After noticing that a possible hole occurs at x = 1 and using polynomial long division on the numerator you should
get:
f (x) = (6x2 − x − 2) · x−1
x−1
A hole occurs at x = 1 which turns out to be the point (1, 3) because 6 · 12 − 1 − 2 = 3.
The y-intercept always occurs where x = 0 which turns out to be the point (0, -2) because f (0) = −2.
To find the x-intercepts you need to factor the remaining part of the function:
(2x + 1)(3x − 2)
Thus the zeroes (x-intercepts) are x = − 12 , 32 .
231
1.36. Zeroes of Rational Functions
www.ck12.org
Example 5
Identify the zeroes and holes of the following rational function.
f (x) =
2(x+1)(x+1)(x+1)
2(x+1)
The hole occurs at x = −1 which turns out to be a double zero. The hole still wins so the point (-1, 0) is a hole. There
are no zeroes. The constant 2 in front of the numerator and the denominator serves to illustrate the fact that constant
scalars do not impact the x values of either the zeroes or holes of a function.
Review
Identify the intercepts and holes of each of the following rational functions.
x3 +x2 −10x+8
x−2
3
2 −5x+6
g(x) = 6x −17x
x−3
h(x) = (x+2)(1−x)
x−1
j(x) = (x−4)(x+2)(x+2)
x+2
x(x−3)(x−4)(x+4)(x+4)(x+2)
k(x) =
(x−3)(x+4)
f (x) = x(x+1)(x+1)(x−1)
(x−1)(x+1)
x3 −x2 −x+1
g(x) = x2 −1
2
h(x) = 4−x
x−2
1. f (x) =
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Create a function with holes at x = 3, 5, 9 and zeroes at x = 1, 2.
Create a function with holes at x = −1, 4 and zeroes at x = 1.
Create a function with holes at x = 0, 5 and zeroes at x = 2, 3.
Create a function with holes at x = −3, 5 and zeroes at x = 4.
Create a function with holes at x = −2, 6 and zeroes at x = 0, 3.
Create a function with holes at x = 1, 5 and zeroes at x = 0, 6.
Create a function with holes at x = 2, 7 and zeroes at x = 3.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.8.
232
www.ck12.org
Chapter 1. Unit 1
1.37 Graphing when the Degrees of the
Numerator and Denominator are Different
Learning Objectives
Here you’ll learn how to graph rational functions where the degrees of the numerator and denominator are not the
same.
1
Xerxes says that the function y = 4xx−2
2 +7 , has a horizontal asymptote of y = 4 , Yolanda says the function has no
horizontal asymptote, Zeb says that it does have a horizontal asymptote but it’s at y = 0. Which one of them is
correct?
Graphing Rational Functions
In this concept we will touch on the different possibilities for the remaining types of rational functions. You will
need to use your graphing calculator throughout this concept to ensure your sketches are correct.
Let’s graph y =
x+3
2x2 +11x−6
and find all asymptotes, intercepts, and the domain and range.
In this problem the degree of the numerator is less than the degree of the denominator. Whenever this happens the
horizontal asymptote will be y = 0, or the x-axis. Now, even though the x-axis is the horizontal asymptote, there will
still be a zero at x = −3 (solving the numerator for x and setting it equal to zero). The vertical asymptotes will be
the solutions to 2x2 + 11x − 6 = 0. Factoring this quadratic, we have (2x − 1)(x + 6) = 0 and the solutions are x = 12
and −6. The y-intercept is 0, − 12 . At this point, we can plug our function into the graphing calculator to get the
general shape.
Because the middle portion crosses over the horizontal asymptote, the range will be all real numbers. The domain is
x ∈ R; x = −6; x 6= 12 .
233
1.37. Graphing when the Degrees of the Numerator and Denominator are Different
www.ck12.org
Be careful when graphing any rational function. This function does not look like the graph to the left in a TI-83/84.
This is because the calculator does not have the ability to draw the asymptotes separately and wants to make the
function continuous. Make sure to double-check the table (2nd → GRAPH) to find where the function is undefined.
Now, let’s graph f (x) =
x2 +7x−30
x+5
and find all asymptotes, intercepts, and the domain and range.
In this problem the degree of the numerator is greater than the degree of the denominator. When this happens,
there is no horizontal asymptote. Instead there is a slant asymptote. Recall that this function represents division.
20
If we were to divide x2 + 7x − 30 by x + 5, the answer would be x + 2 − x+5
. The slant asymptote would be the
answer, minus the remainder. Therefore,for this problem the slant asymptote is y = x + 2. Everything else is the
2
same. The y-intercept is −30
5 → (0, −6) and the x-intercepts are the solutions to the numerator, x + 7x − 30 = 0 →
(x + 10)(x − 3) → x = −10, 3. There is a vertical asymptote at x = −5. At this point, you can either test a few points
to see where the branches are or use your graphing calculator.
The domain would be all real numbers; x 6= −5. Because of the slant asymptote, there are no restrictions on the
range. It is all real numbers.
Finally, let’s graph y =
x−6
3x2 −16x−12
and find the asymptotes and intercepts.
Because the degree of the numerator is less than the degree of the denominator, there will be a horizontal asymptote
along the x-axis. Next, let’s find the vertical asymptotes by factoring the denominator; (x − 6)(3x + 2). Notice that
the denominator has a factor of (x − 6), which is the entirety of the numerator. That means there will be a hole at
x = 6.
234
www.ck12.org
Chapter 1. Unit 1
1
x−6
will be the same as y = 3x+2
except with a hole at x = 6. There is no
Therefore, the graph of y = 3x2 −16x−12
2
x-intercept, the vertical asymptote is at x = − 3 and the y-intercept is 0, 21 .
Recap
For a rational function; f (x) =
p(x)
q(x)
=
am xm +...+a0
bn xn +...+b0
1. If m, then there is a horizontal asymptote at y = 0.
2. If m = n, then there is a horizontal asymptote at y = abmn (ratio of the leading coefficients).
3. If m > n, then there is a slant asymptote at y = (am xm + . . . + a0 ) ÷ (bn xn + . . . + b0 ) without the remainder. In
this concept, we will only have functions where m is one greater than n.
Examples
Example 1
Earlier, you were asked to determine which student is correct.
The degree of the numerator x − 2 is less than the degree of the denominator 4x2 + 7. We know that whenever this
happens the horizontal asymptote will be y = 0, or the x-axis.
Therefore, Zeb is correct.
Graph the following functions. Find any intercepts and asymptotes.
Example 2
y=
3x+5
2x2 +9x+20
x-intercept: − 53 , 0 , y-intercept: 0, 41
horizontal asymptote: y = 0
235
1.37. Graphing when the Degrees of the Numerator and Denominator are Different
vertical asymptotes: none
Example 3
f (x) =
x2 +4x+4
x2 −3x−4
x-intercept: (−2, 0), y-intercept: (0, −1)
horizontal asymptote: y = 1
vertical asymptotes: x = 4 and x = −1
236
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
Example 4
g(x) =
x2 −16
x+3
x-intercepts: (−4, 0) and (4, 0)
y-intercept: 0, − 16
3
slant asymptote: y = x − 3
vertical asymptotes: x = −3
Example 5
y=
2x+3
6x2 −x−15
x-intercepts: none, hole at x = − 32
y-intercept: 0, − 15
horizontal asymptote: y = 0
vertical asymptote: x =
5
3
237
1.37. Graphing when the Degrees of the Numerator and Denominator are Different
www.ck12.org
Review
Find all asymptotes of the following functions.
1. y =
2. y =
x−2
x2 +6x+8
2
x −4
x+5
x2
x−3
3. y =
4. Find the x-intercepts of the function in #2.
5. Find the x-intercepts of the function in #3.
Graph the following functions. Find any intercepts, asymptote and holes.
x+1
x2 −x−12
2
f (x) = x +3x−10
x−3
x−7
y = 2x2 −11x−21
2 −2
g(x) = 2x
3x+5
2
y = x +x−30
x+6
2 +x−30
f (x) = 2x3x−5x
2 −4x+3
x3 −2x2 −3x
y = x2 −5x+6
f (x) = x22x+5
+5x−6
2 +3x+4
g(x) = −x2x−6
6. y =
7.
8.
9.
10.
11.
12.
13.
14.
15. Determine the slant asymptote of y =
Can you explain your results?
3x2 −x−10
3x+5 .
Now, graph this function. Is there really a slant asymptote?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 9.6.
238
www.ck12.org
Chapter 1. Unit 1
1.38 Graphing when the Degrees of the
Numerator and Denominator are the Same
Learning Objectives
Here you’ll learn how to graph rational functions when the numerator and denominator have the same degree.
4
+5
Darnell says that the function y = 2x
has two vertical asymptotes, Barb says it has only one, and Aruna says it
x4 −16
has four. Which one of them is correct?
Graphing Rational Functions
1
+ k, where x = h and y = k are the asymptotes. In this
We have already graphed functions in the form y = x−h
concept, we will extend graphing rational functions when both the denominator and numerator are linear or both
quadratic. So, there will be no “k” term in this concept. Let’s go through some practice problems to determine any
patterns in graphing this type of rational function.
Let’s graph f (x) =
2x−1
x+4
and find asymptotes, x and y intercepts, domain and range.
To find the vertical asymptote, it is the same as before, the value that makes the denominator zero. In this case,
x = −4. Also the same is how to find the x and y intercepts.
y-intercept (when x = 0): y =
2·0−1
0+4
= − 41
x-intercept (when y = 0):
2x − 1
x+4
0 = 2x − 1
0=
1 = 2x
1
=x
2
When solving for the x-intercept, to get the denominator out, we multiplied both sides by x + 4. But, when we
multiply anything by 0, it remains 0. Therefore, to find the x-intercept, we only need to set the numerator equal to
zero and solve for x.
The last thing to find is the horizontal asymptote. We know that the function is positive, so the branches will be in
the first and third quadrants. Let’s make a table.
TABLE 1.29:
x
−13
−7
−5
−3
−1
0
2
5
14
y
3
5
11
−7
−1
−0.25
0.5
1
1.5
239
1.38. Graphing when the Degrees of the Numerator and Denominator are the Same
www.ck12.org
It looks like the horizontal asymptote is y = 2 because both branches seem to approach 2 as x gets larger, both
positive and negative. If we plug in x = 86, y = 1.9 and when x = −94, y = 2.1. As you can see, even when x is very
large, the function is still approaching 2.
Looking back at the original equation, f (x) = 2x−1
x+4 , extract the leading coefficients and leave them numerator over
2
denominator, 1 . This is the horizontal asymptote. We can generalize this pattern for all rational functions. When
the degree of the numerator is equal to the degree of the denominator, the horizontal asymptote is the ratio of the
leading coefficients.
Finally, the domain is all real numbers; x 6= −4 and the range is all real numbers; y 6= 2.
Now, let’s graph y =
3x2 +10
x2 −1
and find the asymptotes, intercepts, domain, and range.
From the previous problem above, we can conclude that the horizontal asymptote is at y = 3. Because the denominator is squared, there will be two vertical asymptotes because x2 − 1 factors to (x − 1)(x + 1). Therefore, the vertical
asymptotes are x = 1 and x = −1. As for the intercepts, there are no x-intercepts because there is no real solution for
10
3x2 + 10 = 0. Solving for the y-intercept, we have y = −1
= −10.
At this point, put the equation in your calculator to see the general shape. To graph this function using a TI-83 or 84,
∧ 2+10)
enter the function into Y = like this: (3x
(x∧ 2−1) and press GRAPH. You will need to expand the window to include
the bottom portion of the graph. The final graph is below.
240
www.ck12.org
Chapter 1. Unit 1
The domain is still all real numbers except the vertical asymptotes. For this function, that would be all real numbers;
x 6= −1, x 6= 1.
The range is a bit harder to find. Notice the gap in the range from the horizontal asymptote and the y-intercept.
Therefore, the range is (−∞, −10] ∪ (3, ∞).
The notation above is one way to write a range of numbers called interval notation and was already introduced. The
∪ symbol means “union.” Notice that −∞ and ∞ are not included in the range.
In general, rational functions with quadratics in the denominator are split into six regions and have branches in three
of them, like the problem above. However, there are cases when there are no zeros or vertical asymptotes and those
look very different. You should always graph the function in a graphing calculator after you find the critical values
and make as accurate a sketch as you can.
Finally, let’s graph f (x) =
x2 −8x+12
x2 −x−6
and find the intercepts, asymptotes, domain and range.
Let’s factor the numerator and denominator to find the intercepts and vertical asymptotes.
f (x) =
x2 −8x+12
x2 +x−6
=
(x−6)(x−2)
(x+3)(x−2)
Notice that the numerator and denominator both have a factor of (x − 2). When this happens, a hole is created
because x = 2 is both a zero and an asymptote. Therefore, x = 2 is a hole and neither a zero nor an asymptote.
241
1.38. Graphing when the Degrees of the Numerator and Denominator are the Same
www.ck12.org
There is a vertical asymptote at x = −3 and a zero at x = 6. The horizontal asymptote is at y = 1. The graph of
2
x−6
f (x) = x x−8x+12
2 −x−6 will look like the graph of f (x) = x+3 , but with a hole at x = 2. A hole is not part of the domain.
−4
2
And, the output value that corresponds with the hole is not part of the range. In this problem, f (2) = 2−6
2+3 = 6 = − 3
is not part of the range. If you were to graph this function on your graphing calculator, the calculator would not show
there is a hole.
The domain is x ∈ R; x 6= 2, −3 and the range is y ∈ R; y 6= 1, − 23 .
Examples
Example 1
Earlier, you were asked to determine which student is correct.
The vertical asymptote(s) occur(s) when the denominator of the function equals zero. For the function y =
the denominator equals zero when x4 − 16 = 0.
2x4 +5
,
x4 −16
x4 − 16 = 0
x4 = 16
x = 2 or x = −2
Therefore, there are two vertical asymptotes and Darnell is correct.
Graph the following functions. Find all intercepts, asymptotes, the domain and range. Double-check your
answers with a graphing calculator.
Example 2
y=
242
4x−5
2x+7
www.ck12.org
5
5
y-intercept: y = −5
7 = − 7 , x-intercept: 0 = 4x − 5 → x = 4 , horizontal asymptote: y =
2x + 7 = 0 → x = − 72 , domain: R; x 6= − 72 , range: R; y 6= 2
Chapter 1. Unit 1
4
2
= 2, vertical asymptote:
Example 3
f (x) =
x2 −9
x2 +1
2
y-intercept: y = −9
1 = −9, x-intercepts: 0 = x − 9 → x = ±3, horizontal asymptote: y = 1, vertical asymptote: none,
domain: R, range: R; y 6= 1
Special Note: When there are no vertical asymptotes and the numerator and denominator are both quadratics, this
is the general shape. It could also be reflected over the horizontal asymptote.
243
1.38. Graphing when the Degrees of the Numerator and Denominator are the Same
www.ck12.org
Example 4
y=
2x2 +7x+3
x2 +3x+2
y-intercept: 0, 32 , x-intercepts: (−3, 0) and − 21 , 0 , horizontal asymptote: y = 2, vertical asymptotes: x = −2, x =
−1.
domain: R; x 6= −1, −2
range: y ∈ (−∞, 2.1] ∪ [12, ∞)
Example 5
y=
x2 −4
2x2 −5x+2
horizontal asymptote: y = 12 , y-intercept: (0, −2)
vertical asymptotes: x = 21 , x-intercept: (−2, 0)
hole: x = 2, f (2) =
domain: R; x 6= 12 , 2
range: R; y 6= 21 , 43
244
4
3
www.ck12.org
Chapter 1. Unit 1
Review
1.
2.
3.
4.
5.
What are the vertical and horizontal asymptotes for y =
What is the domain of this function?
What is the range of this function?
Are there any x-intercepts? If so, what are they?
Is there a y-intercept? If so, what is it?
x−2
x+7 ?
Graph the following rational functions. Write down the equations of the asymptotes, the domain and range, x and y
intercepts and identify any holes.
x+3
x−5
5x+2
x−4
3−x
2x+10
2 +5x+6
y = xx2 −8x+12
2 +4
y = 2xx2 +x−3
2x2 −x−10
y = 3x
2 +10x+8
x2 −4
y = x2 +3x−10
2
y = 6x4x−7x−3
2 −1
x3 −8
y = x3 +x2 −4x−4
1
Graph y = x−2
+3
6. y =
7. y =
8. y =
9.
10.
11.
12.
13.
14.
15.
and y =
3x−5
x−2
on the same set of axes. Compare the two. What do you notice? Explain your
results.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 9.5.
245
1.39. Analysis of Rational Functions
www.ck12.org
1.39 Analysis of Rational Functions
Learning Objectives
Here you will explore rational functions in a bit more detail. By graphing rational functions using a graphing
calculator or by identifying asymptotes, end behavior, and transformations, you can often get a good idea of the
important information from the function.
Consider the following situation:
Students in your school are currently required to demonstrate their understanding of their classwork at a level of
75% or higher in order to move on to more advanced material.
The amount of homework time T required for each student based on understanding level of p% is given by: T =
(18p)
(100−p)
If the school administration decides to raise the minimum level of understanding to 82%, how will this affect the
students’ homework time?
This is an example of a rational function in a real-world situation.
Analyzing Rational Functions
Finding Vertical Asymptotes and Breaks
Recall that rational functions are defined as r(x) =
p(x)
q(x)
where p(x) and q(x) are polynomials.
To find vertical asymptotes and breaks in the domain of a rational function, set the denominator equal to zero and
p(x)
solve for x. Given r(x) = q(x)
, set q(x) = 0 and solve for x.
Note that some rational functions have vertical asymptotes and others do not. Some rational functions have a break
in the function, but no vertical asymptote. This usually happens when one term in the numerator cancels with one
term in the denominator.
Evaluating End Behavior
The end behavior of a rational function can often be identified by the horizontal asymptote. That is, as the values
of x get very large or very small, the graph of the rational function will approach (but not reach) the horizontal
asymptote.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187341
246
www.ck12.org
Chapter 1. Unit 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187412
Examples
Example 1
Earlier, you were asked a question about students’ homework time.
The number of minutes of homework time T required for each student based on understanding level of p’% is given
by: T = (18p)/(100 − p) If the school administration decides to raise the minimum level of understanding to 82%,
how will this affect the students’ homework time?’
The current time required is: (18(75))/(100 − 75) ==>54min
By substituting in the increased level: (18(82))/(100 − 82) ==>82min
The increase in required proficiency would result in an average of 28mins of added study time per student.
Example 2
Consider the following rational function. Find all restrictions on the domain and asymptotes.
f (x) =
x2 + 2x − 35
x−5
Factoring the numerator
(x − 5)(x + 7)
x+7
(x − 5)
(x
+
7)
f (x) =
7
x+
f (x) =
Canceling
f (x) = x + 7, x 6= 5
Notice that there is no asymptote in this function, but rather a break in the graph at x = 5.
247
1.39. Analysis of Rational Functions
www.ck12.org
Example 3
Find the restrictions on the domain of
h(x) =
3x
x2 − 25
Setting the denominator equal to 0,
x2 − 25 = 0
x2 = 25
x = ±5
Thus, the domain of h(x) is the set all real numbers x with the restriction x 6= ±5. h(x) has two vertical asymptotes,
one at x = 5 and one at x = −5.
Note for graphing using technology problems: If you do not have a graphing calculator, there are a number of
excellent free apps, and an excellent free online calculator here: Desmos Online Graphing Calculator .
Example 4
(Graphing calculator exercise)
1
on the window [-10,10] X [-10,10]. (This means {XMIN}=-10, {XMAX}=10, {YMIN}=-10,
Graph f (x) = x−3
and {YMAX}=10).
Notice:
1. When graphing a rational function by entering the function in the Y = screen, remember that you need to use
parenthesis to group the numerator and denominator of the rational function.
2. Vertical asymptotes are sometimes graphed as vertical lines.
3. Graphs of rational functions can be difficult to interpret if the window settings are not chosen carefully.
y=
248
1
x−3
showing vertical line at x = 3
www.ck12.org
Chapter 1. Unit 1
f (x) is undefined and has a vertical asymptote at x = 3, but the way the graphing calculator draws the graph, it shows
a vertical line at x = 3. One way to “fix” this problem is to press MODE and select the option “Dot” rather than
“Connected”. However, dot graphs can be hard to interpret as well.
Example 5
Solve the equation and find any points of discontinuity using technology: f (x) =
6
x−2
4
+ x+7
.
Input the function into the grapher, don’t forget to use "(" ")" to group polynomials. Your function should look
something like this on your input screen (at least until you tell the calc to process the function): 6/(x - 2) + 4/(x + 7).
Note that some calculators require the " y = " or " f (x) = " and some do not.
Once you are sure you have the information entered correctly, press "calculate", or the equivalent. The graph should
look like:
There are asymptotes at x = 2 and x = 7 and y = 0.
Example 6
Using technology, find all intercepts, asymptotes, and discontinuities, and graph:
2x3 +14x2 −9x+6
.
x+3
Input the function with care, it should look something like: (2x3 + 14x2 − 9x + 6)/(x + 3) before you press the calc
or view button.
The graph should look like:
249
1.39. Analysis of Rational Functions
www.ck12.org
The intercepts are at (apx) (−7.64, 0) and (exactly) (0, 2).
There is a vertical asymptote at x = 3 and a curved asymptote described by: y = 2x2 + 8x − 33.
Example 7
Using technology, simplify, find discontinuities and limitations, then graph:
Input the function accurately. The graph should look like:
The function simplifies to:
3(2x2 +7x+3)
4(x2 − 14 )
There are asymptotes at x =
There is a hole at y = −3 43 .
250
1
2
→
3(2x+1)(x+3)
4(x− 21 )(x+ 12 )
and at y = 12 .
6x2 +21x+9
.
4x2 −1
www.ck12.org
Chapter 1. Unit 1
Review
For questions 1 - 5, factor the numerator and denominator, then set the denominator equal to zero and solve to find
restrictions on the domain.
x2 +3x−10
x−2
2
f (x) = x +2x−24
x−4
2
f (x) = x −12x+32
x−4
21
x2 + 20
y = x+ 4
5
2
y = x +13x+42
x+7
1. y =
2.
3.
4.
5.
For questions 6 - 15, input the function into your graphing tool carefully and accurately. Record any asymptotes or
holes and record x and y intercepts. Finally either copy and print or sketch the image of the graphed function.
x3 +5x2 +3x+7
x−1
2
y = 9x x+6
3
2 +2x+8
f (x) = −8x −8x
x+2
3
2 −7x+1
f (x) = 5x −9x
x2 −4
3 +2x2 +5x+2
y = (−2x
(x−2)(x+7)
3 +2x2 +7
y = 4x(x+2)
2
7x3 +2x2 −7x−3
f (x) =
x3
−6x3 +8x2 +7
f (x) =
x2
−5x2 −2x−5
y = x2 +2
y = xx−1
3 −2
6. y =
7.
8.
9.
10.
11.
12.
13.
14.
15.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.8.
251
1.40. Solving Rational Equations
www.ck12.org
1.40 Solving Rational Equations
Here you will extend your knowledge of linear and quadratic equations to rational equations in general. You will
gain insight as to what extraneous solutions are and how to identify them.
The techniques for solving rational equations are extensions of techniques you already know. Recall that when
there are fractions in an equation you can multiply through by the denominator to clear the fraction. The same
technique helps turn rational expressions into polynomials that you already know how to solve. When you multiply
by a constant there is no problem, but when you multiply through by a value that varies and could possibly be zero
interesting things happen.
Since every equation is trivially true when both sides are multiplied by zero, how do you account for this when
solving rational equations?
Finding Solutions to Rational Equations
The first step in solving rational equations is to transform the equation into a polynomial equation. This is
accomplished by clearing the fraction which means multiplying the entire equation by the common denominator
of all the rational expressions. Then you should solve using what you already know. The last thing to check once
you have the solutions is that they do not make the denominators of any part of the equation equal to zero when
substituted back into the original equation. If so, that solution is called extraneous and is a “fake” solution that was
introduced when both sides of the equation were multiplied by a number that happened to be zero.
Take the following rational equation:
5
x − x+3
= 12
To find the solutions of the equation, first multiply all parts of the equation by (x + 3), the common denominator,
and then simplify.
x(x + 3) − 5 = 12(x + 3)
2
x + 3x − 5 − 12x − 36 = 0
x2 − 9x − 41 = 0
p
(−9)2 − 4 · 1 · (−41)
x=
2·1
√
9±7 5
x=
2
−(−9) ±
The only potential extraneous solution would have been -3 since that is the number that makes the denominator of
the original equation zero. Therefore, both answers are possible.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/60820
252
www.ck12.org
Chapter 1. Unit 1
Examples
Example 1
Earlier, you were asked to account for the extra solutions introduced when both sides of an equation are multiplied
by a variable. In order to deal with the possible extra solutions, you must check each solution to see if it makes the
denominator of any fraction in the original equation zero. If it does, it is called an extraneous solution.
Example 2
Solve the following rational equation
3x
x+4
1
2
− x+2
= − x2 +6x+8
Multiply each part of the equation by the common denominator of x2 + 6x + 8 = (x + 2)(x + 4).
1
−2
3x
−
=
(x + 2)(x + 4)
(x + 2)(x + 4)
x+4 x+2
(x + 2)(x + 4)
3x(x + 2) − (x + 4) = −2
3x2 + 6x − x − 2 = 0
3x2 + 5x − 2 = 0
(3x − 1)(x + 2) = 0
1
x = , −2
3
Note that -2 is an extraneous solution. The only actual solution is x = 31 .
Example 3
Solve the following rational equation for y.
x = 2 + 2+1 1
y+1
This question can be done multiple ways. You can use the clearing fractions technique twice.
"
#
1
1
1
2+
x = 2+
2
+
1
y+1
y+1
2 + y+1
x
1
2x +
= 2 2+
+1
y+1
y+1
x
2
2x +
= 4+
+1
y+1
y+1
x
2
(y + 1) 2x +
= 5+
(y + 1)
y+1
y+1
2x(y + 1) + x = 5(y + 1) + 2
2xy + 2x + x = 5y + 5 + 2
Now just get the y variable to one side of the equation and everything else to the other side.
253
1.40. Solving Rational Equations
www.ck12.org
2xy − 5y = −3x + 7
y(2x − 5) = −3x + 7
−3x + 7
y=
2x − 5
Example 4
Solve the following rational equation.
3x
x−5
+4 = x
3x
x−5
+4 = x
3x + 4x − 20 = x2 − 5x
0 = x2 − 12x + 20
0 = (x − 2)(x − 10)
x = 2, 10
Neither solution is extraneous.
Example 5
In electrical circuits, resistance can be solved for using rational expressions. This is an electric circuit diagram with
three resistors. The first resistor R1 is run in series to the other two resistors R2 and R3 which are run in parallel. If
the total resistance R is 100 ohms and R1 and R3 are each 22 ohms, what is the resistance of R2 ?
The equation of value is:
R = R1 +
254
R2 R3
R2 + R3
www.ck12.org
Chapter 1. Unit 1
R3
R = R1 + RR22+R
3
100 = 22 +
x · 22
x + 22
78(x + 22) = 22x
78x + 1716 = 22x
56x = −1716
x = −30.65
A follow up question would be to ask whether or not ohms can be negative which is beyond the scope of this text.
Review
Solve the following rational equations. Identify any extraneous solutions.
1.
2.
3.
4.
5.
6.
7.
8.
2x−4
16
x = x
4
x
x+1 − x+1 = 2
5
2
x+3 + x−3 = 1
3
5
x−4 − x+4 = 6
x
6
x+1 − x+2 = 4
x
4
x−4 − x−4 = 8
4x
x−2 + 3 = 1
−2x
x+1 + 6 = −x
255
1.40. Solving Rational Equations
1
9. x+2
+ 1 = −2x
−6x−3
10. x+1 − 3 = −4x
3
6
11. x+3
x − x+3 = x2 +3x
2
8
12. x−4
x − x−4 = x2 −4x
2
12
13. x+6
x − x+6 = x2 +6x
3
x+5
14. x − x+5 = x215
+5x
15. Explain what it means for a solution to be extraneous.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.6.
256
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
1.41 Clearing Denominators in Rational
Equations
Learning Objectives
Here you’ll learn how to clear denominators in order to solve rational equations.
Suppose that you had 24 quarts of juice to equally distribute to each member of your class. However, 10 members
of your class were absent, so the amount of juice distributed to each person increased by 15 of a quart. How many
people are in your class? What equation could you set up to find the answer to this question?
Solving by Clearing Denominators
When a rational equation has several terms, it may not be possible to use the method of cross-products. A second
method to solve rational equations is to clear the fractions by multiplying the entire equation by the least common
multiple of the denominators.
Let’s solve the following equations by clearing the fractions:
1. 5x − 1x = 4
Start by clearing the denominators. In this case, there is only one denominator, x. This means that x is the
LCM of all the denominators, since there is only one denominator. Simply multiply the whole equation by
that denominator, x. Then solve for x using methods learned previously.
1
=4
x
5x2 − 1 = 4x
5x −
Clear the denominator.
This is quadratic, so make one side equal to zero.
5x2 − 4x − 1 = 0
Factor.
(5x + 1)(x − 1) = 0
1
x = − or x = 1
5
Use the Zero Product Property to solve for the variable.
2.
3
x+2
4
− x−5
=
2
x2 −3x−10
Factor all denominators and find the least common multiple.
3
4
2
−
−
x + 2 x − 5 (x + 2)(x − 5)
LCM = (x + 2)(x − 5)
Multiply all terms in the equation by the LCM and cancel the common terms.
257
1.41. Clearing Denominators in Rational Equations
www.ck12.org
3
4
2
− (x + 2)(x − 5) ·
= (x + 2)(x − 5) ·
x+2
x−5
(x + 2)(x − 5)
4
3
2
(
(
·
− 5) ·
((
− (x + 2)
(x
−
5)
=(
(x
+
2)(x
(x(
+(
2)(x
− 5) ·
(
(
((
x+2
x−5
(x(
+(
2)(x
− 5)
(
(x + 2)(x − 5) ·
Now solve and simplify.
3(x − 5) − 4(x + 2) = 2
3x − 15 − 4x − 8 = 2
x = −25 Answer
Check your answer.
4
3
4
3
−
=
−
= 0.003
x + 2 x − 5 −25 + 2 −25 − 5
2
2
=
= 0.003
2
2
x − 3x − 10 (−25) − 3(−25) − 10
Now, let’s solve the following real-world problem:
A group of friends decided to pool their money together and buy a birthday gift that cost $200. Later 12 of the
friends decided not to participate any more. This meant that each person paid $15 more than the original share. How
many people were in the group to start?
Let x = the number of friends in the original group
TABLE 1.30:
Original group
Later group
Number of People
x
x − 12
Gift Price
200
200
Share Amount
200
x
200
x−12
Since each person’s share went up by $15 after 12 people refused to pay, we write the equation:
200
200
=
+ 15
x − 12
x
Solve by clearing the fractions. Don’t forget to check!
258
www.ck12.org
Chapter 1. Unit 1
LCM = x(x − 12)
200
200
x(x − 12) ·
= x(x − 12) ·
+ x(x − 12) · 15
x − 12
x
· 200 = x(x − 12) · 200 + x(x − 12) · 15
x
(x
−
12)
12
x−
x
200x = 200(x − 12) + 15x(x − 12)
200x = 200x − 2400 + 15x2 − 180x
0 = 15x2 − 180x − 2400
x = 20, x = −8
The answer is 20 people. We discard the negative solution since it does not make sense in the context of this problem.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/180449
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79447
Examples
Example 1
Earlier, you were asked about distributing 24 quarts of juice equally among members of your class. Remember that
10 members of your class were absent, so the amount of juice distributed to each person increased by 15 of a quart.
How many people are in your class? What equation could you set up to find the answer to this question?
Let x = the number of students in your class.
TABLE 1.31:
259
1.41. Clearing Denominators in Rational Equations
www.ck12.org
TABLE 1.31: (continued)
Total class size
Class size day of
Class size
Total Amount of Juice
x
x-10
24
24
Since each person’s share went up by
1
5
Amount of Juice Per Person
24/x
25/x-10
of a quart, we write the equation:
24
24 1
=
+
x − 10
x
5
Solve by clearing fractions.
LCM = x(x − 10)
24
24
1
x(x − 10) ·
= x(x − 10) · + x(x − 10) ·
x − 10
x
5
24
1
24
·
= x(x − 10) · + x(x − 10) ·
x
(x
−
10)
x − 10
5
x
1
24x = 24(x − 10) + x(x − 10)
5
1 2
24x = 24x − 240 + x − 2x
5
x = 40, x = −30
We discard the answer of -30 because a negative number does not make sense in the context of this problem. Your
class consists of 40 people.
Solve
1
x−1
1
− x+1
= 1x .
Since each denominator cannot be a multiple of another denominator, or in other words, the denominators do not
share common factors, the LCM is the product of the 3 denominators:
LCM = (x − 1)(x + 1)(x).
1
1
1
−
=
x−1 x+1 x
1
1
1
Multiply by the LCM. (x − 1)(x + 1)(x) ·
− (x − 1)(x + 1)(x) ·
= (x − 1)(x + 1)(x) ·
x−1
x+1
x
1
1
1
Simplify by cancelling. (x
−
1)(x + 1)(x) · − (x − 1)
(x
+
1)(x) · = (x − 1)(x + 1)(x) ·
x −
1
x +
1
x
(x + 1)(x) − (x − 1)(x) = (x − 1)(x + 1)
Simplify by distributing.
x2 + x − x2 + x = x2 − 1
Simplify by combining like terms.
2x = x2 − 1
Arrange so that one side is equal to zero.
This quadratic equation cannot be factored, so use the quadratic formula to get:
√
x = 1 ± 2.
260
0 = x2 − 2x − 1
www.ck12.org
Chapter 1. Unit 1
Review
Solve the following equations.
1. x + 1x = 2
1
2. −3 + x+1
= 2x
x
3. 1x − x−2
=2
3
2
4. 2x−1 + x+4
=2
2x
x
5. x−1 − 3x+4 = 3
x−4
6. x+1
x−1 + x+4 = 3
x
x
1
7. x−2
+ x+3
= x2 +x−6
2
8. x2 +4x+3
= 2 + x−2
x+3
1
1
9. x+5
− x−5
= 1−x
x+5
1
1
x
+ x−6
= x+6
10. x2 −36
2x
1
2
11. 3x+3
− 4x+4
= x+1
−x
1
12. x−2
+ 3x−1
x+4 = x2 +2x−8
13. Juan jogs a certain distance and then walks a certain distance. When he jogs he averages seven miles per hour.
When he walks, he averages 3.5 miles per hour. If he walks and jogs a total of six miles in a total of seven
hours, how far does he jog and how far does he walk?
14. A boat travels 60 miles downstream in the same time as it takes it to travel 40 miles upstream. The boat’s
speed in still water is 20 miles/hour. Find the speed of the current.
15. Paul leaves San Diego driving at 50 miles/hour. Two hours later, his mother realizes that he forgot something
and drives in the same direction at 70 miles/hour. How long does it take her to catch up to Paul?
16. On a trip, an airplane flies at a steady speed against the wind. On the return trip the airplane flies with the
wind. The airplane takes the same amount of time to fly 300 miles against the wind as it takes to fly 420 miles
with the wind. The wind is blowing at 30 miles/hour. What is the speed of the airplane when there is no wind?
17. A debt of $420 is shared equally by a group of friends. When five of the friends decide not to pay, the share
of the other friends goes up by $25. How many friends were in the group originally?
18. A non-profit organization collected $2,250 in equal donations from their members to share the cost of improving a park. If there were thirty more members, then each member could contribute $20 less. How many
members does this organization have?
Review (Answers)
To see the Review answers, open this PDF file and look for section 12.9.
261
1.42. Unit 1 Test Review
1.42 Unit 1 Test Review
Your unit 1 test will focus on polynomial and rational functions.
262
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
263
1.42. Unit 1 Test Review
264
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
265
1.42. Unit 1 Test Review
266
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
267
1.42. Unit 1 Test Review
Solutions:
268
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
269
1.42. Unit 1 Test Review
270
www.ck12.org
www.ck12.org
Chapter 1. Unit 1
271
1.43. References
www.ck12.org
1.43 References
1. Laura Guerin;CK-12 Foundation. http://www.meritnation.com/img/study_content/lp/1/11/11/162/616/1154
/1365/LP_1.11.1.2.2.2_ANK_KSB_SNK_html_m7c070b53.png;http://classconnection.s3.amazonaws.com/753
/flashcards/449753/png/screen_shot_2011-09-21_at_8.07.32_pm1316650094429.png . CC BY-NC
2. Laura Guerin;CK-12 Foundation. http://www.meritnation.com/img/study_content/lp/1/11/11/162/616/1154
/1365/LP_1.11.1.2.2.2_ANK_KSB_SNK_html_m7c070b53.png;http://classconnection.s3.amazonaws.com/753
/flashcards/449753/png/screen_shot_2011-09-21_at_8.07.32_pm1316650094429.png . CC BY-NC
3. . https://flic.kr/p/7Thh9c .
4. . https://flic.kr/p/7Thh9c .
5. . . CC BY-NC
6. . . CC BY-NC
7. . . CC BY-NC
8. . . CC BY-NC
9. . . CC BY-NC
10. . . CC BY-NC
11. . . CC BY-NC
12. . . CC BY-NC
13. . . CC BY-NC
14. . . CC BY-NC
15. . . CC BY-NC
16. . . CC BY-NC
17. . . CC BY-NC
18. CK-12. CK-12 . CC BY-NC
19. CK-12. CK-12 . CC BY-NC
20. CK-12. CK-12 . CC BY-NC
21. CK-12. CK-12 . CC BY-NC
22. CK-12. CK-12 . CC BY-NC
23. CK-12. CK-12 . CC BY-NC
24. CK-12. CK-12 . CC BY-NC
25. CK-12. CK-12 . CC BY-NC
26. CK-12;Paula Evans. CK-12;Graph created by Desmos . CC BY-NC
27. CK-12;Paula Evans. CK-12;Graph created by Desmos . CC BY-NC
28. CK-12;Paula Evans. CK-12;Graph created by Desmos . CC BY-NC
29. CK-12;Paula Evans. CK-12;Graph created by Desmos . CC BY-NC
30. CK-12. CK-12 . CC BY-NC
31. CK-12. CK-12 . CC BY-NC
32. CK-12 Foundation;European Environment Agency. http://www.eea.europa.eu/data-and-maps/figures/atmos
pheric-concentration-of-co2-ppm-1;CK-12 Foundation . CCSA
33. CK-12 Foundation. http://colouringbook.org/art/svg/coloring-book/roller-coaster-coloring-book-colouring-co
loring-book-colouring-book-colouringbook-org-art-clip-art-clipart-clipartist-net-openclipart-org-scalable-vector
-graphics-svg/ . CCSA
34. CK-12 Foundation. . CCSA
35. CK-12 Foundation. . CCSA
36. CK-12 Foundation. . CCSA
37. CK-12 Foundation. . CCSA
38. CK-12 Foundation. . CCSA
272
www.ck12.org
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
55.
56.
57.
58.
59.
60.
61.
62.
63.
64.
65.
66.
67.
68.
69.
70.
71.
72.
73.
74.
75.
76.
77.
78.
79.
80.
81.
Chapter 1. Unit 1
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. https://www.flickr.com/photos/davidstanleytravel/8227858392/in/photolist-dx4VDs-nKC9
j9-My5FPU-9rfFDj-rXVj6c-G9QvSn-FTuBdG-NVoCS3-8qRnga-4ykGxj-jeRZix-Nm65u9-MuS8RM-MtrYpqnRand5-LSSMHM-GjMCi9-aaGbZM-opao7s-2Cact-FbjHE8-5nGfWn-pniMjM-6C6LQc-xzgo7p-4jAyjp-dQ
wCx8-Nh11cH-qrnGo4-qq1pzb-j5K9gk-9Pu1uQ-8BxKVf-5J85eG-j5LBAg-cfh4yL-5nLVFj-jeTveb-PSNJ1-K2fT
UM-sngP1k-dBmaU6-hMPbWH-pk7mb-6R8PPi-ab6auH-KzX648-4oyMEc-9NHJGZ-Ny57L3 . CCSA
CK-12 Foundation. https://www.flickr.com/photos/davidstanleytravel/8227858392/in/photolist-dx4VDs-nKC9
j9-My5FPU-9rfFDj-rXVj6c-G9QvSn-FTuBdG-NVoCS3-8qRnga-4ykGxj-jeRZix-Nm65u9-MuS8RM-MtrYpqnRand5-LSSMHM-GjMCi9-aaGbZM-opao7s-2Cact-FbjHE8-5nGfWn-pniMjM-6C6LQc-xzgo7p-4jAyjp-dQ
wCx8-Nh11cH-qrnGo4-qq1pzb-j5K9gk-9Pu1uQ-8BxKVf-5J85eG-j5LBAg-cfh4yL-5nLVFj-jeTveb-PSNJ1-K2fT
UM-sngP1k-dBmaU6-hMPbWH-pk7mb-6R8PPi-ab6auH-KzX648-4oyMEc-9NHJGZ-Ny57L3 . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
. https://commons.wikimedia.org/wiki/File:NM_124_and_US_66_WB_near_Budville_NM.jpg .
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. https://en.wikipedia.org/wiki/File:Us127_east_lansing.jpg . CC BY-NC-SA;CCSA
CK-12 Foundation. . CCSA
CK-12 Foundation. . CC BY-NC-SA;CCSA
. . CC BY-NC-SA
CK-12. CK-12 .
CK-12. CK-12 .
Larame Spence, CK-12 Foundation. CK12.org .
Pete. https://www.flickr.com/photos/comedynose/8882744897/in/ .
. https://www.flickr.com/photos/18521020@N04/16722209219/in/p . CC BY-NC
. https://www.flickr.com/photos/18521020@N04/16722209219/in/p . CC BY-NC
. https://www.flickr.com/photos/18521020@N04/16722209219/in/p . CC BY-NC
. . CC BY-NC
273
1.43. References
www.ck12.org
82. . . CC BY-NC
83. CK-12 Foundation;Desmos. Desmos .
84. NASA’s Glenn Research Center. https://en.wikipedia.org/wiki/File:Boyles_Law_animated.gif;https://www.f
lickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc-4ZY3dc-5hDv2p
-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3-513fUN-bWmBmD
-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6oDf1e-a73VQD-4ZY3
2V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmATZ-6zuyfQ-89kZnn-bWmBe
Z-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
85. NASA’s Glenn Research Center. https://en.wikipedia.org/wiki/File:Boyles_Law_animated.gif;https://www.f
lickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc-4ZY3dc-5hDv2p
-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3-513fUN-bWmBmD
-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6oDf1e-a73VQD-4ZY3
2V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmATZ-6zuyfQ-89kZnn-bWmBe
Z-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
86. NASA’s Glenn Research Center. https://en.wikipedia.org/wiki/File:Boyles_Law_animated.gif;https://www.f
lickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc-4ZY3dc-5hDv2p
-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3-513fUN-bWmBmD
-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6oDf1e-a73VQD-4ZY3
2V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmATZ-6zuyfQ-89kZnn-bWmBe
Z-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
87. NASA’s Glenn Research Center. https://en.wikipedia.org/wiki/File:Boyles_Law_animated.gif;https://www.f
lickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc-4ZY3dc-5hDv2p
-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3-513fUN-bWmBmD
-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6oDf1e-a73VQD-4ZY3
2V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmATZ-6zuyfQ-89kZnn-bWmBe
Z-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
88. NASA’s Glenn Research Center. https://en.wikipedia.org/wiki/File:Boyles_Law_animated.gif;https://www.f
lickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc-4ZY3dc-5hDv2p
-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3-513fUN-bWmBmD
-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6oDf1e-a73VQD-4ZY3
2V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmATZ-6zuyfQ-89kZnn-bWmBe
Z-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
89. . https://www.flickr.com/photos/oddwick/941004365/in/photolist-2r9TPc-cBdwKd-aLq9ek-4ZY5qP-4RPGUc4ZY3dc-5hDv2p-8faGWS-a4vjvY-513fNw-bqmxDX-bWmB5H-513fGG-bWmB2i-zgcjVy-cdHW8L-89peS3513fUN-bWmBmD-7Juwon-4wPEDg-bWmAAD-FDyUp-bWmAn4-89peXW-4ZY4YT-bWmAwt-7ED3V-6
oDf1e-a73VQD-4ZY32V-62XeHW-cdHX63-9agWUA-8WCXQf-bWmArv-97tmxz-89peLh-G7mc9-bWmAT
Z-6zuyfQ-89kZnn-bWmBeZ-4F3zbR-tMiQgG-a4stmF-dojsMU-5EQfEd-6S2puk-bzvP78 .
90. CK-12 Foundation. . CCSA
91. CK-12 Foundation. . CCSA
92. CK-12 Foundation. . CCSA
93. CK-12 Foundation. . CCSA
94. CK-12 Foundation. . CCSA
95. CK-12 Foundation. . CCSA
96. CK-12 Foundation. . CCSA
97. Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
98. Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
274
www.ck12.org
99.
100.
101.
102.
103.
104.
Chapter 1. Unit 1
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
Desmos. https://www.flickr.com/photos/88158121@N00/5264675378/in/photolist-92dQz9-Eu1H8y-dNnzdLgg8Pu-4joH9Q-5EvPtm-4RvqPH-6WffXz-77cJAq-6Cskyq-KGKy88-f4g1D8-bHRJ2k-qABFH-72QBBS-9a
dYtH-5qsLfp-6i74v3-7fbqBV-7DeAxE-fv6wKA-eB2RuH-6nnhEr-5b8kAz-4Rvqix-5TuvAc-4RzBbA-dPKHH9
-7w2emB-5tihMm-5aXaUG-7HE7Vn-5WMHB3-6bACZC-5jXasy-9r9DTQ-5NHzWn-9EvNVu-4RMXjF-5T
uwq4-pnbPHX-6QgwvY-nKCW6m-cgvCnm-cVoUYY-a1zW65-eL4KwY-5kDwF2-t6jh6G-dZ9anM .
CK-12. CK-12 .
CK-12 Foundation. . CCSA
275
www.ck12.org
C HAPTER
2
Unit 2
Chapter Outline
2.1
T RIGONOMETRIC R ATIOS WITH A C ALCULATOR
2.2
I NTRODUCTION TO T RIG I DENTITIES
2.3
M ULTIPLICATION OF P OLYNOMIALS BY B INOMIALS
2.4
D ISTANCE F ORMULA
2.5
R EFERENCE A NGLES AND A NGLES IN THE U NIT C IRCLE
2.6
C ONVERSION BETWEEN D EGREES AND R ADIANS
2.7
R ADIAN M EASURE
2.8
A RC L ENGTH
2.9
A PPLICATIONS OF R ADIAN M EASURE
2.10
T HE U NIT C IRCLE
2.11
T RIGONOMETRIC R ATIOS ON THE U NIT C IRCLE
2.12
R ECIPROCAL T RIGONOMETRIC F UNCTIONS
2.13
C OFUNCTION I DENTITIES AND R EFLECTION
2.14
F INDING E XACT T RIG VALUES USING S UM AND D IFFERENCE F ORMULAS
2.15
S INE S UM AND D IFFERENCE F ORMULAS
2.16
C OSINE S UM AND D IFFERENCE F ORMULAS
2.17
TANGENT S UM AND D IFFERENCE F ORMULAS
2.18
S OLVING T RIG E QUATIONS USING S UM AND D IFFERENCE F ORMULAS
2.19
F INDING E XACT T RIG VALUES USING D OUBLE AND H ALF A NGLE F ORMULAS
2.20
S UM AND D IFFERENCE I DENTITIES
2.21
D OUBLE A NGLE I DENTITIES
2.22
S IMPLIFYING T RIG E XPRESSIONS USING D OUBLE AND H ALF A NGLE F ORMU LAS
276
2.23
D OUBLE , H ALF, AND P OWER R EDUCING I DENTITIES
2.24
S IX T RIGONOMETRIC F UNCTIONS AND R ADIANS
2.25
G RAPHING S INE AND C OSINE
2.26
A MPLITUDE , P ERIOD, AND F REQUENCY
2.27
H ORIZONTAL T RANSLATIONS OR P HASE S HIFTS
2.28
G RAPHING TANGENT
2.29
V ERTICAL T RANSLATIONS
2.30
S INE AND C OSECANT G RAPHS
2.31
C OSINE AND S ECANT G RAPHS
2.32
TANGENT AND C OTANGENT G RAPHS
www.ck12.org
Chapter 2. Unit 2
2.33
P UTTING IT ALL TOGETHER
2.34
T RIGONOMETRIC E QUATIONS U SING FACTORING
2.35
T RIGONOMETRIC E QUATIONS U SING THE Q UADRATIC F ORMULA
2.36
T RIG . W ORD P ROBLEMS AND IROC
2.37
U NIT 2 T EST R EVIEW
2.38
R EFERENCES
277
2.1. Trigonometric Ratios with a Calculator
www.ck12.org
2.1 Trigonometric Ratios with a Calculator
Learning Objectives
Here you’ll learn how to solve for missing sides in right triangles that are not one of the special right triangles.
What if you wanted to find the missing sides of a right triangle with angles of 20◦ and 70◦ and a hypotenuse length
of 10 inches? How could you use trigonometry to help you?
Trigonometric Ratios with a Calculator
The trigonometric ratios are not dependent on the exact side lengths, but the angles. There is one fixed value for
every angle, from 0◦ to 90◦ . Your scientific (or graphing) calculator knows the values of the sine, cosine and tangent
of all of these angles. Depending on your calculator, you should have [SIN], [COS], and [TAN] buttons. Use these
to find the sine, cosine, and tangent of any acute angle. One application of the trigonometric ratios is to use them to
find the missing sides of a right triangle. All you need is one angle, other than the right angle, and one side.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/137588
Calculating Trigonometric Functions with a Calculator
Find the trigonometric value, using your calculator. Round to 4 decimal places.
Depending on your calculator, you enter the degree and then press the trig button or the other way around. Also,
make sure the mode of your calculator is in DEGREES.
a) sin 78◦
sin 78◦ = 0.97815
b) cos 60◦
cos 60◦ = 0.5
c) tan 15◦
tan 15◦ = 0.26795
278
www.ck12.org
Chapter 2. Unit 2
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/1371
Solving for Unknown Values
1. Find the value of each variable. Round your answer to the nearest tenth.
We are given the hypotenuse. Use sine to find b, and cosine to find a. Use your calculator to evaluate the sine and
cosine of the angles.
b
30
30 · sin 22◦ = b
sin 22◦ =
a
30
30 · cos 22◦ = a
cos 22◦ =
b ≈ 11.2
a ≈ 27.8
2. Find the value of each variable. Round your answer to the nearest tenth.
We are given the adjacent leg to 42◦ . To find c, use cosine and use tangent to find d.
ad jacent
9
=
hypotenuse c
◦
c · cos 42 = 9
9
c=
≈ 12.1
cos 42◦
cos 42◦ =
opposite
d
=
ad jacent
9
◦
9 · tan 42 = d
tan 42◦ =
d ≈ 27.0
279
2.1. Trigonometric Ratios with a Calculator
www.ck12.org
Any time you use trigonometric ratios, use only the information that you are given. This will result in the most
accurate answers.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/137589
Earlier Problem Revisited
Use trigonometric ratios to find the missing sides. Round to the nearest tenth.
Find the length of a and b using sine or cosine ratios:
280
www.ck12.org
Chapter 2. Unit 2
a
10
◦
10 · cos 20 = a
a
10
◦
10 · sin 70 = a
a ≈ 9.4
a ≈ 9.4
b
10
10 · sin 20◦ = b
b
10
10 · cos 70◦ = b
b ≈ 3.4
b ≈ 3.4
cos 20◦ =
sin 20◦ =
sin 70◦ =
cos 70◦ =
Examples
Example 1
What is tan 45◦ ?
Using your calculator, you should find that tan 45◦ = 1
Example 2
Find the length of the missing sides and round your answers to the nearest tenth:
.
Use tangent for x and cosine for y.
tan 28◦ =
x
11
11 · tan 28◦ = x
x ≈ 5.8
cos 28◦ =
11
y
11
=y
cos 28◦
y ≈ 12.5
Example 3
Find the length of the missing sides and round your answers to the nearest tenth:
281
2.1. Trigonometric Ratios with a Calculator
www.ck12.org
.
Use tangent for y and cosine for x.
tan 40◦ =
y
16
cos 40◦ =
16 · tan 40◦ = y
y ≈ 13.4
16
x
16
=x
cos 40◦
x ≈ 20.9
Review
Use your calculator to find the value of each trig function below. Round to four decimal places.
1.
2.
3.
4.
5.
6.
7.
sin 24◦
cos 45◦
tan 88◦
sin 43◦
tan 12◦
cos 79◦
sin 82◦
Find the length of the missing sides. Round your answers to the nearest tenth.
8.
9.
10.
11.
12. Find sin 80◦ and cos 10◦ .
282
www.ck12.org
Chapter 2. Unit 2
13. Use your knowledge of where the trigonometric ratios come from to explain your result to the previous
question.
14. Generalize your result to the previous two questions. If sin θ = x, then cos? = x.
15. How are tan θ and tan(90 − θ) related? Explain.
Review (Answers)
To view the Review answers, open this PDF file and look for section 8.7.
283
2.2. Introduction to Trig Identities
www.ck12.org
2.2 Introduction to Trig Identities
Learning Objectives
Here you’ll learn about the basic trigonometric identities and how to use them.
You are given a list of Trig Identities. One of those identities is cos(−θ) = cos θ. Prove this identity without
graphing.
Trigonometric Identities
Trigonometric identities are true for any value of x (as long as the value is in the domain). You have learned about
secant, cosecant, and cotangent, which are all reciprocal functions of sine, cosine and tangent. These functions can
be rewritten as the Reciprocal Identities because they are always true.
Reciprocal Identities: csc θ =
1
sin θ
sec θ =
1
cos θ
cot θ =
1
tan θ
Other identities involve the tangent, variations on the Pythagorean Theorem, phase shifts, and negative angles. We
will discover them in this concept.
We know that tan θ =
opposite
ad jacent .
Let’s show that tan θ =
sin θ
cos θ .
This is the Tangent Identity.
Whenever we are trying to verify, or prove, an identity, we start with the statement we are trying to prove and
sin θ
work towards the desired answer. In this case, we will start with tan θ = cos
θ and show that it is equivalent to
opposite
tan θ = ad jacent . First, rewrite sine and cosine in terms of the ratios of the sides.
tan θ =
=
sin θ
cos θ
opposite
hypotenuse
ad jacent
hypotenuse
Then, rewrite the complex fraction as a division problem and simplify.
opposite
ad jacent
÷
hypotenuse hypotenuse
(
hypotenuse
opposite (
((((
·
=
(
hypotenuse
(((( ad jacent
(
opposite
=
ad jacent
=
We now have what we wanted to prove and we are done. Once you verify an identity, you may use it to verify
other identities.
Now, let’s show that sin2 θ + cos2 θ = 1 is a true identity.
Change the sine and cosine in the equation into the ratios. In this problem, we will use y as the opposite side, x is
the adjacent side, and r is the hypotenuse (or radius), as in the unit circle.
284
www.ck12.org
Chapter 2. Unit 2
sin2 θ + cos2 θ = 1
y 2 x 2
+
=1
r
r
y2 x2
+ =1
r2 r2
y2 + x2
=1
r2
Now, x2 + y2 = r2 from the Pythagorean Theorem. Substitute this in for the numerator of the fraction.
r2
=1
r2
This is one of the Pythagorean Identities and very useful.
Finally, let’s verify that sin π2 − θ = cos θ by using the graphs of the functions.
The function y = sin π2 − x is a phase shift of π2 of the sine curve.
The red function above is y = sin x and the blue is y = cos x. If we were to shift the sine curve π2 , it would overlap
perfectly with the cosine curve, thus proving this Cofunction Identity.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/91979
Examples
Example 1
Earlier, you were asked to prove the identity of cos(−θ) = cos θ without graphing.
First, recall that cos θ = x, where (x, y) is the endpoint of the terminal side of θ on the unit circle.
Now, if we have cos(−θ), what is its endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise
direction, rather than the usual counter-clockwise. If we make this rotation, we see that cos(−θ) = x as well, as
illustrated in the following diagram.
285
2.2. Introduction to Trig Identities
www.ck12.org
We know that x = x, so we can set the two expressions equal to one another.
cos θ = cos(−θ)
We can now flip this identity around to get:
cos(−θ) = cos θ
Example 2
Prove the Pythagorean Identity: 1 + tan2 θ = sec2 θ.
First, let’s use the Tangent Identity and the Reciprocal Identity to change tangent and secant in terms of sine and
cosine.
1 + tan2 θ = sec2 θ
1+
1
sin2 θ
=
cos2 θ cos2 θ
Now, change the 1 into a fraction with a base of cos2 θ and simplify.
sin2 θ
cos2 θ
2
cos θ sin2 θ
+
cos2 θ cos2 θ
cos2 θ + sin2 θ
cos2 θ
1
cos2 θ
1+
1
cos2 θ
1
=
cos2 θ
1
=
cos2 θ
1
=
cos2 θ
=
In the second to last step, we arrived at the original Pythagorean Identity sin2 θ + cos2 θ in the numerator of the
left-hand side. Therefore, we can substitute in 1 for this and the two sides of the equation are the same.
Example 3
Without graphing, show that sin(−θ) = − sin θ.
First, recall that sin θ = y, where (x, y) is the endpoint of the terminal side of θ on the unit circle.
286
www.ck12.org
Chapter 2. Unit 2
Now, if we have sin(−θ), what is it’s endpoint? Well, the negative sign tells us that the angle is rotated in a clockwise
direction, rather than the usual counter-clockwise. By looking at the picture, we see that sin(−θ) = −y. Therefore,
if sin θ = y, then − sin θ = −y and combining the equations, we have sin(−θ) = − sin θ.
Review
1.
2.
3.
4.
5.
6.
θ
Show that cot θ = cos
sin θ .
θ
Show that tan θ = sec
csc θ .
2
2 θ.
Show that 1 + cot θ = csc
Explain why cos π2 − θ = sin θ by using the graphs of the two functions.
Show that sec(−θ) = sec θ.
Explain why tan(−θ) = − tan θ is true, using the Tangent Identity and the other Negative Angle Identities.
Verify the following identities.
7.
8.
9.
10.
11.
12.
cot θ sec θ = csc θ
cos θ
tan θ
cot θ = sec θ
sin θ csc θ = 1
cot(−θ) = − cot θ
tan2 x csc x cos x = 1
sin (−x)
= cos2 x
tan2 x
Show that sin2 θ + cos2 θ = 1 is true for the following values of θ.
13. π4
14. 2π
3
15. − 7π
6
16. Recall that a function is odd if f (−x) = − f (x) and even if f (−x) = f (x). Which of the six trigonometric
functions are odd? Which are even?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.6.
287
2.3. Multiplication of Polynomials by Binomials
www.ck12.org
2.3 Multiplication of Polynomials by Binomials
Learning Objectives
Here you’ll learn how to multiply one polynomial by another and simplify your answer. You’l also learn how to
solve real-world problems using multiplication of polynomials.
Multiplication of Polynomials by Binomials
Let’s start by multiplying two binomials together. A binomial is a polynomial with two terms, so a product of two
binomials will take the form (a + b)(c + d).
We can still use the Distributive Property here if we do it cleverly. First, let’s think of the first set of parentheses as
one term. The Distributive Property says that we can multiply that term by c, multiply it by d, and then add those
two products together: (a + b)(c + d) = (a + b) · c + (a + b) · d.
We can rewrite this expression as c(a + b) + d(a + b). Now let’s look at each half separately. We can apply the
distributive property again to each set of parentheses in turn, and that gives us c(a+b)+d(a+b) = ca+cb+da+db.
What you should notice is that when multiplying any two polynomials, every term in one polynomial is multiplied
by every term in the other polynomial.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133001
Multiplying and Simplifying
1. Multiply and simplify: (2x + 1)(x + 3)
We must multiply each term in the first polynomial by each term in the second polynomial. Let’s try to be systematic
to make sure that we get all the products.
First, multiply the first term in the first set of parentheses by all the terms in the second set of parentheses.
288
www.ck12.org
Chapter 2. Unit 2
Now we’re done with the first term. Next we multiply the second term in the first parentheses by all terms in the
second parentheses and add them to the previous terms.
Now we’re done with the multiplication and we can simplify:
(2x)(x) + (2x)(3) + (1)(x) + (1)(3) = 2x2 + 6x + x + 3 = 2x2 + 7x + 3
This way of multiplying polynomials is called in-line multiplication or horizontal multiplication. Another method
for multiplying polynomials is to use vertical multiplication, similar to the vertical multiplication you learned with
regular numbers.
2. Multiply and simplify:
a) (4x − 5)(x − 20)
With horizontal multiplication this would be
(4x − 5)(x − 20) = (4x)(x) + (4x)(−20) + (−5)(x) + (−5)(−20)
= 4x2 − 80x − 5x + 100
= 4x2 − 85x + 100
To do vertical multiplication instead, we arrange the polynomials on top of each other with like terms in the same
columns:
4x − 5
x − 20
− 80x + 100
2
4x − 5x
4x2 − 85x + 100
Both techniques result in the same answer: 4x2 − 85x + 100. We’ll use vertical multiplication for the other problems.
b) (3x − 2)(3x + 2)
3x − 2
3x + 2
6x − 4
9x2 − 6x
9x2 + 0x − 4
289
2.3. Multiplication of Polynomials by Binomials
www.ck12.org
The answer is 9x2 − 4.
c) (3x2 + 2x − 5)(2x − 3)
It’s better to place the smaller polynomial on the bottom:
3x2 + 2x − 5
2x − 3
− 9x2 − 6x + 15
6x3 + 4x2 − 10x
6x3 − 5x2 − 16x + 15
The answer is 6x3 − 5x2 − 16x + 15.
d) (x2 − 9)(4x4 + 5x2 − 2)
Set up the multiplication vertically and leave gaps for missing powers of x:
4x4 + 5x2 − 2
x2 − 9
− 36x4 − 45x2 + 18
4x6 + 5x4 − 2x2
4x6 − 31x4 − 47x2 + 18
The answer is 4x6 − 31x4 − 47x2 + 18.
Solve Real-World Problems Using Multiplication of Polynomials
In this section, we’ll see how multiplication of polynomials is applied to finding the areas and volumes of geometric
shapes.
Real-World Application: Finding Areas and Volumes
a) Find the areas of the figure:
We use the formula for the area of a rectangle: Area = length × width.
For the big rectangle:
290
www.ck12.org
Chapter 2. Unit 2
Length = b + 3, Width = b + 2
Area = (b + 3)(b + 2)
= b2 + 2b + 3b + 6
= b2 + 5b + 6
b) Find the volumes of the figure:
The volume of this shape = (area of the base)(height).
Area of the base = x(x + 2)
= x2 + 2x
Height = 2x + 1
Volume = (x2 + 2x)(2x + 1)
= 2x3 + x2 + 4x2 + 2x
= 2x3 + 5x2 + 2x
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133002
Examples
Example 1
Find the areas of the figure:
291
2.3. Multiplication of Polynomials by Binomials
www.ck12.org
We could add up the areas of the blue and orange rectangles, but it’s easier to just find the area of the whole big
rectangle and subtract the area of the yellow rectangle.
Area of big rectangle = 20(12) = 240
Area of yellow rectangle = (12 − x)(20 − 2x)
= 240 − 24x − 20x + 2x2
= 240 − 44x + 2x2
= 2x2 − 44x + 240
The desired area is the difference between the two:
Area = 240 − (2x2 − 44x + 240)
= 240 + (−2x2 + 44x − 240)
= 240 − 2x2 + 44x − 240
= −2x2 + 44x
Example 2
Find the volumes of the figure:
292
www.ck12.org
Chapter 2. Unit 2
The volume of this shape = (area of the base)(height).
Area of the base = (4a − 3)(2a + 1)
= 8a2 + 4a − 6a − 3
= 8a2 − 2a − 3
Height = a + 4
Volume = (8a2 − 2a − 3)(a + 4)
Let’s multiply using the vertical method:
8a2 − 2a − 3
a+4
2
32a − 8a − 12
3
8a − 2a2 − 3a
8a3 + 30a2 − 11a − 12
The volume is 8a3 + 30a2 − 11a − 12.
Review
Multiply and simplify.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
(x − 3)(x + 2)
(a + b)(a − 5)
(x + 2)(x2 − 3)
(a2 + 2)(3a2 − 4)
(7x − 2)(9x − 5)
(2x − 1)(2x2 − x + 3)
(3x + 2)(9x2 − 6x + 4)
(a2 + 2a − 3)(a2 − 3a + 4)
3(x − 5)(2x + 7)
5x(x + 4)(2x − 3)
Find the areas of the following figures.
293
2.3. Multiplication of Polynomials by Binomials
11.
12.
Find the volumes of the following figures.
13.
14.
Review (Answers)
To view the Review answers, open this PDF file and look for section 9.4.
294
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.4 Distance Formula
Learning Objectives
Here you’l learn how to use the distance formula to find the distance between two points in the coordinate plane.
You’ll also learn how to find the missing coordinate of a point given its distance from another known point. Finally,
you’ll solve real-world applications using the distance formula.
The Distance Formula
In the last section, we saw how to use the Pythagorean Theorem to find lengths. In this section, you’ll learn how to
use the Pythagorean Theorem to find the distance between two coordinate points.
Finding the Distance Between Two Points
1. Find the distance between points A = (1, 4) and B = (5, 2).
Plot the two points on the coordinate plane.
In order to get from point A = (1, 4) to point B = (5, 2), we need to move 4 units to the right and 2 units down. These
lines make the legs of a right triangle.
To find the distance between A and B we find the value of the hypotenuse, d, using the Pythagorean Theorem.
d 2 = 22 + 42 = 20
√
√
d = 20 = 2 5 = 4.47
2. Find the distance between points C = (2, −1) and D = (−3, −4).
295
2.4. Distance Formula
www.ck12.org
We plot the two points on the graph above.
In order to get from point C to point D, we need to move 3 units down and 5 units to the left.
We find the distance from C to D by finding the length of d with the Pythagorean Theorem.
d 2 = 32 + 52 = 34
√
d = 34 = 5.83
The Distance Formula
The procedure we just used can be generalized by using the Pythagorean Theorem to derive a formula for the distance
between any two points on the coordinate plane.
Let’s find the distance between two general points A = (x1 , y1 ) and B = (x2 , y2 ).
Start by plotting the points on the coordinate plane:
In order to move from point A to point B in the coordinate plane, we move x2 − x1 units to the right and y2 − y1 units
up.
We can find the length d by using the Pythagorean Theorem:
d 2 = (x1 − x2 )2 + (y1 − y2 )2
q
(x1 − x2 )2 + (y1 − y2 )2 . This is called the Distance Formula. More formally:
q
Given any two points (x1 , y1 ) and (x2 , y2 ), the distance between them is d = (x1 − x2 )2 + (y1 − y2 )2 .
Therefore, d =
We can use this formula to find the distance between any two points on the coordinate plane. Notice that the distance
is the same whether you are going from point A to point B or from point B to point A, so it does not matter which
order you plug the points into the distance formula.
Let’s now apply the distance formula to the following examples.
296
www.ck12.org
Chapter 2. Unit 2
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133308
Using the Distance Formula
Plug the values of the two points into the distance formula. Be sure to simplify if possible.
a) (-3, 5) and (4, -2)
q
√
√
√
d = (−3 − 4)2 + (5 − (−2))2 = 49 + 49 = 98 = 7 2
b) (12, 16) and (19, 21)
q
√
√
d = (12 − 19)2 + (16 − 21)2 = 49 + 25 = 74
c) (11.5, 2.3) and (-4.2, -3.9)
q
√
d = (11.5 + 4.2)2 + (2.3 + 3.9)2 = 284.93 = 16.88
Applications Using Distance and Midpoint Formulas
The distance and midpoint formula are useful in geometry situations where we want to find the distance between
two points or the point halfway between two points.
Proving that a Triangle is Isosceles
Plot the points A = (4, −2), B = (5, 5), and C = (−1, 3) and connect them to make a triangle. Show that the triangle
is isosceles.
Let’s start by plotting the three points on the coordinate plane and making a triangle:
297
2.4. Distance Formula
www.ck12.org
We use the distance formula three times to find the lengths of the three sides of the triangle.
q
q
√
√
(4 − 5)2 + (−2 − 5)2 = (−1)2 + (−7)2 = 50 = 5 2
q
q
√
√
BC = (5 + 1)2 + (5 − 3)2 = (6)2 + (2)2 = 40 = 2 10
q
q
√
√
AC = (4 + 1)2 + (−2 − 3)2 = (5)2 + (−5)2 = 50 = 5 2
AB =
Notice that AB = AC, therefore triangle ABC is isosceles.
Real-World Application: Walking Speed
At 8 AM one day, Amir decides to walk in a straight line on the beach. After two hours of making no turns and
traveling at a steady rate, Amir is two miles east and four miles north of his starting point. How far did Amir walk
and what was his walking speed?
Let’s start by plotting Amir’s route on a coordinate graph. We can place his starting point at the origin: A = (0, 0).
Then his ending point will be at B = (2, 4).
298
www.ck12.org
Chapter 2. Unit 2
The distance can be found with the distance formula:
d=
q
q
√
√
(2 − 0)2 + (4 − 0)2 = (2)2 + (4)2 = 4 + 16 = 20
d = 4.47 miles
Since Amir walked 4.47 miles in 2 hours, his speed is s =
4.47 miles
2 hours
= 2.24 mi/h.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133309
Example
Example 1
Find all points on the line y = 2 that are exactly 8 units away from the point (-3, 7).
Let’s make a sketch of the given situation.
Draw line segments from the point (-3, 7) to the line y = 2.
Let k be the missing value of x we are seeking.
299
2.4. Distance Formula
www.ck12.org
Let’s use the distance formula:
Square both sides of the equation:
Therefore:
Use the quadratic formula:
Therefore:
q
8 = (−3 − k)2 + (7 − 2)2
64 = (−3 − k)2 + 25
0 = 9 + 6k + k2 − 39 or 0 = k2 + 6k − 30
√
√
−6 ± 36 + 120 −6 ± 156
k=
=
2
2
k = 3.24 or k = −9.24
The points are (-9.24, 2) and (3.24, 2).
Review
Find the distance between the two points.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
(3, -4) and (6, 0)
(-1, 0) and (4, 2)
(-3, 2) and (6, 2)
(0.5, -2.5) and (4, -4)
(12, -10) and (0, -6)
(-5, -3) and (-2, 11)
(2.3, 4.5) and (-3.4, -5.2)
Find all points having an x−coordinate of -4 whose distance from the point (4, 2) is 10.
Find all points having a y−coordinate of 3 whose distance from the point (-2, 5) is 8.
Find three points that are each 13 units away from the point (3, 2) but do not have an x−coordinate of 3 or a
y−coordinate of 2.
11. Plot the points A = (1, 0), B = (6, 4),C = (9, −2) and D = (−6, −4), E = (−1, 0), F = (2, −6). Prove that
triangles ABC and DEF are congruent.
12. Plot the points A = (4, −3), B = (3, 4),C = (−2, −1), D = (−1, −8). Show that ABCD is a rhombus (all sides
are equal)
300
www.ck12.org
Chapter 2. Unit 2
13. Plot points A = (−5, 3), B = (6, 0),C = (5, 5). Find the length of each side. Show that ABC is a right triangle.
Find its area.
14. Find the area of the circle with center (-5, 4) and a point on the circle (3, 2).
15. Michelle decides to ride her bike one day. First she rides her bike due south for 12 miles and then she rides
in a new direction for a while longer. When she stops Michelle is 2 miles south and 10 miles west from her
starting point. Find the total distance that Michelle covered from her starting point.
Review (Answers)
To view the Review answers, open this PDF file and look for section 11.11.
301
2.5. Reference Angles and Angles in the Unit Circle
www.ck12.org
2.5 Reference Angles and Angles in the Unit
Circle
Learning Objectives
Here you’ll learn the definition of reference angles and how to express angles on the unit circle.
When you walk into math class one day, your teacher has a surprise for the class. You’re going to play series of
games related to the things you’ve been learning about in class. For the first game, your teacher hands each group
of students a spinner with an "x" and "y" axis marked. The game is to see how many angles you identify correctly.
However, in this game, you are supposed to give what is called the "reference angle". You spin your spinner three
times. Each picture below shows one of the spins:
302
www.ck12.org
Chapter 2. Unit 2
Can you correctly identify the reference angles for these pictures?
Reference Angles
Consider the angle 150◦ . If we graph this angle in standard position, we see that the terminal side of this angle is a
reflection of the terminal side of 30◦ , across the y−axis.
303
2.5. Reference Angles and Angles in the Unit Circle
www.ck12.org
Notice that 150◦ makes a 30◦ angle with the negative x−axis. Therefore we say that 30◦ is the reference angle for
150◦ . Formally, the reference angle of an angle in standard position is the angle formed with the closest portion of
the x−axis. Notice that 30◦ is the reference angle for many angles. For example, it is the reference angle for 210◦
and for −30◦ .
In general, identifying the reference angle for an angle will help you determine the values of the trig functions of the
angle.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70413
Identifying Reference Angles
Graph each of the following angles and identify their reference angles.
a. 140◦
140◦ makes a 40◦ angles with the negative x−axis. Therefore the reference angle is 40◦ .
b. 240◦
240◦ makes a 60◦ angle with the negative x−axis. Therefore the reference angle is 60◦
304
www.ck12.org
Chapter 2. Unit 2
c. 380◦
380◦ is a full rotation of 360◦ , plus an additional 20◦ . So this angle is co-terminal with 20◦ , and 20◦ is its reference
angle.
Determining the Value of Trigonometric Functions
1. Find the ordered pair for 240◦ and use it to find the value of sin 240◦ .
√
sin 240◦ = − 2 3
As we found in part b under the question above, the reference angle for 240◦ is 60◦ . The figure below shows 60◦ and
the three other angles in the unit circle that have 60◦ as a reference angle.
305
2.5. Reference Angles and Angles in the Unit Circle
www.ck12.org
◦
The terminal side of the angle
a reflection of the terminal side of 60◦ over √
both axes. So the
240 √represents
3
1
coordinates of the point are − 2 , − 2 . The y−coordinate is the sine value, so sin 240◦ = − 2 3 .
Just as the figure above shows 60◦ and three related angles, we can make similar graphs for 30◦ and 45◦ .
Knowing these ordered pairs will help you find the value of any of the trig functions for these angles.
2. Find the value of cot 300◦
cot 300◦ = − √1
3
Using the graph above, you will find that the ordered pair is
x
y
1
=
2
√
− 3
2
306
= 12 × − √2 = − √1
3
3
√ . Therefore the cotangent value is cot 300◦ =
3
1
2,− 2
www.ck12.org
Chapter 2. Unit 2
We can also use the concept of a reference angle and the ordered pairs we have identified to determine the values of
the trig functions for other angles.
Examples
Example 1
Earlier, you were asked if you can correctly identify the reference angles in the pictures.
Since you know how to measure reference angles now, upon examination of the spinners, you know that the first
angle is 30◦ , the second angle is 45◦ , and the third angle is 60◦ .
Example 2
Graph 210◦ and identify its reference angle.
The graph of 210◦ looks like this:
and since the angle makes a 30◦ angle with the negative "x" axis, the reference angle is 30◦ .
Example 3
Graph 315◦ and identify its reference angle.
The graph of 315◦ looks like this:
307
2.5. Reference Angles and Angles in the Unit Circle
www.ck12.org
and since the angle makes a 45◦ angle with the positive "x" axis, the reference angle is 45◦ .
Example 4
Find the ordered pair for 150◦ and use it to find the value of cos 150◦ .
Since the reference angle is
30◦ ,
√
3
1
we know that the coordinates for the point on the unit circle are − 2 , 2 . This
is the same as the value for 30◦ , except the "x" coordinate is negative instead of positive. Knowing this,
Review
1. Graph 100◦ and identify its reference angle.
2. Graph 200◦ and identify its reference angle.
3. Graph 290◦ and identify its reference angle.
Calculate each value using the unit circle and special right triangles.
4.
5.
6.
7.
308
sin 120◦
cos 120◦
csc 120◦
cos 135◦
www.ck12.org
8.
9.
10.
11.
12.
13.
14.
15.
Chapter 2. Unit 2
sin 135◦
tan 135◦
sin 210◦
cos 210◦
cot 210◦
sin 225◦
cos 225◦
sec 225◦
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.18.
309
2.6. Conversion between Degrees and Radians
www.ck12.org
2.6 Conversion between Degrees and Radians
Learning Objectives
Here you’ll learn how to convert degrees to radians, and vice versa.
You are hard at work in the school science lab when your teacher asks you to turn a knob on a detector you are using
75◦ degrees. Unfortunately, you have been working in radians for a while, and so you’re having trouble remembering
how far to turn the knob. Is there a way to translate the instructions in degrees to radians?
Conversion between Degrees and Radians
Since degrees and radians are different ways of measuring the distance moved around the circumference of a circle,
it is reasonable to suppose that there is a conversion formula between these two units. This formula works for all
degrees and radians. Remember that: π radians = 180◦ . If you divide both sides of this equation by π, you will have
the conversion formula:
radians ×
180
= degrees
π
If we have a degree measure and wish to convert it to radians, then manipulating the equation above gives:
degrees ×
π
= radians
180
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70436
Let’s look at a few problems where we convert between degrees and radians.
1. Convert
11π
3
to degree measure.
From the last section, you should recognize that this angle is a multiple of
this angle, π3 × 11 = 60◦ × 11 = 660◦ .
Here is what it would look like using the formula:
radians × 180
π = degrees
310
π
3
(or 60 degrees), so there are 11, π3 ’s in
www.ck12.org
Chapter 2. Unit 2
2. Convert −120◦ to radian measure. Leave the answer in terms of π.
π
= radians
180
−120◦ π
π
=
−120◦ ×
180
180
degrees ×
and reducing to lowest terms gives us − 2π
3
You could also have noticed that 120 is 2 × 60. Since 60◦ is
and you have the answer, − 2π
3 .
11π
12 radians terms
180
π = degrees
3. Express
radians ×
π
3
radians, then 120 is 2, π3 ’s, or
2π
3 .
Make it negative
of degrees.
π
Note: Sometimes students have trouble remembering if it is 180
π or 180 . It might be helpful to remember that radian
measure is almost always expressed in terms of π. If you want to convert from radians to degrees, you want the π to
cancel out when you multiply, so it must be in the denominator.
Examples
Example 1
Earlier, you were asked is there a way to translate the instructions in degrees to radians.
Since you now know that the conversion for a measurement in degrees to radians is
degrees ×
π
= radians
180
you can find the solution to convert 75◦ to radians:
75◦ ×
π
75π 5π
=
=
180 180 12
Example 2
Convert the following degree measures to radians. All answers should be in terms of π.
240◦ , 270◦ , 315◦ , −210◦ , 120◦
4π 3π 7π
3 , 2 , 4 ,
− 7π
6 ,
2π
3
311
2.6. Conversion between Degrees and Radians
www.ck12.org
Example 3
Convert the following degree measures to radians. All answers should be in terms of π.
15◦ , −450◦ , 72◦ , 720◦ , 330◦
π
12 ,
π
− 5π
2 , 5 , 4π,
11π
6
Example 4
Convert the following radian measures to degrees
7π
π 11π 2π
2 , 5 , 3 , 5π, 2
90◦ , 396◦ , 120◦ , 540◦ ,
630◦
Review
Convert the following degree measures to radians. All answers should be in terms of π.
1.
2.
3.
4.
5.
6.
7.
90◦
360◦
50◦
110◦
495◦
−85◦
−120◦
Convert the following radian measures to degrees.
8. 5π
12
9. 3π
5
10. 8π
15
11. 7π
10
12. 5π
2
13. 3π
14. 7π
2
15. Why do you think there are two different ways to measure angles? When do you think it might be more
convenient to use radians than degrees?
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.2.
312
www.ck12.org
Chapter 2. Unit 2
2.7 Radian Measure
Learning Objectives
Here you’ll learn what radian measure is, and how to find the radian values for common angles on the unit circle.
While working on an experiment in your school science lab, your teacher asks you to turn up a detector by rotating
the knob π2 radians. You are immediately puzzled, since you don’t know what a radian measure is or how far to turn
the knob.
Measure of Radians
Until now, we have used degrees to measure angles. But, what exactly is a degree? A degree is 3601 th of a complete
rotation around a circle. Radians are alternate units used to measure angles in trigonometry. Just as it sounds, a
radian is based on the radius of a circle. One radian (abbreviated rad) is the angle created by bending the radius
length around the arc of a circle. Because a radian is based on an actual part of the circle rather than an arbitrary
division, it is a much more natural unit of angle measure for upper level mathematics.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/176403
What if we were to rotate all the way around the circle? Continuing to add radius lengths, we find that it takes a little
more than 6 of them to complete the rotation.
313
2.7. Radian Measure
www.ck12.org
Recall from geometry that the arc length of a complete rotation is the circumference, where the formula is equal
to 2π times the length of the radius. 2π is approximately 6.28, so the circumference is a little more than 6 radius
lengths. Or, in terms of radian measure, a complete rotation (360 degrees) is 2π radians.
360 degrees = 2π radians
With this as our starting point, we can find the radian measure of other angles. Half of a rotation, or 180 degrees,
must therefore be π radians, and 90 degrees must be 12 π, written π2 .
Extending the radian measure past the first quadrant, the quadrantal angles have been determined, except 270◦ .
Because 270◦ is halfway between 180◦ (π) and 360◦ (2π), it must be 1.5π, usually written 3π
2 .
For the 45◦ angles, the radians are all multiples of π4 .
For example, 135◦ is 3 · 45◦ . Therefore, the radian measure should be 3 · π4 , or
of 45◦ , in radians:
314
3π
4 .
Here are the rest of the multiples
www.ck12.org
Chapter 2. Unit 2
Notice that the additional angles in the drawing all have reference angles of 45 degrees and their radian measures are
all multiples of π4 . All of the even multiples are the quadrantal angles and are reduced, just like any other fraction.
Let’s do some problems that involve radian measures.
1. Find the radian measure of these angles.
TABLE 2.1:
Angle in Degrees
90
45
30
Angle in Radians
π
2
Because 45 is half of 90, half of 12 π is 14 π. 30 is one-third of a right angle, so multiplying gives:
π 1 π
× =
2 3 6
and because 60 is twice as large as 30:
2×
π 2π π
=
=
6
6
3
Here is the completed table:
TABLE 2.2:
Angle in Degrees
90
45
30
Angle in Radians
π
2
π
4
π
6
315
2.7. Radian Measure
www.ck12.org
There is a formula to convert between radians and degrees that you may already have discovered while doing this
example. However, many angles that are commonly used can be found easily from the values in this table. For
example, most students find it easy to remember 30 and 60. 30 is π over 6 and 60 is π over 3. Knowing these angles,
you can find any of the special angles that have reference angles of 30 and 60 because they will all have the same
denominators. The same is true of multiples of π4 (45 degrees) and π2 (90 degrees).
2. Complete the following radian measures by counting in multiples of
π
3
and π6 :
Notice that all of the angles with 60-degree reference angles are multiples of π3 , and all of those with 30-degree
reference angles are multiples of π6 . Counting in these terms based on this pattern, rather than converting back to
degrees, will help you better understand radians.
3. Find the radian measure of these angles.
TABLE 2.3:
Angle in Degrees
120
180
240
270
316
Angle in Radians
2π
3
www.ck12.org
Chapter 2. Unit 2
TABLE 2.3: (continued)
Angle in Degrees
300
Angle in Radians
Because 30 is one-third of a right angle, multiplying gives:
π 1 π
× =
2 3 6
adding this to the known value for ninety degrees of π2 :
π π 3π π 4π 2π
+ =
+ =
=
2 6
6
6
6
3
Here is the completed table:
TABLE 2.4:
Angle in Degrees
120
180
240
300
Angle in Radians
2π
3
π
4π
3
5π
3
Examples
Example 1
Earlier, you were given a problem about rotating the knob.
Since 45◦ = π4 rad, then 2 × π4 =
of the knob.
π
2
= 2 × 45◦ . Therefore, a turn of
π
2
is equal to 90◦ , which is
1
4
of a complete rotation
Example 2
Give the radian measure of 60◦
30 is one-third of a right angle. This means that since 90◦ = π2 , then 30◦ = π6 . Therefore, multiplying gives:
π
6
×2 =
π
3
Example 3
Give the radian measure of 75◦
15 is one-sixth of a right triangle. This means that since 90◦ = π2 , then 15◦ =
π
12
×5 =
π
12 .
Therefore, multiplying gives:
5π
12
317
2.7. Radian Measure
www.ck12.org
Example 4
Give the radian measure of 180◦
Since 90◦ = π2 , then 180◦ =
2π
2
=π
Review
Find the radian measure of each angle.
1.
2.
3.
4.
5.
6.
90◦
120◦
300◦
330◦
−45◦
135◦
Find the degree measure of each angle.
7. 3π
2
8. 5π
4
9. 7π
6
10. π6
11. 5π
3
12. π
π
13. Explain why if you are given an angle in degrees and you multiply it by 180
you will get the same angle in
radians.
14. Explain why if you are given an angle in radians and you multiply it by 180
π you will get the same angle in
degrees.
15. Explain in your own words why it makes sense that there are 2π radians in a circle.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.1.
318
www.ck12.org
Chapter 2. Unit 2
2.8 Arc Length
Learning Objectives
Here you will learn how to find the length of an arc and how to measure angles in radians.
A radian is another unit of measurement for angles just like degrees. Radians are especially convenient for measuring
angles that have to do with circles. There are 360◦ in a circle and 90◦ in a right angle. How many radians are in a
circle and how many radians make up a right angle?
Arc Length
Recall that a portion of a circle is called an arc. One way to measure an arc is with degrees. The measure of an arc
c = 70◦ and mGH
d = 70◦ .
is equal to the measure of its corresponding central angle. Below, mDC
When you measure an arc in degrees, it tells you the relative size of the arc compared to the whole circle. It does
not tell you anything about the absolute size of the arc or the circle it came from. Both arcs above have the same
d is physically longer, due to circle E being bigger.
measure, but GH
This leads to another way of describing the size of an arc. Arc length measures the distance (in units such as inches
d has a greater arc length than
or centimeters) along a circle between the endpoints that define the arc. Above, GH
c Because the radius of a circle is what determines the circle’s size, the length of an arc intercepted by a given
DC.
angle will be directly proportional to the radius of the circle. You will derive this fact in Examples A and B.
Arc length suggests a new way of measuring angles in circles. Previously, you have measured all angles in degrees,
but you can also measure angles in radians. 1 radian is the angle that creates an arc that has a length equal to
the radius. Below, the arc has a length equal to the radius. The angle that is created is 1 radian.
319
2.8. Arc Length
www.ck12.org
You can use the formula for the circumference of a circle to show that there are 2π radians in a circle (You will
justify this in the Examples). This means that:
2π radians = 360◦
Therefore, you have the following conversions:
180
degrees (≈ 57.3◦ )
π
π radians (≈ 0.17 radians)
1 degree =
180
1 radian =
Note that if a given angle has no units, it is assumed to be in radians.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/73428
A sector is a portion of a filled circle bounded by two radii and an arc (a “wedge” of a circle).
Let’s do a problem involving sector similarity.
Show that two sectors with the same central angle are similar by using transformations to explain why the two shaded
sectors below are similar.
320
www.ck12.org
Chapter 2. Unit 2
Two shapes are similar if a similarity transformation exists between them. Draw a vector from point A to point E.
Translate the red sector along the vector.
Rotate the image so that EC0 lies on EH. Because angles are preserved with translations and rotations, the image of
ED0 will lie on EG.
Dilate the image about point E by a factor of
EH
AC .
321
2.8. Arc Length
www.ck12.org
The two sectors are similar because a sequence of rigid transformations followed by a dilation carried one sector to
the other.
Next, let’s look at a problem involving arc length.
Explain why the length of arc s is equal to θr, where θ is the central angle in radians and r is the radius of the circle.
In a circle of radius 1, the measure of a central angle in radians will be equal to the length of the intercepted arc.
This is because the number of radians equals the number of radii that make up the arc.
If a sector has radius r, it is similar to a sector of radius 1 with the same central angle (as shown in the previous
problem about sectors).
Because the sectors are similar, corresponding lengths are proportional:
r
s
=
1 x
s = xr
322
www.ck12.org
Chapter 2. Unit 2
x must be equal to the measure of θ in radians. Therefore:
s = θr
Why does this make sense? Remember that if θ is in radians, then θ is equal to the number of radii that fit around
the arc. The number of radii that fit around the arc multiplied by the length of the radius will equal the length of the
arc.
Finally, lets look at a problem where we measure arc length
d and the length of GH.
d
Find mGH
d = 106◦ . To find the length of the arc, multiply the radius (6 in) by the measure of the central angle in radians.
mGH
π
π
radians. This means that 106◦ = 106 · 180
≈ 1.85 radians.
Remember that 1 degree = 180
Now you can find the length of the arc:
s = θr
s = (1.85)(6)
s ≈ 11.1 in
Don’t forget that your angle must be in radians in order to use the formula s = θr!
Examples
Example 1
Earlier, you were asked about radians in cir
There are 360◦ in a circle and 90◦ in a right angle. How many radians are in a circle and how many radians make up
a right angle?
There are 2π radians in a circle. A right angle is 90◦ and 90◦ is
angle.
1
4
of a circle, so there are
2π
4
=
π
2
radians in a right
323
2.8. Arc Length
www.ck12.org
Example 2
Explain why a circle has 2π radians.
The circumference of a circle with radius 1 is 2π(1) = 2π. Therefore, 2π radii fit around a circle with radius 1. All
circles are similar, so 2π radii must fit around all circles. 1 radian is the angle that creates an arc that has a length
equal to the radius, so 2π radians is the angle that creates an arc with a length equal to 2π radii. Therefore, a circle
is 2π radians because 2π radii fit around any circle.
Example 3
How many radians are there in 150◦ ?
π
Remember that 1 degree = 180
radians. This means that 150◦ = 150 ·
π
180
≈ 2.62 radians.
Example 4
c
Find the length of CD.
c is s = (2.62)(4) ≈ 10.48 cm.
s = θr and θ = 150◦ = 2.62 radians. The length of CD
Review
1. What’s the difference between finding the measure of an arc and the length of an arc?
2. What is the relationship between a radian and a radius?
3. How are radians and degrees related?
4. Explain why the two shaded sectors below are similar.
324
www.ck12.org
Chapter 2. Unit 2
5. Justify why the length of arc s is equal to θr, where θ is the central angle in radians and r is the radius of the
circle.
Convert each angle measured in degrees to an angle measured in radians. Leave answers in terms of π.
6. 180◦
7. 360◦
8. 90◦
9. 60◦
10. 30◦
c in each circle.
Find the measure in degrees and length in centimeters of CD
11.
12.
325
2.8. Arc Length
13.
14.
326
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
15. Explain how to find the length of an arc when given the central angle in radians. How does this compare to the
process of finding the length of an arc when given the central angle in degrees?
Review (Answers)
To see the Review answers, open this PDF file and look for section 8.9.
327
2.9. Applications of Radian Measure
www.ck12.org
2.9 Applications of Radian Measure
Learning Objectives
•
•
•
•
Solve problems involving angles of rotation using radian measure.
Calculate the length of an arc and the area of a sector.
Approximate the length of a chord given the central angle and radius.
Solve problems about angular speed.
Rotations
Example 1: The hands of a clock show 11:20. Express the obtuse angle formed by the hour and minute hands in
radian measure.
Solution: The following diagram shows the location of the hands at the specified time.
Because there are 12 increments on a clock, the angle between each hour marking on the clock is (or ). So, the angle
between the 12 and the 4 is (or ). Because the rotation from 12 to 4 is one-third of a complete rotation, it seems
reasonable to assume that the hour hand is moving continuously and has therefore moved one-third of the distance
between the 11 and the 12. This means that the angle between the hour hand and the 12 is two-thirds of the distance
between the 11 and the 12. So, , and the total measure of the angle is therefore .
Length of Arc
The length of an arc on a circle depends on both the angle of rotation and the radius length of the circle. If you recall
from the last lesson, the measure of an angle in radians is defined as the length of the arc cut off by one radius length.
What if the radius is 4 cm? Then, the length of the half-circle arc would be multiplied by the radius length, or cm in
length.
328
www.ck12.org
Chapter 2. Unit 2
This results in a formula that can be used to calculate the length of any arc.
where is the length of the arc, is the radius, and is the measure of the angle in radians.
Solving this equation for will give us a formula for finding the radian measure given the arc length and the radius
length:
Example 2: The free-throw line on an NCAA basketball court is 12 ft wide. In international competition, it is only
about 11.81 ft. How much longer is the half circle above the free-throw line on the NCAA court?
Solution: Find both arc lengths.
So the answer is approximately
This is approximately 0.3 ft, or about 3.6 inches longer.
Example 3: Two connected gears are rotating. The smaller gear has a radius of 4 inches and the larger gear’s radius
is 7 inches. What is the angle through which the larger gear has rotated when the smaller gear has made one complete
rotation?
329
2.9. Applications of Radian Measure
www.ck12.org
Solution: Because the blue gear performs one complete rotation, the length of the arc traveled is:
So, an arc length on the larger circle would form an angle as follows:
So the angle is approximately 3.6 radians.
Area of a Sector
One of the most common geometric formulas is the area of a circle:
In terms of angle rotation, this is the area created by radians.
A half-circle, or radian rotation would create a section, or sector of the circle equal to half the area or:
So an angle of 1 radian would define an area of a sector equal to:
From this we can determine the area of the sector created by any angle, radians, to be:
Example 4: Crops are often grown using a technique called center pivot irrigation that results in circular shaped
fields.
Here is a satellite image taken over fields in Kansas that use this type of irrigation system.
If the irrigation pipe is 450 m in length, what is the area that can be irrigated after a rotation of radians?
330
www.ck12.org
Chapter 2. Unit 2
Solution: Using the formula:
The area is approximately 212,058 square meters.
Length of a Chord
You may recall from your Geometry studies that a chord is a segment that begins and ends on a circle.
is a chord in the circle.
We can calculate the length of any chord if we know the angle measure and the length of the radius. Because each
endpoint of the chord is on the circle, the distance from the center to and is the same as the radius length.
331
2.9. Applications of Radian Measure
www.ck12.org
Next, if we bisect the angle, the angle bisector must be perpendicular to the chord and bisect it (we will leave the
proof of this to your Geometry class). This forms a right triangle.
We can now use a simple sine ratio to find half the chord, called here, and double the result to find the length of the
chord.
So the length of the chord is:
Example 5: Find the length of the chord of a circle with radius 8 cm and a central angle of . Approximate your
answer to the nearest mm.
Solution: We must first convert the angle measure to radians:
Using the formula, half of the chord length should be the radius of the circle times the sine of half the angle.
Multiply this result by 2.
332
www.ck12.org
Chapter 2. Unit 2
So, the length of the arc is approximately 13.1 cm.
Angular Velocity
What about objects that are traveling on a circular path? Do you remember playing on a merry-go-round when you
were younger?
If two people are riding on the outer edge, their velocities should be the same. But, what if one person is close to the
center and the other person is on the edge? They are on the same object, but their speed is actually not the same.
Look at the following drawing.
Imagine the point on the larger circle is the person on the edge of the merry-go-round and the point on the smaller
circle is the person towards the middle. If the merry-go-round spins exactly once, then both individuals will also
make one complete revolution in the same amount of time.
However, it is obvious that the person in the center did not travel nearly as far. The circumference (and of course the
radius) of that circle is much smaller and therefore the person who traveled a greater distance in the same amount
of time is actually traveling faster, even though they are on the same object. So the person on the edge has a greater
linear velocity (recall that linear velocity is found using ). If you have ever actually ridden on a merry-go-round, you
know this already because it is much more fun to be on the edge than in the center! But, there is something about
the two individuals traveling around that is the same. They will both cover the same rotation in the same period of
time. This type of speed, measuring the angle of rotation over a given amount of time is called the angular velocity.
The formula for angular velocity is:
is the last letter in the Greek alphabet, omega, and is commonly used as the symbol for angular velocity. is the angle
of rotation expressed in radian measure, and is the time to complete the rotation.
333
2.9. Applications of Radian Measure
www.ck12.org
In this drawing, is exactly one radian, or the length of the radius bent around the circle. If it took point exactly 2
seconds to rotate through the angle, the angular velocity of would be:
In order to know the linear speed of the particle, we would have to know the actual distance, that is, the length of
the radius. Let’s say that the radius is 5 cm.
If linear velocity is then, or 2.5 cm per second.
If the angle were not exactly 1 radian, then the distance traveled by the point on the circle is the length of the arc, ,
or, the radius length times the measure of the angle in radians.
Substituting into the formula for linear velocity gives: or .
Look back at the formula for angular velocity. Substituting gives the following relationship between linear and
angular velocity, . So, the linear velocity is equal to the radius times the angular velocity.
Remember in a unit circle, the radius is 1 unit, so in this case the linear velocity is the same as the angular velocity.
Here, the distance traveled around the circle is the same for a given unit of time as the angle of rotation, measured
in radians.
Example 6: Lindsay and Megan are riding on a Merry-go-round. Megan is standing 2.5 feet from the center and
Lindsay is riding on the outside edge 7 feet from the center. It takes them 6 seconds to complete a rotation. Calculate
the linear and angular velocity of each girl.
Solution: We are told that it takes 6 seconds to complete a rotation. A complete rotation is the same as radians. So
the angular velocity is:
radians per second, which is slightly more than 1 (about 1.05), radian per second. Because both girls cover the
same angle of rotation in the same amount of time, their angular speed is the same. In this case they rotate through
approximately 60 degrees of the circle every second.
As we discussed previously, their linear velocities are different. Using the formula, Megan’s linear velocity is:
Lindsay’s linear velocity is:
334
www.ck12.org
Chapter 2. Unit 2
Points to Consider
• What is the difference between finding arc length and the area of a sector?
• What is the difference between linear velocity and angular velocity?
• How are linear and angular velocity related?
Review Questions
1. The following Figure ?? shows a 24-hour clock.
a. What is the angle between each number of the clock expressed in:
a. exact radian measure in terms of ?
b. to the nearest tenth of a radian?
c. in degree measure?
b. Estimate the measure of the angle between the hands at the time shown in:
a. to the nearest whole degree
b. in radian measure in terms of
2. The following picture is a window of a building on the campus of Princeton University in Princeton, New
Jersey.
a. What is the exact radian measure in terms of between two consecutive circular dots on the small circle
in the center of the window?
335
2.9. Applications of Radian Measure
www.ck12.org
b. If the radius of this circle is about 0.5 m, what is the length of the arc between the centers of each
consecutive dot? Round your answer to the nearest cm.
3. Now look at the next larger circle in the window.
a. Find the exact radian measure in terms of between two consecutive dots in this window.
b. The radius of the glass portion of this window is approximately 1.20 m. Calculate an estimate of the
length of the highlighted chord to the nearest cm. Explain the reasoning behind your solution.
4. The state championship game is to be held at Ray Diaz Memorial Arena. The seating forms a perfect circle
around the court. The principal of Archimedes High School is sent the following diagram showing the seating
allotted to the students at her school.
336
www.ck12.org
Chapter 2. Unit 2
It is 55 ft from the center of the court to the beginning of the stands and 110 ft from the center to the end.
Calculate the approximate number of square feet each of the following groups has been granted:
a. the students from Archimedes.
b. general admission.
c. the press and officials.
5. Doris and Lois go for a ride on a carousel. Doris rides on one of the outside horses and Lois rides on one of
the smaller horses near the center. Lois’ horse is 3 m from the center of the carousel, and Doris’ horse is 7
m farther away from the center than Lois’. When the carousel starts, it takes them 12 seconds to complete a
rotation.
a. Calculate the linear velocity of each girl.
b. Calculate the angular velocity of the horses on the carousel.
6. The Large Hadron Collider near Geneva, Switzerland began operation in 2008 and is designed to perform
experiments that physicists hope will provide important information about the underlying structure of the
universe. The LHC is circular with a circumference of approximately 27,000 m. Protons will be accelerated
to a speed that is very close to the speed of light ( meters per second).
a. How long does it take a proton to make a complete rotation around the collider?
b. What is the approximate (to the nearest meter per second) angular speed of a proton traveling around the
collider?
c. Approximately how many times would a proton travel around the collider in one full second?
337
2.10. The Unit Circle
www.ck12.org
2.10 The Unit Circle
Here you will use your knowledge of basic triangle trigonometry to identify key points and angles around a circle of
radius one centered at the origin. The unit circle is a circle of radius one, centered at the origin, that summarizes all
the 30-60-90 and 45-45-90 triangle relationships
that exist. When memorized, it is extremely useful for evaluating
expressions like cos(135◦ ) or sin − 5π
.
It
also
helps
to produce the parent graphs of sine and cosine.
3
How can you use the unit circle to evaluate cos(135◦ ) and sin − 5π
3 ?
The Unit Circle
You already know how to translate between degrees and radians and the triangle ratios for 30-60-90 and 45-4590 right triangles. In order to be ready to completely fill in and memorize a unit circle, two triangles need to be
worked out. Start by finding the side lengths of a 30-60-90 triangle and a 45-45-90 triangle each with hypotenuse
equal to 1.
TABLE 2.5:
30◦
x
1
2
60√◦
x√ 3
3
90◦
2x
1
2
TABLE 2.6:
45◦
x√
2
45◦
x√
2
2
2
90√◦
x 2
1
This is enough information to fill out the important points in the first quadrant of the unit circle. The values of the
x and y coordinates for each of the key points are shown below. Remember that the x and y coordinates come from
the lengths of the legs of the special right triangles, as shown specifically for the 30◦ angle. Always remember to
measure the angle from the positive portion of the x-axis.
338
www.ck12.org
Chapter 2. Unit 2
Knowing the first quadrant well is the key to knowing the entire unit circle. Every other point on the unit circle can
be found using logic and this quadrant, so there is no need to memorize the whole circle.
To use your knowledge of the first quadrant of the unit circle to identify the angles and important points of the
second quadrant, notice that the heights are mirrored and equal which correspond to the y values. The x values are
all negative.
339
2.10. The Unit Circle
www.ck12.org
There is a pattern in the heights of the points in the first quadrant that can help you remember the points.
√ √
Notice that the heights of the points in the first quadrant are the y-coordinates: 0, 12 , 2 2 , 2 3 , 1
√ √ √ √
When rewritten, the pattern becomes clear: 02 , 2 1 , 2 2 , 2 3 , 2 4 .
The three points in the
the most often mixed up. This pattern illustrates how they increase in size from
√ middle are √
2
small 1 , to medium
, to large 3 . When you fill in the unit circle, look for the heights that are small, medium
2
2
2
and large and this will tell
√ you
√ were each value should go. Notice that the heights for these five points in the second
2
1
quadrant are also 0, ,
, 3 , 1.
2
2
2
This technique also works
√ for√the widths.
√ √ This can make memorizing the 16 points of the unit circle a matter of
1
2
0
logic and the pattern: ,
,
, 3, 4.
2
2
2
2
2
One last item to note is that coterminal angles are sets of angles such as 90º, 450º, and -270º that start at the positive
x-axis and end at the same terminal side. Since coterminal angles end at identical points along the unit circle,
trigonometric expressions involving coterminal angles are equivalent: sin (90) = sin (450) = sin (−270).
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61156
340
www.ck12.org
Chapter 2. Unit 2
Examples
Example 1
Earlier, you were asked how you can use the unit circle to evaluate cos (135) and sin − 5π
3 . The x value of a point
along the unit circle corresponds to the cosine of the angle. The y value of a point corresponds to the sine of the
angle. When the angles and points are memorized, simply recall the x or y coordinate. If the construction of the unit
circle is understood, it becomes easier to determine the coordinates.
When evaluating cos(135◦ ) your thought process should be something like this:
√ √ √
◦
You know 135 goes with the point − 2 2 , 2 2 and cosine is the x portion. So, cos(135◦ ) = − 2 2 .
When evaluating sin − 5π
3 your thought process should be something like this:
√ √
3
1
5π
=
−
You know − 3 goes with the point 2 , − 2 3 and sine is the y portion. So, sin − 5π
3
2
Example 2
Evaluate cos 60◦ using the unit circle and right triangle trigonometry. What is the connection between the x
coordinate of the point and the cosine of the angle?
√ ◦
The point on the unit circle for 60 is 12 , 2 3 and the point is one unit from the origin. This can be represented
as a 30-60-90 triangle.
Since cosine is adjacent over hypotenuse, cosine turns out to be exactly the x coordinate 12 .
341
2.10. The Unit Circle
www.ck12.org
Example 3
Using knowledge of the first quadrant of the unit circle, identify the angles and important points of the third quadrant.
Both the x values and y values are negative and their respective coordinates correspond to those of the other
quadrants.
Example 4
For each of the six trigonometric functions, identify the quadrants where they are positive and the quadrants where
they are negative.
In quadrant I, the hypotenuse, adjacent and opposite side are all positive. Thus all 6 trigonometric functions are
positive.
In quadrant II the hypotenuse and opposite sides are positive and the adjacent side is negative. This means that
every trigonometric expression involving an adjacent side is negative. Sine and its reciprocal cosecant are the only
two trigonometric functions that do not refer to the adjacent side which makes them the only positive ones.
In quadrant III only the hypotenuse is positive. Thus the only trigonometric functions that are positive are tangent
and its reciprocal cotangent because these functions refer to both adjacent and opposite sides which will both be
negative.
In quadrant IV the hypotenuse and the adjacent sides are positive while the opposite side is negative. This means
that only cosine and its reciprocal secant are positive.
A mnemonic device to remember which trigonometric functions are positive and which trigonometric functions are
negative is “All Students Take Calculus.” All refers to all the trigonometric functions are positive in quadrant I. The
letter S refers to sine and its reciprocal cosecant that are positive in quadrant II. The letter T refers to tangent and its
reciprocal cotangent that are positive in quadrant III. The letter C refers to cosine and its reciprocal secant that are
positive in quadrant IV.
342
www.ck12.org
Chapter 2. Unit 2
Example 5
Evaluate the following trigonometric expressions using the unit circle.
1. sin π2
sin π2 = 1
2. cos 210◦
√
cos 210◦ = − 2 3
3. tan 315◦
tan 315◦ = −1
4. cot 270◦
cot 270◦ = 0
5. sec 11π
6
sec 11π
6 =
1
cos 11π
6
= √2 =
3
2
√
3
3
6. csc − 5π
4
csc − 5π
4 =
1
sin − 5π
4
√
= − √2 = − 2
2
Review
Name the angle between 0◦ and 360◦ that is coterminal with. . .
1. −20◦
2. 475◦
3. −220◦
4. 690◦
5. −45◦
Use your knowledge of the unit circle to help determine whether each of the following trigonometric expressions is
positive or negative.
6. tan 143◦
7. cos π3
8. sin 362◦
9. csc 3π
4
Use your knowledge of the unit circle to evaluate each of the following trigonometric expressions.
10. cos 120◦
343
2.10. The Unit Circle
11. sec π3
12. tan 225◦
13. cot 120◦
14. sin 11π
6
15. csc 240◦
√
16. Find sin θ and tan θ if cos θ = 2 3 and cot θ > 0.
17. Find cos θ and tan θ if sin θ = − 12 and sec θ < 0.
18. Draw the complete unit circle (all four quadrants) and label the important points.
Review (Answers)
To see the Review answers, open this PDF file and look for section 5.1.
344
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.11 Trigonometric Ratios on the Unit Circle
Here you’ll learn how to determine exact value of trigonometric ratios for multiples of 0◦ , 30◦ and 45◦ (or 0, π6 , π4
radians).
What are the exact values of the following trigonometric functions?
a. cos 495◦
b. tan 5π
3
Trigonometric Ratios on the Unit Circle
√
Recall special right triangles from Geometry. In a (30◦ − 60◦ − 90◦ ) triangle, the sides are in the ratio 1 : 3 : 2.
√
In an isosceles triangle (45◦ − 45◦ − 90◦ ), the congruent sides and the hypotenuse are in the ratio 1 : 1 : 2.
In a (30◦ − 60◦ − 90◦ ) triangle, the sides are in the ratio 1 :
√
3 : 2.
Now let’s make the hypotenuse equal to 1 in each of the triangles so we’ll be able to put them inside the unit circle.
Using the appropriate ratios, the new side lengths are:
345
2.11. Trigonometric Ratios on the Unit Circle
www.ck12.org
Using these triangles, we can evaluate sine, cosine and tangent for each of the angle measures.
√
2
sin 45 =
2
√
2
cos 45◦ =
2
◦
◦
tan 45 = 1
√
3
sin 60 =
2
1
cos 60◦ =
2
√
◦
◦
tan 60 =
3
2
1
2
√
= 3
1
2√
3
cos 30◦ =
2
sin 30◦ =
◦
tan 30 =
1
2
√
3
2
√
3
=
2
These triangles can now fit inside the unit circle.
Putting together the trigonometric ratios and the coordinates of the points on the circle, which represent the lengths of
the legs of the triangles, √
(∆x, ∆y), we can see that each point is actually (cos θ, sin θ), where θ is the reference angle.
◦
For example, sin 60 = 3 is the y - coordinate of the point on the unit circle in the triangle with reference angle
2
60◦ . By reflecting these triangles across the axes and finding the points on the axes, we can find the trigonometric
ratios of all multiples of 0◦ , 30◦ and 45◦ (or 0, π6 , π4 radians).
346
www.ck12.org
Chapter 2. Unit 2
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/99159
Let’s solve the following problems using the unit circle.
1. Find sin 3π
2 .
3π
Find 3π
2 on the unit circle and the corresponding point is (0, −1). Since each point on the unit circle is (cos θ, sin θ), sin 2 =
−1.
2. Find tan 7π
6 .
This time we need to look at the ratio
√
− 12
7π
1
√
√
tan 6 =
=
= 33 .
3
− 23
sin θ
cos θ .
We can use the unit circle to find sin
7π
6
= − 21
and cos
7π
6
√
= − 2 3 . Now,
347
2.11. Trigonometric Ratios on the Unit Circle
www.ck12.org
Quadrants in a Unit Circle
Another way to approach these exact value problems is to use the reference angles and the special right triangles.
The benefit of this method is that there is no need to memorize the entire unit circle. If you memorize the special
right triangles, can determine reference angles and know where the ratios are positive and negative you can put the
pieces together to get the ratios. Looking at the unit circle above, we see that all of the ratios are positive in Quadrant
I, sine is the only positive ratio in Quadrant II, tangent is the only positive ratio in Quadrant III and cosine is the only
positive ratio in Quadrant IV.
Keeping this diagram in mind will help you remember where cosine, sine and tangent are positive and negative. You
can also use the pneumonic device - All Students Take Calculus, or ASTC, to recall which is positive (all the others
would be negative) in which quadrant.
The coordinates on the vertices will help you determine the ratios for the multiples of 90◦ or π2 .
Now, let’s find the exact values for the following trigonometric functions using the alternative method.
1. cos 120◦
First, we need to determine in which quadrant the angles lies. Since 120◦ is between 90◦ and 180◦ it will lie in
Quadrant II. Next, find the reference angle. Since we are in QII, we will subtract from 180◦ to get 60◦ . We can use
the reference angle to find the ratio, cos 60◦ = 12 . Since we are in QII where only sine is positive, cos 120◦ = − 12 .
2. sin 5π
3
This time we will need to work in terms of radians but the process is the same. The angle 5π
3√lies in QIV and the
√
3
π
π
5π
reference angle is . This means that our ratio will be negative. Since sin =
, sin = − 3 .
3
3
2
3
2
3. tan 7π
2
3π
The angle 7π
2 represents more than one entire revolution and it is equivalent to 2π + 2 . Since our angle is a multiple
π
sin θ
7π
−1
of 2 we are looking at an angle on an axis. In this case, the point is (0, −1). Because tan θ = cos
θ , tan 2 = 0 , which
7π
is undefined. Thus, tan 2 is undefined.
Examples
Example 1
Earlier, you were asked to find the exact values of the following trigonometric ratios.
348
www.ck12.org
Chapter 2. Unit 2
a. cos 495◦
First, we need to determine in which quadrant the angle lies. Since 495◦ − 360◦ = 135◦ is between 90◦ and 180◦ it
will lie in Quadrant II. Next, find the reference angle. Since√
we are in QII, we will subtract from 180◦ to get 45◦ .
We can use the reference angle to find the ratio, cos 45◦ = 2 2 . Since we are in QII where only sine is positive,
√
cos 495◦ = − 2 2 .
b. tan 5π
3
In problem #2 above we established that the angle 5π
QIV and the reference angle is π3 . This means that the
3 lies in√
√
tangent ratio will be negative. Since tan π3 = 3, tan 5π
3 = − 3.
Find the exact trigonometric ratios. You may use either method.
Example 2
cos 7π
3
7π
3
has a reference angle of
π
3
in QI. cos π3 =
1
2
1
and since cosine is positive in QI, cos 7π
3 = 2.
Example 3
tan 9π
2
9π
2
is coterminal to
π
2
which has coordinates (0, 1). So tan 9π
2 =
sin 9π
2
cos 9π
2
=
1
0
which is undefined.
Example 4
sin 405◦
405◦
has a reference angle of
45◦
in QI.
sin 45◦
√
√
2
◦
= 2 and since sine is positive in QI, sin 405 = 2 2 .
Example 5
tan 11π
6
11π
6
is coterminal to
π
6
in QIV. tan π6 =
√
√
3 and since tangent is negative in QIV, tan 11π = − 3 .
3
6
3
Example 6
cos 2π
3
2π
3
is coterminal to
π
3
in QII. cos π3 =
1
2
1
and since cosine is negative in QII, cos 2π
3 = 2.
Review
Find the exact values for the following trigonometric functions.
1. sin 3π
4
349
2.11. Trigonometric Ratios on the Unit Circle
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
cos 3π
2
tan 300◦
sin 150◦
cos 4π
3
tan π
cos − 15π
4
sin 225◦
tan 7π
6
sin 315◦
cos 450◦ sin − 7π
2
cos 17π
6
tan 270◦
sin(−210◦ )
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 13.7.
350
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.12 Reciprocal Trigonometric Functions
Here you’ll learn about the ratios of the reciprocal trigonometric ratios cosecant, secant and cotangent for angles that
are multiples of 0◦ , 30◦ and 45◦ (or 0, π6 , π4 radians) without a calculator and evaluate the reciprocal trigonometric
functions for all other angles using the calculator.
A ladder propped up against a house forms an angle of 30◦ with the ground. What is the secant of this angle?
Reciprocal Trigonometric Functions
Each of the trigonometric ratios has a reciprocal function associated with it as shown below.
1
H
sin θ = csc θ, so csc θ = O (hypotenuse over opposite)
The reciprocal of cosine is secant: cos1 θ = sec θ, so sec θ = HA (hypotenuse over adjacent)
The reciprocal of tangent is cotangent: tan1 θ = cot θ, so cot θ = OA (adjacent over opposite)
Let’s use a calculator to evaluate sec 2π
5 .
The reciprocal of sine is cosecant:
First, be sure that your calculator is in radian mode. To check/change the mode, press the MODE button and make
sure RADIAN is highlighted. If it is not, use the arrow keys to move the cursor to RADIANS and press enter to
select RADIAN as the mode. Now we are ready to use the calculator to evaluate the reciprocal trig function. Since
the calculator does not have a button for secant, however, we must utilize the reciprocal relationship between cosine
and secant:
Since sec θ =
1
2π
1
, sec
=
= 3.2361.
cos θ
5
cos 2π
5
Now, let’s use a calculator to evaluate cot 100◦ .
This time we will need to be in degree mode. After the mode has been changed we can use the reciprocal of
cotangent, which is tangent, to evaluate as shown:
Since cot θ =
1
1
, cot 100◦ =
≈ −0.1763.
tan θ
tan 100◦
Finally, let’s find the exact value of csc 5π
3 without using a calculator and give our answer in exact form.
The reciprocal of cosecant is sine so we will first find sin 5π
3 Using either the unit circle or the alternative method,
√
3
◦
we can determine that sin 5π
3 is − 2 using a 60 reference angle in the fourth quadrant. Now, find its reciprocal:
√
1√
= − √2 = − 2 3 3 .
3
− 23
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/182867
351
2.12. Reciprocal Trigonometric Functions
www.ck12.org
Examples
Example 1
Earlier, you were asked to find the secant of the angle of the ladder propped up against the house.
The secant is the reciprocal of the cosine. So to find sec 30◦ , use the cosine.
√
◦
cos 30 = 2 3 .
√
2 3
2
◦
√
Therefore, sec 30 =
= 3 .
3
Use your calculator to evaluate the following reciprocal trigonometric functions.
Example 2
csc 7π
8
csc 7π
8 =
1
sin 7π
8
= 2.6131
Example 3
cot 85◦
cot 85◦ =
1
tan 85◦
= 0.0875
Evaluate the following without using a calculator. Give all answers in exact form.
Example 4
sec 225◦
sec 225◦ is the reciprocal of cot 225◦ , a 45◦ reference
√ angle in quadrant three where cosine is negative. Because
1
1
◦
◦
◦
√
√
, cos 225 = −
, and sec 225 = − 2.
cos 45 =
2
2
Example 5
csc 5π
6
5π
π
◦
csc 5π
6 is the reciprocal of sin 6 , a 6 or 30 reference angle in the second quadrant where sine is positive. Because
1
5π
sin π6 = 12 , sin 5π
6 = 2 , and csc 6 = 2.
Review
Use your calculator to evaluate the reciprocal trigonometric functions. Round your answers to four decimal places.
1. csc 95◦
2. cot 278◦
352
www.ck12.org
3.
4.
5.
6.
7.
8.
Chapter 2. Unit 2
sec 14π
5
cot(−245◦ )
sec 6π
7
csc 23π
13
cot 333◦
csc 9π
5
Evaluate the following trigonometric functions without using a calculator. Give your answers exactly.
9.
10.
11.
12.
13.
14.
15.
16.
sec 5π
6
csc − 3π
2
cot 225◦
sec 11π
3
csc 7π
6
sec 270◦
cot 5π
3
csc 315◦
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 13.8.
353
2.13. Cofunction Identities and Reflection
www.ck12.org
2.13 Cofunction Identities and Reflection
Learning Objectives
Here you’ll learn about the four cofunction identities and how to apply them to solve for the values of trig functions.
While toying with a triangular puzzle piece, you start practicing your math skills to see what you can find out about it.
You realize one of the interior angles of the puzzle piece is 30◦ , and decide to compute the trig functions associated
with this angle. You immediately want to compute the cosine of the angle, but can only remember the values of your
sine functions.
Is there a way to use this knowledge of sine functions to help you in your computation of the cosine function for
30◦ ?
Cofunction Identities and Reflection
In a right triangle, you can apply what are called "cofunction identities". These are called cofunction identities
because the functions have common values. These identities are summarized below.
sin θ = cos(90◦ − θ)
cos θ = sin(90◦ − θ)
tan θ = cot(90◦ − θ)
cot θ = tan(90◦ − θ)
Let’s take a look at some problems involving cofunction identities and reflection.
1. Find the value of cos 120◦ .
Because this angle has a reference angle of 60◦ , the answer is cos 120◦ = − 12 .
2. Find the value of cos(−120◦ ).
Because this angle has a reference angle of 60◦ , the answer is cos(−120◦ ) = cos 240◦ = − 21 .
3. Find the value of sin 135◦ .
Because this angle has a reference angle of 45◦ , the answer is sin 135◦ =
√
2
2
Examples
Example 1
Earlier, you were asked if there is a way to use your knowledge of sine functions to help you in your computation of
the cosine function.
Since you now know the cofunction relationships, you can use your knowledge of sine functions to help you with
the cosine computation:
√
cos 30◦ = sin(90◦ − 30◦ ) = sin(60◦ ) = 2 3
354
www.ck12.org
Chapter 2. Unit 2
Example 2
Find the value of sin 45◦ using a cofunction identity.
The sine of
45◦
is equal to
cos(90◦ − 45◦ )
=
cos 45◦
√
= 22 .
Example 3
Find the value of cos 45◦ using a cofunction identity.
The cosine of 45◦ is equal to sin(90◦ − 45◦ ) = sin 45◦ =
√
2.
2
Example 4
Find the value of cos 60◦ using a cofunction identity.
The cosine of 60◦ is equal to sin(90◦ − 60◦ ) = sin 30◦ = .5.
Review
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Find a value for θ for which sin θ = cos 15◦ is true.
Find a value for θ for which cos θ = sin 55◦ is true.
Find a value for θ for which tan θ = cot 80◦ is true.
Find a value for θ for which cot θ = tan 30◦ is true.
Use cofunction identities to help you write the expression tan 255◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression sin 120◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression cos 310◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression cot 260◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression cos 280◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression tan 60◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression sin 100◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression cos 70◦ as the function of an acute angle of measure
less than 45◦ .
Use cofunction identities to help you write the expression cot 240◦ as the function of an acute angle of measure
less than 45◦ .
Use a right triangle to prove that sin θ = cos(90◦ − θ).
Use the sine and cosine cofunction identities to prove that tan(90◦ − θ) = cot θ.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.24.
355
2.14. Finding Exact Trig Values using Sum and Difference Formulas
www.ck12.org
2.14 Finding Exact Trig Values using Sum and
Difference Formulas
Learning Objectives
Here you’ll use the sum and difference formulas to find exact values of angles other than the critical angles.
You measure an angle with your protractor to be 165◦ . How could you find the exact sine of this angle without using
a calculator?
Sum and Difference Formulas
√
√
You know that sin 30◦ = 21 , cos 135◦ = − 2 2 , tan 300◦ = − 3, etc... from the special right triangles. In this concept,
we will learn how to find the exact values of the trig functions for angles other than these multiples of 30◦ , 45◦ , and
60◦ . Using the Sum and Difference Formulas, we can find these exact trig values.
Sum and Difference Formulas
sin(a ± b) = sin a cos b ± cos a sin b
cos(a ± b) = cos a cos b ∓ sin a sin b
tan a ± tan b
tan(a ± b) =
1 ∓ tan a tan b
Let’s find the following exact values using the Sum and Difference Formulas.
1. sin 75◦
This is an example of where we can use the sine sum formula from above, sin(a + b) = sin a cos b + cos a sin b, where
a = 45◦ and b = 30◦ .
sin 75◦ = sin(45◦ + 30◦ )
= sin 45◦ cos 30◦ + cos 45◦ sin 30◦
√
√ √
2
3
2 1
=
·
+
·
2
2
2 2
√
√
6+ 2
=
4
In general, sin(a + b) 6= sin a + sin b and similar statements can be made for the other sum and difference formulas.
2. cos 11π
12
For this problem, we could use either the sum or difference cosine formula,
the sum formula.
356
11π
12
=
2π
3
+ π4 or
11π
12
=
7π
6
− π4 . Let’s use
www.ck12.org
Chapter 2. Unit 2
11π
2π π
cos
= cos
+
12
3
4
2π
π
2π
π
= cos cos − sin sin
3√ 4 √ √3
4
2
3
2
1
−
·
=− ·
2√ 2 √ 2
2
2+ 6
=−
4
π
3. tan − 12
This angle is the difference between
π
4
and π3 .
tan
This angle is also the same as
answer.
23π
12 .
π
4
−
tan π4 − tan π3
π
=
3
1 + tan π4 tan π3
√
1− 3
√
=
1+ 3
You could have also used this value and done tan
π
4
+ 5π
3 and arrived at the same
Examples
Example 1
Earlier, you were asked to find the exact value of sin 165◦ without using the calculator.
We can use the sine sum formula, sin(a + b) = sin a cos b + cos a sin b, where a = 120◦ and b = 45◦ .
sin 165◦ = sin(120◦ + 45◦ )
= sin 120◦ cos 45◦ + cos 120◦ sin 45◦
√
√ √
3
2 −1
2
=
·
+
·
2
2
2
2
√
√
6− 2
=
4
Example 2
Find the exact value of cos 15◦ .
357
2.14. Finding Exact Trig Values using Sum and Difference Formulas
www.ck12.org
cos 15◦ = cos(45◦ − 30◦ )
= cos 45◦ cos 30◦ + sin 45◦ sin 30◦
√ √
√
2
3
2 1
=
·
+
·
2
2
2 2
√
√
6+ 2
=
4
Example 3
Find the exact value of tan 255◦ .
tan(210◦ + 45◦ ) =
=
tan 210◦ + tan 45◦
1 − tan 210◦ tan 45◦
√
3
1
3 +
√
1 − 33
=
√
3+3
3√
3− 3
3
√
3+3
√
=
3− 3
Review
Find the exact value of the following trig functions.
sin 15◦
cos 5π
12
tan 345◦
cos(−255◦ )
sin 13π
12
sin 17π
12
cos 15◦
tan(−15◦ )
sin 345◦
Now, use sin 15◦ from #1, and find sin 345◦ . Do you arrive at the same answer? Why or why not?
Using cos 15◦ from #7, find cos 165◦ . What is another way you could find cos 165◦ ?
Describe any patterns you see between the sine, cosine, and tangent of these “new” angles.
Using your calculator, find the sin 142◦ . Now, use the sum formula and your calculator to find the sin 142◦
using 83◦ and 59◦ .
14. Use the sine difference formula to find sin 142◦ with any two angles you choose. Do you arrive at the same
answer? Why or why not?
15. Challenge Using sin(a+b) = sin a cos b+cos a sin b and cos(a+b) = cos a cos b−sin a sin b, show that tan(a+
tan a+tan b
b) = 1−tan
a tan b .
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.12.
358
www.ck12.org
Chapter 2. Unit 2
2.15 Sine Sum and Difference Formulas
Learning Objectives
Here you’ll learn to rewrite sine functions with addition or subtraction in their arguments in a more easily solvable
form.
You’ve gotten quite good at knowing the values of trig functions. So much so that you and your friends play a game
before class everyday to see who can get the most trig functions of different angles correct. However, your friend
Jane keeps getting the trig functions of more angles right. You’re amazed by her memory, until she smiles one day
and tells you that she’s been fooling you all this time.
"What you do you mean?" you say.
"I have a trick that lets me calculate more functions in my mind by breaking them down into sums of angles." she
replies.
You’re really surprised by this. And all this time you thought she just had an amazing memory!
"Here, let me show you," she says. She takes a piece of paper out and writes down:
sin 7π
12
"This looks like an unusual value to remember for a trig function. So I have a special rule that helps me to evaluate
it by breaking it into a sum of different numbers."
Sine Sum and Difference Formulas
Our goal here is to figure out a formula that lets you break down a the sine of a sum of two angles (or a difference of
two angles) into a simpler formula that lets you use the sine of only one argument in each term.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143490
To find
sin(a + b)
:
359
2.15. Sine Sum and Difference Formulas
www.ck12.org
hπ
i
− (a + b)
h2 π
i
= cos
−a −b
π2 π
= cos
− a cos b + sin
− a sin b
2
2
= sin a cos b + cos a sin b
sin(a + b) = cos
Set θ = a + b
Distribute the negative
Difference Formula for cosines
Co-function Identities
In conclusion, sin(a + b) = sin a cos b + cos a sin b, which is the sum formula for sine.
To obtain the identity for sin(a − b):
sin(a − b) = sin[a + (−b)]
= sin a cos(−b) + cos a sin(−b)
sin(a − b) = sin a cos b − cos a sin b
Use the sine sum formula
Use cos(−b) = cos b, and sin(−b) = − sin b
In conclusion, sin(a − b) = sin a cos b − cos a sin b, so, this is the difference formula for sine.
Using the Sine Sum and Difference Formula
1. Find the exact value of sin 5π
12
Recall that there are multiple angles that add or subtract to equal any angle. Choose whichever formula that you feel
more comfortable with.
5π
3π 2π
sin
= sin
+
12
12 12
3π
2π
3π
2π
= sin cos
+ cos sin
12
√ 12 √ 12 √ 12
5π
2
3
2 1
=
×
+
×
sin
12
2
2
2
√2
√
6+ 2
=
4
2. Given sin α =
sin(α + β).
12
13 ,
where α is in Quadrant II, and sin β = 53 , where β is in Quadrant I, find the exact value of
To find the exact value of sin(α + β), here we use sin(α + β) = sin α cos β + cos α sin β. The values of sin α and sin β
are known, however the values of cos α and cos β need to be found.
Use sin2 α + cos2 α = 1, to find the values of each of the missing cosine values.
12 2
2
2
For cos a : sin2 α + cos2 α = 1, substituting sin α = 12
+ cos2 α = 144
13 transforms to 13
169 + cos α = 1 or cos α =
25
5
5
169 cos α = ± 13 , however, since α is in Quadrant II, the cosine is negative, cos α = − 13 .
2
9
For cos β use sin2 β + cos2 β = 1 and substitute sin β = 53 , 35 + cos2 β = 25
+ cos2 β = 1 or cos2 β = 16
25 and cos β =
4
4
± 5 and since β is in Quadrant I, cos β = 5
Now the sum formula for the sine of two angles can be found:
360
www.ck12.org
Chapter 2. Unit 2
5
3
48 15
12 4
× + −
× or
−
sin(α + β) =
13 5
13
5
65 65
33
sin(α + β) =
65
3. Find the exact value of sin 15◦
Recall that there are multiple angles that add or subtract to equal any angle. Choose whichever formula that you feel
more comfortable with.
sin 15◦ = sin (45◦ − 30◦ )
= sin 45◦ cos 30◦ + cos 45◦ sin 30◦
sin 15◦ = (.707) × (.866) + (.707) × (.5)
= (.612262) × (.3535)
= .2164
Examples
Example 1
Earlier, you were given a problem about using the sine sum formula.
With the sine sum formula, you can break the sine into easier to calculate quantities:
7π
4π 3π
sin
= sin
+
12
12 12
π π
= sin
+
3 4
π
π
π
π
= sin( ) cos( ) + cos( )sin( )
3
4
3
4
√ ! √ ! √ !
3
2
1
2
=
+
2
2
2
2
√
√
6
2
=
+
4
4
√
√
6+ 2
=
4
Example 2
Find the exact value for sin 345◦
sin 345◦ = sin(300◦ + 45◦ ) = sin 300◦ cos 45◦ + cos 300◦ sin 45◦
√
√
√
√
√
√ √
3
2 1
2
6
2
2− 6
=−
·
+ ·
=−
+
=
2
2
2 2
4
4
4
361
2.15. Sine Sum and Difference Formulas
www.ck12.org
Example 3
Find the exact value for sin 17π
12
sin
9π 8π
17π
2π
3π
2π
3π 2π
3π
= sin
+
+
+ cos sin
= sin
= sin cos
12
12 12
4
3
4
3
4
3
√ √
√
√
√
√
√
1
2
2
3
2
6 − 2− 6
· (− ) + −
·
=−
−
=
=
2
2
2
2
4
4
4
Example 4
5
If sin y = − 13
, y is in quad III, and sin z = 45 , z is in quad II find sin(y + z)
5
If sin y = − 13
and in Quadrant III, then cosine is also negative. By the Pythagorean Theorem, the second leg is
4
2
2
12(5 + b = 132 ), so cos y = − 12
13 . If the sin z = 5 and in Quadrant II, then the cosine is also negative. By the
2
2
Pythagorean Theorem, the second leg is 3(4 + b = 52 ), so cos = − 35 . To find sin(y + z), plug this information into
the sine sum formula.
sin(y + z) = sin y cos z + cos y sin z
5
3
12 4 15 48
33
= − ·− +− · =
−
=−
13
5
13 5 65 65
65
Review
Find the exact value for each sine expression.
1.
2.
3.
4.
5.
sin 75◦
sin 105◦
sin 165◦
sin 255◦
sin −15◦
Write each expression as the sine of an angle.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
sin 46◦ cos 20◦ + cos 46◦ sin 20◦
sin 3x cos 2x − cos 3x sin 2x
sin 54◦ cos 12◦ + cos 54◦ sin 12◦
sin 29◦ cos 10◦ − cos 29◦ sin 10◦
sin 4y cos 3y + cos 4y sin 2y
Prove that sin(x − π2 ) = − cos(x)
Suppose that x, y, and z are the three angles of a triangle. Prove that sin(x + y) = sin(z)
Prove that sin( π2 − x) = cos(x)
Prove that sin(x + π) = − sin(x)
Prove that sin(x − y) + sin(x + y) = 2 sin(x) cos(y)
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.7.
362
www.ck12.org
Chapter 2. Unit 2
2.16 Cosine Sum and Difference Formulas
Learning Objectives
Here you’ll learn to rewrite cosine functions with addition or subtraction in their arguments in a more easily solvable
form.
While playing a board game with friends, you are using a spinner like this one:
When you tap the spinner with your hand, it rotates 110◦ . However, at that moment, someone taps the game board
and the spinner moves back a little to 80◦ . One of your friends, who is a grade above you in math, starts talking to
you about trig functions.
"Do you think you can calculate the cosine of the difference between those angles?" he asks.
"Hmm," you reply. "Sure. I think it’s just cos(110◦ − 80◦ ) = cos 30◦ ."
Your friend smiles. "Are you sure?" he asks.
You realize you aren’t sure at all. Can you solve this problem? Read this lesson, and by the end you’ll be able to
calculate the cosine of the difference of the angles.
Formulas for the Sum and Difference of Cosines
When thinking about how to calculate values for trig functions, it is natural to consider what the value is for the trig
function of a difference of two angles.
363
2.16. Cosine Sum and Difference Formulas
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143489
For example, is cos 15◦ = cos(45◦ − 30◦ )? Upon appearance, yes, it is. This section explores how to find an
expression that would equal cos(45◦ − 30◦ ). To simplify this, let the two given angles be a and b where 0 < b < a <
2π.
Begin with the unit circle and place the angles a and b in standard position as shown in Figure A. Point Pt1 lies on
the terminal side of b, so its coordinates are (cos b, sin b) and Point Pt2 lies on the terminal side of a so its coordinates
are (cos a, sin a). Place the a − b in standard position, as shown in Figure B. The point A has coordinates (1, 0) and
the Pt3 is on the terminal side of the angle a − b, so its coordinates are (cos[a − b], sin[a − b]).
364
www.ck12.org
Chapter 2. Unit 2
Triangles OP1 P2 in figure A and Triangle OAP3 in figure B are congruent. (Two sides and the included angle, a − b,
are equal). Therefore the unknown side of each triangle must also be equal. That is: d (A, P3 ) = d (P1 , P2 )
Applying the distance formula to the triangles in Figures A and B and setting them equal to each other:
q
q
2
2
[cos(a − b) − 1] + [sin(a − b) − 0] = (cos a − cos b)2 + (sin a − sin b)2
Square both sides to eliminate the square root.
[cos(a − b) − 1]2 + [sin(a − b) − 0]2 = (cos a − cos b)2 + (sin a − sin b)2
FOIL all four squared expressions and simplify.
cos2 (a − b) − 2 cos(a − b) + 1 + sin2 (a − b) = cos2 a − 2 cos a cos b + cos2 b + sin2 a − 2 sin a sin b + sin2 b
2
2
2
sin2 (a − b) + cos2 (a − b) −2 cos(a − b) + 1 = |sin2 a +
{zcos a} −2 cos a cos b + |sin b +
{zcos b} −2 sin a sin b
|
{z
}
1 − 2 cos(a − b) + 1 = 1 − 2 cos a cos b + 1 − 2 sin a sin b
2 − 2 cos(a − b) = 2 − 2 cos a cos b − 2 sin a sin b
−2 cos(a − b) = −2 cos a cos b − 2 sin a sin b
cos(a − b) = cos a cos b + sin a sin b
In cos(a − b) = cos a cos b + sin a sin b, the difference formula for cosine, you can substitute a − (−b) = a + b to
obtain: cos(a + b) = cos[a − (−b)] or cos a cos(−b) + sin a sin(−b). since cos(−b) = cos b and sin(−b) = − sin b,
then cos(a + b) = cos a cos b − sin a sin b, which is the sum formula for cosine.
Using the Cosine Difference Formula
1. Find an equivalent form of cos
π
2
− θ using the cosine difference formula.
365
2.16. Cosine Sum and Difference Formulas
www.ck12.org
π
π
− θ = cos cos θ + sin sin θ
2
2
2
π
π
π
cos
− θ = 0 × cos θ + 1 × sin θ, substitute cos = 0 and sin = 1
2
2
2
π
cos
− θ = sin θ
2
cos
π
We know that is a true identity because of our understanding of the sine and cosine curves, which are a phase shift
of π2 off from each other.
The cosine formulas can also be used to find exact values of cosine that we weren’t able to find before, such as
15◦ = (45◦ − 30◦ ), 75◦ = (45◦ + 30◦ ), among others.
2. Find the exact value of cos 15◦
Use the difference formula where a = 45◦ and b = 30◦ .
cos(45◦ − 30◦ ) = cos 45◦ cos 30◦ + sin 45◦ sin 30◦
√
√
√
2
3
2 1
◦
cos 15 =
×
+
×
2
2
2
√2
√
6+ 2
cos 15◦ =
4
3. Find the exact value of cos 5π
12 , in radians.
π
π
π
3π
cos 5π
12 = cos 4 + 6 , notice that 4 = 12 and
π
6
=
2π
12
π
π
π
π
π
= cos cos − sin sin
4 6
6
√ 4 √6 √ 4
π
π
π
π
2
3
2 1
cos cos − sin sin =
×
−
×
4
6
4
6
2
2
2
√2
√
6− 2
=
4
cos
π
+
Examples
Example 1
Earlier, you were asked if you can calculate the cosine of the difference between the two angles.
It would seem that your friend was having some fun with you, since he figured you didn’t know the cosine difference
formula. But now, with this formula in hand, you can readily solve for the difference of the two angles:
cos(110◦ − 80◦ )
= (cos 110◦ )(cos 80◦ ) + (sin 110◦ )(sin 80◦ )
= (−.342)(.174) + (.9397)(.9848)
= −.0595 + .9254 = .8659
Therefore,
cos(110◦ − 80◦ ) = .8659
366
www.ck12.org
Chapter 2. Unit 2
Example 2
Find the exact value for cos 5π
12
π π
2π 3π
π
π
π
π
5π
cos
= cos
+
= cos
+
= cos cos − sin sin
12
12 12
6 4
6
4
6
4
√
√
√
√
√
√ √
3
2 1
2
6
2
6− 2
·
− ·
=
−
=
=
2
2
2 2
4
4
4
Example 3
Find the exact value for cos 7π
12
π π
7π
4π 3π
π
π
π
π
cos
= cos
+
= cos
+
= cos cos − sin sin
12
12 12
3 4
3
4
3
4
√
√
√
√
√
√ √
2
3
2
2
6
2− 6
1
−
·
=
−
=
= ·
2 2
2
2
4
4
4
Example 4
Find the exact value for cos 345◦
cos 345◦ = cos(315◦ + 30◦ ) = cos 315◦ cos 30◦ − sin 315◦ sin 30◦
√
√
√ √
√
2
3
2 1
6+ 2
·
− (−
)· =
=
2
2
2
2
4
Review
Find the exact value for each cosine expression.
1.
2.
3.
4.
5.
cos 75◦
cos 105◦
cos 165◦
cos 255◦
cos −15◦
Write each expression as the cosine of an angle.
cos 96◦ cos 20◦ + sin 96◦ sin 20◦
cos 4x cos 3x − sin 4x sin 3x
cos 37◦ cos 12◦ + sin 37◦ sin 12◦
cos 59◦ cos 10◦ − sin 59◦ sin 10◦
cos 5y cos 2y + sin 5y sin 2y
√
π
11. Prove that cos(x − 4 ) = 2 2 (cos(x) + sin(x))
12. If cos(x) cos(y) = sin(x) sin(y), then what does cos(x + y) equal?
13. Prove that cos(x − π2 ) = sin(x)
14. Use the fact that cos( π2 − x) = sin(x) (shown in examples), to show that sin( π2 − x) = cos(x).
15. Prove that cos(x − y) + cos(x + y) = 2 cos(x) cos(y).
6.
7.
8.
9.
10.
367
2.16. Cosine Sum and Difference Formulas
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.6.
368
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.17 Tangent Sum and Difference Formulas
Learning Objectives
Here you’ll learn to rewrite tangent functions with addition or subtraction in their arguments in a more easily solvable
form.
Suppose you were given two angles and asked to find the tangent of the difference of them. For example, can you
compute:
tan(120◦ − 40◦ )
Would you just subtract the angles and then take the tangent of the result? Or is something more complicated
required to solve this problem? Keep reading, and by the end of this lesson, you’ll be able to calculate trig functions
like the one above.
Tangent Sum and Difference Formulas
In this lesson, we want to find a formula that will make computing the tangent of a sum of arguments or a difference
of arguments easier. As first, it may seem that you should just add (or subtract) the arguments and take the tangent
of the result. However, it’s not quite that easy.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143491
To find the sum formula for tangent:
sin(a + b)
cos(a + b)
sin a cos b + sin b cos a
=
cos a cos b − sin a sin b
tan(a + b) =
=
=
=
tan(a + b) =
Using tan θ =
sin θ
cos θ
Substituting the sum formulas for sine and cosine
sin a cos b+sin b cos a
cos a cos b
cos a cos b−sin a sin b
cos a cos b
sin a cos b
sin b cos a
cos a cos b + cos a cos b
cos a cos b
sin a sin b
cos a cos b − cos a cos b
sin a
sin b
cos a + cos b
sin a sin b
1 − cos
a cos b
Reduce each of the fractions
tan a + tan b
1 − tan a tan b
Sum formula for tangent
Divide both the numerator and the denominator by cos a cos b
Substitute
sin θ
= tan θ
cos θ
369
2.17. Tangent Sum and Difference Formulas
In conclusion, tan(a + b) =
tangent:
tan a+tan b
1−tan a tan b .
www.ck12.org
Substituting −b for b in the above results in the difference formula for
tan(a − b) =
tan a − tan b
1 + tan a tan b
Using the Tangent Difference Formula
1. Find the exact value of tan 285◦ .
Use the difference formula for tangent, with 285◦ = 330◦ − 45◦
tan(330◦ − 45◦ ) =
tan 330◦ − tan 45◦
1 + tan 330◦ tan 45◦
√
3
3 √− 1
1 − 33 · 1
√
−3 − 3
√
=
=
3− 3
√
√
−3 − 3 3 + 3
√ ·
√
=
3− 3 3+ 3
√
−9 − 6 3 − 3
=
9−3
√
−12 − 6 3
=
6√
= −2 − 3
−
To verify this on the calculator, tan 285◦ = −3.732 and −2 −
√
3 = −3.732.
2. Verify the tangent difference formula by finding tan 6π
6 , since this should be equal to tan π = 0.
7π
π
Use the difference formula for tangent, with tan 6π
6 = tan( 6 − 6 )
tan(
π
tan 7π
7π π
6 − tan 6
− )=
π
6
6
1 + tan 7π
6 tan 6
=
=
√
√
2
− 62
6 √
√
1 − 62 · 62
=
0
2
1 − 36
0
34
36
=0
3. Find the exact value of tan 165◦ .
Use the difference formula for tangent, with 165◦ = 225◦ − 60◦
tan 225◦ − tan 60◦
1 + tan 225◦ tan 60◦
√
1− 3
√ =1
=
1−1· 3
tan(225◦ − 60◦ ) =
370
www.ck12.org
Chapter 2. Unit 2
Examples
Example 1
Earlier, you were asked to find tan(120◦ − 40◦ ).
You can use the tangent difference formula:
tan(a − b) =
tan a − tan b
1 + tan a tan b
to help solve this. Substituting in known quantities:
tan(120◦ − 40◦ ) =
tan 120◦ − tan 40◦
−1.732 − .839
−2.571
=
=
= 5.674
◦
◦
1 + (tan 120 )(tan 40 ) 1 + (−1.732)(.839) −.453148
Example 2
Find the exact value for tan 75◦
tan 75◦ = tan(45◦ + 30◦ )
tan 45◦ + tan 30◦
=
1 − tan 45◦ tan 30◦
=
√
3
3√
1 − 1 · 33
1+
=
√
3+ 3
3√
3− 3
3
√
√
3+ 3 3+ 3
√ ·
√
=
3− 3 3+ 3
√
√
9 + 6 3 + 3 12 + 6 3
=
=
9−3
6
√
= 2+ 3
Example 3
Simplify tan(π + θ)
tan(π + θ) =
tan π+tan θ
1−tan π tan θ
=
tan θ
1
= tan θ
Example 4
Find the exact value for tan 15◦
371
2.17. Tangent Sum and Difference Formulas
www.ck12.org
tan 15◦ = tan(45◦ − 30◦ )
tan 45◦ − tan 30◦
=
1 + tan 45◦ tan 30◦
=
√
3
3√
1 + 1 · 33
1−
=
√
3− 3
3√
3+ 3
3
√
√
3− 3 3− 3
√ ·
√
=
3+ 3 3− 3
√
√
9 + 6 3 + 3 12 + 6 3
=
=
9−3
6
√
= 2+ 3
Review
Find the exact value for each tangent expression.
1.
2.
3.
4.
5.
tan 5π
12
tan 11π
12
tan −165◦
tan 255◦
tan −15◦
Write each expression as the tangent of an angle.
6.
7.
8.
9.
10.
11.
tan 15◦ +tan 42◦
1−tan 15◦ tan 42◦
tan 65◦ −tan 12◦
1+tan 65◦ tan 12◦
tan 10◦ +tan 50◦
1−tan 10◦ tan 50◦
tan 2y+tan 4
1−tan 2 tan 4y
tan x−tan 3x
1+tan x tan 3x
tan 2x−tan y
1+tan 2x tan y
12. Prove that tan(x + π4 ) = 1+tan(x)
1−tan(x)
13. Prove that tan(x − π2 ) = − cot(x)
14. Prove that tan( π2 − x) = cot(x)
15. Prove that tan(x + y) tan(x − y) =
tan2 (x)−tan2 (y)
1−tan2 (x) tan2 (y)
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.8.
372
www.ck12.org
Chapter 2. Unit 2
2.18 Solving Trig Equations using Sum and
Difference Formulas
Learning Objectives
Here you’ll solve trig equations using the sum and difference formulas.
As Agent Trigonometry, you are given a piece of the puzzle: sin( π2 − x) = −1. What is the value of x?
Solving Trigonometric Functions
We can use the sum and difference formulas to solve trigonometric equations. For this concept, we will only find
solutions in the interval 0 ≤ x < 2π.
Let’s solve the following functions using the sum and difference formulas.
1. cos(x − π) =
√
2
2
Use the formula to simplify the left-hand side and then solve for x.
√
2
cos(x − π) =
2
√
2
cos x cos π + sin x sin π =
2
√
2
− cos x =
2√
2
cos x = −
2
The cosine negative in the 2nd and 3rd quadrants. x =
2. sin x + π4 + 1 = sin
π
4
−x
3π
4
and
5π
4 .
π
π
sin x +
+ 1 = sin
−x
4
4
π
π
π
π
sin x cos + cos x sin + 1 = sin cos x − cos sin x
√4
√4
√4
√ 4
2
2
2
2
sin x ·
+ cos x ·
+1 =
. cos x −
· sin x
2
2
2
√2
2 sin x = −1
√
1
2
sin x = − √ = −
2
2
In the interval, x =
5π
4
and
7π
4 .
373
2.18. Solving Trig Equations using Sum and Difference Formulas
www.ck12.org
3. 2 sin x + π3 = tan π3
π
π
= tan
2 sin x +
3
3
π √
π
2 sin x cos + cos x sin
= 3
3
3
√
1
3 √
2 sin x · + 2 cos x ·
= 3
2
√ 2
√
sin x + 3 cos x = 3
√
sin x = 3(1 − cos x)
sin2 x = 3(1 − 2 cos x + cos2 x)
1 − cos2 x = 3 − 6 cos x + 3 cos2 x
square both sides
substitute sin2 x = 1 − cos2 x
0 = 4 cos2 x − 6 cos x + 2
0 = 2 cos2 x − 3 cos x + 1
At this point, we can factor the equation to be (2 cos x − 1)(cos x − 1) = 0. cos x = 12 , and 1, so x = 0, π3 , 5π
3 . Be
5π
careful with these answers. When we check these solutions it turns out that 3 does not work.
5π π
2 sin
+
3
3
Therefore,
5π
3
π
= tan
3
√
2 sin 2π = 3
√
0 6= 3
is an extraneous solution.
Examples
Example 1
Earlier, you were asked to find the value of x from the equation sin( π2 − x) = −1.
First, simplify the expression sin( π2 − x) as:
π
π
π
sin( − x) = sin cos x − cos sin x
2
2
2
= 1 · cos x − 0 · sin x
= cosx
So what you’re now looking for is the value of x where cos x = −1.
The cosine of 180◦ is equal to −1.
Solve the following equations in the interval 0 ≤ x < 2π.
Example 2
cos(2π − x) =
374
1
2
www.ck12.org
Chapter 2. Unit 2
1
2
1
cos 2π cos x + sin 2π sin x =
2
1
cos x =
2
5π
π
x = and
3
3
cos(2π − x) =
Example 3
π
6
sin
− x + 1 = sin x + π6
π
− x + 1 = sin x +
6
6
π
π
π
π
sin cos x − cos sin x + 1 = sin x cos + cos x sin
6
6
6
√6
√
1
1
3
3
cos x −
sin x + 1 =
sin x + cos x
2
2
2
2
√
1 = 3 sin x
1
√ = sin x
3
1
−1
√
x = sin
= 0.6155 and 2.5261 rad
3
sin
π
Example 4
cos
π
2
+ x = tan π4
π
+ x = tan
2
4
π
π
cos cos x − sin sin x = 1
2
2
− sin x = 1
cos
π
sin x = −1
3π
x=
2
Review
Solve the following trig equations in the interval 0 ≤ x < 2π.
1.
2.
3.
4.
√
sin(x − π) = − 2 2
cos(2π + x)
= −1
π
tan x + 4 = 1
sin π2 − x = 12
375
2.18. Solving Trig Equations using Sum and Difference Formulas
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
3π
sin x + 3π
4 + sin x − 4 = 1
sin x + π6 = − sin x − π6
cos x + π6 = cos x − π6 + 1
cos x + π3 + cos x − π3 = 1
tan(x + π) + 2 sin(x + π)
=0
tan(x + π)+ cos x + π2 = 0
tan x + π4 = tan x − π4
2π
sin x − 5π
3 − sin x − 3 = 0
4 sin(x + π) − 2 = 2 cos x + π2
1 + 2 cos(x − π) + cos x = 0
Real Life Application The height, h (in feet), of twopeople in different seats on a Ferris wheel can be
modeled by h1 = 50 cos 3t + 46 and h2 = 50 cos 3 t − 3π
4 + 46 where t is the time (in minutes). When are the
two people at the same height?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.14.
376
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.19 Finding Exact Trig Values using Double
and Half Angle Formulas
Learning Objectives
Here you’ll use the half and double angle formulas to find exact values of angles other than the critical angles.
You want to find the exact value of tan 3π
8 . How could you find this value without using a calculator?
Double Angle and Half Angle Formulas
In this concept, we will learn how to find the exact values of the trig functions for
angles that are half or double of
a
other angles. Here we will introduce the Double-Angle (2a) and Half-Angle 2 Formulas.
Double-Angle and Half-Angle Formulas
cos 2a = cos2 a − sin2 a
sin 2a = 2 sin a cos a
2 tan a
tan 2a =
1 − tan2 a
= 2 cos2 a − 1
a
2
a
cos
2
sin
= 1 − sin2 a
r
1 − cos a
=±
2
r
1 + cos a
=±
2
tan
The signs of sin a2 and cos 2a depend on which quadrant
solve for the exact value.
a
2
a 1 − cos a
=
2
sin a
sin a
=
1 + cos a
lies in. For cos 2a and tan 2a any formula can be used to
Let’s find the exact value of cos π8 .
π
8
is half of
π
4
and in the first quadrant.
1 π
·
cos
2 4
r
=
1 + cos π4
2
s
1+
2
=
√
2
2
√
1 2+ 2
=
·
2
2
p
√
2+ 2
=
2
s
Now, let’s find the exact value of sin 2a if cos a = − 54 and
3π
2
≤ a < 2π.
377
2.19. Finding Exact Trig Values using Double and Half Angle Formulas
To use the sine double-angle formula, we also need to find sin a, which would be
www.ck12.org
3
5
because a is in the 4th quadrant.
sin 2a = 2 sin a cos a
4
3
= 2· ·−
5
5
24
=−
25
Finally, let’s find the exact value of tan 2a for a from the previous problem.
Use tan a =
sin a
cos a
=
3
5
− 45
= − 43 to solve for tan 2a.
2 · − 34
− 32
1− −4
7
16
tan 2a =
=
3 2
3 16
24
=− ·
=−
2 7
7
Examples
Example 1
Earlier, you were asked to find the value of tan 3π
8 without a calculator.
3π
8
a
= 12 · 3π
4 so we can use the formula tan 2 =
sin a
1+cos a
tan
for a =
sin 3π
3π
4
=
8
1 + cos 3π
4
=
If we simplify this expression, we get
3π
4
√
2
2 √
1 + −2 2
√
2 + 1.
Example 2
Find the exact value of cos − 5π
8 .
rd
− 5π
8 is in the 3 quadrant.
s
1 + cos − 5π
5π 1
5π
1
5π
4
− =
−
→ cos
−
=−
8
2
4
2
4
2
s
s
√
p
√
√
1 − 22
1 2− 2
2− 2
=−
=
·
=
2
2
2
2
Example 3
Given the function cos a =
378
4
7
and 0 ≤ a < π2 , find sin 2a.
www.ck12.org
Chapter 2. Unit 2
First, find sin a. 42 + y2 = 72 → y =
√
√
sin 2a = 2 · 733 · 74 = 8 4933
√
√
33, so sin a = 733
Example 4
Given the function cos a =
4
7
and 0 ≤ a < π2 , find tan 2a .
You can use either tan a2 formula.
√
3
7
3
a 1 − 47
33
tan = √ = · √ = √ =
33
2
7
11
33
33
7
Review
Find the exact value of the following angles.
1.
2.
3.
4.
5.
6.
7.
8.
sin 105◦
tan π8
cos 5π
12
cos 165◦
sin 3π
8
π
tan − 12
sin 11π
8
cos 19π
12
5
13
and
3π
2
8
11
and
π
2
The cos a =
9.
10.
11.
12.
sin 2a
cos 2a
tan a2
cos 2a
The sin a =
13.
14.
15.
16.
≤ a < 2π. Find:
≤ a < π. Find:
tan 2a
sin 2a
cos 2a
sin 2a
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.15.
379
2.20. Sum and Difference Identities
www.ck12.org
2.20 Sum and Difference Identities
Here you will add six identities to your toolbox: the sum and difference identities for sine, cosine and tangent. You
will use these identities along with previous identities for proofs and simplifying expressions.
With your knowledge of special angles like the sine and cosine of 30◦ and 45◦ , you can find the sine and cosine of
15◦ , the difference of 45◦ and 30◦ , and 75◦ , the sum of 45◦ and 30◦ . Using what you know about the unit circle and
the sum and difference identities, how do you determine sin 15◦ and sin 75◦ ?
Sum and Difference Identities
First look at the derivation of the cosine difference identity:
cos(α − β) = cos α cos β + sin α sin β
Start by drawing two arbitrary angles α and β. In the image above α is the angle in red and β is the angle in
blue. The difference α − β is noted in black as θ. The reason why there are two pictures is because the image on
the right has the same angle θ in a rotated position. This will be useful to work with because the length of AB will
be the same as the length of CD.
AB = CD
q
q
2
2
(cos α − cos β) + (sin α − sin β) = (cos θ − 1)2 + (sin θ − 0)2
(cos α − cos β)2 + (sin α − sin β)2 = (cos θ − 1)2 + (sin θ − 0)2
(cos α)2 − 2 cos α cos β + (cos β)2 + (sin α)2 − 2 sin α sin β + (sin β)2 = (cos θ − 1)2 + (sin θ)2
2 − 2 cos α cos β − 2 sin α sin β = (cos θ)2 − 2 cos θ + 1 + (sin θ)2
2 − 2 cos α cos β − 2 sin α sin β = 1 − 2 cos θ + 1
−2 cos α cos β − 2 sin α sin β = −2 cos θ
cos α cos β + sin α sin β = cos θ
= cos(α − β)
380
www.ck12.org
Chapter 2. Unit 2
You can use this identity to prove the cosine of a sum identity. First, start with the cosine of a difference and make
a substitution. Then use the odd-even identity.
cos α cos β + sin α sin β = cos(α − β)
Let γ = −β
cos α cos(−γ) + sin α sin(−γ) = cos(α + γ)
cos α cos γ − sin α sin γ = cos(α + γ)
The proofs for sine and tangent are left to the videos and examples. They are listed here for your reference. Cotangent, secant and cosecant are excluded because you can use reciprocal identities to get those once you have sine,
cosine and tangent.
Summary
• cos(α ± β) = cos α cos β ∓ sin α sin β
• sin(α ± β) = sin α cos β ± cos α sin β
sin(α±β)
tan α±tan β
• tan(α ± β) = cos(α±β)
= 1∓tan
α tan β
The order of the plus or minus signs is important because for cosine of a sum, the negative sign is used on the other
side of the identity. This is the opposite of sine of a sum, where a positive sign is used on the other side of the
identity.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61321
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/109807
381
2.20. Sum and Difference Identities
www.ck12.org
Examples
Example 1
Earlier, you were asked to evaluate sin 15◦ and sin 75◦ exactly without a calculator. To do this you need to use the
sine of a difference and sine of a sum.
√
2
sin(45 − 30 ) = sin 45 cos 30 − cos 45 sin 30 =
·
2
√
2
·
sin(45◦ + 30◦ ) = sin 45◦ cos 30◦ + cos 45◦ sin 30◦ =
2
◦
◦
◦
◦
◦
◦
√
3
−
2
√
3
+
2
√
√
√
2 1
6− 2
· =
2 2
√
√ 4 √
2 1
6+ 2
· =
2 2
4
Example 2
Find the exact value of tan 15◦ without using a calculator.
tan 15◦
=
tan(45◦ − 30◦ )
=
tan 45◦ −tan 30◦
1+tan 45◦ tan 30◦
=
√
3
3
√
1+1· 3 3
1−
√
=
3− √3
3+
3
A final solution will not have a radical in the denominator. In this case multiplying through by the conjugate of the
denominator will eliminate the radical.
√
√
(3 − 3) · (3 − 3)
√
√
=
(3 + 3) · (3 − 3)
√
(3 − 3)2
=
9−3
√
(3 − 3)2
=
6
Example 3
Evaluate the expression exactly without using a calculator.
cos 50◦ cos 5◦ + sin 50◦ sin 5◦
Once you know the general form of the sum and difference identities then you will recognize this as cosine of a
difference.
√
◦
◦
◦
◦
◦
◦
◦
cos 50 cos 5 + sin 50 sin 5 = cos(50 − 5 ) = cos 45 = 2 2
Example 4
Use a sum or difference identity to find an exact value of cot
5π
12
Start with the definition of cotangent as the inverse of tangent.
382
.
www.ck12.org
Chapter 2. Unit 2
Example 5
Prove the following identity:
sin(x−y)
sin(x+y)
=
tan x−tan y
tan x+tan y
Here are the steps:
sin(x − y)
sin(x + y)
sin x cos y − cos x sin y
sin x cos y + cos x sin y
1
sin x cos y − cos x sin y cos x·cos y
·
1
sin x cos y + cos x sin y
cos x·cos y
y
sin x cos
cos
x sin y
−
cos x·
cos x·cos y
cos y
x sin y
y
sin x cos
cos
+
x·cos y
y
cos x·
cos
cos
=
tan x − tan y
tan x + tan y
=
=
=
tan x − tan y
=
tan x + tan y
Review
Find the exact value for each expression by using a sum or difference identity.
1. cos 75◦
2. cos 105◦
3. cos 165◦
4. sin 105◦
5. sec 105◦
6. tan 75◦
7. Prove the sine of a sum identity.
8. Prove the tangent of a sum identity.
9. Prove the tangent of a difference identity.
10. Evaluate without a calculator: cos 50◦ cos 10◦ − sin 50◦ sin 10◦ .
11. Evaluate without a calculator: sin 35◦ cos 5◦ − cos 35◦ sin 5◦ .
12. Evaluate without a calculator: sin 55◦ cos 5◦ + cos 55◦ sin 5◦ .
13. If cos α cos β = sin α sin β, then what does cos(α + β) equal?
x
14. Prove that tan x + π4 = 1+tan
1−tan x .
15. Prove that sin(x + π) = − sin x.
383
2.20. Sum and Difference Identities
Review (Answers)
To see the Review answers, open this PDF file and look for section 6.3.
384
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.21 Double Angle Identities
Learning Objectives
Here you’ll learn the double angle identities and how to use them to rewrite trig equations into a more easily solvable
form.
Finding the values for trig functions is pretty familiar to you by now. The trig functions of some particular angles
may even seem obvious, since you’ve worked with them so many times. In some cases, you might be able to use
this knowledge to your benefit to make calculating the values of some trig equations easier. For example, if someone
asked you to evaluate
cos 120◦
without consulting a table of trig values, could you do it?
You might notice right away that this is equal to four times 30◦ . Can this help you? Read this lessons, and at its
conclusion you’ll know how to use certain formulas to simplify multiples of familiar angles to solve problems.
Double Angle Identities
Here we’ll start with the sum and difference formulas for sine, cosine, and tangent. We can use these identities to
help derive a new formula for when we are given a trig function that has twice a given angle as the argument.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143660
For example, sin(2θ). This way, if we are given θ and are asked to find sin(2θ), we can use our new double angle
identity to help simplify the problem. Let’s start with the derivation of the double angle identities.
One of the formulas for calculating the sum of two angles is:
sin(α + β) = sin α cos β + cos α sin β
If α and β are both the same angle in the above formula, then
sin(α + α) = sin α cos α + cos α sin α
sin 2α = 2 sin α cos α
This is the double angle formula for the sine function. The same procedure can be used in the sum formula for
cosine, start with the sum angle formula:
385
2.21. Double Angle Identities
www.ck12.org
cos(α + β) = cos α cos β − sin α sin β
If α and β are both the same angle in the above formula, then
cos(α + α) = cos α cos α − sin α sin α
cos 2α = cos2 α − sin2 α
This is one of the double angle formulas for the cosine function. Two more formulas can be derived by using the
Pythagorean Identity, sin2 α + cos2 α = 1.
sin2 α = 1 − cos2 α and likewise cos2 α = 1 − sin2 α
Using sin2 α = 1 − cos2 α :
Using cos2 α = 1 − sin2 α :
cos 2α = cos2 α − sin2 α
cos 2α = cos2 α − sin2 α
= cos2 α − (1 − cos2 α)
= (1 − sin2 α) − sin2 α
= cos2 α − 1 + cos2 α
= 1 − sin2 α − sin2 α
= 2 cos2 α − 1
= 1 − 2 sin2 α
Therefore, the double angle formulas for cos 2α are:
cos 2α = cos2 α − sin2 α
cos 2α = 2 cos2 α − 1
cos 2α = 1 − 2 sin2 α
Finally, we can calculate the double angle formula for tangent, using the tangent sum formula:
tan(α + β) =
tan α + tan β
1 − tan α tan β
If α and β are both the same angle in the above formula, then
tan α + tan α
1 − tan α tan α
2 tan α
tan 2α =
1 − tan2 α
tan(α + α) =
We can use these formulas to help simplify calculations of trig functions of certain arguments.
Let’s look at a few problems involving double angle identities.
1. If sin a =
5
13
and a is in Quadrant II, find sin 2a, cos 2a, and tan 2a.
To use sin 2a = 2 sin a cos a, the value of cos a must be found first.
386
www.ck12.org
Chapter 2. Unit 2
= cos2 a + sin2 a = 1
2
5
2
=1
= cos a +
13
25
= cos2 a +
=1
169
12
144
, cos a = ±
= cos2 a =
169
13
.
However since a is in Quadrant II, cos a is negative or cos a = − 12
13 .
5
12
120
sin 2a = 2 sin a cos a = 2
× −
= sin 2a = −
13
13
169
For cos 2a, use cos(2a) = cos2 a − sin2 a
2
12 2
5
144 − 25
cos(2a) = −
−
or
13
13
169
119
cos(2a) =
169
For tan 2a, use tan 2a =
2 tan a
.
1−tan2 a
From above, tan a =
tan(2a) =
2 · −5
12
1−
=
−5 2
12
5
13
− 12
13
5
.
= − 12
−5
6
25
1 − 144
=
−5
6
119
144
5 144
120
=− ·
=−
6 119
119
2. Find cos 4θ.
Think of cos 4θ as cos(2θ + 2θ).
cos 4θ = cos(2θ + 2θ) = cos 2θ cos 2θ − sin 2θ sin 2θ = cos2 2θ − sin2 2θ
Now, use the double angle formulas for both sine and cosine. For cosine, you can pick which formula you would
like to use. In general, because we are proving a cosine identity, stay with cosine.
= (2 cos2 θ − 1)2 − (2 sin θ cos θ)2
= 4 cos4 θ − 4 cos2 θ + 1 − 4 sin2 θ cos2 θ
= 4 cos4 θ − 4 cos2 θ + 1 − 4(1 − cos2 θ) cos2 θ
= 4 cos4 θ − 4 cos2 θ + 1 − 4 cos2 θ + 4 cos4 θ
= 8 cos4 θ − 8 cos2 θ + 1
3. Solve the trigonometric equation sin 2x = sin x such that (−π ≤ x < π)
387
2.21. Double Angle Identities
www.ck12.org
Using the sine double angle formula:
sin 2x = sin x
2 sin x cos x = sin x
2 sin x cos x − sin x = 0
sin x(2 cos x − 1) = 0



&

y
2 cos x − 1 = 0
2 cos x = 1
sin x = 0
x = 0, −π
cos x =
1
2
π π
x = ,−
3 3
Examples
Example 1
Earlier, you were asked to solve cos 120◦ .
You can simplify this into a familiar angle:
cos(2 × 60◦ )
And then apply the double angle identity:
cos(2 × 60◦ ) = 2 cos2 60◦ − 1
= (2)(cos 60◦ )(cos 60◦ ) − 1
1 1
= (2)( )( ) − 1
2 2
1
=−
2
Example 2
4
5 and x is in Quad II, find the exact values of cos 2x, sin 2x and tan 2x
If sin x = 45 and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem,
p
side is 3(b = 52 − 42 ). So, cos x = − 35 and tan x = − 43 . Using this, we can find sin 2x, cos 2x, and tan 2x.
If sin x =
388
the third
www.ck12.org
Chapter 2. Unit 2
2 tan x
1 − tan2 x
2 · − 43
=
2
1 − − 43
cos 2x = 1 − sin2 x
2
4
= 1−2·
5
= 1−2·
sin 2x = 2 sin x cos x
4
3
= 2· ·−
5
5
24
=−
25
= 1−
=−
tan 2x =
16
25
− 38
7
8
= − ÷−
3
9
1−
8
9
= − ·−
3
7
24
=
7
=
32
25
7
25
16
9
Example 3
Find the exact value of cos2 15◦ − sin2 15◦
This is one of the forms for cos 2x.
cos2 15◦ − sin2 15◦ = cos(15◦ · 2)
= cos 30◦
√
3
=
2
Example 4
Verify the identity: cos 3θ = 4 cos3 θ − 3 cos θ
Step 1: Use the cosine sum formula
cos 3θ = 4 cos3 θ − 3 cos θ
cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ
Step 2: Use double angle formulas for cos 2θ and sin 2θ
= (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ
Step 3: Distribute and simplify.
= 2 cos3 θ − cos θ − 2 sin2 θ cos θ
= − cos θ(−2 cos2 θ + 2 sin2 θ + 1)
= − cos θ[−2 cos2 θ + 2(1 − cos2 θ) + 1]
→ Substitute 1 − cos2 θ for sin2 θ
= − cos θ[−2 cos2 θ + 2 − 2 cos2 θ + 1]
= − cos θ(−4 cos2 θ + 3)
= 4 cos3 θ − 3 cos θ
389
2.21. Double Angle Identities
Review
Simplify each expression so that it is in terms of sin(x) and cos(x).
1.
2.
3.
4.
sin 2x + cos x
sin 2x + cos 2x
sin 3x + cos 2x
sin 2x + cos 3x
Solve each equation on the interval [0, 2π).
5.
6.
7.
8.
9.
sin(2x) = 2 sin(x)
cos(2x) = sin(x)
sin(2x) − tan(x) = 0
cos2 (x) + cos(x) = cos(2x)
cos(2x) = cos(x)
Simplify each expression so that only one calculation would be needed in order to evaluate.
10.
11.
12.
13.
14.
15.
2 cos2 (15◦ ) − 1
2 sin(25◦ ) cos(25◦ )
1 − 2 sin2 (35◦ )
cos2 (60◦ ) − sin2 (60◦ )
2 sin(125◦ ) cos(125◦ )
1 − 2 sin2 (32◦ )
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.10.
390
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.22 Simplifying Trig Expressions using Double
and Half Angle Formulas
Learning Objectives
Here you’ll use the half and double angle formulas to simplify more complicated expressions.
As Agent Trigonometry, you are given the following cryptic clue. How could you simplify this clue?
tan 2x
tanx
1+tan x
Simplifying Trigonometric Expressions
We can also use the double-angle and half-angle formulas to simplify trigonometric expressions.
Let’s simplify
cos 2x
sin x cos x .
Use cos 2a = cos2 a − sin2 a and then factor.
cos 2x
cos2 x − sin2 x
=
sin x cos x
sin x + cos x
(
((
(sin
(cos x − sin x)(
(cos
x+
x)
((
=
(
(
sin(x(
+(
cos x
(
= cos x − sin x
Now, let’s find the formula for sin 3x.
You will need to use the sum formula and the double-angle formula. sin 3x = sin(2x + x)
sin 3x = sin(2x + x)
= sin 2x cos x + cos 2x sin x
= 2 sin x cos x cos x + sin x(2 cos2 x − 1)
= 2 sin x cos2 x + 2 sin x cos2 x − sin x
= 4 sin x cos2 x − sin x
= sin x(4 cos2 x − 1)
Finally, let’s verify the identity cos x + 2 sin2 2x = 1.
Simplify the left-hand side use the half-angle formula.
391
2.22. Simplifying Trig Expressions using Double and Half Angle Formulas
x
cos x + 2 sin2
2
!2
r
1 − cos x
cos x + 2
2
1 − cos x
2
cos x + 1 − cos x
cos x + 2 ·
1
Examples
Example 1
Earlier, you were asked to simplify
Use tan 2a =
2 tan a
1−tan2 a
tan 2x
tanx
1+tan x
.
and then factor.
2 tan x 1 + tan x
·
1 − tan2 x
tanx
2 tan x
1 + tan x
2
=
·
=
(1 + tan x)(1 − tan x)
tanx
1 − tan x
tan 2x
tanx
1+tan x
=
Example 2
Simplify
sin 2x
sin x
=
sin 2x
sin x .
2 sin x cos x
sin x
= 2 cos x
Example 3
Verify cos x + 2 cos2 2x = 1 + 2 cos x.
x
cos x + 2 cos2 = 1 + 2 cos x
2
r
2
1 + cos x
cos x + 2
=
2
cos x + 1 + cos x =
1 + 2 cos x =
Review
Simplify the following expressions.
1.
2.
392
√
2 + 2 cos x cos 2x
cos 2x
cos2 x
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
3. tan 2x(1 + tan x)
4. cos 2x − 3 sin2 x
2x
5. 1+cos
cot x
6. (1 + cos x)2 tan 2x
Verify the following identities.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
sin x
cot 2x = 1−cos
x
sin x
1−cos x
=
1+cos x
sin x
sin 2x
=
tan
x
1+cos 2x
2
(sin x + cos x) = 1 + sin 2x
sin x tan 2x + 2 cos x = 2 cos2 2x
cot x + tan x = 2 csc 2x
cos 3x = 4 cos3 x − 3 cos x
cos 3x = cos3 x − 3 sin2 x cos x
sin 2x − tan x = tan x cos 2x
cos4 x − sin4 x = cos 2x
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.16.
393
2.23. Double, Half, and Power Reducing Identities
www.ck12.org
2.23 Double, Half, and Power Reducing
Identities
Here you will prove and use the double, half, and power reducing identities.
These identities are significantly more involved and less intuitive than previous identities. By practicing and working
with these advanced identities, your toolbox and fluency substituting and proving on your own will increase. Each
identity in this concept is named aptly. Double angles work on finding sin 80◦ if you already know sin 40◦ . Half
angles allow you to find sin 15◦ if you already know sin 30◦ . Power reducing identities allow you to find sin2 15◦ if
you know the sine and cosine of 30◦ .
What is sin2 15◦ ?
Double Angle, Half Angle, and Power Reducing Identities
Double Angle Identities
The double angle identities are proved by applying the sum and difference identities. They are left as review
problems. These are the double angle identities.
• sin 2x = 2 sin x cos x
• cos 2x = cos2 x − sin2 x
2 tan x
• tan 2x = 1−tan
2x
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61324
Half Angle Identities
The half angle identities are a rewritten version of the power reducing identities. The proofs are left as review
problems.
r
1 − cos x
r 2
1 + cos x
• cos 2x = ±
2
r
1
−
cos
x
• tan 2x = ±
1 + cos x
x
2
• sin = ±
394
www.ck12.org
Chapter 2. Unit 2
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61326
Power Reducing Identities
The power reducing identities allow you to write a trigonometric function that is squared in terms of smaller
powers. The proofs are left as examples and review problems.
2x
• sin2 x = 1−cos
2
2x
• cos2 x = 1+cos
2
1−cos 2x
2
• tan x = 1+cos 2x
Power reducing identities are most useful when you are asked to rewrite expressions such as sin4 x as an expression
without powers greater than one. While sin x · sin x · sin x · sin x does technically simplify this expression as necessary,
you should try to get the terms to sum together not multiply together.
sin4 x = (sin2 x)2
1 − cos 2x 2
=
2
1 − 2 cos 2x + cos2 2x
=
4
1 + cos 4x
1
1 − 2 cos 2x +
=
4
2
Examples
Example 1
Earlier, you were asked to find sin2 15◦ . In order to fully identify sin2 15◦ you need to use the power reducing formula.
1 − cos 2x
2
√
1 − cos 30◦ 1
3
2
◦
sin 15 =
= −
2
2
4
√
2− 3
=
4
sin2 x =
Example 2
Write the following expression with only sin x and cos x : sin 2x + cos 3x.
395
2.23. Double, Half, and Power Reducing Identities
www.ck12.org
sin 2x + cos 3x = 2 sin x cos x + cos(2x + x)
= 2 sin x cos x + cos 2x cos x − sin 2x sin x
= 2 sin x cos x + (cos2 x − sin2 x) cos x − (2 sin x cos x) sin x
= 2 sin x cos x + cos3 x − sin2 x cos x − 2 sin2 x cos x
= 2 sin x cos x + cos3 x − 3 sin2 x cos x
Example 3
Use half angles to find an exact value of tan 22.5◦ without using a calculator.
r
1 − cos x
x
tan 2 = ±
1 + cos x
◦
45◦
r
v
u
u1−
t
=
±
1 + cos 45◦
1+
1 − cos 45◦
=±
2
s
√
(2 − 2)2
=±
2
tan 22.5 = tan
√
2
2
√
2
2
v
u2
u −
= ±t 2
2
2+
√
2
2
√
2
2
s
=±
√
2− 2
√
2+ 2
Example 4
Prove the power reducing identity for sine.
sin2 x =
1−cos 2x
2
Using the double angle identity for cosine:
cos 2x = cos2 x − sin2 x
cos 2x = (1 − sin2 x) − sin2 x
cos 2x = 1 − 2 sin2 x
This expression is an equivalent expression to the double angle identity and is often considered an alternate form.
Example 5
Simplify the following identity: sin4 x − cos4 x.
Here are the steps:
sin4 x − cos4 x = (sin2 x − cos2 x)(sin2 x + cos2 x)
= −(cos2 x − sin2 x)
= − cos 2x
396
www.ck12.org
Chapter 2. Unit 2
Review
Prove the following identities.
1. sin 2x = 2 sin x cos x
2. cos 2x = cos2 x − sin2 x
3. tan 2x =
2 tan x
1−tan2 x
4. cos2 x =
1+cos 2x
2
1−cos 2x
1+cos 2x
5. tan2 x =
r
1 − cos x
2
r
1 + cos x
7. cos 2x = ±
2
r
1 − cos x
8. tan 2x = ±
1 + cos x
x
2
6. sin = ±
9. csc 2x = 12 csc x sec x
10. cot 2x =
cot2 x−1
2 cot x
Find the value of each expression using half angle identities.
11. tan 15◦
12. tan 22.5◦
13. sec 22.5◦
14. Show that tan 2x =
1−cos x
sin x .
15. Using your knowledge from the answer to question 14, show that tan 2x =
sin x
1+cos x .
Review (Answers)
To see the Review answers, open this PDF file and look for section 6.4.
397
2.24. Six Trigonometric Functions and Radians
www.ck12.org
2.24 Six Trigonometric Functions and Radians
Learning Objectives
Here you’ll learn what the values of trig functions are when angles are expressed in radians.
While working in your math class one day, you are given a sheet of values in radians and asked to find the various
trigonometric functions of them, such as sine, cosine, and tangent. The first question asks you to find the sin π6 . You
are about to start converting the measurements in radians into degrees when you wonder if it might be possible to
just take the values of the functions directly.
Do you think this is possible? As it turns out, it is indeed possible to apply trig functions to measurements in radians.
Here you’ll learn to do just that.
Trigonometric Functions and Radians
Even though you are used to performing the trig functions on degrees, they still will work on radians. The only
difference is the way the problem looks. If you see sin π6 , that is still sin 30◦ and the answer is still 12 .
Most scientific and graphing calculators have a MODE setting that will allow you to either convert between the two,
or to find approximations for trig functions using either measure. It is important that if you are using your calculator
to estimate a trig function that you know which mode you are using. Look at the following screen:
If you entered this expecting to find the sine of 30 degrees you would realize that something is wrong because the
answer should be 12 . In fact, as you may have suspected, the calculator is interpreting this as 30 radians. In this case,
changing the mode to degrees and recalculating will give the expected result.
398
www.ck12.org
Chapter 2. Unit 2
Scientific calculators will usually have a 3-letter display that shows either DEG or RAD to tell you which mode the
calculator is in.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/182422
Let’s take a look at a few example problems.
1. Find tan 3π
4 .
If needed, convert
3π
4
to degrees. Doing this, we find that it is 135◦ . So, this is tan 135◦ , which is -1.
2. Find the value of cos 11π
6 .
If needed, convert
11π
6
to degrees. Doing this, we find that it is
330◦ .
So, this is
cos 330◦ ,
√
which is 2 3 .
3. Convert 1 radian to degree measure.
Many students get so used to using π in radian measure that they incorrectly think that 1 radian means 1π radians.
While it is more convenient and common to express radian measure in terms of π, don’t lose sight of the fact that π
radians is a number. It specifies an angle created by a rotation of approximately 3.14 radius lengths. So 1 radian is a
rotation created by an arc that is only a single radius in length.
radians ×
180
= degrees
π
So 1 radian would be 180
π degrees. Using any scientific or graphing calculator will give a reasonable approximation
for this degree measure, approximately 57.3◦ .
Examples
Example 1
Earlier, you were asked to find sin π6 .
As you have learned in this section, the sin π6 is the same as sin 30◦ , which equals 21 . You could find this either by
converting π6 to degrees, or by using your calculator with angles entered in radians.
399
2.24. Six Trigonometric Functions and Radians
www.ck12.org
Example 2
Using a calculator, find the approximate degree measure (to the nearest tenth) of the angle expressed in radians:
6π
7
154.3◦
Example 3
Using a calculator, find the approximate degree measure (to the nearest tenth) of the angle expressed in radians:
20π
11
327.3◦
Example 4
Gina wanted to calculate the sin 210◦ and got the following answer on her calculator:
Fortunately, Kylie saw her answer and told her that it was obviously incorrect.
1. Write the correct answer, in simplest radical form.
2. Explain what she did wrong.
The correct answer is − 12 . Her calculator was is the wrong mode and she calculated the sine of 210 radians.
Review
Using a calculator, find the approximate degree measure (to the nearest tenth) of the angle expressed in radians.
1.
2.
3.
4.
5.
6.
7.
4π
7
5π
6
8π
11
5π
3
8π
3
7π
4
12π
5
Find the value of each using your calculator.
8. sin 3π
2
9. cos π2
400
www.ck12.org
10.
11.
12.
13.
14.
15.
Chapter 2. Unit 2
tan π6
sin 5π
6
tan 4π
3
cot 7π
3
sec 11π
6
Do you think radians will always be written in terms of π? Is it possible to have, for example, exactly 2
radians?
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.3.
401
2.25. Graphing Sine and Cosine
www.ck12.org
2.25 Graphing Sine and Cosine
Learning Objectives
Here you’ll learn how to graph and stretch the sine and cosine functions.
Your mission, should you choose to accept it, as Agent Trigonometry is to graph the function y = 2 cos x. What are
the minimum and maximum of your graph?
Graphing Sine and Cosine
In this concept, we will take the unit circle and graph it on the Cartesian plane.
To do this, we are going to “unravel” the unit circle. Recall that for the unit circle the coordinates are (cos θ, sin θ)
where θ is the central angle. To graph, y = sin x rewrite the coordinates as (x, sin x) where x is the central angle, in
radians. Below we expanded the sine coordinates for 3π
4 .
Notice that the curve ranges from 1 to -1. The maximum value is 1, which is at x = π2 . The minimum value is -1 at
x = 3π
2 . This “height” of the sine function is called the amplitude. The amplitude is the absolute value of average
between the highest and lowest points on the curve.
Now, look at the domain. It seems that, if we had continued the curve, it would repeat. This means that the sine
curve is periodic. Look back at the unit circle, the sine value changes until it reaches 2π. After 2π, the sine values
repeat. Therefore, the curve above will repeat every 2π units, making the period 2π. The domain is all real numbers.
Similarly, when we expand the cosine curve, y = cos x, from the unit circle, we have:
402
www.ck12.org
Chapter 2. Unit 2
Notice that the range is also between 1 and -1 and the domain will be all real numbers. The cosine curve is also
periodic, with a period of 2π. If we draw the graph past 2π, it would look like:
Comparing y = sin x and y = cos x (below), we see that the curves are almost identical, except that the sine curve
starts at y = 0 and the cosine curve starts at y = 1.
If we shift either curve
is called a phase shift.
π
2
units to the left or right, they will overlap. Any horizontal shift of a trigonometric function
Let’s identify the highlighted points on y = sin x and y = cos x below.
403
2.25. Graphing Sine and Cosine
www.ck12.org
√
For eachpoint, think
about what the sine or cosine value is at those values. For point A, sin = 2 2 , therefore the
√ point is π , 2 . For point B, we have to work backwards because it is not exactly on a vertical line, but it is on a
π
4
4
2
horizontal one. When is sin x = − 12 ? When
x=
11π 1
option. Therefore, the point is 6 , 2 .
7π
6
or
11π
6 .
By looking at point B’s location, we know it is the second
For the cosine curve, point C is the same as point A because the sine and cosine for π4 is the same. As for point D, we
4π
use the same logic as we did for point B. When does cos x = − 12? When x = 2π
3 or 3 . Again, looking at the location
4π 1
of point D, we know it is the second option. The point is 3 , 2 .
Amplitude
In addition to graphing y = sin x and y = cos x, we can stretch the graphs by placing a number in front of the sine or
cosine, such as y = a sin x or y = a cos x. |a| is the amplitude of the curve.
Let’s graph y = 3 sin x over two periods.
Start with the basic sine curve. Recall that one period of the parent graph, y = sin x, is 2π. Therefore, two periods will
be 4π. The 3 indicates that the range will now be from 3 to -3 and the curve will be stretched so that the maximum
is 3 and the minimum is -3. The red curve is y = 3 sin x.
404
www.ck12.org
Chapter 2. Unit 2
Notice that the x-intercepts are the same as the parent graph. Typically, when we graph a trigonometric function, we
always show two full periods of the function to indicate that it does repeat.
Now, let’s graph y = 12 cos x over two periods.
Now, the amplitude will be
1
2
and the function will be “smooshed” rather than stretched.
Finally, let’s graph y = − sin x over two periods.
The last two problems dealt with changing a and a was positive. Now, a is negative. Just like with other functions,
when the leading coefficient is negative, the function is reflected over the x-axis. y = − sin x is in red.
Examples
Example 1
Earlier, you were asked to find the minimum and maximum of the graph y = 2 cos x.
The 2 in front of the cosine function indicates that the range will now be from 2 to -2 and the curve will be stretched
so that the maximum is 2 and the minimum is -2.
Example 2
Is the point
5π 1
6 ,2
on y = sin x? How do you know?
Substitute in the point for x and y and see if the equation holds true.
405
2.25. Graphing Sine and Cosine
1
2
= sin
5π
6
This is true, so
5π 1
6 ,2
is on the graph.
Graph the following functions for two full periods.
Example 3
y = 6 cos x
Stretch the cosine curve so that the maximum is 6 and the minimum is -6.
Example 4
y = −3 cos x
The graph is reflected over the x-axis and stretched so that the amplitude is 3.
406
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
Example 5
y = 32 sin x
The fraction is equivalent to 1.5, making 1.5 the amplitude.
Review
1. Determine the exact value of each point on y = sin x or y = cos x.
407
2.25. Graphing Sine and Cosine
www.ck12.org
2. List all the points in the interval [0, 4π] where
sin x =5πcos
x. Use the graph from #1 to help you.
π
3. Draw from y = sin x from [0, 2π]. Find f 3 and f 3 . Plot these values on the curve.
For questions 4-12, graph the sine or cosine curve over two periods.
y = 2 sin x
y = −5 cos x
y = 41 cos x
y = − 23 sin x
y = 4 sin x
y = −1.5 cos x
y = 35 cos x
y = 10 sin x
y = −7.2 sin x
Graph y = sin x and y = cos x on the same set of axes. How many units would you have to shift the sine curve
(to the left or right) so that it perfectly overlaps the cosine curve?
14. Graph y = sin x and y = − cos x on the same set of axes. How many units would you have to shift the sine
curve (to the left or right) so that it perfectly overlaps y = − cos x?
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
Write the equation for each sine or cosine curve below. a > 0 for both questions.
15.
408
www.ck12.org
Chapter 2. Unit 2
16.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.1.
409
2.26. Amplitude, Period, and Frequency
www.ck12.org
2.26 Amplitude, Period, and Frequency
Learning Objectives
Here you’ll learn how to solve problems that involve both the amplitude and period of a trig function.
You are working in science lab one afternoon when your teacher asks you to do a little more advanced work with
her on sound. Excited to help, you readily agree. She gives you a device that graphs sound waves as they come in
through a microphone. She then gives you a "baseline" graph of what the sound wave’s graph would look like:
She then asks you to plot the sound wave she’s about to generate. However, she tells you that the sound wave will
be twice as loud and twice as high in pitch as the baseline sound wave she gave you.
Can you determine how large the graph needs to be to plot the new sound wave? What about the spacing of numbers
on the "x" axis?
Amplitude and Period
In other lessons you have dealt with how find the amplitude of a wave, or the period of a wave. Here we’ll take a few
minutes to work problems that involve both the amplitude and period, giving us two variables to work with when
thinking about sinusoidal equations.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70562
410
www.ck12.org
Chapter 2. Unit 2
Finding the Period, Amplitude, and Frequency
1. Find the period, amplitude and frequency of y = 2 cos 12 x and sketch a graph from 0 to 2π.
This is a cosine graph that has been stretched both vertically and horizontally. It will now reach up to 2 and down to
-2. The frequency is 12 and to see a complete period we would need to graph the interval [0, 4π]. Since we are only
going out to 2π, we will only see half of a wave. A complete cosine wave looks like this:
This means that this half needs to be stretched out so it finishes at 2π, which means that at π the graph should cross
the x−axis:
The final sketch would look like this:
amplitude = 2, frequency = 12 , period =
2π
1
2
= 4π
2. Identify the period, amplitude, frequency, and equation of the following sinusoid:
411
2.26. Amplitude, Period, and Frequency
www.ck12.org
The amplitude is 1.5. Notice that the units on the x−axis are not labeled in terms of π. This appears to be a sine
wave because the y−intercept is 0.
One wave appears to complete in 1 unit (not 1π units!), so the period is 1. If one wave is completed in 1 unit, how
many waves will be in 2π units? In previous problems, you were given the frequency and asked to find the period
using the following relationship:
p=
2π
B
Where B is the frequency and p is the period. With just a little bit of algebra, we can transform this formula and
solve it for B:
p=
2π
2π
→ Bp = 2π → B =
B
p
Therefore, the frequency is:
B=
2π
= 2π
1
If we were to graph this out to 2π we would see 2π (or a little more than 6) complete waves.
Replacing these values in the equation gives: f (x) = 1.5 sin 2πx.
3. Find the period, amplitude and frequency of y = 3 sin 2x and sketch a graph from 0 to 6π.
This is a sine graph that has been stretched both vertically and horizontally. It will now reach up to 3 and down to
-3. The frequency is 2 and so we will see the wave repeat twice over the interval from 0 to 2π.
412
www.ck12.org
amplitude = 3, frequency = 2, period =
Chapter 2. Unit 2
2π
2
=π
Examples
Example 1
Earlier, you were asked can you determine how large the graphs needs to be to plot the new sound wave.
You know that the amplitude of the wave is the maximum height it makes above zero. You also know that the
frequency is the number of cycles in a second. The scale of the graph you make should be able to take into account a
maximum height of the wave that has been doubled, as well as a frequency that is twice as high. Your graph should
look like this:
Example 2
Identify the amplitude, period, and frequency of y = cos 2x
period: π, amplitude: 1, frequency: 2
Example 3
Identify the amplitude, period, and frequency of y = 3 sin x
413
2.26. Amplitude, Period, and Frequency
period: 2π, amplitude: 3, frequency: 1
Example 4
Identify the amplitude, period, and frequency of y = 2 sin πx
period: 2, amplitude: 2, frequency: π
Review
Find the period, amplitude, and frequency of the following functions.
1.
2.
3.
4.
5.
6.
y = 2 sin(3x)
y = 5 cos( 34 x)
y = 3 cos(2x)
y = −2 sin( 21 x)
y = − sin(2x)
y = 21 cos(4x)
Identify the equation of each of the following graphs.
7.
8.
414
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
9.
10.
Graph each of the following functions from 0 to 2π.
11.
12.
13.
14.
15.
16.
17.
18.
y = 2 cos(4x)
y = 3 sin( 54 x)
y = − cos(2x)
y = −2 sin( 21 x)
y = 4 sec(3x)
y = 21 cos(3x)
y = 4 tan(3x)
y = 21 csc(3x)
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.16.
415
2.27. Horizontal Translations or Phase Shifts
www.ck12.org
2.27 Horizontal Translations or Phase Shifts
Learning Objectives
Here you’ll learn how to express horizontal translations of graphs algebraically.
You are working on a graphing project in your math class, where you are supposed to graph several functions. Things
seem to be going well, until you realize that there is a bold, vertical line three units to the left of where you placed
your "y" axis! As it turns out, you’ve accidentally shifted your entire graph. You didn’t notice that your instructor
had placed a bold line where the "y" axis was supposed to be. And now, all of the points for your graph of the cosine
function are three points farther to the right than they are supposed to be along the "x" axis.
You might be able to keep all of your work, if you can find a way to rewrite the equation so that it takes into account
the change in your graph.
Can you think of a way to rewrite the function so that the graph is correct the way you plotted it?
Horizontal Translations
Horizontal translations involve placing a constant inside the argument of the trig function being plotted. If we return
to the example of the parabola, y = x2 , what change would you make to the equation to have it move to the right or
left? Many students guess that if you move the graph vertically by adding to the y−value, then we should add to the
x−value in order to translate horizontally. This is correct, but the graph itself behaves in the opposite way than what
you may think.
Here is the graph of y = (x + 2)2 .
Notice that adding 2 to the x−value shifted the graph 2 units to the left, or in the negative direction.
416
www.ck12.org
Chapter 2. Unit 2
To compare, the graph y = (x − 2)2 moves the graph 2 units to the right or in the positive direction.
We will use the letter C to represent the horizontal shift value. Therefore, subtracting C from the x−value will shift
the graph to the right and adding C will shift the graph C units to the left.
Adding to our previous equations, we now have y = D ± sin(x ±C) and y = D ± cos(x ±C) where D is the vertical
translation and C is the opposite sign of the horizontal shift.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70572
Sketching Graphs
1. Sketch y = sin x − π2
This is a sine wave that has been translated
π
2
units to the right.
417
2.27. Horizontal Translations or Phase Shifts
www.ck12.org
Horizontal translations are also referred to as phase shifts. Two waves that are identical, but have been moved
horizontally are said to be “out of phase” with each other. Remember that cosine and sine are really the same waves
with this phase variation.
y = sin x can be thought of as a cosine wave shifted horizontally to the right by
Alternatively, we could also think of cosine as a sine wave that has been shifted
2. Draw a sketch of y = 1 + cos(x − π)
418
π
2
π
2
radians.
radians to the left.
www.ck12.org
Chapter 2. Unit 2
This is a cosine curve that has been translated up 1 unit and π units to the right. It may help you to use the quadrant
angles to draw these sketches. Plot the points of y = cos x at 0, π2 , π, 3π
2 , 2π (as well as the negatives), and then
translate those points before drawing the translated curve. The blue curve below is the final answer.
Graphing Functions
Graph y = −2 + sin x + 3π
2
This is a sine curve that has been translated 2 units down and moved
quadrant angles on y = sin x and translate them down 2 units.
Then, take that result and shift it
3π
2
3π
2
radians to the left. Again, start with the
to the left. The blue graph is the final answer.
419
2.27. Horizontal Translations or Phase Shifts
www.ck12.org
Examples
Example 1
Earlier, you were asked if you can think of a way to rewrite the function so that the graph is correct the way you
plotted it.
As you’ve now seen by reading this Concept, it is possible to shift an entire graph to the left or the right by changing
the argument of the graph.
So in this case, you can keep your graph by changing the function to y = cos(x − 3)
Example 2
Draw a sketch of y = 3 + cos(x − π2 )
As we’ve seen, the 3 shifts the graph vertically 3 units, while the − π2 shifts the graph to the right by
Example 3
Draw a sketch of y = sin(x + π4 )
420
π
2
units.
www.ck12.org
The
π
4
Chapter 2. Unit 2
shifts the graph to the left by π4 .
Example 4
Draw a sketch of y = 2 + cos(x + 2π)
The 2 added to the function shifts the graph up by 2 units, and the 2π added in the argument of the function brings
the function back to where it started, so the cosine graph isn’t shifted horizontally at all.
Review
Graph each of the following functions.
1.
2.
3.
4.
5.
6.
y = cos(x − π2 )
y = sin(x + 3π
2 )
y = cos(x + π4 )
y = cos(x − 3π
4 )
y = −1 + cos(x − π4 )
y = 1 + sin(x + π2 )
421
2.27. Horizontal Translations or Phase Shifts
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
y = −2 + cos(x + π4 )
y = 3 + cos(x − 3π
2 )
y = −4 + sec(x − π4 )
y = 3 + csc(x − π2 )
y = 2 + tan(x + π4 )
y = −3 + cot(x − 3π
2 )
3π
y = 1 + cos(x − 4 )
y = 5 + sec(x + π2 )
y = −1 + csc(x + π4 )
y = 3 + tan(x − 3π
2 )
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.13.
422
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.28 Graphing Tangent
Learning Objectives
Here you’ll learn how to graph a tangent function.
Your mission, should you choose to accept it, as Agent Trigonometry is to find the period and the zeros of the
function y = 12 tan 4x.
Graph of a Tangent Function
The graph of the tangent function is very different from the sine and cosine functions. First, recall that the tangent
opposite
ratio is tan θ = hypotenuse . In radians, the coordinate for the tangent function would be (θ, tan θ)
x
θ
0
y
tan θ
0
π
√6
3
3
π
4
1
π
3
√
3
π
2
und.
2π
3
√
− 3
3π
4
−1
5π
√6
3
−
3
π
0
After π, the y-values repeat, making the tangent function periodic with a period of π.
The red portion of the graph represents the coordinates in the table above. Repeating this portion, we get the entire
π π
3π
tangent graph. Notice that there are vertical asymptotes at x = − 3π
2 , − 2 , 2 and 2 . If we were to extend the graph out
in either direction, there would continue to be vertical asymptotes at the odd multiples of π2 . Therefore, the domain
is all real numbers, x 6= nπ ± π2 , where n is an integer. The range would be all real numbers. Just like with sine and
cosine functions, you can change the amplitude, phase shift, and vertical shift.
423
2.28. Graphing Tangent
www.ck12.org
The standard form of the equation is y = a tan b(x − h) + k where a, b, h, and k are the same as they are for the other
trigonometric functions. For simplicity, we will not address phase shifts (k) in this concept.
Let’s graph y = 3 tan x + 1 from [−2π, 2π] and state the domain and range.
First, the amplitude is 3, which means each y-value will be tripled. Then, we will shift the function up one unit.
Notice that the vertical asymptotes did not change. The period of this function is still π. Therefore, if we were to
change the period of a tangent function, we would use a different formula than what we used for sine and cosine. To
π
change the period of a tangent function, use the formula |b|
.
The domain will be all real numbers, except where the asymptotes occur. Therefore, the domain of this function will
be x ∈ R, x ∈
/ nπ ± π2 . The range is all real numbers.
Now, let’s graph y = − tan 2π from [0, 2π], state the domain and range, and find all zeros within this domain.
The period of this tangent function will be
424
π
2
and the curves will be reflected over the x-axis.
www.ck12.org
Chapter 2. Unit 2
5π 7π π
π
The domain is all real numbers, x ∈
/ π4 , 3π
4 , 4 , 4 , 4 ± 2 n where n is any integer. The range is all real numbers. To
find the zeros, set y = 0.
0 = − tan 2x
0 = tan 2x
2x = tan−1 0 = 0, π, 2π, 3π, 4π
3π
π
x = 0, , π, , 2π
2
2
Finally, let’s graph y = 41 tan 14 x from [0, 4π] and state the domain and range.
This function has a period of π1 = 4π. The domain is all real numbers, except 2π, 6π, 10π, 2π ± 4πn, where n is any
4
integer. The range is all real numbers.
Examples
Example 1
Earlier, you were asked to find the period and zeros of the function y = 12 tan 4x.
The period is π4 .
The zeros are where y is zero.
425
2.28. Graphing Tangent
www.ck12.org
0=
1
tan 4x
2
0 = tan 4x
4x = tan−1 0 = 0, π, 2π, 3π
1
x = (0, π, 2π, 3π)
4
π π 3
x = 0, , , π
4 2 4
Example 2
Find the period of the function y = −4 tan 23 x.
The period is
π
3
2
= π · 23 =
2π
3 .
Example 3
Find the zeros of the function from Example 2, from [0, 2π].
The zeros are where y is zero.
3
0 = −4 tan x
2
3
0 = tan x
2
3
x = tan−1 0 = 0, π, 2π, 3π
2
2
x = (0, π, 2π, 3π)
3
2π 4π
x = 0, , , 2π
3 3
Example 4
Find the equation of the tangent function with an amplitude of 8 and a period of 6π.
The general equation is y = a tan bx. We know that a = 8. Let’s use the period to solve for the frequency, or b.
π
= 6π
b
π
1
b=
=
6π 6
The equation is y = 8 tan 16 x.
Review
Graph the following tangent functions over [0, 4π]. Determine the period, domain, and range.
1. y = 2 tan x
426
www.ck12.org
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
Chapter 2. Unit 2
y = − 13 tan x
y = − tan 3x
y = 4 tan 2x
y = 21 tan 4x
y = − tan 21 x
y = 4 + tan x
y = −3 + tan 3x
y = 1 + 23 tan 12 x
Find the zeros of the function from #1.
Find the zeros of the function from #3.
Find the zeros of the function from #5.
Write the equation of the tangent function, in the form y = a tan bx, with the given amplitude and period.
13.
14.
15.
16.
Amplitude: 3 Period: 3π
2
Amplitude: 14 Period: 2π
Amplitude: -2.5 Period: 8
Challenge Graph y = 2 tan 31 x + π4 − 1 over [0, 6π]. Determine the domain and period.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.5.
427
2.29. Vertical Translations
www.ck12.org
2.29 Vertical Translations
Learning Objectives
Here you’ll learn how to express vertical translations of graphs algebraically.
You are working on a graphing project in your math class, where you are supposed to graph several functions. You
are working on graphing a cosine function, and things seem to be going well, until you realize that there is a bold,
horizontal line two units above where you placed your "x" axis! As it turns out, you’ve accidentally shifted your
entire graph. You didn’t notice that your instructor had placed a bold line where the "x" axis was supposed to be.
And now, all of the points for your graph of the cosine function are two points lower than they are supposed to be
along the "y" axis.
You might be able to keep all of your work, if you can find a way to rewrite the equation so that it takes into account
the change in your graph.
Can you think of a way to rewrite the function so that the graph is correct the way you plotted it?
Vertical Translations
When you first learned about vertical translations in a coordinate grid, you started with simple shapes. Here is a
rectangle:
To translate this rectangle vertically, move all points and lines up by a specified number of units. We do this by
adjusting the y−coordinate of the points. So to translate this rectangle 5 units up, add 5 to every y−coordinate.
428
www.ck12.org
Chapter 2. Unit 2
This process worked the same way for functions. Since the value of a function corresponds to the y−value on its
graph, to move a function up 5 units, we would increase the value of the function by 5. Therefore, to translate y = x2
up five units, you would increase the y−value by 5. Because y is equal to x2 , then the equation y = x2 + 5, will show
this translation.
429
2.29. Vertical Translations
www.ck12.org
Hence, for any graph, adding a constant to the equation will move it up, and subtracting a constant will move it
down. From this, we can conclude that the graphs of y = sin x and y = cos x will follow the same rules. That is, the
graph of y = sin(x) + 2 will be the same as y = sin x, only it will be translated, or shifted, 2 units up.
To avoid confusion, this translation is usually written in front of the function: y = 2 + sin x.
Various texts use different notation, but we will use D as the constant for vertical translations. This would lead to
the following equations: y = D ± sin x and y = D ± cos x where D is the vertical translation. D can be positive or
negative.
Another way to think of this is to view sine or cosine curves “wrapped” around a horizontal line. For y = sin x and
y = cos x, the graphs are wrapped around the x−axis, or the horizontal line, y = 0.
For y = 3 + sin x, we know the curve is translated up 3 units. In this context, think of the sine curve as being
“wrapped” around the line, y = 3.
Either method works for the translation of a sine or cosine curve. Pick the thought process that works best for you.
430
www.ck12.org
Chapter 2. Unit 2
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70572
Let’s take a look at a few problems involving vertical translations.
1. Find the minimum and maximum of y = −6 + cos x
This is a cosine wave that has been shifted down 6 units, or is now wrapped around the line y = −6. Because the
graph still rises and falls one unit in either direction, the cosine curve will extend one unit above the “wrapping line”
and one unit below it. The minimum is −7 and the maximum is−5.
2. Graph y = 4 + cos x.
This will be the basic cosine curve, shifted up 4 units.
3. Find the minimum and maximum of y = sin x + 3
This is a sine wave that has been shifted up 3 units, so now instead of going up and down around the ’x’ axis, it will
go up and down around the line y = 3. Since the sine function rises and falls one unit in each direction, the new
minimum is 2 and the new maximum is 4.
Examples
Example 1
Earlier, you were asked if you can think of a way to rewrite the function so that the graph is correct the way you
plotted it.
Since you now know how to shift a graph vertically by adding or removing a constant after the function, you can
keep your graph by changing the equation to y = cos(x) − 2
Example 2
Which of the following is true for the equation: y = sin x − π2
431
2.29. Vertical Translations
www.ck12.org
The minimum value is 0.
The maximum value is 3.
The y−intercept is -2.
The y−intercept is -1.
This is the same graph as y = cos(x).
"This is the same graph as y = cos(x)." is the answer to this question, since the
it the same as a cosine graph.
π
2
is a shift to the graph which makes
Example 3
Which of the following is true for the equation: y = 1 + sin x
The minimum value is 0.
The maximum value is 3.
The y−intercept is -2.
The y−intercept is -1.
This is the same graph as y = cos(x).
"The minimum value is 0." is the answer to this question, since this graph is the same as a regular sine graph, which
ranges from -1 to 1, but shifted upward one unit on the "y" axis, so it ranges from 0 to 2.
Example 4
Which of the following is true for the equation: y = 2 + cos x
The minimum value is 0.
The maximum value is 3.
The y−intercept is -2.
The y−intercept is -1.
This is the same graph as y = cos(x).
"The maximum value is 3." is the answer to this question, since a cosine graph (which normally ranges between -1
and 1) shifted upward by two units. Therefore its new range is from 1 to 3.
Review
Use vertical translations to graph each of the following functions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
432
y = x3 + 4
y = x2 − 3
y = sin(x) − 4
y = cos(x) + 7
y = sec(x) − 3
y = tan(x) + 2
y = 3 + sin(x)
y = cos(x) + 1
y = 6 + sec(x)
www.ck12.org
Chapter 2. Unit 2
10. y = tan(x) − 4
Find the minimum and maximum value of each of the following functions.
11.
12.
13.
14.
15.
16.
17.
18.
y = sin(x) + 6
y = cos(x) − 1
y = sin(x) − 4
y = −3 + cos(x)
y = 2 + cos(x)
Give an example of a sine function with a y-intercept of 6.
Give an example of a cosine function with a maximum of -1.
Give an example of a sine function with a minimum of 0.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 2.12.
433
2.30. Sine and Cosecant Graphs
www.ck12.org
2.30 Sine and Cosecant Graphs
Learning Objectives
Here you’ll learn how to draw the graphs of the sine and cosecant functions.
Imagine for a moment that you have a clock that has only one hand - that rotates counterclockwise! However, the
hand is very slim all the way until the tip, where there is a ball on the end. In fact, the hand is so slim you won’t
notice it. You only notice the ball on the end of the rotating hand. This hand is rotating faster than normal. Here is a
picture of the clock:
Consider what it would be like if you put a light next to the clock and let the shadow of the hands fall on the far wall.
What pattern would that shadow trace out? If you think about it, you might realize that the shadow would make an
up and down motion, over and over as the hand of the clock rotated. Now imagine that instead of a wall, there was
a large piece of paper for the shadow to fall on. And wherever the shadow fell, there would be a mark on the paper.
Finally, imagine moving the paper as the clock rotates. Can you imagine sort of pattern this would trace out?
Sine and Cosecant Graphs
By now, you have become very familiar with the specific values of sine, cosine, and tangents for certain angles of
rotation around the coordinate grid. In mathematics, we can often learn a lot by looking at how one quantity changes
as we consistently vary another. We will be looking at the sine value as a function of the angle of rotation around
434
www.ck12.org
Chapter 2. Unit 2
the coordinate plane. We refer to any such function as a circular function, because they can be defined using the
unit circle. Recall from earlier sections that the sine of an angle in standard position is the ratio of yr , where y is the
y−coordinate of any point on the circle and r is the distance from the origin to that point.
Because the ratios are the same for a given angle, regardless of the length of the radius r, we can use the unit circle
as a basis for all calculations.
The denominator is now 1, so we have the simpler expression, sin x = y. The advantage to this is that we can use the
y−coordinate of the point on the unit circle to trace the value of sin θ through a complete rotation. Imagine if we
start at 0 and then rotate counter-clockwise through gradually increasing angles. Since the y−coordinate is the sine
value, watch the height of the point as you rotate.
Through Quadrant I that height gets larger, starting at 0, increasing quickly at first, then slower until the angle reaches
90◦ , at which point, the height is at its maximum value, 1.
435
2.30. Sine and Cosecant Graphs
436
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
As you rotate into the second quadrant, the height starts to decrease towards zero.
When you start to rotate into the third and fourth quadrants, the length of the segment increases, but this time in a
negative direction, growing to -1 at 270◦ and heading back toward 0 at 360◦ .
437
2.30. Sine and Cosecant Graphs
www.ck12.org
After one complete rotation, even though the angle continues to increase, the sine values will repeat themselves. The
same would have been true if we chose to rotate clockwise to investigate negative angles, and this is why the sine
function is a periodic function. The period is 2π because that is the angle measure before the sine of the angle will
repeat its values.
Let’s translate this circular motion into a graph of the sine value vs. the angle of rotation. The following sequence
of pictures demonstrates the connection. These pictures plot (θ, sin θ) on the coordinate plane as (x, y).
438
www.ck12.org
Chapter 2. Unit 2
After we rotate around the circle once, the values start repeating. Therefore, the sine curve, or “wave,” also continues
to repeat. The easiest way to sketch a sine curve is to plot the points for the quadrant angles. The value of sin θ goes
from 0 to 1 to 0 to -1 and back to 0. Graphed along a horizontal axis, it would look like this:
Filling in the gaps in between and allowing for multiple rotations as well as negative angles results in the graph of
y = sin x where x is any angle of rotation, in radians.
439
2.30. Sine and Cosecant Graphs
www.ck12.org
As we have already mentioned, sin x has a period of 2π. You should also note that the y−values never go above 1 or
below -1, so the range of a sine curve is {−1 ≤ y ≤ 1}. Because angles can be any value and will continue to rotate
around the circle infinitely, there is no restriction on the angle x, so the domain of sin x is all reals.
Cosecant is the reciprocal of sine, or 1y . Therefore, whenever the sine is zero, the cosecant is going to have a vertical
asymptote because it will be undefined. It also has the same sign as the sine function in the same quadrants. Here is
the graph.
The period of the function is 2π, just like sine. The domain of the function is all real numbers, except multiples of
π {. . . − 2π, −π, 0, π, 2π . . .}. The range is all real numbers greater than or equal to 1, as well as all real numbers less
than or equal to -1. Notice that the range is everything except where sine is defined (other than the points at the top
and bottom of the sine curve).
440
www.ck12.org
Chapter 2. Unit 2
Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs
at 1 and -1. Many students are reminded of parabolas when they look at the half-period of the cosecant graph. While
they are similar in that they each have a local minimum or maximum and they have the same beginning and ending
behavior, the comparisons end there. Parabolas are not restricted by asymptotes, whereas the cosecant curve is.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143444
Graphing Functions
Graph the following functions:
441
2.30. Sine and Cosecant Graphs
www.ck12.org
1. g(x) = 21 sin (3x).
As you can see from the graph, the 12 in front of the function reduces the function’s height, while the 3 inside the
argument of the function makes the function "squished" along the "x" axis.
2. f (x) = 31 csc
1
2x
.
3. f (x) = 5 sin 2(x + π3 ) .
442
www.ck12.org
Chapter 2. Unit 2
Examples
Example 1
Earlier, you were asked what pattern the shadow would trace out.
As you have seen in this lesson, the shadow of a light applied vertically to a rotating clock hand would trace out a sine
graph. The graph would begin at zero when the hand is lying flat along the positive "x" axis. It would then increase
until the hand was vertical. It would then decrease until the rotating hand was pointing straight down. Finally, the
graph would increase again to zero when the hand is returning to the positive "x" axis.
Example 2
Graph g(x) = 5 csc
1
4 (x + π
).
Example 3
Determine the function creating this graph:
443
2.30. Sine and Cosecant Graphs
www.ck12.org
This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical shift is -1. The
3π
2π
period is the difference between the two given x−values, 7π
4 − 4 = π, so the frequency
is ππ =2. The horizontal
3π
shift incorporates the frequency, so in y = csc x the corresponding x−value to 4 , 0 is 2 , 1 . The difference
π
3π
2π
π
π
π
between the x−valuesis 3π
4 − 2 = 4 − 4 = 4 and then multiply it by the frequency, 2 · 4 = 2 . The equation is
π
y = −1 + csc 2(x − 2 ).
Example 4
Graph h(x) = 3 sin
1
π
2 (x + 2
).
Review
Graph each of the following functions.
1.
2.
3.
4.
5.
6.
7.
444
f (x) = sin(x).
h(x) = sin(2x).
k(x) = sin(2x + π).
m(x) = 2 sin(2x + π).
g(x) = 2 sin(2x + π) + 2.
f (x) = csc(x).
h(x) = csc(2x).
www.ck12.org
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
Chapter 2. Unit 2
k(x) = csc(2x + π).
m(x) = 2 csc(2x + π).
g(x) = 2 csc(2x + π) + 2.
h(x) = sin(3x).
k(x) = sin(3x + π2 ).
m(x) = 3 sin(3x + π2 ).
g(x) = 3 sin(3x + π2 ) + 3.
h(x) = csc(3x).
k(x) = csc(3x + 3π
2 ).
m(x) = 4 csc(3x + 3π
2 ).
g(x) = 4 csc(3x + 3π
2 ) − 3.
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.9.
445
2.31. Cosine and Secant Graphs
www.ck12.org
2.31 Cosine and Secant Graphs
Learning Objectives
Here you’ll learn to draw the graphs of the cosine and secant functions.
Imagine for a moment that you have a clock that has only one hand - that rotates counterclockwise!. However, the
hand is very slim all the way until the tip, where there is a ball on the end. In fact, the hand is so slim you won’t
notice it. You only notice the ball on the end of the rotating hand. This hand is rotating faster than normal.
Consider what it would be like if you put a light above the clock and let the shadow of the hands fall on the wall
under the clock. What pattern would that shadow trace out? If you think about it, you might realize that the shadow
would make an left and right motion, over and over as the hand of the clock rotated. Now imagine that instead of
a wall, there was a large piece of paper for the shadow to fall on. And wherever the shadow fell, there would be
a mark on the paper. Finally, imagine moving the paper as the clock rotates. Can you imagine sort of pattern this
would trace out?
446
www.ck12.org
Chapter 2. Unit 2
Cosine and Secant Graphs
If you have read other Trigonometry sections in this course, you may have learned that sine and cosine are very
closely related. The cosine of an angle is the same as the sine of its complementary angle. So, it should not be a
surprise that sine and cosine waves are very similar in that they are both periodic with a period of 2π, a range from
-1 to 1, and a domain of all real angles.
The cosine of an angle is the ratio of xr , so in the unit circle, the cosine is the x−coordinate of the point of rotation.
If we trace the x−coordinate through a rotation, notice the change in the distance of cos x starts at one. The
x−coordinate at 0◦ is 1 and the x−coordinate for 90◦ is 0, so the cosine value is decreasing from 1 to 0 through
the 1st quadrant.
447
2.31. Cosine and Secant Graphs
www.ck12.org
Here is a sequence of rotations. Compare the x− coordinate of the point of rotation with the height of the point as it
traces along the horizontal. These pictures plot (θ, cos θ) on the coordinate plane as (x, y).
448
www.ck12.org
Chapter 2. Unit 2
Plotting the quadrant angles and filling in the in-between values shows the graph of y = cos x
The graph of y = cos x has a period of 2π. The range of a cosine curve is {−1 ≤ y ≤ 1} and the domain of cos x
is all reals. If you’ve studied the sine function, you may notice that the shape of the curve is exactly the same, but
shifted by π2 .
449
2.31. Cosine and Secant Graphs
www.ck12.org
Secant is the reciprocal of cosine, or 1x . Therefore, whenever the cosine is zero, the secant is going to have a vertical
asymptote because it will be undefined. It also has the same sign as the cosine function in the same quadrants. Here
is the graph.
The period of the function is 2π, just like cosine. The domain of the function is all real numbers, except multiples
π
3π 5π
of π starting at π2 . . . . −π
2 , 2 , 0, 2 , 2 . . . . The range is all real numbers greater than or equal to 1 as well as all real
numbers less than or equal to -1. Notice that the range is everything except where cosine is defined (other than the
tops and bottoms of the cosine curve).
450
www.ck12.org
Chapter 2. Unit 2
Notice again the reciprocal relationships at 0 and the asymptotes. Also look at the intersection points of the graphs
at 1 and -1. Again, this graph looks parabolic, but it is not.
451
2.31. Cosine and Secant Graphs
www.ck12.org
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/70557
Sketch the graph
Sketch a graph of h(x) = 5 + 12 sec 4x over the interval [0, 2π].
If you compare this example to f (x) = sec x, it will be translated 5 units up, with an amplitude of
of 4. This means in our interval of 0 to 2π, there will be 4 secant curves.
452
1
2
and a frequency
www.ck12.org
Chapter 2. Unit 2
Find the equation for the graph below.
First of all, this could be either a secant or cosecant function. Let’s say this is a secant function.
Secant usually
intersects the y−axis at (0, 1) at a minimum. Now, that corresponding minimum is π2 , −2 . Because there is no
amplitude change, we can say that the vertical shift is the difference between the two y−values, -3. It looks like
π
8π
there is a phase shift and a period change. From minimum to minimum is one period, which is 9π
2 − 2 = 2 = 4π
2π
1
and B = 4π = 2 . Lastly, we need to find the horizontal shift. Since secant usually intersects the y−axis at (0, 1) at
a minimum, and now the corresponding minimum is π2 , −2 , we can say that the horizontal shift is the difference
between the two x−values, π2 . Therefore, our equation is f (x) = −3 + sec 12 (x − π2 ).
Graph the function h(x) = 2 − 3 cos 4x
Examples
Example 1
Earlier, you were asked what would the shadow trace out.
453
2.31. Cosine and Secant Graphs
www.ck12.org
As you have learned in this section, a light shining down on the rotating hand would create a shadow in the pattern
of a cosine function, starting at a maximum value as the hand is lying along the "x" axis, going through zero to a
maximum negative value when the hand is lying along the negative "y" axis. It would then begin to increase until it
returned to a maximum value when the rotating hand was again lying along the positive "x" axis.
Example 2
Graph y = −2 + 21 sec(4(x − 1)).
Example 3
Determine the function creating this graph:
This could be either a secant or cosecant function. We will use a cosecant model. First, the vertical shift is -1. The
3π
period is the difference between the two given x−values, 7π
is 2π
4 − 4 = π, so the frequency
π =2. The horizontal
3π
shift incorporates the frequency, so in y = csc x the corresponding x−value to 4 , 0 is π2 , 1 . The difference
π
3π
2π
π
π
π
between the x−valuesis 3π
4 − 2 = 4 − 4 = 4 and then multiply it by the frequency, 2 · 4 = 2 . The equation is
π
y = −1 + csc 2(x − 2 ).
Example 4
Graph h(x) = 31 cos 2x
454
www.ck12.org
Chapter 2. Unit 2
Review
Graph each of the following functions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
f (x) = cos(x).
h(x) = cos(2x).
k(x) = cos(2x + π).
m(x) = −2 cos(2x + π).
g(x) = −2 cos(2x + π) + 1.
f (x) = sec(x).
h(x) = sec(3x).
k(x) = sec(3x + π).
m(x) = 2 sec(3x + π).
g(x) = 3 + 2 sec(3x + π).
h(x) = cos( 2x ).
k(x) = cos( 2x + π2 ).
m(x) = 2 cos( 2x + π2 ).
g(x) = 2 cos( 2x + π2 ) − 3.
h(x) = sec( 4x ).
k(x) = sec( 4x + 3π
2 ).
x
m(x) = −3 sec( 4 + 3π
2 ).
g(x) = 2 − 3 sec( 4x + 3π
2 ).
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.10.
455
2.32. Tangent and Cotangent Graphs
www.ck12.org
2.32 Tangent and Cotangent Graphs
Learning Objectives
Here you’ll learn to draw the graphs of the tangent and cotangent functions.
What if your instructor gave you a set of graphs like these:
and asked you to identify which were the graphs of the tangent and cotangent functions?
Tangent and Cotangent Graphs
The name of the tangent function comes from the tangent line of a circle. This is a line that is perpendicular to the
radius at a point on the circle so that the line touches the circle at exactly one point.
If we extend angle θ through the unit circle so that it intersects with the tangent line, tan θ will be equal to the length
of the red segment. Below, this segment is labeled the "tangent segment".
456
www.ck12.org
Chapter 2. Unit 2
Why? The dashed segment is 1 because it is the radius of the unit circle. Recall that in general, tan θ = xy . So here,
tangent segment
= tangent segment
tan θ =
1
As the value of θ increases, the value of tan θ changes. As we rotate through the first quadrant, the value of tan θ will
increase very slowly at first and then more rapidly.
457
2.32. Tangent and Cotangent Graphs
458
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
As we get very close to the y−axis the segment gets infinitely large, until when the angle really hits 90◦ , at which
point the extension of the angle and the tangent line will actually be parallel and therefore never intersect.
459
2.32. Tangent and Cotangent Graphs
www.ck12.org
This means there is no finite length of the tangent segment, or the tangent segment is infinitely large.
Let’s translate this portion of the graph onto the coordinate plane. Plot (θ, tan θ) as (x, y).
460
www.ck12.org
Chapter 2. Unit 2
461
2.32. Tangent and Cotangent Graphs
www.ck12.org
In fact as we get infinitely close to 90◦ , the tangent value increases without bound, until when we actually reach
90◦ , at which point the tangent is undefined. Recall there are some angles (90◦ and 270◦ , for example) for which the
tangent is not defined. Therefore, at these points, there are going to be vertical asymptotes.
Rotating past 90◦ , the intersection of the extension of the angle and the tangent line is actually below the x−axis.
This fits nicely with what we know about the tangent for a 2nd quadrant angle being negative. At first, it will have
very large negative values, but as the angle rotates, the segment gets shorter, reaches 0, then crosses back into the
positive numbers as the angle enters the 3rd quadrant. The segment will again get infinitely large as it approaches
270◦ . After being undefined at 270◦ , the angle crosses into the 4th quadrant and once again changes from being
infinitely negative, to approaching zero as we complete a full rotation.
462
www.ck12.org
Chapter 2. Unit 2
The graph y = tan x over several rotations would look like this:
463
2.32. Tangent and Cotangent Graphs
www.ck12.org
Notice the x−axis is measured in radians. Our asymptotes occur every π radians, starting at π2 . The period of the
π
graph is therefore π radians. The domain is all reals except for the asymptotes at π2 , 3π
2 , − 2 , etc. and the range is all
real numbers.
Cotangent is the reciprocal of tangent, xy , so it would make sense that where ever the tangent had an asymptote, now
the cotangent will be zero. The opposite of this is also true. When the tangent is zero, now the cotangent will have
an asymptote. The shape of the curve is generally the same, so the graph looks like this:
464
www.ck12.org
Chapter 2. Unit 2
465
2.32. Tangent and Cotangent Graphs
www.ck12.org
When you overlap the two functions, notice that the graphs consistently intersect at 1 and -1. These are the angles
that have 45◦ as reference angles, which always have tangents and cotangents equal to 1 or -1. It makes sense that
1 and -1 are the only values for which a function and it’s reciprocal are the same. Keep this in mind as we look at
cosecant and secant compared to their reciprocals of sine and cosine.
The cotangent function has a domain of all real angles except multiples of π {. . . − 2π, −π, 0, π, 2π . . .} The range is
all real numbers.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/143445
Sketching Graphs
1. Sketch the graph of g(x) = −2 + cot 31 x over the interval [0, 6π]
Starting with y = cot x, g(x) would be shifted down two and frequency is 13 , which means the period would be 3π,
instead of π. So, in our interval of [0, 6π] there would be two complete repetitions. The red graph is y = cot x.
2. Sketch the graph of y = −3 tan x − π4 over the interval [−π, 2π]
If you compare this graph to y = tan x, it will be stretched and flipped. It will also have a phase shift of
The red graph is y = tan x.
466
π
4
to the right.
www.ck12.org
Chapter 2. Unit 2
3. Sketch the graph of h(x) = 4 tan x + π2 + 3 over the interval [0, 2π]
The constant in front of the tangent function will cause the graph to be stretched. It will also have a phase shift of π2
to the left. Finally, the graph will be shifted up three. Here you can see both graphs, where the red graph is y = tan x.
Examples
Example 1
Earlier, you were asked to identify which graphs are tangent and cotangent.
As you can tell after completing this section, when presented with the graphs:
1
467
2.32. Tangent and Cotangent Graphs
4
468
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
The tangent and cotangent graphs are the third and sixth graphs.
Example 2
Graph y = −1 + 31 cot 2x
Example 3
Graph f (x) = 4 + tan(0.5(x − π))
Example 4
Graph y = −2 tan 2x
469
2.32. Tangent and Cotangent Graphs
Review
Graph each of the following functions
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
f (x) = tan(x)
h(x) = tan(2x)
k(x) = tan(2x + π)
m(x) = − tan(2x + π)
g(x) = − tan(2x + π) + 3
f (x) = cot(x)
h(x) = cot(2x)
k(x) = cot(2x + π)
m(x) = 3 cot(2x + π)
g(x) = −2 + 3 cot(2x + π)
h(x) = tan( 2x )
k(x) = tan( 2x + π4 )
m(x) = 3 tan( 2x + π4 )
g(x) = 3 tan( 2x + π4 ) − 1
h(x) = cot( 2x )
k(x) = cot( 2x + 3π
2 )
m(x) = −3 cot( 2x + 3π
2 )
g(x) = 2 − 3 cot( 2x + 3π
2 )
Review (Answers)
To see the Review answers, open this PDF file and look for section 2.11.
470
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.33 Putting it all Together
Learning Objectives
To graph sine and cosine functions where the amplitude is changed and horizontal and vertical shifts.
Your mission, should you choose to accept it, as Agent Trigonometry is to the find the domain and the range of the
function y = 12 sin(x + 2) − 3.
Graphing Functions
Here you’ll change the amplitude, the horizontal shifts, vertical shifts, and reflections.
Let’s graph y = 4 sin x − π4 and find the domain and range.
First, stretch the curve so that the amplitude is 4, making the maximums and minimums 4 and -4. Then, shift the
curve π4 units to the right.
As for the domain, it is all real numbers because the sine curve is periodic and infinite. The range will be from the
maximum to the minimum; y ∈ [−4, 4].
Now, let’s graph y = −2 cos(x − 1) + 1 and find the domain and range.
The -2 indicates the cosine curve is flipped and stretched so that the amplitude is 2. Then, move the curve up one
unit and to the right one unit.
The domain is all real numbers and the range is y ∈ [−1, 3].
471
2.33. Putting it all Together
www.ck12.org
Finally, let’s find the equation of the sine curve below.
First, let’s find the amplitude. The range is from 1 to -5, which is a total distance of 6. Divided by 2, we find that the
amplitude is 3. Halfway between 1 and -5 is 1+(−5)
= −2, so that is our vertical shift. Lastly, we need to find the
5
horizontal shift. The easiest way to do this is to superimpose the curve y = 3 sin(x) − 2 over this curve and determine
the movement from one maximum to the closest maximum of this curve.
472
www.ck12.org
π
2
Subtracting
π
2
− π6 =
3π
6
Chapter 2. Unit 2
and π6 , we have:
− π6 =
2π
6
=
π
3
Making the equation y = 3 sin x + π3 − 2.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/182870
Examples
Example 1
Earlier, you were asked to find the domain and range of the function y = 12 sin(x + 2) − 3.
The 12 indicates the sine curve is smooshed so that the amplitude is 21 . Then, move the curve down two units and to
the left three units.
The domain is all real numbers and the range is y ∈ [− 52 , − 23 ].
For Examples 2 3, graph the functions. State the domain and range. Show two full periods.
Example 2
y = −2 sin x − π2
The domain is all real numbers and the range is y ∈ [−2, 2].
473
2.33. Putting it all Together
Example 3
y = 13 cos(x + 1) − 2
The domain is all real numbers and the range is y ∈ −2 13 , −1 23 .
Example 4
Write one sine equation and one cosine equation for the curve below.
474
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
The amplitude and vertical shift is the same, whether the equation is a sine or cosine curve. The vertical shift is
-2 because that is the number that is halfway between the maximum and minimum. The difference between the
maximum and minimum is 1, so the amplitude is half of that, or 12 . As a sine curve, the function is y = −2 + 21 sin x.
As a cosine curve, there will be a shift of π2 , y = 12 cos x − π2 − 2.
Review
Determine if the following statements are true or false.
1. To change a cosine curve into a sine curve, shift the curve π2 units.
2. For any given sine or cosine graph, there are infinitely many possible equations that can be written to represent
the curve.
3. The amplitude is the same as the maximum value of the sine or cosine curve.
4. The horizontal shift is always in terms of π.
5. The domain of any sine or cosine function is always all real numbers.
Graph the following sine or cosine functions such that x ∈ [−2π, 2π]. State the domain and range.
6.
7.
8.
9.
10.
11.
y = sin x + π4 + 1
y = 2 − 3 cos x y = 43 sin x − 2π
3
y = −5 sin(x − 3)− 2
y = 2 cos x + 5π
6 − 1.5
y = −2.8 cos(x − 8) + 4
Use the graph below to answer questions 12-15.
475
2.33. Putting it all Together
www.ck12.org
12. Write a sine equation for the function where the amplitude is positive.
13. Write a cosine equation for the function where the amplitude is positive.
14. How often does a sine or cosine curve repeat itself? How can you use this to help you write different equations
for the same graph?
15. Write a second sine and cosine equation with different horizontal shifts.
Use the graph below to answer questions 16-20.
16.
17.
18.
19.
20.
Write a sine equation for the function where the amplitude is positive.
Write a cosine equation for the function where the amplitude is positive.
Write a sine equation for the function where the amplitude is negative.
Write a cosine equation for the function where the amplitude is negative.
Describe the similarities and differences between the four equations from questions 16-19.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 14.3.
476
www.ck12.org
Chapter 2. Unit 2
2.34 Trigonometric Equations Using Factoring
Learning Objectives
Here you’ll learn how to factor trig equations and then solve them using the factored form.
Solving trig equations is an important process in mathematics. Quite often you’ll see powers of trigonometric
functions and be asked to solve for the values of the variable which make the equation true. For example, suppose
you were given the trig equation
2 sin x cos x = cos x
Trigonometric Equations Using Factoring
You have no doubt had experience with factoring. You have probably factored equations when looking for the
possible values of some variable, such as "x". It might interest you to find out that you can use the same factoring
method for more than just a variable that is a number. You can factor trigonometric equations to find the possible
values the function can take to satisfy an equation.
Algebraic skills like factoring and substitution that are used to solve various equations are very useful when solving
trigonometric equations. As with algebraic expressions, one must be careful to avoid dividing by zero during these
maneuvers.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/182926
Solving for Unknown Values
1. Solve 2 sin2 x − 3 sin x + 1 = 0 for 0 < x ≤ 2π.
477
2.34. Trigonometric Equations Using Factoring
www.ck12.org
2 sin2 x − 3 sin x + 1 = 0
Factor this like a quadratic equation
(2 sin x − 1)(sin x − 1) = 0
↓
&
2 sin x − 1 = 0
or sin x − 1 = 0
2 sin x = 1
1
sin x =
2
π
5π
x = and x =
6
6
sin x = 1
π
x=
2
2. Solve 2 tan x sin x + 2 sin x = tan x + 1 for all values of x.
Pull out sin x
There is a common factor of (tan x + 1)
Think of the −(tan x + 1) as (−1)(tan x + 1), which is why there is a −1 behind the 2 sin x.
3. Solve 2 sin2 x + 3 sin x − 2 = 0 for all x, [0, π].
2 sin2 x + 3 sin x − 2 = 0 → Factor like a quadratic
(2 sin x − 1)(sin x + 2) = 0
.
&
2 sin x − 1 = 0
sin x + 2 = 0
1
sin x = −2
sin x =
2
π
5π
x = and x =
There is no solution because the range of sin x is [−1, 1].
6
6
Some trigonometric equations have no solutions. This means that there is no replacement for the variable that will
result in a true expression.
Examples
Example 1
Earlier, you were asked to solve this:
478
www.ck12.org
Chapter 2. Unit 2
2 sin x cos x = cos x
Subtract cos x from both sides and factor it out of the equation:
2 sin x cos x − cos x = 0
cos x(2 sin x − 1) = 0
Now set each factor equal to zero and solve. The first is cos x:
cos x = 0
π 3π
x= ,
2 2
And now for the other term:
2 sin x − 1 = 0
1
sin x =
2
π 5π
x= ,
6 6
Example 2
Solve the trigonometric equation 4 sin x cos x + 2 cos x − 2 sin x − 1 = 0 such that 0 ≤ x < 2π.
Use factoring by grouping.
2 sin x + 1 = 0
or
2 sin x = −1
1
sin x = −
2
7π 11π
x= ,
6 6
2 cos x − 1 = 0
2 cos x = 1
1
cos x =
2
π 5π
x= ,
3 3
479
2.34. Trigonometric Equations Using Factoring
www.ck12.org
Example 3
Solve tan2 x = 3 tan x for x over [0, π].
tan2 x = 3 tan x
tan2 x − 3 tan x = 0
tan x(tan x − 3) = 0
tan x = 0
x = 0, π
or
tan x = 3
x = 1.25
Example 4
Find all the solutions for the trigonometric equation 2 sin2 4x − 3 cos 4x = 0 over the interval [0, 2π).
2 sin2 4x − 3 cos 4x = 0
x
x
2 1 − cos2
− 3 cos = 0
4
4
x
2 x
2 − 2 cos − 3 cos = 0
4
4
x
2 x
2 cos + 3 cos − 2 = 0
4 4 x
x
2 cos − 1 cos + 2 = 0
4
4
.
&
x
x
2 cos − 1 = 0 or cos + 2 = 0
4
4
x
x
2 cos = 1
cos = −2
4
4
x 1
cos =
4 2
x π
5π
=
or
4 3
3
4π
20π
x=
or
3
3
20π
3
is eliminated as a solution because it is outside of the range and cos 4x = −2 will not generate any solutions
because −2 is outside of the range of cosine. Therefore, the only solution is 4π
3 .
Review
Solve each equation for x over the interval [0, 2π).
1. cos2 (x) + 2 cos(x) + 1 = 0
2. 1 − 2 sin(x) + sin2 (x) = 0
3. 2 cos(x) sin(x) − cos(x) = 0
480
www.ck12.org
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Chapter 2. Unit 2
sin(x) tan2 (x) − sin(x) = 0
sec2 (x) = 4
sin2 (x) − 2 sin(x) = 0
3 sin(x) = 2 cos2 (x)
2 sin2 (x) + 3 sin(x) = 2
tan(x) sin2 (x) = tan(x)
2 sin2 (x) + sin(x) = 1
2 cos(x) tan(x) − tan(x) = 0
sin2 (x) + sin(x) = 2
tan(x)(2 cos2 (x) + 3 cos(x) − 2) = 0
sin2 (x) + 1 = 2 sin(x)
2 cos2 (x) − 3 cos(x) = 2
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.4.
481
2.35. Trigonometric Equations Using the Quadratic Formula
www.ck12.org
2.35 Trigonometric Equations Using the
Quadratic Formula
Learning Objectives
Here you’ll learn how to use the quadratic equation to find solutions of trig functions.
Solving equations is a fundamental part of mathematics. Being able to find which values of a variable fit an equation
allows us to determine all sorts of interesting behavior, both in math and in the sciences. Solving trig equations for
angles that satisfy the equation is one application of mathematical methods for solving equations. Suppose someone
gave you the following equation:
3 sin2 θ + 8 sin θ − 3 = 0
Quadratic Functions with Trigonometric Equations
When solving quadratic equations that do not factor, the quadratic formula is often used.
Remember that the quadratic equation is:
ax2 + bx + c = 0 (where a, b, and c are constants)
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/176568
In this situation, you can use the quadratic formula to find out what values of "x" satisfy the equation.
The same method can be applied when solving trigonometric equations that do not factor. The values for a is the
numerical coefficient of the function’s squared term, b is the numerical coefficient of the function term that is to the
first power and c is a constant. The formula will result in two answers and both will have to be evaluated within the
designated interval.
Solving for Unknown Values
1. Solve 3 cot2 x − 3 cot x = 1 for exact values of x over the interval [0, 2π].
3 cot2 x − 3 cot x = 1
3 cot2 x − 3 cot x − 1 = 0
The equation will not factor. Use the quadratic formula for cot x, a = 3, b = −3, c = −1.
482
www.ck12.org
Chapter 2. Unit 2
√
b2 − 4ac
cot x =
2a p
−(−3) ± (−3)2 − 4(3)(−1)
cot x =
2(3)
√
3 ± 9 + 12
cot x =
√6
3 + 21
cot x =
or
6
3 + 4.5826
cot x =
6
cot x = 1.2638
1
tan x =
1.2638
x = 0.6694, 3.81099
−b ±
√
21
cot x =
6
3 − 4.5826
cot x =
6
cot x = −0.2638
1
tan x =
−0.2638
x = 1.8287, 4.9703
3−
2. Solve −5 cos2 x + 9 sin x + 3 = 0 for values of x over the interval [0, 2π].
Change cos2 x to 1 − sin2 x from the Pythagorean Identity.
−5 cos2 x + 9 sin x + 3 = 0
−5(1 − sin2 x) + 9 sin x + 3 = 0
−5 + 5 sin2 x + 9 sin x + 3 = 0
5 sin2 x + 9 sin x − 2 = 0
p
92 − 4(5)(−2)
sin x =
2(5)
√
−9 ± 81 + 40
sin x =
10
√
−9 ± 121
sin x =
10
−9 + 11
−9 − 11
and sin x =
sin x =
10
10
1
sin x = and − 2
5
−1
sin (0.2) and sin−1 (−2)
−9 ±
x ≈ .201 rad and π − .201 ≈ 2.941
This is the only solutions for x since −2 is not in the range of values.
3. Solve 3 sin2 x − 6 sin x − 2 = 0 for values of x over the interval [0, 2π].
3 sin2 x − 6 sin x − 2 = 0
483
2.35. Trigonometric Equations Using the Quadratic Formula
www.ck12.org
p
(−6)2 − 4(3)(−2)
sin x =
2(3)
√
6 ± 36 − 24
sin x =
√6
6 ± 12
sin x =
6
6 + 3.46
6 − 3.46
sin x =
and sin x =
10
10
sin x = .946 and .254
6±
sin−1 (0.946) and sin−1 (0.254)
x ≈ 71.08 deg and ≈ 14.71 deg
Examples
Example 1
Earlier, you were asked to solve an equation.
The original equation to solve was:
3 sin2 θ + 8 sin θ − 3 = 0
Using the quadratic formula, with a = 3, b = 8, c = −3, we get:
p
√
√
64 − (4)(3)(−3) −8± 100
−8±
−b± b2 − 4ac
sin θ =
=
=
=
2a
6
6
−8±10
6
= 31 or − 3
The solution of -3 is ignored because sine can’t take that value, however:
sin−1 13 = 19.471◦
Example 2
Solve sin2 x − 2 sin x − 3 = 0 for x over [0, π].
You can factor this one like a quadratic.
sin2 x − 2 sin x − 3 = 0
(sin x − 3)(sin x + 1) = 0
sin x − 3 = 0
sin x + 1 = 0
sin x = 3
or
x = sin−1 (3)
For this problem the only solution is
3π
2
because sine cannot be 3 (it is not in the range).
Example 3
Solve tan2 x + tan x − 2 = 0 for values of x over the interval − π2 , π2 .
484
sin x = −1
3π
x=
2
www.ck12.org
Chapter 2. Unit 2
tan2 x + tan x − 2 = 0
p
12 − 4(1)(−2)
= tan x
2
√
−1 ± 1 + 8
= tan x
2
−1 ± 3
= tan x
2
tan x = −2 or
−1 ±
1
tan x = 1 when x = π4 , in the interval − π2 , π2
tan x = −2 when x = −1.107 rad
Example 4
Solve the trigonometric equation such that 5 cos2 θ − 6 sin θ = 0 over the interval [0, 2π].
5 cos2 θ − 6 sin θ = 0 over the interval [0, 2π].
5 1 − sin2 x − 6 sin x = 0
−5 sin2 x − 6 sin x + 5 = 0
5 sin2 x + 6 sin x − 5 = 0
p
−6 ± 62 − 4(5)(−5)
= sin x
2(5)
√
−6 ± 36 + 100
= sin x
10 √
−6 ± 136
= sin x
10 √
−6 ± 2 34
= sin x
10 √
−3 ± 34
= sin x
5
x = sin
−1
−3+
√ √ 34 or sin−1 −3− 34 x = 0.6018 rad or 2.5398 rad from the first expression, the second
5
5
expression will not yield any answers because it is out the range of sine.
Review
Solve each equation using the quadratic formula.
1. 3x2 + 10x + 2 = 0
2. 5x2 + 10x + 2 = 0
3. 2x2 + 6x − 5 = 0
Use the quadratic formula to solve each quadratic equation over the interval [0, 2π).
485
2.35. Trigonometric Equations Using the Quadratic Formula
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
3 cos2 (x) + 10 cos(x) + 2 = 0
5 sin2 (x) + 10 sin(x) + 2 = 0
2 sin2 (x) + 6 sin(x) − 5 = 0
6 cos2 (x) − 5 cos(x) − 21 = 0
9 tan2 (x) − 42 tan(x) + 49 = 0
sin2 (x) + 3 sin(x) = 5
3 cos2 (x) − 4 sin(x) = 0
−2 cos2 (x) + 4 sin(x) = 0
tan2 (x) + tan(x) = 3
cot2 (x) + 5 tan(x) + 14 = 0
sin2 (x) + sin(x) = 1
What type of sine or cosine equations have no solution?
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.5.
486
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.36 Trig. Word Problems and IROC
Homework Practice:
487
2.36. Trig. Word Problems and IROC
488
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
489
2.36. Trig. Word Problems and IROC
Solutions:
490
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
2.37 Unit 2 Test Review
491
2.37. Unit 2 Test Review
492
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
493
2.37. Unit 2 Test Review
494
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
495
2.37. Unit 2 Test Review
Solutions:
496
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
497
2.37. Unit 2 Test Review
498
www.ck12.org
www.ck12.org
Chapter 2. Unit 2
499
2.38. References
www.ck12.org
2.38 References
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
33.
34.
35.
36.
37.
38.
39.
40.
41.
42.
500
CK-12. CK-12 .
CK-12;Gustavb. CK-12;Wikimedia .
CK-12;Gustavb. CK-12;Wikimedia .
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
. . CC BY-NC-SA
Courtesy of the USDA. http://commons.wikimedia.org/wiki/File:PivotIrrigationOnCotton.jpg . Public Domain
Courtesy of the NASA. http://commons.wikimedia.org/wiki/File:Crops_Kansas_AST_20010624.jpg . Public
Domain
User:Johne7236/Wikipedia. http://commons.wikimedia.org/wiki/File:W.F._Mangels_Kiddie_Galloping_Hor
se_Carrousel.jpg . Public Domain
Flickr:vxla. http://www.flickr.com/photos/14812197@N00/8727193240 . CC BY 2.0
CK-12. CK-12 . CC BY-NC-SA
CK-12 Foundation;CK-12;Gustavb. CK-12;Wikimedia . CCSA;CC BY-NC-SA
CK-12 Foundation;CK-12;Gustavb. CK-12;Wikimedia . CCSA;CC BY-NC-SA
CK-12 Foundation;CK-12;Gustavb. CK-12;Wikimedia . CCSA
CK-12 Foundation. . CCSA;CC BY-NC-SA
ck12. ck12.org .
ck12. ck12.org .
ck12. ck12.org .
ck12. ck12.org .
CK-12 Foundation;CK-12. CK-12 . CCSA
CK-12. CK-12 .
CK-12. CK-12 .
CK-12. CK-12 .
CK-12. CK-12 .
CK-12. CK-12 .
. CK-12 .
. CK-12 .
. CK-12 .
. CK-12 .
. CK-12 .
www.ck12.org
43.
44.
45.
46.
47.
48.
49.
Chapter 2. Unit 2
. CK-12 .
. CK-12 .
CK-12 Foundation. CK12.org .
. CK12.org .
. CK12.org .
. CK12.org .
. CK12.org .
501
www.ck12.org
C HAPTER
3
Chapter Outline
502
3.1
G RAPHS OF E XPONENTIAL F UNCTIONS
3.2
L AWS OF E XPONENTS FOR R EAL N UMBERS
3.3
Z ERO AND N EGATIVE E XPONENTS
3.4
I NVERSE F UNCTIONS
3.5
G RAPHS OF L OGS AS THE I NVERSE OF E XPONENTIAL F UNCTIONS
3.6
D EFINING L OGARITHMS
3.7
G RAPHS OF L OGARITHMIC F UNCTIONS
3.8
G RAPHING L OGARITHMIC F UNCTIONS
3.9
C HANGE OF BASE
3.10
PH
3.11
I NTENSITY AND L OUDNESS OF S OUND
3.12
E ARTHQUAKE M AGNITUDE S CALES
3.13
A PPLICATIONS
3.14
S IMPLIFYING R ADICALS
3.15
S OLVING E XPONENTIAL E QUATIONS
3.16
E XPONENTIAL E QUATIONS
3.17
H ALF L IFE
3.18
P RODUCT AND Q UOTIENT P ROPERTIES OF L OGARITHMS
3.19
P ROPERTIES OF L OGS
3.20
P OWER P ROPERTY OF L OGARITHMS
3.21
S OLVING L OGARITHMIC E QUATIONS
3.22
L OGARITHMIC M ODELS
3.23
U NIT 3 T EST R EVIEW
3.24
R EFERENCES
Unit 3
www.ck12.org
Chapter 3. Unit 3
3.1 Graphs of Exponential Functions
Learning Objectives
Here you’ll learn how to graph exponential functions and how to compare the graphs of exponential functions on the
same coordinate axes.
Graphs of Exponential Functions
A colony of bacteria has a population of three thousand at noon on Monday. During the next week, the colony’s
population doubles every day. What is the population of the bacteria colony just before midnight on Saturday?
At first glance, this seems like a problem you could solve using a geometric sequence. And you could, if the bacteria
population doubled all at once every day; since it doubled every day for five days, the final population would be
3000 times 25 .
But bacteria don’t reproduce all at once; their population grows slowly over the course of an entire day. So how do
we figure out the population after five and a half days?
Exponential Functions
Exponential functions are a lot like geometrical sequences. The main difference between them is that a geometric
sequence is discrete while an exponential function is continuous.
Discrete means that the sequence has values only at distinct points (the 1st term, 2nd term, etc.)
Continuous means that the function has values for all possible values of x. The integers are included, but also all
the numbers in between.
The problem with the bacteria is an example of a continuous function. Here’s an example of a discrete function:
An ant walks past several stacks of Lego blocks. There is one block in the first stack, 3 blocks in the 2nd stack and
9 blocks in the 3rd stack. In fact, in each successive stack there are triple the number of blocks than in the previous
stack.
In this example, each stack has a distinct number of blocks and the next stack is made by adding a certain number of
whole pieces all at once. More importantly, however, there are no values of the sequence between the stacks. You
can’t ask how high the stack is between the 2nd and 3rd stack, as no stack exists at that position!
As a result of this difference, we use a geometric series to describe quantities that have values at discrete points, and
we use exponential functions to describe quantities that have values that change continuously.
When we graph an exponential function, we draw the graph with a solid curve to show that the function has values
at any time during the day. On the other hand, when we graph a geometric sequence, we draw discrete points to
signify that the sequence only has value at those points but not in between.
Here are graphs for the two examples above:
503
3.1. Graphs of Exponential Functions
www.ck12.org
The formula for an exponential function is similar to the formula for finding the terms in a geometric sequence. An
exponential function takes the form
y = A · bx
where A is the starting amount and b is the amount by which the total is multiplied every time. For example, the
bacteria problem above would have the equation y = 3000 · 2x .
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133172
Compare Graphs of Exponential Functions
Let’s graph a few exponential functions and see what happens as we change the constants in the formula. The basic
shape of the exponential function should stay the same—but it may become steeper or shallower depending on the
constants we are using.
First, let’s see what happens when we change the value of A.
Compare the graphs of y = 2x and y = 3 · 2x .
Let’s make a table of values for both functions.
TABLE 3.1:
x
-3
-2
504
y = 2x
1
8
1
4
y = 3 · 2x
y = 3 · 2−3 = 3 · 213 =
y = 3 · 2−2 = 3 · 212 =
3
8
3
4
www.ck12.org
Chapter 3. Unit 3
TABLE 3.1: (continued)
x
-1
0
1
2
3
y = 2x
y = 3 · 2x
y = 3 · 2−1 = 3 · 211 = 23
y = 3 · 20 = 3
y = 3 · 21 = 6
y = 3 · 22 = 3 · 4 = 12
y = 3 · 23 = 3 · 8 = 24
1
2
1
2
4
8
Now let’s use this table to graph the functions.
We can see that the function y = 3 · 2x is bigger than the function y = 2x . In both functions, the value of y doubles
every time x increases by one. However, y = 3 · 2x “starts” with a value of 3, while y = 2x “starts” with a value of 1,
so it makes sense that y = 3 · 2x would be bigger as its values of y keep getting doubled.
Similarly, if the starting value of A is smaller, the values of the entire function will be smaller.
Comparing Graphs
Compare the graphs of y = 2x and y = 13 · 2x .
Let’s make a table of values for both functions.
TABLE 3.2:
x
-3
-2
-1
0
1
2
3
y = 2x
1
8
1
4
1
2
1
2
4
8
y=
y=
y=
y=
y=
y=
y=
y=
1
3
1
3
1
3
1
3
1
3
1
3
1
3
1
3
· 2x
· 2−3 = 13 · 213 =
· 2−2 = 13 · 212 =
· 2−1 = 13 · 211 =
· 20 = 31
· 21 = 32
· 22 = 31 · 4 = 43
· 23 = 31 · 8 = 83
1
24
1
12
1
6
Now let’s use this table to graph the functions.
505
3.1. Graphs of Exponential Functions
www.ck12.org
As we expected, the exponential function y = 13 · 2x is smaller than the exponential function y = 2x .
So what happens if the starting value of A is negative? Let’s find out.
Example C
Graph the exponential function y = −5 · 2x .
Solution
Let’s make a table of values:
TABLE 3.3:
x
-2
-1
0
1
2
3
Now let’s graph the function:
506
y = −5 · 2x
− 54
− 52
-5
-10
-20
-40
www.ck12.org
Chapter 3. Unit 3
This result shouldn’t be unexpected. Since the starting value is negative and keeps doubling over time, it makes
sense that the value of y gets farther from zero, but in a negative direction. The graph is basically just like the graph
of y = 5 · 2x , only mirror-reversed about the x−axis.
Now, let’s compare exponential functions whose bases (b) are different.
Graphing Multiple Functions
Graph the following exponential functions on the same graph: y = 2x , y = 3x , y = 5x , y = 10x .
First we’ll make a table of values for all four functions.
TABLE 3.4:
x
-2
-1
0
1
2
3
y = 2x
y = 3x
y = 5x
y = 10x
1
4
1
2
1
9
1
3
1
25
1
5
1
100
1
10
1
2
4
8
1
3
9
27
1
5
25
125
1
10
100
1000
Now let’s graph the functions.
507
3.1. Graphs of Exponential Functions
www.ck12.org
Notice that for x = 0, all four functions equal 1. They all “start out” at the same point, but the ones with higher
values for b grow faster when x is positive—and also shrink faster when x is negative.
Finally, let’s explore what happens for values of b that are less than 1.
Example E
Graph the exponential function y = 5 ·
1 x
2 .
Solution
Let’s start by making a table of values. (Remember that a fraction to a negative power is equivalent to its reciprocal
to the same positive power.)
TABLE 3.5:
x
-3
-2
-1
0
1
2
Now let’s graph the function.
508
y = 5·
y = 5·
y = 5·
y = 5·
y = 5·
y = 5·
y = 5·
1 x
2
1 −3
= 5 · 23 =
2
1 −2
= 5 · 22 =
2
1 −1
= 5 · 21 =
2
1 0
2 = 5·1 = 5
1 1
5
2 = 2
1 2
= 54
2
40
20
10
www.ck12.org
Chapter 3. Unit 3
This graph looks very different than the graphs from the previous example! What’s going on here?
When we raise a number greater than 1 to the power of x, it gets bigger as x gets bigger. But when we raise a number
smaller than 1 to the power of x, it gets smaller as x gets bigger—as you can see from the table of values above. This
makes sense because multiplying any number by a quantity less than 1 always makes it smaller.
So, when the base b of an exponential function is between 0 and 1, the graph is like an ordinary exponential graph,
only decreasing instead of increasing. Graphs like this represent exponential decay instead of exponential growth.
Exponential decay functions are used to describe quantities that decrease over a period of time.
When b can be written as a fraction, we can use the Property of Negative Exponents to write the function in a
x
different form. For instance, y = 5 · 12 is equivalent to 5 · 2−x . These two forms are both commonly used, so it’s
important to know that they are equivalent.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/133173
Examples
Example 1
Graph the exponential function y = 8 · 3−x
a.) Here is our table of values and the graph of the function.
TABLE 3.6:
x
-3
y = 8 · 3−x
y = 8 · 3−(−3) = 8 · 33 = 216
509
3.1. Graphs of Exponential Functions
www.ck12.org
TABLE 3.6: (continued)
y = 8 · 3−x
y = 8 · 3−(−2) = 8 · 32 = 72
y = 8 · 3−(−1) = 8 · 31 = 24
y = 8 · 30 = 8
y = 8 · 3−1 = 83
y = 8 · 3−2 = 89
x
-2
-1
0
1
2
Example 2
Graph the functions y = 4x and y = 4−x on the same coordinate axes.
Here is the table of values for the two functions. Looking at the values in the table, we can see that the two functions
are “backwards” of each other, in the sense that the values for the two functions are reciprocals.
TABLE 3.7:
x
-3
-2
-1
0
1
2
3
y = 4x
1
y = 4−3 = 64
1
−2
y = 4 = 16
y = 4−1 = 14
y = 40 = 1
y = 41 = 4
y = 42 = 16
y = 43 = 64
y = 4−x
y = 4−(−3) = 64
y = 4−(−2) = 16
y = 4−(−1) = 4
y = 40 = 1
y = 4−1 = 14
1
y = 4−2 = 16
1
−3
y = 4 = 64
Here is the graph of the two functions. Notice that the two functions are mirror images of each other if the mirror is
placed vertically on the y−axis.
510
www.ck12.org
Chapter 3. Unit 3
In the next lesson, you’ll see how exponential growth and decay functions can be used to represent situations in the
real world.
Review
Graph the following exponential functions by making a table of values.
1.
2.
3.
4.
y = 3x
y = 5 · 3x
y = 40 · 4x
y = 3 · 10x
Graph the following exponential functions.
5.
6.
7.
8.
9.
10.
11.
12.
x
y = 51
x
y = 4 · 23
y = 3−x
y = 43 · 6−x
Which two of the eight graphs above are mirror images of each other?
What function would produce a graph that is the mirror image of the one in problem 4?
How else might you write the exponential function in problem 5?
How else might you write the function in problem 6?
Solve the following problems.
13. A chain letter is sent out to 10 people telling everyone to make 10 copies of the letter and send each one to a
new person.
a. Assume that everyone who receives the letter sends it to ten new people and that each cycle takes a week.
How many people receive the letter on the sixth week?
b. What if everyone only sends the letter to 9 new people? How many people will then get letters on the
sixth week?
14. Nadia received $200 for her 10th birthday. If she saves it in a bank account with 7.5% interest compounded
yearly, how much money will she have in the bank by her 21st birthday?
511
3.1. Graphs of Exponential Functions
Review (Answers)
To view the Review answers, open this PDF file and look for section 8.11.
R
512
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.2 Laws of Exponents for Real Numbers
Students should check the below link for an interactive understanding of the exponents.
• http://www.mathsisfun.com/algebra/exponent-laws.html
Multimedia link
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/160490
Watch the video at: https://www.youtube.com/watch?v=byV55vUjDPI
513
3.3. Zero and Negative Exponents
www.ck12.org
3.3 Zero and Negative Exponents
Here you’ll learn how to work with zero and negative exponents.
How can you use the quotient rules for exponents to understand the meaning of a zero or negative exponent?
Watch This
Khan Academy Negative Exponent Intuition
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/59342
Guidance
Zero Exponent
Recall that
am
= am−n
an
. If m = n, then the following would be true:
am
= am−n = a0
an
33
= 33−3 = 30
33
However, any quantity divided by itself is equal to one. Therefore,
general:
a0 = 1 if a 6= 0.
Note that if a = 0, 00 is not defined.
Negative Exponents
42 × 4−2 = 42+(−2) = 40 = 1
Therefore:
514
33
33
= 1 which means 30 = 1. This is true in
www.ck12.org
Chapter 3. Unit 3
42 × 4−2 = 1
42 × 4−2
1
= 2
2
4
4
2
−2
4 ×4
1
= 2
2
4
4
4−2 =
Divide both sides by 42 .
Simplify the equation.
1
42
This is true in general and creates the following laws for negative exponents:
•
a−m =
1
am
•
1
a−m
= am
These laws for negative exponents can be expressed in many ways:
• If a term has a negative exponent, write it as 1 over the term with a positive exponent. For example: a−m = a1m
1
and a−m
= am
−2
2
• If a term has a negative exponent, write the reciprocal with a positive exponent. For example: 23
= 32
−m
and a−m = a 1 = a1m
• If the term is a factor in the numerator with a negative exponent, write it in the denominator with a positive
n
exponent. For example: 3x−3 y = 3y
and a−m bn = a1m (bn ) = abm
x3
• If the term is a factor in the denominator with a negative exponent, write it in the numerator with a positive
3
n
m
exponent. For example: x2x−2 = 2x3 (x2 ) and ab−m = bn aa = bn am
These ways for understanding negative exponents provide shortcuts for arriving at solutions without doing tedious
calculations. The results will be the same.
Example A
Evaluate the following using the laws of exponents.
3 −2
4
Solution:
There are two methods that can be used to evaluate the expression.
Method 1: Apply the negative exponent rule
a−m =
1
am
515
3.3. Zero and Negative Exponents
−2
3
=
4
1
3 2
Write the expression with a positive exponent by applying a−m =
4
1
1
= 32
2
3
1
=
1
1
1
.
am
a n
b
=
an
bn
=
an
bn
Evaluate the powers.
9
16
= 1÷
9
16
Apply the law of exponents for raising a quotient to a power.
42
4
32
42
www.ck12.org
9
16
Divide
9
16 16
= 1×
=
16
9
9
−2
3
16
=
4
9
1÷
Method 2: Apply the shortcut and write the reciprocal with a positive exponent.
−2 2
3
4
=
4
3
2
4
42
= 2
3
3
42 16
=
32
9
−2
3
16
=
4
9
Write the reciprocal with a positive exponent.
Apply the law of exponents for raising a quotient to a power.
a n
b
Simplify.
Applying the shortcut facilitates the process for obtaining the solution.
Example B
State the following using only positive exponents: (If possible, use shortcuts)
i) y−6
−3
ii) ab
iii)
x5
y−4
iv) a2 × a−5
Solutions:
i)
y−6
y−6 =
516
Write the expression with a positive exponent by applying
1
y6
a−m =
1
.
am
www.ck12.org
Chapter 3. Unit 3
ii)
a −3
Write the reciprocal with a positive exponent.
b
a −3 b 3
=
b
a
3
b
b3
= 3
a
a
a −3 b3
= 3
b
a
a n
Apply the law of exponents for raising a quotient to a power.
b
=
an
bn
iii)
x5
y−4
Apply the negative exponent rule.
x5
= x5
y−4
y4
1
1
a−m
= am
Simplify.
x5
= x 5 y4
y−4
iv)
a2 × a−5
Apply the product rule for exponents am × an = am+n .
a2 × a−5 = a2+(−5)
Simplify.
a2+(−5) = a−3
Write the expression with a positive exponent by applying a−m =
a−3 =
1
.
am
1
a3
a2 × a−5 =
1
a3
Example C
Evaluate the following:
7−2 +7−1
7−3 +7−4
Solution:
There are two methods that can be used to evaluate the problem.
Method 1: Work with the terms in the problem in exponential form.
Numerator:
517
3.3. Zero and Negative Exponents
7−2 =
www.ck12.org
1
1
and 7−1 =
2
7
7
Apply the definition a−m =
1 1
+
72 7 1 1 7
+
72 7 7
7
1+7
8
1
+ 2= 2 = 2
2
7
7
7
7
1
am
A common denominator is needed to add the fractions.
Multiply
1
7
by to obtain the common denominator of 72
7
7
Add the fractions.
Denominator:
1
1
and 7−4 = 4
3
7
7
1
1
+
73 74
7 1
1
+ 4
3
7 7
7
1
1+7
8
7
+ 4= 4 = 4
4
7
7
7
7
7−3 =
Apply the definition a−m =
1
am
A common denominator is needed to add the fractions.
Multiply
1
7
by to obtain the common denominator of 74
3
7
7
Add the fractions.
Numerator and Denominator:
8
8
÷ 4
2
7
7
8 74
×
72
8
74 74
8
× = 2 = 72 = 49
72
7
8
Divide the numerator by the denominator.
Multiply by the reciprocal.
Simplify.
7−2 + 7−1
= 49
7−3 + 7−4
Method 2: Multiply the numerator and the denominator by 74 . This will change all negative exponents to
positive exponents. Apply the product rule for exponents and work with the terms in exponential form.
7−2 + 7−1
7−3 + 7−4
4 −2
7 7 + 7−1
74 7−3 + 7−4
72 + 73
71 + 70
49 + 343 392
=
= 49
7+1
8
Apply the distributive property with the product rule for exponents.
Evaluate the numerator and the denominator.
7−2 + 7−1
= 49
7−3 + 7−4
Whichever method is used, the result is the same.
518
www.ck12.org
Chapter 3. Unit 3
Concept Problem Revisited
m
By the quotient rule for exponents, xxm = xm−m = x0 . Since anything divided by itself is equal to 1 (besides 0),
Therefore, x0 = 1 as long as x 6= 0.
Also by the quotient rule for exponents,
you would have
x2
x5
=
x·x
x·x·x·x·x
=
1
.
x3
x2
x5
xm
xm
= 1.
= x2−5 = x−3 . If you were to expand and reduce the original expression
Therefore, x−3 =
1
.
x3
This generalizes to x−a =
1
xa .
Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers.
Some examples are: In the expression 25 , ’2’ is the base. In the expression (−3y)4 , ’−3y’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times
to multiply the base times itself. Some examples are:
In the expression 25 , ’5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: 25 = 2 × 2 ×
2×2×2
In the expression (−3y)4 , ’4’ is the exponent. It means to multiply −3y times itself 4 times as shown here:
(−3y)4 = −3y × −3y × −3y × −3y.
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents
when dealing with exponential expressions.
Guided Practice
1. Use the laws of exponents to simplify the following: (−3x2 )3 (9x4 y)−2
2. Rewrite the following using only positive exponents. (x2 y−1 − 1)2
3. Use the laws of exponents to evaluate the following: [5−4 × (25)3 ]2
Answers:
1.
519
3.3. Zero and Negative Exponents
www.ck12.org
(−3x2 )3 (9x4 y)−2
Apply the laws of exponents (am )n = amn and a−m =
1
am
1
(−3x ) (9x y) = (−3 x )
Simplify and apply (ab)n = an bn
(9x4 y)2
1
1
3 6
6
(−3 x )
= −27x
Simplify.
(9x4 y)2
(92 x8 y2 )
1
am
−27x6
− 27x6
Simplify and apply the quotient rule for exponents n = am−n .
=
2
8
2
8
2
(9 x y )
81x y
a
2 3
4
−2
3 6
−27x6
1x−2
=
−
81x8 y2
3y2
Apply the negative exponent rule a−m =
(−3x2 )3 (9x4 y)−2 = −
1
am
1
3x2 y2
2.
(x2 y−1 − 1)2
Begin by expanding the binomial and using the FOIL m
to apply the product rule for exponents am × an = am+n
(x2 y−1 − 1)(x2 y−1 − 1) = (x2+2 y−1+(−1) − 1x2 y−1 − 1x2 y−1 + 1) Simplify.
(x2+2 y−1+(−1) − 1x2 y−1 − 1x2 y−1 + 1) = (x4 y−2 − 2x2 y−1 + 1)
4
2x2
x
4 −2
2 −1
−
+1
(x y − 2x y + 1) =
y2
y
4
2x2
x
2 −1
2
(x y − 1) =
−
+1
y2
y
Apply the negative exponent rule a−m =
3.
[5−4 × (25)3 ]2
[5
−4
3 2
× (25) ] = [5
Try to do this one by applying the laws of exponents.
−4
2 3 2
× (5 ) ]
[5−4 × (52 )3 ]2 = [5−4 × 56 ]2
[5−4 × 56 ]2 = (52 )2
(52 )2 = 54
54 = 625
[5−4 × (25)3 ]2 = 54 = 625
Practice
Evaluate each of the following expressions:
520
1
.
am
www.ck12.org
Chapter 3. Unit 3
2 0
3
−2
− 25
(−3)−3
1. −
2.
3.
−2
4. 6 × 12
5. 7−4 × 74
Simplify the following:
6.
7.
8.
9.
10.
(2−1 − 2−2 )2
(40 + 4−1 )−1
(3−1 − 2−1 )−2
(x−1 + y−1 )2
1
1 −1
− 6−1
60
Rewrite the following using positive exponents only. Simplify where possible.
11. (4wx−2 y3 z−4 )3
2 3 −2
12. da−2bbcc −6
13. (x2 − 1)(x−2 + 2)
14. m4 (m2 + m − 5m−2 )
−2 −y−2
15. xx−1 +y
−1
521
3.4. Inverse Functions
www.ck12.org
3.4 Inverse Functions
Learning Objectives
Here you’ll learn how to find the inverse of a relation and function.
2
A planet’s maximum distance from the sun (in astronomical units) is given by the formula d = p 3 , were p is the
period (in years) of the planet’s orbit around the sun. What is the inverse of this function?
Inverse Functions
By now, you are probably familiar with the term “inverse”. Multiplication and division are inverses of each other.
More examples are addition and subtraction and the square and square root. We are going to extend this idea to
functions. An inverse relation maps the output values to the input values to create another relation. In other words,
we switch the x and y values. The domain of the original relation becomes the range of the inverse relation and the
range of the original relation becomes the domain of the inverse relation.
Let’s find the inverse mapping of S = {(6, −1), (−2, −5), (−3, 4), (0, 3), (2, 2)}.
Here, we will find the inverse of this relation by mapping it over the line y = x. As was stated above in the definition,
the inverse relation switched the domain and range of the original function. So, the inverse of this relation, S, is S−1
(said “s inverse”) and will flip all the x and y values.
S−1 = {(−1, 6), (−5, −2), (4, −3), (3, 0), (2, 2)}
If we plot the two relations on the x − y plane, we have:
The blue points are all the points in S and the red points are all the points in S−1 . Notice that the points in S−1 are a
reflection of the points in S over the line, y = x. All inverses have this property.
522
www.ck12.org
Chapter 3. Unit 3
If we were to fold the graph on y = x, each inverse point S−1 should lie on the original point from S. The point (2, 2)
lies on this line, so it has no reflection. Any value on this line will remain the same.
Domain of S: x ∈ {6, −2, −3, 0, 2}
Range of S: y ∈ {−1, −5, 4, 3, 2}
Domain of S0 : x ∈ {−1, −5, 4, 3, 2}
Range of S0 : y ∈ {6, −2, −3, 0, 2}
By looking at the domains and ranges of S and S−1 , we see that they are both functions (no x-values repeat). When
the inverse of a function is also a function, we say that the original function is a one-to-one function. Each value
maps one unique value onto another unique value.
Now, let’s find the inverse of f (x) = 32 x − 1.
This is a linear function. Let’s solve by doing a little investigation. First, draw the line along with y = x on the same
set of axes.
Notice the points on the function (blue line). Map these points over y = x by switching their x and y values. You
could also fold the graph along y = x and trace the reflection.
523
3.4. Inverse Functions
www.ck12.org
The red line in the graph to the right is the inverse of f (x) = 23 x − 1. Using slope triangles between (-1, 0) and (1,
3), we see that the slope is 32 . Use (-1, 0) to find the y-intercept.
3
f −1 (x) = x + b
2
3
0 = (−1) + b
2
3
=b
2
The equation of the inverse, read “ f inverse”, is f −1 (x) = 32 x + 32 .
You may have noticed that the slopes of f and f −1 are reciprocals of each other. This will always be the case for
linear functions. Also, the x-intercept of f becomes the y-intercept of f −1 and vice versa.
Alternate Method: There is also an algebraic approach to finding the inverse of any function. Let’s repeat this
example using algebra.
Step 1: Change f (x) to y.
y = 23 x − 1
Step 2: Switch the x and y. Change y to y−1 for the inverse.
x = 32 y−1 − 1
Step 3: Solve for y0 .
2
x = y−1 − 1
3 3
3
2 −1
(x + 1) = ·
y
2
2
3
3
3
x + = y−1
2
2
524
www.ck12.org
Chapter 3. Unit 3
The algebraic method will work for any type of function.
√
Finally, let’s determine if g(x) = x − 2 and f (x) = x2 + 2 are inverses of each other.
There are two different ways to determine if two functions are inverses of each other. The first, is to find f −1 and
g−1 and see if f −1 = g and g−1 = f .
x=
p
y−1 − 2
x2 = y−1 − 2
x2 + 2 = y−1 = g−1 (x)
x = (y−1 )2 + 2
and
x − 2 = (y−1 )2
√
± x − 2 = y−1 = f −1 (x)
Notice the ± sign in front of the square root for f −1 . That means that g−1 is
√
√
x − 2 and − x − 2.
Therefore, f −1 is not really a function because it fails the vertical line test. However, if you were to take each
part separately, individually, they are functions. You can also think about reflecting f (x) over y = x. It would be a
parabola on its side, which is not a function.
The inverse of g would then be only half of the parabola, see below. Despite the restrictions on the domains, f and
g are inverses of each other.
525
3.4. Inverse Functions
www.ck12.org
Alternate Method: The second, and easier, way to determine if two functions are inverses of each other is to use
composition. If f ◦ g = g ◦ f = x, then f and g are inverses of each other. Think about it; if everything cancels
out and all that remains is x, each operation within the functions are opposites, making the functions “opposites” or
inverses of each other.
q
(x2 + 2) − 2
√
= x2
f ◦g =
g◦ f =
√
2
x−2 +2
= x−2+2
and
=x
=x
Because f ◦ g = g ◦ f = x, f and g are inverses of each other. Both f ◦ g = x and g ◦ f = x in order for f and g to be
inverses of each other.
Examples
Example 1
2
Earlier, you were asked to find the inverse of the function d = p 3 .
2
In the function d = p 3 , d is the equivalent of y and p is the equivalent of x. So rewrite the equation and follow the
step-by-step process illustrated above.
2
y = x3 ,
Switch the x and y. Change y to y−1 for the inverse.
2
x = (y−1 ) 3
Solve for y0 .
2
x = (y−1 ) 3
3
2 3
x 2 = (y−1 ) 3 · 2
3
x 2 = y−1
526
www.ck12.org
Chapter 3. Unit 3
3
Now replace y and x with d and p. The inverse d is p 2 .
Example 2
Find the inverse of g(x) = − 34 x + 12 algebraically.
3
y = − x + 12
4
3
x = − y−1 + 12
4
3
x − 12 = − y−1
4
4
− (x − 12) = y−1
3
4
g−1 (x) = − x + 16
3
Example 3
Find the inverse of f (x) = 2x3 + 5 algebraically. Is the inverse a function?
y = 2x3 + 5
x = 2(y−1 )3 + 5
x − 5 = 2(y−1 )3
x−5
= (y−1 )3
2
r
3 x−5
−1
f (x) =
2
Yes, f −1 is a function. Plot in your graphing calculator if you are unsure and see if it passes the vertical line test.
Example 4
Determine if h(x) = 4x4 − 7 and j(x) =
1
4
√
4
x − 7 are inverses of each other using compositions.
First, find h( j(x)).
4
1√
4
h( j(x)) = 4
x+7 −7
4
4
1
= 4·
x+7−7
4
1
= x
64
Because h( j(x)) 6= x, we know that h and j are not inverses of each other. Therefore, there is no point to find j(h(x)).
527
3.4. Inverse Functions
www.ck12.org
Review
Write the inverses of the following functions. State whether or not the inverse is a function.
1. (2, 3), (−4, 8), (−5, 9), (1, 1)
2. (9, −6), (8, −5), (7, 3), (4, 3)
Find the inverses of the following functions algebraically. Note any restrictions to the domain of the inverse
functions.
3.
4.
5.
6.
7.
8.
f (x) = 6x − 9
1
f (x) = 4x+3
√
f (x) = x + 7
f (x) = x2 + 5
f (x) = x√3 − 11
5
f (x) = x + 16
Determine whether f and g are inverses of each other by checking to see whether finding f ◦ g = x or g ◦ f = x. You
do not need to show both.
9.
10.
11.
12.
f (x) = 23 x − 14 and g(x) = 32 x + 21
f (x) = x+5
8 and g(x) = 8x + 5
√
3
3
f (x) = 3x − 7 and g(x) = x3 − 7
x
9x
f (x) = x−9
, x 6= 9 and g(x) = x−1
Find the inverses of the following functions algebraically. Note any restrictions to the domain of the inverse
functions. These problems are a little trickier as you will need to factor out the y variable to solve. Use the example
below as a guide.
f (x) =
3x+13
2x−11
Example:
• x = 3y+13
2y−11 First, switch x and y
• 2xy − 11x = 3y + 13 Multiply both sides by 2y − 11 to eliminate the fraction
• 2xy − 3y = 11x + 13 Now rearrange the terms to get both terms with y in them on one side and everything else
on the other side
• y(2x − 3) = 11x + 13 Factor out the y
• y = 11x+13
2x−3 Finally, Divide both sides by 2x − 3 to isolate y.
So, the inverse of f (x) =
13. f (x) =
14. f (x) =
x+7
x ,x
x
x−8 , x
3x+13
2x−11 , x
6=
11
2
is f −1 (x) =
11x+13
2x−3 , x
6= 32 .
6 0
=
6= 8
Multi-step problem
15. In many countries, the temperature is measured in degrees Celsius. In the US we typically use degrees
Fahrenheit. For travelers, it is helpful to be able to convert from one unit of measure to another. The following
problem will help you learn to do this using an inverse function.
528
www.ck12.org
Chapter 3. Unit 3
a. The temperature at which water freezes will give us one point on a line in which x represents the degrees
in Celsius and y represents the degrees in Fahrenheit. Water freezes at 0 degrees Celsius and 32 degrees
Fahrenheit so the first point is (0, 32). The temperature at which water boils gives us the second point
(100, 212), because water boils at 100 degrees Celsius or 212 degrees Fahrenheit. Use this information
to show that the equation to convert from Celsius to Fahrenheit is y = 95 x + 32 or F = 95 C + 32.
b. Find the inverse of the equation above by solving for C to derive a formula that will allow us to convert
from Fahrenheit to Celsius.
c. Show that your inverse is correct by showing that the composition of the two functions simplifies to
either F or C (depending on which one you put into the other.)
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 7.11.
529
3.5. Graphs of Logs as the Inverse of Exponential Functions
www.ck12.org
3.5 Graphs of Logs as the Inverse of
Exponential Functions
Learning Objectives
Here you will explore the graphs of logarithmic functions, and compare them to the graphs of exponential functions.
You will also see how the common rules for graphing using transformations apply to log graphs.
Graphs of Logs as the Inverse of Exponential Functions
This lesson explores the statement that the log functions are identified as the inverse of exponential functions. These
functions will be compared visually through the graphs of both the logarithmic and exponential functions.
Warm-Up
An inverse of a function f (x) will undo the the operations performed in that function. Later in this section you will
see how logarithms are inverses of exponential functions.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/217487
Example 1: Reflecting Points Across a Diagonal Line
Graph the line y = x on a coordinate plane, as well as the following points: (1,0), (3,0), and (3,2). Now reflect these
three coordinates over the line, and draw/label the new reflected points.
• Solution:
The graph below shows the original points in red and the reflected points in purple. Notice that the original
coordinates reflected over the line y = x result in new coordinates that are: 1) equidistant from the line they
are reflect over, and 2) inverse values from one another; for example (1,0) is the inverse of (0,1).
Work it Out 1: Reflecting an Exponential Function Across a Diagonal Line
Start with the function f (x) = 3x .
530
www.ck12.org
Chapter 3. Unit 3
FIGURE 3.1
Points reflected across the line y=x
a. Create a table of values for x and f (x) where x = -3, -2, -1, 0, 1, 2, and 3
b. Graph the coordinates and connect them to create a curve
c. Determine the location for each coordinate reflected over the line y = x, and connect them with a curve
Discussion
Some values from the first set of coordinates are graphed below (in green), along with the line y = x (in
purple). Recall from Example 1 that when an (x, y) coordinate is reflected over y = x, the inverse coordinate
is (y, x). For example, the inverse of the green point (0, 1) is (1, 0), which is represented with the blue point.
If blue coordinates are the inverses of green coordinates, does this mean the equation for blue line will be the
inverse of the equation for the green line?
Work it Out 2: Graphing Logarithmic Functions
1 1 1
For the function f (x) = log3 x, find f (x) for the following x-values: 27
, 9 , 3 , 1, 2, 3. Then graph these points
and reflect them and their graph across the line y = x. Remember from previous work with logs that if logb (a) =
x, then bx = a.
Discussion
The table below contains some of the identified values for the function f (x) = log3 x
TABLE 3.8:
531
3.5. Graphs of Logs as the Inverse of Exponential Functions
TABLE 3.8: (continued)
x
1
27
1
9
1
3
1
2
3
532
f(x)
-3
0
.631
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
FIGURE 3.2
The blue dot coordinates are the inverse
of the green dot coordinates
Below are graphs of the function f (x) = log3 x and the same function reflected across the line y = x. Which color
curve below is the logarithmic function, and which is its inverse, the exponential function? Explain your
reasoning.
Comparing this graph to the graph from Active Learning 2 above reveals an important relationship between
the functions f (x) = 3x and f (x) = log3 x: they are inverses.
It is often helpful to visualize an exponential function and its inverse on the same graph, for example f (x) =
ax and f (x) = loga x.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/217484
533
3.5. Graphs of Logs as the Inverse of Exponential Functions
www.ck12.org
FIGURE 3.3
Work it Out 3
An equation for an exponential function is y = A(xh ) + k.
a. What is the equation for its inverse, logarithmic function?
b. What is the effect on both the exponential graph as k becomes smaller/larger? Given this, what do you predict
will happen to the inverse logarithmic equation and graph? Why?
c. What is the effect on both the exponential graph as h becomes smaller/larger? Given this, what do you predict
will happen to the inverse logarithmic equation and graph? Why?
Discussion
The log equation would be y = logA (x − k) + h. Notice that k is being subtracted in the exponential equation.
As k grows larger, the graph will shift up on the y-axis, and as k grows smaller, the graph will shift down.
Because the logarithmic graph is a reflection over the line y = x, it will have an inverse reaction as k changes.
Note that this doesn’t mean it will have an opposite reaction, which would mean the log function would shift
up as the exponential function shifted down. Instead, the log function will shift left as the exponential function
shifts down, and vice versa. Use this information to predict what happens as the value of h changes.
534
www.ck12.org
Chapter 3. Unit 3
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/217482
The table below shows the x and y values of the points on an exponential curve. Identify the corresponding
coordinates of the points that would appear on the logarithmic curve. Can you identify the function?
TABLE 3.9:
Point on exponential curve
(-3, 1/8)
(-2, 1/4)
(-1, 1/2)
(0, 1)
(1, 2)
(2, 4)
(3, 8)
7. Sketch the graphs of the functions f (x) =
where they intersect.
Corresponding point on logarithmic curve
(1/8, -3)
#1. (__ , __)
#2. (__ , __)
#3. (__ , __)
#4. (__ , __)
#5. (__ , __)
#6. (__ , __)
1 x
3
and g(x) = log 1 (x)on the same axes. Determine the coordinate
3
For questions 8 and 9, graph the logarithmic function and related exponential function on the same axes.
8. y = log3 (x)
9. y = log5 (x − 2) + 6
10. Explain how changing h impacts both the graph of an exponential function and the logarithmic inverse. Why
do the graphs change in these ways?
535
3.6. Defining Logarithms
www.ck12.org
3.6 Defining Logarithms
Learning Objectives
Here you’ll define and learn how to use logarithms.
You go a concert and you want to know how loud it is in decibels. The decibel level of a sound is found by first
assigning an intensity I0 to a very soft sound, or the threshold. The decibel level can then be measured with the
formula d = 10 · log I0I where I is the intensity of the sound. If the intensity of the concert is 1,000,000,000(I0), what
is its decibel level?
Logarithm
You can probably guess that x = 3 in 2x = 8 and x = 4 in 2x = 16. But, what is x if 2x = 12? Until now, we did
not have an inverse to an exponential function. But, because we have variables in the exponent, we need a way
to get them out of the exponent. We will now introduce the logarithm. A logarithm is defined as the inverse of
an exponential function. It is written logb a = x such that bx = a. Therefore, if 52 = 25 (exponential form), then
log5 25 = 2 (logarithmic form).
There are two special logarithms, or logs. One has base 10, and rather that writing log10 , we just write log. The
other is the natural log, the inverse of the natural number. The natural log has base e and is written ln. This is the
only log that is not written using log.
Let’s rewrite log3 27 = 3 in exponential form.
Use the definition above, also called the “key”.
logb a = x ↔ bx = a
log3 27 = 3 ↔ 33 = 27
Now, let’s find the following.
a. log 1000
log 1000 = x ⇒ 10x = 1000, x = 3.
1
b. log7 49
1
log7 49
= x ⇒ 7x =
1
49 , x
= −2.
c. log 1 (−8)
2
log 1 (−8) = x ⇒
2
negative.
1 x
2
= −8. There is no solution. A positive number when raised to any power will never be
Using the key, we can rearrange all of these in terms of exponents.
There are two special logarithms that you may encounter while writing them into exponential form.
536
www.ck12.org
Chapter 3. Unit 3
The first is logb 1 = 0, because b0 = 1. The second is logb b = 1 because b1 = b · b can be any number except 1.
Finally, let’s use a calculator to find the following logarithms and round our answers to the nearest hundredth.
a. ln 7
Locate the LN button on your calculator. Depending on the brand, you may have to input the number first. For a
TI-83 or 84, press LN, followed by the 7 and ENTER. The answer is 1.95.
b. log 35
The LOG button on the calculator is base 10. Press LOG, 35, ENTER. The answer is 1.54.
c. log5 226
To use the calculator for a base other than 10 or the natural log, you need to use the change of base formula.
Change of Base Formula: loga x =
logb x
logb a ,
such that x, a, and b > 0 and a and b 6= 1.
So, to use this for a calculator, you can use either LN or LOG.
log5 226 =
log 226
log 5
or
ln 226
ln 5
≈ 3.37
In the TI-83 or 84, the keystrokes would be LOG(226)/LOG(5), ENTER.
Examples
Example 1
Earlier, you were asked to find the decibel level of a concert if the intensity is 1,000,000,000(I0).
Plug the given values into the equation d = 10 · log I0I and solve for d.
1, 000, 000, 000(I0)
I0
d = 10 · log 1, 000, 000, 000
d = 10 · log
d = 10 · 9 = 90
Therefore, the decibel level of the concert is 90.
Example 2
Write 62 = 36 in logarithmic form.
Using the key, we have: 62 = 36 → log6 36 = 2.
For Examples 3-5, evaluate the expressions without a calculator.
Change each logarithm into exponential form and solve for x.
Example 3
log 1 16
2
log 1 16 →
2
24 = 16, so
1 x
2
= 16. x must be negative because the answer is not a fraction, like the base.
1 −4
2
= 16. Therefore, log 1 16 = −4.
2
537
3.6. Defining Logarithms
www.ck12.org
Example 4
log 100
log 100 → 10x = 100. x = 2, therefore, log 100 = 2.
Example 5
log64 81
√
1
log64 81 → 64x = 81 . First, 64 = 8, so 64 2 = 8. To make this a fraction, we need to make the power negative.
1
64− 2 = 81 , therefore log64 18 = − 12 .
Example 6
Use the change of base formula to evaluate log8 79 in a calculator.
Rewriting log8 97 using the change of base formula, we have:
−0.12.
log 79
log 8 .
Plugging it into a calculator, we get
log( 97 )
log 8
≈
Review
Convert the following exponential equations to logarithmic equations.
1. 3x = 5
2. ax = b
3. 4(5x ) = 10
Convert the following logarithmic equations to exponential equations.
4. log2 32 = x
5. log 1 x = −2
3
6. loga y = b
Evaluate the following logarithmic expressions without a calculator.
7. log5 25
8. log 1 27
3
1
9. log 10
10. log2 64
Evaluate the following logarithmic expressions using a calculator. You may need to use the Change of Base Formula
for some problems.
11.
12.
13.
14.
15.
538
log 72
ln 8
log2 12
log3 9
log11 32
www.ck12.org
Chapter 3. Unit 3
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.5.
539
3.7. Graphs of Logarithmic Functions
www.ck12.org
3.7 Graphs of Logarithmic Functions
Learning Objectives
Here you will explore the graphs of logarithmic functions, and compare them to the graphs of exponential functions.
You will also see how the common rules for graphing using transformations apply to log graphs.
Logarithmic functions are in many ways similar to exponential functions, and can be applied in similar situations.
Log functions can be used to model radioactive decay (to estimate the age of fossils, for instance), or estimate
the discharge of electricity from a capacitor. Logarithmic functions can be graphed to allow quick and reasonably
accurate estimation of the useable area of a WiFi network, or estimate the minimum length of a filtering pipe in order
to have a specific purity of kerosene.
As you work with the graphs of log functions in this lesson, you will notice that they bear a resemblance to another
family of graphs which you have explored already. At the end of the lesson, we will revisit this fact.
Can you identify what other family the log functions resemble, and explain in your own words why that is the case
before the review?
Graphing Logarithmic Functions
In a previous lesson, log functions were identified as the inverses of exponential functions, in this lesson we explore
that fact visually through the graphs of logarithmic functions.
As you can see below, because the function f (x) = log2 x is the inverse of the function g(x) = 2x , the graphs of these
functions are reflections over the line y = x.
We can verify that the functions are inverses by looking at the graph. For example, the graph of g(x) = 2x contains
the point (1, 2), while the graph of f (x) = log2 x contains the point (2, 1).
540
www.ck12.org
Chapter 3. Unit 3
Also, note that while that the graph of g(x) = 2x is asymptotic to the x-axis, the graph of f (x) = log2 x is asymptotic to
the y-axis. This behavior of the graphs gives us a visual interpretation of the restricted range of g and the restricted
domain of f.
When graphing log functions, it is important to consider x- values across the domain of the function. In particular,
we should look at the behavior of the graph as it gets closer and closer to the asymptote. Consider f (x) = log2 x for
values of x between 0 and 1.
If x = 1/2, then f (1/2) = log2 (1/2) = -1 because 2−1 = 1/2
If x = 1/4, then f (1/4) = log2 (1/4) = -2 because 2−2 = 1/4
If x = 1/8, then f (1/8) = log2 (1/8) = -3 because 2−3 = 1/8
From these values you can see that if we choose x values that are closer and closer to 0, the y values decrease
(heading towards −∞!). In terms of the graph, these values show us that the graph gets closer and closer to the
y-axis. Formally we say that the vertical asymptote of the graph is x = 0.
Graphing Logarithmic Functions Using Transformations
Consider again the log function f (x) = log2 x. The table below summarizes how we can use the graph of this function
to graph other related functions.
TABLE 3.10:
Equation
g(x) = log2 (x - a), for a > 0
g(x) = log2 (x+a) for a > 0
g(x) = log2 (x) + a for a > 0
g(x) = log2 (x) - a for a > 0
g(x) = alog2 (x) for a > 0
g(x) = -alog2 (x) , for a > 0
g(x) = log2 (-x)
Relationship to f (x) = log2 x
Obtain a graph of g by shifting the
graph of f a units to the right.
Obtain a graph of g by shifting the
graph of f a units to the left.
Obtain a graph of g by shifting the
graph of f up a units.
Obtain a graph of g by shifting the
graph of f down a units.
Obtain a graph of g by vertically
stretching the graph of f by a factor
of a.
Obtain a graph of g by vertically
stretching the graph of f by a factor
of a, and by reflecting the graph
over the x-axis.
Obtain a graph of g by reflecting the
graph of f over the y-axis.
Domain
x>a
x > -a
x>0
x>0
x>0
x>0
x<0
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187462
541
3.7. Graphs of Logarithmic Functions
www.ck12.org
Examples
Example 1
Earlier, you were asked a question about a function family.
Can you identify what other family the log functions resemble, and explain in your own words why that is the case?
By now, I am sure you are quite familiar with the answer, it was repeated several times throughout the lesson.
Logarithmic function graphs are the inverses of exponential function graphs, because the very definition of a
logarithmic function it as the inverse of an exponential function!
Example 2
Graph the function f (x) = log4 x and state the domain and range of the function.
The function f (x) = log4 x is the inverse of the function g(x) = 4x . We can sketch a graph of f (x) by evaluating the
function for several values of x, or by reflecting the graph of g over the line y = x.
If we choose to plot points, it is helpful to organize the points in a table:
TABLE 3.11:
x
1/4
1
4
16
y = log4 x
0
1
2
The graph is asymptotic to the y-axis, so the domain of f is the set of all real numbers that are greater than 0. We
can write this as a set:{x ∈ R|x > 0} . While the graph might look as if it has a horizontal asymptote, it does in fact
continue to rise. The range is R.
542
www.ck12.org
Chapter 3. Unit 3
Example 3
Graph the following functions: f (x) = log2 (x), g(x) = log2 (x) + 3, and h(x) = log2 (x + 3).
The graph below shows these three functions together:
Notice that the location of the 3 in the equation makes a difference! When the 3 is added to log2 x , the shift is
vertical. When the 3 is added to the x, the shift is horizontal. It is also important to remember that adding 3 to the x
is a horizontal shift to the left. This makes sense if you consider the function value when x = -3:
h(-3) = log2 (-3 + 3) = log2 0 = undefined
This is the vertical asymptote!
Note that in order to graph these functions, we evaluated them by investigating specific values of x. If we want to
know what the x value is for a particular y value, we need to solve a logarithmic equation.
Example 4
Graph the function y = log2 (x − 3).
Start by making a table:
TABLE 3.12:
x
3
4
5
6
7
8
y
−∞
0
1
1.585
2
2.32
Since 20 = 1 that means x − 3 = 1 and x = 4. That means that when y = 0, x = 4.
Since 21 = 2 that means x − 3 = 2 and x = 5. That means that when y = 1, x = 5.
The graph looks like:
543
3.7. Graphs of Logarithmic Functions
www.ck12.org
Example 5
Which of the following functions is graphed in the image below? a) y = log4 x, b) y = −log4 x, or c) y = log4 − x?
All three functions are varieties of log4 x, but the image shows the function reflected across the y axis, therefore c)
y = log4 − x is correct.
Example 6
Graph the function y = log3 x.
To graph y = log3 x we will start with a table of values:
TABLE 3.13:
x
3
9
27
81
y
1
2
3
4
Plotting those points and drawing a smooth curve between them gives:
544
www.ck12.org
Chapter 3. Unit 3
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187463
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187456
Review
Identify the domain and range, then sketch the graph.
1.
2.
3.
4.
5.
y = log6 (x − 1) − 5
y = log5 (x − 1) + 3
y = log4 (3x + 11) − 5
y = log6 (3x + 14) + 1
y = log5 (2x + 2) + 5
Look at the graphs below and identify the function that the graph represents from the functions listed below.
a)
b)
c)
d)
e)
f (x) = −log10 x
f (x) = log10 x
f (x) = −log10 (−2x)
f (x) = log10 (−3x)
f (x) = log10 x + 3
545
3.7. Graphs of Logarithmic Functions
6.
7.
8.
546
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
9.
10.
Graph the following logarithmic functions.
11.
12.
13.
14.
y = log10 (2x)
y = log1/2 (x + 2)
y = log3 (2x + 2)
- 19. The table below shows the x and y values of the points on an exponential curve. Switch them and identify
the corresponding coordinates of the points that would appear on the logarithmic curve. Can you identify the
function?
TABLE 3.14:
Point on exponential curve
(-3, 1/8)
(-2, 1/4)
(-1, 1/2)
(0, 1)
(1, 2)
(2, 4)
(3, 8)
Corresponding point on logarithmic curve
(1/8, -3)
14. (__ , __)
15. (__ , __)
16. (__ , __)
17. (__ , __)
18. (__ , __)
19. (__ , __)
Graph logarithmic functions, using the inverse of the related exponential function. Then graph the pair of functions
on the same axes.
20. y = log3 x
21. y = log5 x
547
3.7. Graphs of Logarithmic Functions
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.6.
548
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.8 Graphing Logarithmic Functions
Learning Objectives
Here you’ll learn how to graph a logarithmic function by hand and using a calculator.
Your math homework assignment is to find out which quadrants the graph of the function f (x) = 4 ln(x + 3) falls
in. On the way home, your best friend tells you, "This is the easiest homework assignment ever! All logarithmic
functions fall in Quadrants I and IV." You’re not so sure, so you go home and graph the function as instructed. Your
graph falls in Quadrant I as your friend thought, but instead of Quadrant IV, it also falls in Quadrants II and III.
Which one of you is correct?
Graphing Logarithmic Functions
Now that we are more comfortable with using these functions as inverses, let’s use this idea to graph a logarithmic
function. Recall that functions are inverses of each other when they are mirror images over the line y = x. Therefore,
if we reflect y = bx over y = x, then we will get the graph of y = logb x.
549
3.8. Graphing Logarithmic Functions
www.ck12.org
Recall that an exponential function has a horizontal asymptote. Because the logarithm is its inverse, it will have a
vertical asymptote. The general form of a logarithmic function is f (x) = a logb (x − h) + k and the vertical asymptote
is x = h. The domain is x > h and the range is all real numbers. Lastly, if b > 1, the graph moves up to the right. If
0 < b < 1, the graph moves down to the right.
Let’s graph y = log3 (x − 4) and state the domain and range.
FIGURE 3.4
To graph a logarithmic function without a calculator, start by drawing the vertical asymptote, at x = 4. We know the
graph is going to have the general shape of the first function above. Plot a few points, such as (5, 0), (7, 1), and (13,
2) and connect.
The domain is x > 4 and the range is all real numbers.
Now, let’s determine if (16, 1) is on y = log(x − 6).
550
www.ck12.org
Chapter 3. Unit 3
Plug in the point to the equation to see if it holds true.
1 = log(16 − 6)
1 = log 10
1=1
Yes, this is true, so (16, 1) is on the graph.
Finally, let’s graph f (x) = 2 ln(x + 1).
To graph a natural log, we can use a graphing calculator. Press Y = and enter in the function, Y = 2 ln(x + 1),
GRAPH.
FIGURE 3.5
Examples
Example 1
Earlier, you were asked to determine if your friend was correct.
The vertical asymptote of the function f (x) = 4 ln(x + 3) is x = −3. Since x will approach −3 but never quite reach
it, x can assume some negative values. Hence, the function will fall in Quadrants II and III. Therefore, you are
correct and your friend is wrong.
Example 2
Graph y = log 1 x + 2 in an appropriate window.
4
First, there is a vertical asymptote at x = 0. Now, determine a few easy points, points where the log is easy to find;
such as (1, 2), (4, 1), (8, 0.5), and (16, 0).
To graph a logarithmic function using a TI-83/84, enter the function into Y = and use the Change of Base Formula:
logb x
loga x = log
. The keystrokes would be:
ba
Y=
log(x)
log( 14 )
+ 2, GRAPH
551
3.8. Graphing Logarithmic Functions
www.ck12.org
FIGURE 3.6
To see a table of values, press 2nd → GRAPH.
Example 3
Graph y = − log x using a graphing calculator. Find the domain and range.
The keystrokes are Y = − log(x), GRAPH.
FIGURE 3.7
The domain is x > 0 and the range is all real numbers.
Example 4
Is (-2, 1) on the graph of f (x) = log 1 (x + 4)?
2
Plug (-2, 1) into f (x) = log 1 (x + 4) to see if the equation holds true.
2
552
www.ck12.org
Chapter 3. Unit 3
1 = log 1 (−2 + 4)
2
1 = log 1 2
2
1 6= −1
Therefore, (-2, 1) is not on the graph. However, (-2, -1) is.
Review
Graph the following logarithmic functions without using a calculator. State the equation of the asymptote, the
domain and the range of each function.
1.
2.
3.
4.
5.
6.
y = log5 x
y = log2 (x + 1)
y = log(x) − 4
y = log 1 (x − 1) + 3
3
y = − log 1 (x + 3) − 5
2
y = log4 (2 − x) + 2
Graph the following logarithmic functions using your graphing calculator.
7.
8.
9.
10.
11.
12.
13.
14.
15.
y = ln(x + 6) − 1
y = − ln(x − 1) + 2
y = log(1 − x) + 3
y = log(x + 2) − 4
How would you graph y = log4 x on the graphing calculator? Graph the function.
Graph y = log 3 x on the graphing calculator.
4
Is (3, 8) on the graph of y = log3 (2x − 3) + 7?
Is (9, -2) on the graph of y = log 1 (x − 5) ?
4
Is (4, 5) on the graph of y = 5 log2 (8 − x)?
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.7.
553
3.9. Change of Base
www.ck12.org
3.9 Change of Base
Here you will extend your knowledge of log properties to a simple way to change the base of a logarithm. While it
is possible to change bases by always going back to exponential form, it is more efficient to find out how to change
the base of logarithms in general. Since there are only base e and base 10 logarithms on most calculators, how
would you evaluate an expression like log3 12?
Changing the Base of Logarithms
The change of base property states:
logb x =
loga x
loga b
You can derive this formula by converting logb x to exponential form and then taking the log base x of both
sides. This is shown below.
logb x = y
by = x
loga by = loga x
y loga b = loga x
loga x
y=
loga b
Therefore, logb x =
loga x
loga b .
If you were to evaluate log3 4 using your calculator, you may need to use the change of base formula since some
calculators only have base 10 or base e . The result would be:
log3 4 =
log10 4
log10 3
=
ln 4
ln 3
≈ 1.262
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/61000
Examples
Example 1
Earlier, you were asked how to use a calculator to evaluate an expression like log3 12. In order to evaluate an
expression like log3 12 you have some options on your calculator:
554
www.ck12.org
ln 12
ln 3
=
log 12
log 3
Chapter 3. Unit 3
≈ 2.26
Some graphing calculators also have another option. Press the MATH followed by the A buttons and enter log3 12.
Example 2
Prove the following log identity.
loga b =
1
logb a
loga b =
logx b
logx a
=
1
1
logb a
=
logx a
logx b
Example 3
Simplify to an exact result: (log4 5) · (log3 4) · (log5 81) · (log5 25)
log 5
log 4
4
2
log 4 log 3 log 5
· log
3 · log 5 · log 5 =
log 5
log 4
4 4·log 3 2·log 5
· log
log 3 · log 5 · log 5 = 4 · 2 = 8
Example 4
Evaluate: log2 48 − log4 36
log 48 log 36
−
log 2
log 4
log 48 log 62
=
−
log 2
log 22
log 48 2 · log 6
=
−
log 2
2 · log 2
log 48 − log 6
=
log 2
log 48
6
=
log 2
log 8
=
log 2
log 23
=
log 2
3 · log 2
=
log 2
=3
log2 48 − log4 36 =
Example 5
Given log3 5 ≈ 1.465 find log25 27 without using a log button on the calculator.
log25 27 =
log 33
log 52
= 32 ·
1
5
( log
log 3 )
1
= 32 · log1 5 ≈ 32 · 1.465
= 1.0239
3
555
3.9. Change of Base
Review
Evaluate each expression by changing the base and using your calculator.
1. log6 15
2. log9 12
3. log5 25
Evaluate each expression.
4. log8 (log4 (log3 81))
5. log2 3 · log3 4 · log6 16 · log4 6
6. log 125 · log9 4 · log4 81 · log5 10
7. log5 (5log5 125 )
8. log(log6 (log2 64))
9. 10log100 9
10. (log4 x)(logx 16)
11. log49 495
12. 3 log24 248
13. 4log2 3
Prove the following properties of logarithms.
14. (loga b)(logb c) = loga c
15. (loga b)(logb c)(logc d) = loga d
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.5.
556
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.10 pH
Learning Objectives
• Define pH.
• List pH values of common materials.
Why is grapefruit juice acidic?
Grapefruit juice has a pH of somewhere between 2.9-3.3, depending on the specific product. Excessive exposure to
this juice can cause erosion of tooth enamel and can lead to tooth damage. The acids in grapefruit juice are carbonbased, with citric acid being one of the major constituents. This compound has three ionizable hydrogens on each
molecule which contribute to the relatively low pH of the juice. Another component of grape juice is malic acid,
containing two ionizable hydrogens per molecule.
The pH Scale
Expressing the acidity of a solution by using the molarity of the hydrogen ion is cumbersome because the quantities
are generally very small. Danish scientist Søren Sørenson (1868-1939) proposed an easier system for indicating
the concentration of H+ called the pH scale. The letters pH stand for the power of the hydrogen ion. The pH of a
solution is the negative logarithm of the hydrogen-ion concentration.
pH = -log[H+ ]
In pure water or a neutral solution the [H+ ] = 1.0 × 10−7 M. Substituting into the pH expression:
pH = -log[1.0 × 10−7 ] = -(-7.00) = 7.00
557
3.10. pH
www.ck12.org
The pH of pure water or any neutral solution is thus 7.00. For recording purposes, the numbers to the right of the
decimal point in the pH value are the significant figures. Since 1.0 × 10−7 has two significant figures, the pH can be
reported as 7.00.
A logarithmic scale condenses the range of acidity to numbers that are easy to use. Consider a solution with
[H+ ] = 1.0 × 10−4 M. That is a hydrogen-ion concentration that is 1000 times higher than the concentration in
pure water. The pH of such a solution is 4.00, a difference of just 3 pH units. Notice that when the [H+ ] is written in
scientific notation and the coefficient is 1, the pH is simply the exponent with the sign changed. The pH of a solution
with the [H+ ] = 1 × 10−2 M is 2 and the pH of a solution with the [H+ ] = 1 × 10−10 M is 10.
As we saw earlier, a solution with the [H+ ] higher than 1.0 × 10−7 is acidic, while a solution with the [H+ ] lower
than 1.0 × 10−7 is basic. Consequently, solutions whose pH is less than 7 are acidic, while those with a pH higher
than 7 are basic. Figure 3.8 illustrates this relationship, along with some examples of various solutions.
FIGURE 3.8
The pH values for several common materials.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/185620
558
www.ck12.org
Chapter 3. Unit 3
Summary
• The concept of pH is defined.
• pH values for several common materials are listed.
Review
1. What is one value of using pH instead of molar concentrations?
2. Is coffee an acidic or a basic substance?
3. If a material has a pH of 9.3, is it acidic or basic?
Vocabulary
• pH: The negative logarithm of the hydrogen-ion concentration.
559
3.11. Intensity and Loudness of Sound
www.ck12.org
3.11 Intensity and Loudness of Sound
Learning Objectives
• Define intensity of sound and relate it to loudness.
• Compare decibel levels of different sounds.
• Identify factors that affect sound intensity.
A friend whispers to you in a voice so soft that she has to lean very close so you can hear what she’s saying. Later
that day, your friend shouts to you from across the gymnasium. Now her voice is loud enough for you to hear her
clearly even though she’s several meters away. Obviously, sounds can vary in loudness.
It’s All About Energy
Loudness refers to how loud or soft a sound seems to a listener. The loudness of sound is determined, in turn, by the
intensity of the sound waves. Intensity is a measure of the amount of energy in sound waves. The unit of intensity
is the decibel (dB).
Decibel Levels
The Figure 3.9 shows decibel levels of several different sounds. As decibel levels get higher, sound waves have
greater intensity and sounds are louder. For every 10-decibel increase in the intensity of sound, loudness is 10 times
greater. Therefore, a 30-decibel “quiet” room is 10 times louder than a 20-decibel whisper, and a 40-decibel light
560
www.ck12.org
Chapter 3. Unit 3
rainfall is 100 times louder than the whisper. High-decibel sounds are dangerous. They can damage the ears and
cause loss of hearing.
FIGURE 3.9
Q: How much louder than a 20-decibel whisper is the 60-decibel sound of a vacuum cleaner?
A: The vacuum cleaner is 10,000 times louder than the whisper!
Amplitude and Distance
The intensity of sound waves determines the loudness of sounds, but what determines intensity? Intensity results
from two factors: the amplitude of the sound waves and how far they have traveled from the source of the sound.
• Amplitude is a measure of the size of sound waves. It depends on the amount of energy that started the waves.
Greater amplitude waves have more energy and greater intensity, so they sound louder.
• As sound waves travel farther from their source, the more spread out their energy becomes. You can see how
this works in the Figure 3.10. As distance from the sound source increases, the area covered by the sound
waves increases. The same amount of energy is spread over a greater area, so the intensity and loudness of the
sound is less. This explains why even loud sounds fade away as you move farther from the source.
Q: Why can low-amplitude sounds like whispers be heard only over short distances?
A: The sound waves already have so little energy that spreading them out over a wider area quickly reduces their
intensity below the level of hearing.
The spectrum of sound varies for different musical instruments, which is why they all sound different playing the
same musical note. Observe the Amplitude vs Frequency graph, known as the Fourier Transform, on the top right
graph in the Pan Flute simulation below to learn more:
561
3.11. Intensity and Loudness of Sound
www.ck12.org
FIGURE 3.10
This diagram represents just a small section of the total area of sound waves
spreading out from a source.
Sound
waves actually travel away from the
source in all directions.
SIMULATION
Learn about wave harmonics in a column of air by looking
closely at the sound produced by a pan flute.
URL:
http://www.ck12.org/physics/wave-
speed/simulationint/Pan-Flute
Further Reading
• Hearing and the Ear
• Hearing Loss
Summary
• Loudness refers to how loud or soft a sound seems to a listener. The loudness of sound is determined, in turn,
by the intensity, or amount of energy, in sound waves. The unit of intensity is the decibel (dB).
• As decibel levels get higher, sound waves have greater intensity and sounds are louder. For every 10-decibel
increase in the intensity of sound, loudness is 10 times greater.
• Intensity of sound results from two factors: the amplitude of the sound waves and how far they have traveled
from the source of the sound.
Review
1.
2.
3.
4.
562
Define loudness and intensity of sound. How are the two concepts related?
What is the unit of intensity of sound?
At what decibel level do sounds start to become harmful to the ears and hearing?
Relate amplitude and distance to the intensity and loudness of sound.
www.ck12.org
Chapter 3. Unit 3
Vocabulary
• Loudness: how loud or soft a sound seems to a listener.
• Intensity: amount of energy in sound waves, measured in decibel (dB).
563
3.12. Earthquake Magnitude Scales
www.ck12.org
3.12 Earthquake Magnitude Scales
Learning Objectives
• Describe how scientists express the size and intensity of an earthquake.
• Compare the Richter magnitude scale to the moment magnitude scale.
How do scientists measure earthquakes?
This 8.8 magnitude earthquake in Chile in 2010 caused over 500 deaths and thousands of injuries. Earthquakes and
the damage they cause can be measured in a few different ways. One way is to describe the damage. The other way
is to measure the energy of the quake.
Earthquake Intensity
The ways seismologists measure an earthquake have changed over the decades. Initially, they could only measure
what people felt and saw: the intensity. For this, they used the Mercalli scale.
Mercalli Scale
Early in the 20th century, earthquakes were described in terms of what people felt and the damage that was done to
buildings. The Mercalli intensity scale describes earthquake intensity.
564
www.ck12.org
Chapter 3. Unit 3
There are many problems with the Mercalli scale. The damage from an earthquake is affected by many things. The
type of ground a building sits on is very important to what happens to that building in a quake. Different people
experience an earthquake differently. Using this scale, comparisons between earthquakes were difficult to make. A
new scale was needed.
Earthquake Magnitude
For decades scientists have had equipment that can measure earthquake magnitude. The earthquake magnitude is
the energy released during the quake.
The Richter Magnitude Scale
Charles Richter developed the Richter magnitude scale in 1935. The Richter scale measures the magnitude of an
earthquake’s largest jolt of energy. This is determined by using the height of the waves recorded on a seismograph.
The Richter scale is logarithmic. The magnitudes jump from one level to the next. The height of the largest wave
increases 10 times with each level. So the height of the largest seismic wave of a magnitude 5 quake is 10 times that
of a magnitude 4 quake. A magnitude 5 is 100 times that of a magnitude 3 quake. With each level, thirty times more
energy is released. A difference of two levels on the Richter scale equals 900 times more released energy.
The Richter scale has limitations. A single sharp jolt measures higher on the Richter scale than a very long intense
earthquake. Yet, this is misleading because the longer quake releases more energy. Earthquakes that release more
energy are likely to do more damage. As a result, another scale was needed.
The Moment Magnitude Scale
The moment magnitude scale is the favored method of measuring earthquake magnitudes. It measures the total
energy released by an earthquake. Moment magnitude is calculated by two things. One is the length of the fault
break. The other is the distance the ground moves along the fault. Japan’s Tōhoku earthquake in 2011 had a
magnitude of 9.0 (Figure 3.11).
FIGURE 3.11
Earthquake and tsunami damage in Japan, 2011. The Tōhoku earthquake
had a magnitude of 9.0.
Summary
• Mercalli Intensity Scale depends on many factors other than the amount of energy released in the earthquake.
These factors include the type of rock in the region. They also include the quality of the structures built in the
area.
• The Richter scale is a logarithmic scale that measures the largest jolt of energy released by an earthquake.
565
3.12. Earthquake Magnitude Scales
www.ck12.org
• The moment magnitude scale is a logarithmic scale that measures the total amount of energy released by an
earthquake.
Review
1. Under what circumstances might the Mercalli intensity scale be useful today? Why was it replaced by the
Richter and then the moment magnitude scales?
2. Why do scientists prefer the moment magnitude scale to the Richter scale?
3. In 2011 there was a 5.8 magnitude quake that struck Virginia and a 9.0 quake that struck Japan. How might a
5.8 quake rate on the Mercalli scale in Virginia? How might a 5.8 magnitude quake rate on the Mercalli scale
in Japan?
Explore More
Use the resource below to answer the questions that follow.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/116784
1.
2.
3.
4.
5.
566
What was the magnitude of the 2010 Haiti earthquake?
What does magnitude mean?
How much larger is each earthquake magnitude?
What is the difference between two magnitudes?
What was the largest earthquake ever measured?
www.ck12.org
Chapter 3. Unit 3
3.13 Applications
Learning objectives
• Work with the decibel system for measuring loudness of sound.
• Work with the Richter scale, which measures the magnitude of earthquakes.
• Work with pH values and concentrations of hydrogen ions.
Introduction
Because logarithms are related to exponential relationships, logarithms are useful for measuring phenomena that
involve very large numbers or very small numbers. In this lesson you will learn about three situations in which a
quantity is measured using logarithms. In each situation, a logarithm is used to simplify measurements of either very
small numbers or very large numbers. We begin with measuring the intensity of sound.
Intensity of sound
Sound intensity is measured using a logarithmic scale. The intensity of a sound wave is measured in Watts per square
meter, or W/m2 . Our hearing threshold (or the minimum intensity we can hear at a frequency of 1000 Hz), is 2.5 ×
1012 W/m2 . The intensity of sound is often measured using the decibel (dB) system. We can think of this system as
a function. The input of the function is the intensity of the sound, and the output is some number of decibels. The
decibel is a dimensionless unit; however, because decibels are used in common and scientific discussions of sound,
the values of the scale have become familiar to people.
We can calculate the decibel measure as follows:
Intensity level (dB) = 10log
h
intensity of sound in W /m2
i
.937×10−12W /m2
An intensity of .937 × 10−12 W/m2 corresponds to 0 decibels:
10log
h
.937×10−12W /m2
.937×10−12W /m2
i
= 10log1 = 10(0) = 0.
Note: The sound equivalent to 0 decibels is approximately the lowest sound that humans can hear. If the intensity is
ten times as large, the decibel level is 10:
10log
h
.937×10−11W /m2
.937×10−12W /m2
i
= 10log10 = 10(1) = 10
If the intensity is 100 times as large, the decibel level is 20, and if the intensity is 1000 times as large, the decibel level
is 30. (The scale is created this way in order to correspond to human hearing. We tend to underestimate intensity.)
The threshold for pain caused by sound is 1 W/m2 . This intensity corresponds to about 120 decibels:
567
3.13. Applications
10log
h
1W /m2
W /m2
.937×10−12
www.ck12.org
i
≈ 10 : 12 = 120
Many common phenomena are louder than this. For example, a jet can reach about 140 decibels, and concert can
reach about 150 decibels.
(Source: Ohanian, H.C. (1989) Physics. New York: W.W. Norton & Company.)
For ease of calculation, the equation is often simplified: .937 is rounded to 1:
TABLE 3.15:
Intensity level (dB)
= 10log
h
= 10log
h
intensity of sound in W /m2
i
1×10−12W /m2
intensity of sound in W /m2
i
10−12W /m2
In the example below we will use this simplified equation to answer a question about decibels. (In the review
exercises, you can also use this simplified equation).
Example 1: Verify that a sound of intensity 100 times that of a sound of 0 dB corresponds to 20 dB.
100×10−12
Solution: dB = 10log
= 10log(100) = 10(2) = 20.
10−12
Intensity and magnitude of earthquakes
An earthquake occurs when energy is released from within the earth, often caused by movement along fault lines. An
earthquake can be measured in terms of its intensity, or its magnitude. Intensity refers to the effect of the earthquake,
which depends on location with respect to the epicenter of the quake. Intensity and magnitude are not the same
thing.
As mentioned in lesson 3, the magnitude of an earthquake is measured using logarithms. In 1935, scientist Charles
Richter developed this scale in order to compare the “size” of earthquakes. You can think of Richter scale as a
function in which the input is the amplitude of a seismic wave, as measured by a seismograph, and the output is a
magnitude. However, there is more than one way to calculate the magnitude of an earthquake because earthquakes
produce two different kinds of waves that can be measured for amplitude. The calculations are further complicated
by the need for a correction factor, which is a function of the distance between the epicenter and the location of the
seismograph.
Given these complexities, seismologists may use different formulas, depending on the conditions of a specific
earthquake. This is done so that the measurement of the magnitude of a specific earthquake is consistent with
Richter’s original definition. Source: http://earthquake.usgs.gov/learning/topics/richter.php
Even without a specific formula, we can use the Richter scale to compare the size of earthquakes. For example, the
1906 San Francisco earthquake had a magnitude of about 7.7. The 1989 Loma Prieta earthquake had a magnitude
of about 6.9. (The epicenter of the quake was near Loma Prieta peak in the Santa Cruz mountains, south of San
Francisco.) Because the Richter scale is logarithmic, this means that the 1906 quake was six times as strong as the
1989 quake:
107.7
106.9
= 107.7−6.9 = 10.8 ≈ 6.3
This kind of calculation explains why magnitudes are reported using a whole number and a decimal. In fact, a
decimal difference makes a big difference in the size of the earthquake, as shown below and in the review exercises
568
www.ck12.org
Chapter 3. Unit 3
Example 2: An earthquake has a magnitude of 3.5. A second earthquake is 100 times as strong. What is the
magnitude of the second earthquake?
Solution: The second earthquake is 100 times as strong as the earthquake of magnitude 3.5. This means that if the
magnitude of the second earthquake is x, then:
10x
103.5
= 100
10x−3.5 = 100 = 102
x − 3.5 = 2
x = 5.5
So the magnitude of the second earthquake is 5.5.
The pH scale
If you have studied chemistry, you may have learned about acids and bases. An acid is a substance that produces
hydrogen ions when added to water. A hydrogen ion is a positively charged atom of hydrogen, written as H+ . A
base is a substance that produces hydroxide ions (OH − ) when added to water. Acids and bases play important roles
in everyday life, including within the human body. For example, our stomachs produce acids in order to breakdown
foods. However, for people who suffer from gastric reflux, acids travel up to and can damage the esophagus.
Substances that are bases are often used in cleaners, but a strong base is dangerous: it can burn your skin.
To measure the concentration of an acid or a base in a substance, we use the pH scale, which was invented in the
early 1900’s by a Danish scientist named Soren Sorenson. The pH of a substance depends on the concentration of
H+ , which is written with the symbol [H+ ].
pH = - log [H+ ]
(Note: concentration is usually measured in moles per liter. A mole is 6.02 × 1023 units. Here, it would be 6.02 ×
1023 hydrogen ions.)
For example, the concentration of H+ in stomach acid is about 1 × 10 −1 . So the pH of stomach acid is -log (10−1 )
= -(-1) = 1. The pH scale ranges from 0 to 14. A substance with a low pH is an acid. A substance with a high pH is
a base. A substance with a pH in the middle of the scale is considered to be neutral.
Example 3: The pH of ammonia is 11. What is the concentration of H+ ?
Solution: pH = - log [H+ ]. If we substitute 11 for pH we can solve for H+ :
11 = −log[H + ]
−11 = log[H + ]
+
10−11 = 10log[H ]
10−11 = H +
Lesson Summary
In this lesson we have looked at three examples of logarithmic scales. In the case of the decibel system, using a
logarithm has produced a simple way of categorizing the intensity of sound. The Richter scale allows us to compare
earthquakes. And, the pH scale allows us to categorize acids and bases. In each case, a logarithm helps us work with
large or small numbers, in order to more easily understand the quantities involved in certain real world phenomena.
569
3.13. Applications
www.ck12.org
Points to Consider
1. How are the decibel system and the Richter scale the same, and how are they different?
2. What other phenomena might be modeled using a logarithmic scale?
Review Questions
1. Verify that a sound of intensity 1000 times that of a sound of 0 dB corresponds to 30 dB.
2. Calculate the decibel level of a sound with intensity 10−8 W/m2 .
3. The 2004 Indian Ocean earthquake was recorded to have a magnitude of about 9.5. In 1960, an earthquake in
Chile was recorded to have a magnitude of 9.1. How much stronger was the 2004 Indian Ocean quake?
4. Two earthquakes of the same magnitude do not necessarily cause the same amount of destruction. How is that
possible?
5. The concentration of H+ in pure water is 1 × 10 −7 . What is the pH?
6. The pH of normal human blood is 7.4. What is the concentration of H+
Solutions:
1. dB = 10 log(1000) = 10(3) = 30 dB
2. 40 dB
3. 2.51 times
4. Different grounds and materials used to create buildings.
5. 7
6. 10−7.4 = H +
Vocabulary
Acid
An acid is a substance that produces hydrogen ions when added to water.
Amplitude
The amplitude of a wave is the distance from its highest (or lowest) point to its center.
Base
A base is a substance that produces hydroxide ions (OH − ) when added to water
Decibel
A decibel is a unitless measure of the intensity of sound.
Mole
6.02 × 1023 units of a substance.
570
www.ck12.org
Chapter 3. Unit 3
Seismograph
A seismograph is a device used to measure the amplitude of earthquakes.
571
3.14. Simplifying Radicals
www.ck12.org
3.14 Simplifying Radicals
Learning Objectives
Here you’ll learn how to simplify expressions containing radicals.
Simplifying Radicals
In algebra, you learned how to simplify radicals. Let’s review it here. Some key points to remember:
1. One way to simplify a radical is to factor out the perfect squares (see Example A).
2. When
√ adding
√ radicals, you can only combine radicals with the same number underneath it. For example,
2 5 + 3 6 cannot be combined, because 5 and 6 are not the same number (see Example B).
3. To multiply two radicals, multiply what is under the radicals and what is in front (see Example B).
4. To divide radicals, you need to simplify the denominator, which means multiplying the top and bottom of the
fraction by the radical in the denominator (see Example C).
What if you were asked to find the sum of
could add them?
√
√
32 and 3 8? How could you combine these two terms so that you
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/136644
Examples
Example 1
√
Simplify the radical: 75.
√
√
√
75 = 25 · 3 = 5 3
Example 2
√
√
Simplify the radicals: 2 5 + 3 80.
√
√
√
√
√
√
√
2 5 + 3 80 = 2 5 + 3( 16 · 5) = 2 5 + (3 · 4) 5 = 14 5
572
www.ck12.org
Chapter 3. Unit 3
Example 3
Simplify the radicals. (For each radical, find the square number(s) that are factors).
a.
√
50
√
√
√
50 = 25 · 2 = 5 2
b.
√
27
√
√
√
27 = 9 · 3 = 3 3
c.
√
272
√
√
√
272 = 16 · 17 = 4 17
Example 4
Simplify the radicals.
√
√
a. 2 10 + 160
Simplify
√
√
√
√
√
√
√
√
160 before adding: 2 10 + 160 = 2 10 + 16 · 10 = 2 10 + 4 10 = 6 10
√
√
b. 5 6 · 4 18
√
√
√
√
√
√
√
5 6 · 4 18 = 5 · 4 6 · 18 = 20 108 = 20 36 · 3 = 20 · 6 3 = 120 3
c.
√
√
8 · 12 2
√
√
√
√
8 · 12 2 = 12 8 · 2 = 12 16 = 12 · 4 = 48
d.
√ 2
5 2
√ 2
√ 2
√
5 2 = 52
2 = 25 · 2 = 50 → the and the 2 cancel each other out
Example 5
Divide and simplify the radicals. (Rewrite all division problems like a fraction).
√
√
a. 4 6 ÷ 3
573
3.14. Simplifying Radicals
√
b. √30
8
√
√
√
√
√
√
√30 · √8 = √240 = 168 · 15 = 4 815 = 215
8
8
64
√
c.
8 √2
6
7
√
√
√
√
√
7
4 14
4 14
8 14
√
·
= 6·7 = 3·7 = 21
6 7
7
Notice, we do not really “divide” radicals, but get them out of the denominator of a fraction.
8 √2
Review
Simplify the radicals.
√
1. √
48 √
2. 2√ 5 + 20
3. 24 √ 2
4. 6 3
√ √
5. 8 8 · 10
√
2
6. 2 30
√
7. √320
8. 4√ 5
6
9. √12
10
√
21√ 5
10.
9 15
Review (Answers)
To see the Review answers, open this PDF file and look for section 8.1.
574
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.15 Solving Exponential Equations
Learning Objectives
Here you’ll learn how to solve exponential equations.
"I’m thinking of a number," you tell your best friend. "The number I’m thinking of satisfies the equation 4x+1 = 256.
What number are you thinking of?
Exponential Equations
Until now, we have only solved pretty basic exponential equations. But, what happens when the power is not easily
found? We must use logarithms, followed by the Power Property to solve for the exponent.
Let’s solve the following more complicated exponential equations.
1. 6x = 49 (Round your answer to the nearest three decimal places)
To solve this exponential equation, let’s take the logarithm of both sides. The easiest logs to use are either ln (the
natural log), or log (log, base 10). We will use the natural log.
6x = 49
ln 6x = ln 49
x ln 6 = ln 49
ln 49
≈ 2.172
x=
ln 6
2. 10x−3 = 1003x+11
Change 100 into a power of 10.
10x−3 = 102(3x+11)
x − 3 = 6x + 22
−25 = 5x
−5 = x
3. 82x−3 − 4 = 5
Add 4 to both sides and then take the log of both sides.
575
3.15. Solving Exponential Equations
www.ck12.org
82x−3 − 4 = 5
82x−3 = 9
log 82x−3 = log 9
(2x − 3) log 8 = log 9
log 9
2x − 3 =
log 8
log 9
2x = 3 +
log 8
log 9
3
≈ 2.56
x= +
2 2 log 8
Notice that we did not find the numeric value of log 9 or log 8 until the very end. This will ensure that we have the
most accurate answer.
Examples
Example 1
Earlier, you were asked to determine what number you are thinking of if the number satisfies the equation 4x+1 =
256.
We can rewrite the equation 4x+1 = 256 as 22(x+1) = 28 and solve for x.
22(x+1) = 28
22x+2 = 28
2x + 2 = 8
x=3
Therefore, you’re thinking of the number 3.
Example 2
Solve: 4x−8 = 16.
Change 16 to 42 and set the exponents equal to each other.
4x−8 = 16
4x−8 = 42
x−8 = 2
x = 10
Example 3
Solve: 2(7)3x+1 = 48.
576
www.ck12.org
Chapter 3. Unit 3
Divide both sides by 2 and then take the log of both sides.
2(7)3x+1 = 48
73x+1 = 24
ln 73x+1 = ln 24
(3x + 1) ln 7 = ln 24
ln 24
3x + 1 =
ln 7
ln 24
ln 7
1 ln 24
≈ 0.211
x=− +
3 3 ln 7
3x = −1 +
Example 4
Solve: 23 · 5x+2 + 9 = 21.
Subtract 9 from both sides and multiply both sides by 32 . Then, take the log of both sides.
2 x+2
· 5 + 9 = 21
3
2 x+2
·5
= 12
3
5x+2 = 18
(x + 2) log 5 = log 18
log 18
x=
− 2 ≈ −0.204
log 5
Review
Use logarithms and a calculator to solve the following equations for x. Round answers to three decimal places.
1.
2.
3.
4.
5.
6.
7.
8.
5x = 65
7x = 75
2x = 90
3x−2 = 43
6x+1 + 3 = 13
6(113x−2 ) = 216
8 + 132x−5 = 35
1
x−3 − 5 = 14
2 ·7
Solve the following exponential equations without a calculator.
9.
10.
11.
12.
13.
4x = 8
9x−2 = 27
52x+1 = 125
93 = 34x−6
7(2x−3 ) = 56
577
3.15. Solving Exponential Equations
14. 16x · 4x+1 = 32x+1
15. 33x+5 = 3 · 9x+3
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.10.
578
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.16 Exponential Equations
Learning Objectives
Here you’ll learn how to solve basic equations that contain exponents.
The following exponential equation is one in which the variable appears in the exponent.
√
9x+1 = 27
How can you solve this type of equation where you can’t isolate the variable?
Exponential Equations
When an equation has exponents, sometimes the variable will be in the exponent and sometimes it won’t. There are
different strategies for solving each type of equation.
• When the variable is in the exponent: Rewrite each side of the equation so that the bases of the exponent are
the same. Then, create a new equation where you set the exponents equal to each other and solve.
• When the variable is not in the exponent: Manipulate the equation so the exponent is no longer there. Or,
rewrite each side of the equation so that both sides have the same exponent. Then, create a new equation
where you set the bases equal to each other and solve.
Let’s solve the following exponential equations:
1. 25x−3 =
1 3x+18
5
The variable appears in the exponent. Write both sides of the equation as a power of 5.
(52 )x−3 = (5−1 )3x+18
Apply the law of exponents for raising a power to a power
(am )n = amn
579
3.16. Exponential Equations
www.ck12.org
52(x−3) = 5−1(3x+18)
Simplify the exponents.
52x−6 = 5−3x−18
The bases are the same so the exponents are equal quantities.
2x − 6 = −3x − 18
Set the exponents equal to each other and solve the equation.
2x − 6+6 = −3x − 18+6
2x = −3x−12
2x+3x = −3x+3x − 12
5x = −12
5x −12
=
5
5
5x −12
=
5
5
−12
x=
5
1
2. 4(x − 2) 2 = 16
The variable appears in the base.
1
4(x − 2) 2 = 16
Divide both sides of the equation by 4.
1
2
4(x − 2)
16
=
4
4
4
1
4(x − 2) 2
16
=
4
4
1
(x − 2) 2 = 4
h
i
1 2
(x − 2) 2
1
(x − 2) 2 ×2 = (4)1×2
1
Multiply the exponents on each side of the equation by the reciprocal of .
2
Apply the law of exponents (am )n = amn .
Simplify the exponents.
(x − 2)1 = 42
x − 2 = 16
Solve the equation.
x − 2+2 = 16+2
x = 18
x = 18
2
3. (2x − 4) 3 =
√
3
9
The variable appears in the base.
580
www.ck12.org
Chapter 3. Unit 3
2
(2x − 4) 3 =
√
3
9
2
m
Apply a n =
1
(2x − 4) 3 = (9) 3
2
3
(2x − 4) = (32 )
2
Write 9 as a power of 3.
1
3
Apply the law of exponents (am )n = amn to the right side of the equation.
1
(2x − 4) 3 = (3)2× 3
2
3
(2x − 4) = (3)
2
3
2x − 4 = 3
√
√ m
n
am = n a m, n ∈ N to the right side of the equation.
Simplify the exponents.
The exponents are equal so the bases are equal quantities.
Solve the equation.
2x − 4+4 = 3+4
2x = 7
2x 7
=
2
2
2x 7
=
2
2
7
x=
2
Examples
Example 1
Earlier, √
you were asked how to solve a type of equation where you cannot isolate the variable as in the case of
9x+1 = 27.
To begin, write each side
√ of the equation with a common base. Both 9 and 27 can be written as a power of ‘3’.
Therefore, (32 )x+1 = 33 .
Apply
(am )n = amn
to the left side of the equation. 32x+2 =
√
33
Express the right side of the equation in exponential form and apply
(am )n = amn
1
32x+2 = (33 ) 2
3
32x+2 = 3 2
Now that the bases are the same, then the exponents are equal quantities.
2x + 2 =
3
2
Solve the equation.
2(2x + 2) = 2 32 Multiply both sides of the equation by ‘2’. Simplify and solve.
581
3.16. Exponential Equations
www.ck12.org
3
4x + 4 = 2
2
4x + 4 = 3
4x + 4−4 = 3−4
4x −1
=
4
4
1
x=−
4
Example 2
Use the laws of exponents to solve the following exponential equation: 271−x =
27
1−x
2−x
1
=
9
(33 )1−x = (3−2 )2−x
3 1−x
(3 )
−2 2−x
= (3 )
(3)3(1−x) = (3)−2(2−x)
3
3−3x
=3
−4+2x
3 − 3x = −4 + 2x
1 2−x
9
The variable appears in the exponent.
Write each side of the equation as a power of 3.
Apply the law of exponents (am )n = amn .
Simplify the exponents.
The bases are the same so the exponents are equal quantities.
Solve the equation.
3−3 − 3x = −4−3 + 2x
− 3x = −7 + 2x
− 3x−2x = −7x + 2x−2x
−5x = −7
−5x −7
=
5
5
−5x 7
=
5
−5
7
x=
5
Example 3
1
1
Use the laws of exponents to solve the following exponential equation: (x − 3) 2 = (25) 4
582
www.ck12.org
Chapter 3. Unit 3
1
1
1
2
(x − 3) = (52 )
1
4
Write 25 as a power of 2.
1
2
1
4
Apply the law of exponents (am )n = amn to the right side of the equation.
(x − 3) 2 = (25) 4
(x − 3) = (52 )
1
The variable appears in the base.
1
(x − 3) 2 = (5)2× 4
1
2
(x − 3) = (5)
2
4
1
1
Simplify the exponents.
(x − 3) 2 = (5) 2
The exponents are equal so the bases are equal quantities.
x−3 = 5
Solve the equation.
x − 3+3 = 5+3
x=8
x=8
Example 4
Use the laws of exponents to solve
(8x−4 )(2x )(42x+3 )
32x
= 16.
583
3.16. Exponential Equations
(8x−4 )(2x )(42x+3 )
= 16
32x
[(23 )x−4 ](2x )[(22 )2x+3 ]
= 24
(25 )x
[(23 )x−4 ](2x )[(22 )2x+3 ]
= 24
(25 )x
[(2)3(x−4) ](2x )[(2)2(2x+3) ]
= 24
(2)5(x)
[(2)3x−12 ](2x )[(2)4x+6 ]
= 24
25x
[23x−12 ](2x )[24x+6 ]
= 24
25x
[23x+x+4x−12+6 ]
= 24
25x
28x−6
= 24
25x
www.ck12.org
The variable appears in the exponent.
Write all bases as a power of 2. Write 16 as a power of 2.
Apply the law of exponents (am )n = amn .
Simplify the exponents.
Apply the law of exponents am × an = am+n .
Simplify the exponents.
Apply the laws of exponents
28x−6−5x = 24
Simplify the exponents.
23x−6 = 24
The bases are the same so the exponents are equal quantities.
3x − 6 = 4
Solve the equation.
3x − 6+6 = 4+6
3x = 10
3x 10
=
3
3
3x 10
=
3
3
10
x=
3
Review
Use the laws of exponents to solve the following exponential equations:
1.
2.
3.
4.
5.
6.
7.
√
3
23x−1 = 16
2x+5
36x−2 = 16
1
6(x − 4) 3 = 18
2
(3x − 2) 5 √= 4
36x+1 = √ 6
3
35x−1 = √
9 x
4
2x−1
9
=
27
3
8. (3x − 2) 2 = 8
5
9. (x + 1)− 2 = 32
√ 4x
10.
3
= 27x−3
√
3
11. 43x−1 = 32
2
2
12. (x + 2) 3 = (27) 9
584
am
= am−n .
an
www.ck12.org
Chapter 3. Unit 3
13. (2x−3 )(8x ) = 32
1
1
14. (x − 2) 2 = 9 4
1 2x−7
15. 8x+12 = 16
Review (Answers)
To see the Review answers, open this PDF file and look for section 6.8 for questions 1 to 15. The solution for #16
is:
585
3.17. Half Life
www.ck12.org
3.17 Half Life
Learning Objectives
• Explain what is meant by half life.
• Solve problems involving radioactive decay.
Some of the matter on Earth is unstable and undergoing nuclear decay. Alpha decay is the emission of a helium
nucleus and causes the product to have an atomic number 2 lower than the original and an atomic mass number 4
lower than the original. Beta minus decay is the emission of an electron, causing the product to have an atomic
number 1 greater than the original. Beta plus decay is the emission of a positron, causing the product to have an
atomic number 1 lower than the original.
When an atomic nucleus decays, it does so by releasing one or more particles. The atom often (but not always)
turns into a different element during the decay process. The amount of radiation given off by a certain sample of
radioactive material depends on the amount of material, how quickly it decays, and the nature of the decay product.
Big, rapidly decaying samples are most dangerous.
The measure of how quickly a nucleus decays is given by the half-life of the nucleus. One half-life is the amount of
time it will take for half of the radioactive material to decay.
The type of atom is determined by the atomic number (i.e. the number of protons). The atomic mass of an atom is
approximately the number of protons plus the number of neutrons. Typically, the atomic mass listed in a periodic
table is an average, weighted by the natural abundances of different isotopes. The atomic mass number in a nuclear
decay process is conserved. This means that you will have the same total atomic mass number on both sides of the
equation. Charge is also conserved in a nuclear process.
It is impossible to predict when an individual atom will decay; one can only predict the probability. However, it
is possible to predict when a portion of a macroscopic sample will decay extremely accurately because the sample
contains a vast number of atoms.
586
www.ck12.org
Chapter 3. Unit 3
Key Equations
1 t
M = M0 ( ) tH
2
The amount of mass M of the substance surviving after an original M0 nuclei decay for time t with a half-life of tH .
t = tH
ln MM0
ln 12
The amount of time t it takes a set of nuclei to decay to a specified amount.
N = N0 exp−λt
The number N of nuclei surviving after an original N0 nuclei decay for time t with a half-life of tH .
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/356
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/357
Interactive Simulation
587
3.17. Half Life
www.ck12.org
SIMULATION
Learn about how radiocarbon dating works and how
anthropologists can use this method to figure out who long
ago people lived.
URL:
http://www.ck12.org/physics/isotopes-and-nuclear-
stability/simulationint/Radiocarbon-Dating
Review
1. After 6 seconds, the mass of a sample of radioactive material has reduced from 100 grams to 25 grams. Its
half-life must be
a.
b.
c.
d.
e.
1s
2s
3s
4s
6s
2. For any radioactive material, when does its half-life,
a.
b.
c.
d.
e.
First decrease and then increase?
First increase and then decrease?
Increase with time?
Decrease with time?
Stay the same?
3. If the half-life of a substance is 5 seconds, it ceases to be radioactive (i.e. it ceases emitting particles), . . .
a.
b.
c.
d.
...
...
...
...
after 5 seconds.
after 10 seconds
after 20 seconds.
after a very long time.
4. You have 5 grams of radioactive substance A and 5 grams of radioactive substance B. Both decay by emitting
alpha-radiation, and you know that the higher the number of alpha-particles emitted in a given amount of time,
the more dangerous the sample is. Substance A has a short half-life (around 4 days or so) and substance B has
a longer half-life (around 10 months or so).
a. Which substance is more dangerous right now? Explain.
b. Which substance will be more dangerous in two years? Explain.
5. A certain radioactive material has a half-life of 8 minutes. Suppose you have a large sample of this material,
containing 1025 atoms.
a.
b.
c.
d.
How many atoms decay in the first 8 minutes?
Does this strike you as a dangerous release of radiation? Explain.
How many atoms decay in the second 8 minutes?
What is the ratio of the number of atoms that decay in the first 8 minutes to the number of atoms that
decay in the second 8 minutes?
e. How long would you have to wait until the decay rate drops to 1% of its value in the first 8 minutes?
6. Hidden in your devious secret laboratory are 5.0 grams of radioactive substance A and 5.0 grams of radioactive
substance B. Both emit alpha-radiation. Quick tests determine that substance A has a half-life of 4.2 days and
substance B has a half-life of 310 days.
588
www.ck12.org
Chapter 3. Unit 3
a. How many grams of substance A and how many grams of substance B will you have after waiting 30
days?
b. Which sample (A or B) is more dangerous at this point (i.e., after the 30 days have passed)?
7. The half-life of a certain radioactive material is 4 years. After 24 years, how much of a 75 g sample of this
material will remain?
8. The half life of 53 Ti is 33.0 seconds. You begin with 1000 g of 53 Ti. How much is left after 99.0 seconds?
9. You want to determine the half-life of a radioactive substance. At the moment you start your stopwatch, the
radioactive substance has a mass of 10 g. After 2.0 minutes, the radioactive substance has 0.5 grams left.
What is its half-life?
10. There are two equal amounts of radioactive material. One has a short half-life and the other has a very long
half-life. If you measured the decay rates coming from each sample, which would you expect to have a higher
decay rate? Why?
Review (Answers)
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
c
e
d
a. Substance A decays faster than B b. Substance B because there is more material left to decay.
a. 5 × 1024 atoms b. Decay of a lot of atoms in a short period of time c. 2.5 × 1024 atoms d. 12 e. 26.6 minutes
a. Substance B = 4.6 g and substance A = 0.035 g b. substance B
1.2 g
125 g
0.46 minutes
The one with the short half life, because half life is the rate of decay.
589
3.18. Product and Quotient Properties of Logarithms
www.ck12.org
3.18 Product and Quotient Properties of
Logarithms
Learning Objectives
Here you’ll use and apply the product and quotient properties of logarithms.
Your friend Robbie works as a server at a pizza parlor. You and two of your friends go to the restaurant and order a
pizza. You ask Robbie to bring you separate checks so you can split the cost of the pizza. Instead of bringing you
three checks, Robbie brings you one with the total log3 162 − log3 2. "This is how much each of you owes," he says
as he drops the bill on the table. How much do each of you owe?
Product and Quotient Properties of Logarithms
Just like exponents, logarithms have special properties, or shortcuts, that can be applied when simplifying expressions. In this lesson, we will address two of these properties.
Let’s simplify logb x + logb y.
First, notice that these logs have the same base. If they do not, then the properties do not apply.
logb x = m and logb y = n, then bm = x and bn = y.
Now, multiply the latter two equations together.
bm · bn = xy
bm+n = xy
Recall, that when two exponents with the same base are multiplied, we can add the exponents. Now, reapply the
logarithm to this equation.
bm+n = xy → logb xy = m + n
Recall that m = logb x and n = logb y, therefore logb xy = logb x + logb y.
This is the Product Property of Logarithms.
Now, let’s expand log12 4y.
Applying the Product Property from the previous problem, we have:
log12 4y = log12 4 + log12 y
Finally, let’s simplify log3 15 − log3 5.
As you might expect, the Quotient Property of Logarithms is logb xy = logb x − logb y (proof in the Review section).
Therefore, the answer is:
15
5
= log3 3
log3 15 − log3 5 = log3
=1
590
www.ck12.org
Chapter 3. Unit 3
Examples
Example 1
Earlier, you were asked to find the amount that each of you owes.
If you rewrite log3 162 − log3 2 as log3 162
2 , you get log3 81.
34 = 81 so you each owe $4.
Example 2
Simplify the following expression: log7 8 + log7 x2 + log7 3y.
Combine all the logs together using the Product Property.
log7 8 + log7 x2 + log7 3y = log7 8x2 3y
= log7 24x2 y
Example 3
Simplify the following expression: log y − log 20 + log 8x.
Use both the Product and Quotient Property to condense.
y
· 8x
20
2xy
= log
5
log y − log 20 + log 8x = log
Example 4
Simplify the following expression: log2 32 − log2 z.
Be careful; you do not have to use either rule here, just the definition of a logarithm.
log2 32 − log2 z = 5 − log2 z
Example 5
.
Simplify the following expression: log8 16x
y2
When expanding a log, do the division first and then break the numerator apart further.
log8
16x
= log8 16x − log8 y2
y2
= log8 16 + log8 x − log8 y2
4
= + log8 x − log8 y2
3
To determine log8 16, use the definition and powers of 2: 8n = 16 → 23n = 24 → 3n = 4 → n = 34 .
591
3.18. Product and Quotient Properties of Logarithms
www.ck12.org
Review
Simplify the following logarithmic expressions.
1.
2.
3.
4.
5.
6.
log3 6 + log3 y − log3 4
log 12 − log x + log y2
log6 x2 − log6 x − log6 y
ln 8 + ln 6 − ln 12
ln 7 − ln 14 + ln 10
log11 22 + log11 5 − log11 55
Expand the following logarithmic functions.
7. log6 (5x)
8. log3(abc)
2
9. log ab
10. log9 xy
5
11. log 2x
y2
12. log 8x
15 5
13. log4 9y
14. Write an algebraic proof of the Quotient Property. Start with the expression loga x − loga y and the equations
loga x = m and loga y = n in your proof. Refer to the proof of the Product Property in the first practice
problem as a guide for your proof.
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.8.
592
www.ck12.org
Chapter 3. Unit 3
3.19 Properties of Logs
Here you will be introduced to logarithmic expressions and will learn how they can be combined using properties of
arithmetic.
Log functions are inverses of exponential functions. This means the domain of one is the range of the other. This is
extremely helpful when solving an equation and the unknown is in an exponent. Before solving equations, you must
be able to simplify expressions containing logs. The rules of exponents are applied, but in non-obvious ways. In
order to get a conceptual handle on the properties of logs, it may be helpful to continually ask, what does a log
expression represent? For example, what does log10 1, 000 represent?
Log Properties
Exponential and logarithmic expressions have the same 3 components. They are each written in a different way
so that a different variable is isolated. The following two equations are equivalent to one another.
bx = a ↔ logb a = x
The exponential equation on the left is read “b to the power x is a.” The logarithmic equation on the right is read
“log base b of a is x”.
The two most common bases for logs are 10 and e. At the PreCalculus level log by itself implies log base 10 and
ln implies base e. ln is called the natural log. One important restriction for all log functions is that they must have
strictly positive numbers in their arguments. So, if you press log -2 or log 0 on your calculator, it will give an error.
There are three basic properties of logs that correlate to properties of exponents.
Addition/Multiplication
logb x + logb y = logb (x · y)
bw+z = bw · bz
Subtraction/Division
logb x − logb y = logb xy
bw−z =
bw
bz
Exponentiation
logb (xn ) = n · logb x
(bw )n = bw·n
There are also a few standard results that should be memorized and should serve as baseline reference tools.
• logb 1 = 0
• logb b = 1
• logb (bx ) = x
593
3.19. Properties of Logs
www.ck12.org
• blogb x = x
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/60998
Example
Example 1
Earlier you were asked what log10 1, 000 represents. A log expression represents an exponent. The expression
log10 1, 000 represents the number 3.
log10 1000 = log10 103 = 3
The reason to keep this in mind is that it can solidify the properties of logs. For example, adding exponents implies
bases are multiplied. Thus adding logs means the bases of the exponents are multiplied.
Example 2
Write the expression as a logarithm of a single argument.
log2 12 + log4 6 − log2 24
Note that the center expression is of a different base. First change it to base 2 by switching back to exponential
form.
log4 6 = x ↔ 4x = 6
22x = 6 ↔ log2 6 = 2x
1
1
x = log2 6 = log2 6 2
2
Thus the expression with the same base is:
1
2
log2 12 + log2 6 − log2 24 = log2
= log2
594
√ !
12 · 6
24
√ !
6
2
www.ck12.org
Chapter 3. Unit 3
Example 3
Prove the following log identity:
loga b =
1
logb a
Start by letting the left side of the equation be equal to x. Then, rewrite in exponential form, manipulate, and rewrite
back in logarithmic form until you get the expression from the left side of the equation.
loga b = x
ax = b
1
a = bx
1
logb a = logb b x =
x=
Therefore,
1
logb a
1
x
1
logb a
= loga b because both expressions are equal to x.
Example 4
Rewrite the following expression under a single log.
e
+ 2x · ln 5
4x
e
+ ln(52x )
= ln
4x
e · 52x
= ln
4x
ln e − ln 4x + 2(eln x · ln 5) = ln
Example 5
True or false:
(log3 4x) · (log3 5y) = log3 (4x + 5y)
False. It is true that the log of a product is the sum of logs. It is not true that the product of logs is the log of a sum.
Review
Decide whether each of the following statements is true or false. Explain.
x
x
1. log
=
log
log y
y
2. (log x)n = n log x
3. log x + log y = log xy
Rewrite each of the following expressions under a single log and simplify.
4. log 4x + log(2x + 4)
5. 5 log x + log x
595
3.19. Properties of Logs
6. 4 log2 x + 12 log2 9 − log2 y
7. 6 log3 z2 + 14 log3 y8 − 2 log3 z4 y
Expand the expression as much as possible.
3
8. log4 2x5
2
9. ln 4xy
15
2 3
10. log x (yz)
3
Translate from exponential form to logarithmic form.
11. 2x+1 + 4 = 14
Translate from logarithmic form to exponential form.
12. log2 (x − 1) = 12
Prove the following properties of logarithms.
13. logbn x = 1n logb x
14. logbn xn = logb x
15. log 1
b
1
x
= logb x
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.4.
596
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.20 Power Property of Logarithms
Learning Objectives
Here you’ll use the Power Property of logarithms.
The hypotenuse of a right triangle has a length of log3 278 . How long is the triangle’s hypotenuse?
Power Property
The last property of logs is the Power Property.
logb x = y
Using the definition of a log, we have by = x. Now, raise both sides to the n power.
(by )n = xn
bny = xn
Let’s convert this back to a log with base b, logb xn = ny. Substituting for y, we have logb xn = n logb x.
Therefore, the Power Property says that if there is an exponent within a logarithm, we can pull it out in front of the
logarithm.
Let’s use the Power Property to expand the following logarithms.
1. log6 17x5
To expand this log, we need to use the Product Property and the Power Property.
log6 17x5 = log6 17 + log6 x5
= log6 17 + 5 log6 x
2. ln
4
2x
y3
We will need to use all three properties to expand this problem. Because the expression within the natural log is in
parenthesis, start with moving the 4th power to the front of the log.
2x
ln 3
y
4
= 4 ln
2x
y3
= 4(ln 2x − ln y3 )
= 4(ln 2 + ln x − 3 ln y)
= 4 ln 2 + 4 ln x − 12 ln y
597
3.20. Power Property of Logarithms
www.ck12.org
Depending on how your teacher would like your answer, you can evaluate 4 ln 2 ≈ 2.77, making the final answer
2.77 + 4 ln x − 12 ln y.
Now, let’s condense log 9 − 4 log 5 − 4 log x + 2 log 7 + 2 log y.
This is the opposite of the previous two problems. Start with the Power Property.
log 9 − 4 log 5 − 4 log x + 2 log 7 + 2 log y
log 9 − log 54 − log x4 + log 72 + log y2
Now, start changing things to division and multiplication within one log.
2 2
y
log 9·7
54 x4
Lastly, combine like terms.
2
441y
log 625x
4
Examples
Example 1
Earlier, you were asked to find the length of the triangle’s hypotenuse.
We can rewrite log3 278 and 8 log3 27 and solve.
8 log3 27
= 8·3
= 24
Therefore, the triangle’s hypotenuse is 24 units long.
Example 2
Expand the following expression: ln x3 .
The only thing to do here is apply the Power Property: 3 ln x.
Example 3
2
x y
Expand the following expression: log16 32z
5.
Let’s start with using the Quotient Property.
log16
x2 y
= log16 x2 y − log16 32z5
32z5
Now, apply the Product Property, followed by the Power Property.
598
www.ck12.org
Chapter 3. Unit 3
= log16 x2 + log16 y − log16 32 + log16 z5
5
= 2 log16 x + log16 y − − 5 log16 z
4
Simplify log16 32 → 16n = 32 → 24n = 25 and solve for n. Also, notice that we put parenthesis around the second
log once it was expanded to ensure that the z5 would also be subtracted (because it was in the denominator of the
original expression).
Example 4
Expand the following expression: log(5c4 )2 .
For this problem, you will need to apply the Power Property twice.
log(5c4 )2 = 2 log 5c4
= 2(log 5 + log c4 )
= 2(log 5 + 4 log c)
= 2 log 5 + 8 log c
Important Note: You can write this particular log several different ways. Equivalent logs are: log 25+8 log c, log 25+
log c8 and log 25c8 . Because of these properties, there are several different ways to write one logarithm.
Example 5
Condense into one log: ln 5 − 7 ln x4 + 2 ln y.
To condense this expression into one log, you will need to use all three properties.
ln 5 − 7 ln x4 + 2 ln y = ln 5 − ln x28 + ln y2
= ln
5y2
x28
Important Note: If the problem was ln 5 − (7 ln x4 + 2 ln y), then the answer would have been ln x285y2 . But, because
there are no parentheses, the y2 is in the numerator.
Review
Expand the following logarithmic expressions.
1. log7 y2
2. log12 5z2
3. log4 (9x)3
2
4. log 3x
y
3 2
5. log8 xz4y
599
3.20. Power Property of Logarithms
4 2
6. log5 25x
y
−2
7. ln 6x
y3
5 −2 6
8. ln e yx3
Condense the following logarithmic expressions.
9.
10.
11.
12.
13.
14.
15.
6 log x
2 log6 x + 5 log6 y
3(log x − log y)
1
2 log(x + 1) − 3 log y
4 log2 y + 13 log2 x3
1
5 [10 log2 (x − 3) + log2 32 −log2 y]
4 21 log3 y − 13 log3 x − log3 z
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.9.
600
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
3.21 Solving Logarithmic Equations
Learning Objectives
Here you’ll learn how to solve a logarithmic equation with any base.
"I’m thinking of a number," you tell your best friend. "The number I’m thinking of satisfies the equation log 10x2 −
log x = 3." What number are you thinking of?
Solving Logarithm Equations
A logarithmic equation has the variable within the log. To solve a logarithmic equation, you will need to use the
inverse property, blogb x = x, to cancel out the log.
Let’s solve the following logarithmic equations.
1. log2 (x + 5) = 9
There are two different ways to solve this equation. The first is to use the definition of a logarithm.
log2 (x + 5) = 9
29 = x + 5
512 = x + 5
507 = x
The second way to solve this equation is to put everything into the exponent of a 2, and then use the inverse property.
2log2 (x+5) = 29
x + 5 = 512
x = 507
Make sure to check your answers for logarithmic equations. There can be times when you get an extraneous
solution.To check, plug in 507 for x : log2 (507 + 5) = 9 → log2 512 = 9
2. 3 ln(−x) − 5 = 10
First, add 5 to both sides and then divide by 3 to isolate the natural log.
3 ln(−x) − 5 = 10
3 ln(−x) = 15
ln(−x) = 5
601
3.21. Solving Logarithmic Equations
www.ck12.org
Recall that the inverse of the natural log is the natural number. Therefore, everything needs to be put into the
exponent of e in order to get rid of the log.
eln(−x) = e5
−x = e5
x = −e5 ≈ −148.41
Checking the answer, we have 3 ln(−(−e5 )) − 5 = 10 → 3 ln e5 − 5 = 10 → 3 · 5 − 5 = 10
3. log 5x + log(x − 1) = 2
Condense the left-hand side using the Product Property.
log 5x + log(x − 1) = 2
log[5x(x − 1)] = 2
log(5x2 − 5x) = 2
Now, put everything in the exponent of 10 and solve for x.
10log(5x
2 −5x)
= 102
5x2 − 5x = 100
5x2 − 5x − 100 = 0
x2 − x − 20 = 0
(x − 5)(x + 4) = 0
x = 5, −4
Now, check both answers.
log 5(5) + log(5 − 1) = 2
log 25 + log 4 = 2
log 5(−4) + log((−4) − 1) = 2
log(−20) + log(−5) = 2
log 100 = 2
-4 is an extraneous solution. In the step log(−20) + log(−5) = 2, we cannot take the log of a negative number,
therefore -4 is not a solution. 5 is the only solution.
Examples
Example 1
Earlier, you were asked to determine what number you are thinking of if the number satisfies the equation log 10x2 −
log x = 3.
2
We can rewrite log 10x2 − log x = 3 as log 10x
x = 3 and solve for x.
602
www.ck12.org
Chapter 3. Unit 3
10x2
=3
x
log 10x = 3
log
10log 10x = 103
10x = 1000
x = 100
Therefore, the number you are thinking of is 100.
Example 2
Solve: 9 + 2 log3 x = 23.
Isolate the log and put everything in the exponent of 3.
9 + 2 log3 x = 23
2 log3 x = 14
log3 x = 7
x = 37 = 2187
Example 3
Solve: ln(x − 1) − ln(x + 1) = 8.
Condense the left-hand side using the Quotient Rule and put everything in the exponent of e.
ln(x − 1) − ln(x + 1) = 8
x−1
ln
=8
x+1
x−1
= e8
x+1
x − 1 = (x + 1)e8
x − 1 = xe8 + e8
x − xe8 = 1 + e8
x(1 − e8 ) = 1 + e8
x=
1 + e8
≈ −1.0007
1 − e8
Checking our answer, we get ln(−1.0007 − 1) − ln(−1.0007 + 1) = 8, which does not work because you cannot take
the log of a negative number. Therefore, there is no solution for this equation.
Example 4
Solve: 21 log5 (2x + 5) = 2.
603
3.21. Solving Logarithmic Equations
www.ck12.org
Multiply both sides by 2 and put everything in the exponent of a 5.
1
log (2x + 5) = 2
2 5
log5 (2x + 5) = 4
5log5 (2x+5) = 54
2x + 5 = 625
2x = 620
x = 310
Review
Use properties of logarithms and a calculator to solve the following equations for x. Round answers to three decimal
places and check for extraneous solutions.
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
log2 x = 15
log12 x = 2.5
log9 (x − 5) = 2
log7 (2x + 3) = 3
8 ln(3 − x) = 5
4 log3 3x − log3 x = 5
log(x + 5) + log x = log 14
2 ln x − ln x = 0
3 log3 (x − 5) = 3
2
3 log3 x = 2
5 log 2x − 3 log 1x = log 8
2 ln xe+2 − ln x = 10
2 log6 x + 1 = log6 (5x + 4)
2 log 1 x + 2 = log 1 (x + 10)
2
2
3 log 2 x − log 2 27 = log 2 8
3
3
3
Answers for Review Problems
To see the Review answers, open this PDF file and look for section 8.11.
604
www.ck12.org
Chapter 3. Unit 3
3.22 Logarithmic Models
Learning Objectives
Here you will explore more complex log functions and real-world applications of logarithmic functions.
In prior lessons, you used an exponential model to predict the population of a town based on a constant growth rate
such as 6% per year.
In the real world however, populations often do not just grow continuously and without limit. A town originally
founded near a convenient water source may grow very quickly at first, but the expansion will slow dramatically
as houses and businesses run out of room near the water source, and need to begin transporting water further and
further away.
How can a situation like this be modeled with an equation?
Logarithmic Models
In a prior lesson, we considered the solutions of simple log equations. Now we return to that topic and explore some
more complex examples. Solving more complicated log equations can be less difficult than you might think, by
using our knowledge of log properties.
For example, consider the equation log2 (x) + log2 (x - 2) = 3. We can solve this equation using a log property.
TABLE 3.16:
log2 (x) + log2 (x - 2) = 3
log2 (x(x - 2)) = 3
log2 (x2 - 2x) = 3 ⇒
23 = x2 - 2x
x2 - 2x - 8 = 0
(x - 4) (x + 2) = 0
x = -2, 4
logb x + logb y = logb (xy)
write the equation in exponential form.
Solve the resulting quadratic
The resulting quadratic has two solutions. However, only x = 4 is a solution to our original equation, as log2 (-2) is
undefined. We refer to x = -2 as an extraneous solution.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187518
605
3.22. Logarithmic Models
www.ck12.org
Examples
Example 1
Earlier, you were asked how a situation could be modeled with an equation.
A population that increases continuously at a constant rate may be modeled with an exponential function.
A population that increases rapidly and then levels off may be modeled with a logarithmic function.
Example 2
Solve each equation.
a. log (x + 2) + log 3 = 2
TABLE 3.17:
log (3(x + 2)) = 2
log (3x + 6) = 2
102 = 3x + 6
100 = 3x + 6
3x = 94
x = 94/3
logb x + logb y = logb (xy)
Simplify the expression 3(x+2)
Write the log expression in exponential form
Solve the linear equation
b. ln (x + 2) - ln (x) = 1
TABLE 3.18:
x+2
=
x
x+2
1
e = x
ln
1
ex = x + 2
ex − x = 2
x(e − 1) = 2
2
x = e−1
logb x − logb y = logb
x
y
Write the log expression in exponential form.
Multiply both sides by x.
Factor out x.
Isolate x.
The solution above is an exact solution. If we want a decimal approximation, we can use a calculator to find that x
≈ 1.16. We can also use a graphing calculator to find an approximate solution. Consider again the equation ln (x +
2) - ln (x) = 1. We can solve this equation by solving a system:
(
y = ln(x + 2) − ln(x)
y=1
If you graph the system on your graphing calculator, you should see that the curve and the horizontal line intersection
at one point. Using the INTERSECT function on the CALC menu (press <2nd>[CALC]), you should find that the x
coordinate of the intersection point is approximately 1.16. This method will allow you to find approximate solutions
for more complicated log equations.
Example 3
Use a graphing calculator to solve each equation:
606
www.ck12.org
Chapter 3. Unit 3
a. log(5 - x) + 1 = log x
The graphs of y = log (5 - x) + 1 and y = log x intersect at x ≈ 4.5454545.
Therefore the solution of the equation is x ≈ 4.54.
b. log2 (3x + 8) + 1 = log3 (10 - x)
First, in order to graph the equations, you must rewrite them in terms of a common log or a natural log. The resulting
+ 1 and y = log(10−x)
equations are: y = log(3x+8)
log2
log3 .
The graphs of these functions intersect at x ≈ -1.87. This value is the approximate solution to the equation.
Example 4
Consider population growth:
TABLE 3.19:
Year
1
5
10
20
30
40
Population
2000
4200
6500
8800
10500
12500
If we plot this data, we see that the growth is not quite linear, and it is not exponential either.
We can find a logarithmic function to model this data. First enter the data in the table in L1 and L2. Then press STAT
to get to the CALC menu. This time choose option 9. You should get the function y = 930.4954615 + 2780.218173
ln x. If you view the graph and the data points together, as described in the Technology Note above, you will see
that the graph of the function does not touch the data points, but models the general trend of the data.
Note about technology: you can also do this using an Excel spreadsheet. Enter the data in a worksheet, and create
a scatterplot by inserting a chart. After you create the chart, from the chart menu, choose “add trendline.” You will
then be able to choose the type of function. Note that if you want to use a logarithmic function, the domain of your
607
3.22. Logarithmic Models
www.ck12.org
data set must be positive numbers. The chart menu will actually not allow you to choose a logarithmic trendline if
your data include zero or negative x values. See below:
Example 5
Solve for x : log2 x − log2 (x − 4) = 12.
To solve log2 x − log2 (x − 4) = 12:
x
= 12 : Using logx y − logx z = logx yz
log2 x−4
212 =
x
x−4
4, 096 =
: Write in exponential form
x
x−4
: With a calculator
4, 096x − 16384 = x : Multiply both sides by x − 4
4, 095x = 16, 384 : Simplify
x = 4 : Divide
Example 6
Biologists use the formula n = k · logA to estimate the number of species n that live in a given area A by multiplying
by a constant k which changes by location. If a particular rain forest has a constant k of 943 how many species would
be estimated to live in an area of 950km2 ?
To find the number of species in an area of 950km2 :
n = 943 · log950 : Substitute the given k and A values
n = 943 · 2.977 : With a calculator
n = 2, 807
Therefore 2,807 species would likely live in the area.
608
www.ck12.org
Chapter 3. Unit 3
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187513
Review
Express 1-7 in exponential form:
1
1. log12 1728
= −3
2. log216 6 = 13
3. log 1 19 = 3
3
4.
5.
6.
7.
1
log 1 16
=2
4
log5 125 = 3
log15 225 = 2
log25 5 = 21
For questions 8-13, solve for x.
8.
9.
10.
11.
12.
13.
logx 64 = 2
log3 6561 = x
log5 x = 4
logx 27 = 3
log2 x = 6
log4 64 = x
For questions 14-19, solve for x.
14. 4log( 5x ) + log( 625
4 ) = 2logx
log5 125
15. log5 z + log5 x = 72
16. logp = 2−logp
logp
17. 2logx − 2log(x + 1) = 0
18. log(25 − z3 ) − 3log(4 − z) = 0
3)
19. log(35−y
log(5−y) = 3
Review (Answers)
To see the Review answers, open this PDF file and look for section 3.10.
609
3.23. Unit 3 Test Review
3.23 Unit 3 Test Review
Unit 3 Test Review
610
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
611
3.23. Unit 3 Test Review
612
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
613
3.23. Unit 3 Test Review
614
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
Solutions:
615
3.23. Unit 3 Test Review
616
www.ck12.org
www.ck12.org
Chapter 3. Unit 3
617
3.24. References
www.ck12.org
3.24 References
1. CK-12 Foundation. https://www.flickr.com/photos/oceanyamaha/4579317995/in/photolist-7YEcGx-VpLS3VosTt-r842zw-vmJczY-55pFUb-5EtzBt-aiq7Pb-eZHA5e-7xLNk4-6y4noo-b8GLcX-7YHqZQ-7cH5UJ-rsUq4n
-561KZR-dDPyUD-2rEXy-2rEXw-e1rZr-7cuFp-6U9BCi-2sbYk-2rEXv-2sbYj-oWbwC1-2sbYh-48Ytsy-dDPz
aB-2rEXx-8Y88F4-6HMSTc-5FpUBB-oSciLy-7xM9B8-8LSV1p-8LSUL8-7y5eCX-dg9tw8-2sd89-2sbYm-VpLJ
u-2rEXu-cHECdq-cPcUsm-5UcCCx-67azBF-cPcVe3-cPcUBh-7M5Nqr .
2. CK-12 Foundation. https://www.flickr.com/photos/oceanyamaha/4579317995/in/photolist-7YEcGx-VpLS3VosTt-r842zw-vmJczY-55pFUb-5EtzBt-aiq7Pb-eZHA5e-7xLNk4-6y4noo-b8GLcX-7YHqZQ-7cH5UJ-rsUq4n
-561KZR-dDPyUD-2rEXy-2rEXw-e1rZr-7cuFp-6U9BCi-2sbYk-2rEXv-2sbYj-oWbwC1-2sbYh-48Ytsy-dDPz
aB-2rEXx-8Y88F4-6HMSTc-5FpUBB-oSciLy-7xM9B8-8LSV1p-8LSUL8-7y5eCX-dg9tw8-2sd89-2sbYm-VpLJ
u-2rEXu-cHECdq-cPcUsm-5UcCCx-67azBF-cPcVe3-cPcUBh-7M5Nqr .
3. CK-12 Foundation. https://www.flickr.com/photos/oceanyamaha/4579317995/in/photolist-7YEcGx-VpLS3VosTt-r842zw-vmJczY-55pFUb-5EtzBt-aiq7Pb-eZHA5e-7xLNk4-6y4noo-b8GLcX-7YHqZQ-7cH5UJ-rsUq4n
-561KZR-dDPyUD-2rEXy-2rEXw-e1rZr-7cuFp-6U9BCi-2sbYk-2rEXv-2sbYj-oWbwC1-2sbYh-48Ytsy-dDPz
aB-2rEXx-8Y88F4-6HMSTc-5FpUBB-oSciLy-7xM9B8-8LSV1p-8LSUL8-7y5eCX-dg9tw8-2sd89-2sbYm-VpLJ
u-2rEXu-cHECdq-cPcUsm-5UcCCx-67azBF-cPcVe3-cPcUBh-7M5Nqr .
4. CK-12 Foundation. https://www.flickr.com/photos/oceanyamaha/4579317995/in/photolist-7YEcGx-VpLS3VosTt-r842zw-vmJczY-55pFUb-5EtzBt-aiq7Pb-eZHA5e-7xLNk4-6y4noo-b8GLcX-7YHqZQ-7cH5UJ-rsUq4n
-561KZR-dDPyUD-2rEXy-2rEXw-e1rZr-7cuFp-6U9BCi-2sbYk-2rEXv-2sbYj-oWbwC1-2sbYh-48Ytsy-dDPz
aB-2rEXx-8Y88F4-6HMSTc-5FpUBB-oSciLy-7xM9B8-8LSV1p-8LSUL8-7y5eCX-dg9tw8-2sd89-2sbYm-VpLJ
u-2rEXu-cHECdq-cPcUsm-5UcCCx-67azBF-cPcVe3-cPcUBh-7M5Nqr .
5. CK-12. CK-12 .
6. CK-12. CK-12 .
7. CK-12 Foundation. .
8. Courtesy of Renee Comet, National Cancer Institute. https://commons.wikimedia.org/wiki/File:Grapefruit_%
281%29.jpg . Public Domain
9. CK-12 Foundation - Hana Zavadska and Zachary Wilson;User:Midnightcomm/Wikimedia Commons. http://c
ommons.wikimedia.org/wiki/File:Antacid-L478.jpg . CC-BY-NC-SA 3.0;CC BY 2.5
10. Guitar string photo by Flickr:jar(); illustration by Christopher Auyeung (CK-12 Foundation). Guitar string ph
oto: http://www.flickr.com/photos/jariceiii/5343236121/ . CC BY 2.0;CC BY-NC 3.0
11. Christopher Auyeung;Guitar string photo by Flickr:jar(); illustration by Christopher Auyeung (CK-12 Foundation). CK-12 Foundation;Guitar string photo: http://www.flickr.com/photos/jariceiii/5343236121/;Getty I
mages . CC BY-NC 3.0;CC BY 2.0
12. Christopher Auyeung;Guitar string photo by Flickr:jar(); illustration by Christopher Auyeung (CK-12 Foundation). CK-12 Foundation;Guitar string photo: http://www.flickr.com/photos/jariceiii/5343236121/ . CC
BY-NC 3.0;CC BY 2.0
13. Courtesy of US Navy;User:Crickett/Wikipedia. http://commons.wikimedia.org/wiki/File:Tsunami_damage_north_of_Sendai_1x1_resized.jpg;http://commons.wikimedia.org/wiki/File:Seismogram.gif . Public Domain
14. Courtesy of US Navy;Jodi So;User:Crickett/Wikipedia. http://commons.wikimedia.org/wiki/File:Tsunami_d
amage_north_of_Sendai_1x1_resized.jpg;CK-12 Foundation;http://commons.wikimedia.org/wiki/File:Seismo
gram.gif . Public Domain;CC BY-NC 3.0
618
www.ck12.org
Chapter 4. Unit 4
C HAPTER
4
Unit 4
Chapter Outline
4.1
CK-12 C ALCULUS S TUDY G UIDE - A RITHMETIC O PERATIONS ON F UNCTIONS
4.2
O PERATIONS ON F UNCTIONS
4.3
CK-12 C ALCULUS S TUDY G UIDE - C OMPOSITIONS OF F UNCTIONS
4.4
F UNCTION C OMPOSITION
4.5
C OMPOSITION OF F UNCTIONS
4.6
C OMPOSITION OF F UNCTIONS AND I NVERTIBLE F UNCTIONS
4.7
F UNCTION O PERATIONS
619
4.1. CK-12 Calculus Study Guide - Arithmetic Operations on Functions
www.ck12.org
4.1 CK-12 Calculus Study Guide - Arithmetic
Operations on Functions
Important Points - Arithmetic Operations on Functions
Arithmetic operations on functions involve creating a new function by using a combination of addition, multiplication, and/or division among functions.
Any number of functions can be combined this way, but to illustrate the concept only two functions will be used:
TABLE 4.1:
Notation
( f + g)(x)
(f ·g)(x)
f
g
(x)
Operation
( f + g)(x) = f (x) + g(x)
(f ·g)(x) = f (x) · g(x)
f
g
(x) =
f (x)
g(x)
Example 1
Question:
Let f (x) = x and g(x) = x3 . What is ( f + g)(x)?
Analysis:
( f + g)(x) = f (x) + g(x)
= x + x3
Answer:
( f + g)(x) = x + x3
Below is the graph of the function for reference:
( f + g)(x) = x + x3
620
Example
( f + g)(x) = 1 + x2
(f ·g)(x) = x2
f
g
(x) =
1
x2
www.ck12.org
Chapter 4. Unit 4
Extension:
Whenever you perform addition or subtraction operations among a collection of polynomials, with the maximum
degree of a polynomial being n, then the answer will always be a polynomial of degree less than or equal to n.
Example 2
Question:
Let f (x) = x2 and g(x) = 1. What is ( f − g)(x)?
Analysis:
( f − g)(x) = f (x) − g(x)
= x2 − 1
Answer:
( f − g)(x) = x2 − 1
Below is the graph of the function for reference:
( f − g)(x) = x2 − 1
621
4.1. CK-12 Calculus Study Guide - Arithmetic Operations on Functions
Example 3
Question:
Let f (x) = (x + 1) and g(x) = (x − 1). What is ( f · g)(x)?
Analysis:
( f · g)(x) = f (x) · g(x)
= (x + 1)(x − 1)
= x2 − 1
Answer:
( f · g)(x) = x2 − 1
Note that this answer is the same as the one for Example 2.
Below is the graph of the function for reference:
( f · g)(x) = x2 − 1
622
www.ck12.org
www.ck12.org
Chapter 4. Unit 4
Example 4
Question:
Let f (x) = (x2 − 1) and g(x) = (x + 1). What is
f
g
(x)?
Analysis:
f
g
(x) =
=
Answer:
f
g (x) =
(x2 −1)
(x+1)
f (x)
g(x)
(x2 −1)
(x+1)
for {x ∈ R|x 6= −1}
Below is the graph of the function for reference:
f
(x2 − 1)
(x) =
, {x ∈ R|x 6= −1}
g
(x + 1)
623
4.1. CK-12 Calculus Study Guide - Arithmetic Operations on Functions
www.ck12.org
Extension:
It is especially important to be aware of constraints on the domain for functions of the type
f
g
(x).
Example 5
Question:
What function must g(x) be in order for ( f + g)(x) = ( f − g)(x) = f (x)?
Analysis:
If arithmetic operations on numbers are considered, then (a + b) = (a − b) = a, is satisfied for any a only if b = 0.
In the same way, ( f + g)(x) = ( f − g)(x) = f (x) is satisfied only if g(x) = 0.
Answer:
g(x) = 0
Below is the graph of the function for reference:
g(x) = 0
624
www.ck12.org
Chapter 4. Unit 4
Example 6
Question:
What function must g(x) be in order for ( f · g)(x) =
f
g
(x) = f (x)?
Analysis:
If arithmetic operations on numbers are considered, then (a · b) = ba = a, is satisfied for any a only if b = 1.
In the same way, ( f · g)(x) = gf (x) = f (x) is satisfied only if g(x) = 1.
Answer:
g(x) = 1
Below is the graph of the function for reference:
g(x) = 1
625
4.1. CK-12 Calculus Study Guide - Arithmetic Operations on Functions
Example 7
Question:
Let f (x) = x2 and g(x) = 1. What is f + f · g − gg (x)?
Analysis:
f + f · g − gg (x) =
f (x) + f (x) · g(x) − g(x)
g(x)
= x2 + x2 (1) − 11
= 2x2 − 1
Answer:
f + f · g − gg (x) = 2x2 − 1
Below is the graph of the function for reference:
626
g
(x) = 2x2 − 1
f + f ·g−
g
www.ck12.org
www.ck12.org
Chapter 4. Unit 4
Extension:
It is important to follow order of operations rules when dealing with arithmetic operations of functions.
Example 8
Question:
Let f (x) be any arbitrary function. Let H(x) =


1
1
2

0
x>0
x=0
x<0
. What is ( f · H)(x)?
Analysis:
Note that H(x) is the Heaviside Step Function.
Answer:
Example 9
Question:
Let f (x) = 12 .


1 x > 0
Let g(x) = 0 x = 0


−1 x < 0
.
Let h(x) = 2.
Write H(x), the Heaviside Step Function as the result of arithmetic operations among f (x), g(x), and h(x).
627
4.1. CK-12 Calculus Study Guide - Arithmetic Operations on Functions
www.ck12.org
Analysis:
The most logical course of action is to start with g(x) and see what operations need to be applied to it in order to
obtain H(x). Then this needs to be represented as arithmetic operations of f (x), g(x), and h(x).
H(x) =
=
=
=
1
2 (1 + g(x))
g(x)
1
2+ 2
g(x)
f (x) + h(x)
f + gh (x)
The g(x) used here is known as the sign or signum function, often denoted by sgn(x).
Answer:
H(x) = f + gh (x)
Example 10
Question:
Let f (x) = sin(x) and g(x) = cos(x − b). Find a value of b such that ( f − g)(x) = 0?
Analysis:
( f − g)(x) = f (x) − g(x)
= sin(x) − cos(x − b)
Set this expression to 0 in order to solve for the value of b.
sin(x) − cos(x − b) = 0
sin(x) = cos(x − b)
A value of b such that sin(x) = cos(x − b) occurs for b = π2 .
Answer:
b=
628
π
2
www.ck12.org
Chapter 4. Unit 4
4.2 Operations on Functions
Learning Objectives
Here you will learn how to perform the standard mathematical operations of addition, subtraction, multiplication,
and division on functions. You will also explore the graphs that result from these operations.
Just as numbers can be added, subtracted, multiplied, and divided, so too can functions. Combining functions in this
way can often have surprising results, as the resultant function may not have a graph that appears similar to that of
either input function’s graph.
How can you tell, before completing the entire operation and graphing the result, whether the new function is likely
to resemble one of the input functions? How do you describe combined functions without a graph?
Operations on Functions
Sums and Differences of Functions
Consider the function: f (x) = 48/x + 2x2 .
Notice that the equation has two terms:
The first term: 48/x
The second term: 2x2
Therefore we can think of the function f (x) as the sum of two other functions:
The reciprocal function g(x) = 48/x
The quadratic function b(x) = 2x2
When we add the functions together, we get a new type of graph that resembles both the graphs of g(x) and b(x):
The graph on the right is f (x). The right portion of f (x) resembles the parabola b(x), but is asymptotic to the y-axis.
The left portion of f (x) resembles the left side of g(x), as both functions are asymptotic to the negative y-axis.
There are two points to be stressed here: first, that we can add functions together, and second, that the resulting sum
may be a different kind of function from the original two.
629
4.2. Operations on Functions
www.ck12.org
The sum or difference of a function is more likely to resemble the original two functions if they are from the
same family.
For example, if two functions from the linear function family are added together, the sum function is also a member
of the linear family.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187161
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187162
Examples
Example 1
Earlier, you were asked if you could discover the trick for identifying when a resultant function graph is likely to
resemble the input graphs.
The sum or difference of a function is more likely to resemble the original two functions if they are from the same
family.
In other words, if you are adding or subtracting two quadratic equations, the result is likely to be quadratic, and have
a similar graph.
Example 2
If f (x) = x3 + 2x2 and g(x) = x2 - 5, what is f - g? What does the graph look like?
The difference is: f - g = x3 + 2x2 - (x2 - 5) = x3 + 2x2 - x2 + 5 = x3 + x2 + 5
The graph of the new function, along with f (x), is shown here:
630
www.ck12.org
Chapter 4. Unit 4
Because f (x) and the new function y = x3 + x2 + 5 are both members of the cubic family, they have similar shapes.
To recap: When we add or subtract functions, the resulting sum or difference function may be in the same family as
one or both of the original functions, or it may be a different type of function. The resultant function is more likely
to be in the same family if both of the initial functions are in the same family as each other.
Example 3
Given f (x) = 2x2 and g(x) = x + 1, find r(x) = f (x)/ g(x) and t(x) = g(x)/f (x).
r(x) = f (x)/g(x) = 2x2 /(x + 1). This is a rational function, and does not have a horizontal asymptote. It does, however,
have a vertical asymptote at x = -1, as the domain excludes x = -1.
t(x) = g(x)/ f (x) = (x + 1)/2x2 . This is also a rational function. This function has a horizontal asymptote at y = 0 (the
x-axis), and a vertical asymptote at x = 0 (the y-axis).
631
4.2. Operations on Functions
www.ck12.org
Notice that the graph of this function crosses its asymptote at (-1, 0), but then as x approaches −∞, the function
values approach 0.
In general, if we multiply linear and polynomial functions (quadratics, cubics, and other such functions with higher
exponents, such as y = x4 + 3x2 + 2), we will obtain other polynomial functions. If we divide these kinds of functions,
we will obtain other polynomial functions, or rational functions.
Multiplying and dividing other types of functions may result in more complicated graphs.
Example 4
Given f (x) = 4x2 − 7 and g(x) = 3x2 − 2x + 8, find and graph (use technology) ( f + g)(x).
Step 1: Recall that ( f + g)(x) = f (x) + g(x)
Step 2: Substitute f (x) + g(x) = (4x2 − 7) + (3x2 − 2x + 8)
Step 3: Combine like terms ( f + g)(x) = 7x2 − 2x + 1
So our answer is: ( f + g)(x) = 7x2 − 2x + 1
The graph of f (x) = 7x2 − 2x + 1 looks like:
632
www.ck12.org
Chapter 4. Unit 4
Example 5
Multiply the function by the scalar value - if f (x) = 3x + 10, find 3 · f (x).
To multiply a function by a scalar, multiply each term of the function by the scalar:
Step 1: Substitute: 3 f (x) = 3(3x + 10)
Step 2: Distribute: 3 f (x) = 9x + 30
So our answer is: 3 f (x) = 9x + 30
Example 6
Given f (x) = 3x − 7 and g(x) = 4x + 6, find and graph (use technology) ( f · g)(x).
Step 1: Recall that ( f · g)(x) = f (x) · g(x)
Step 2: Substitute: f (x) · g(x) = (3x − 7)(4x + 6)
Step 3: Distribute (FOIL): ( f · g)(x) = 12x2 + 18x − 28x − 42
Step 4: Combine like terms: ( f · g)(x) = 12x2 − 10x − 42
So our answer is:( f · g)(x) = 12x2 − 10x − 42
The graph of f (x) = 12x2 − 10x − 42 looks like this:
633
4.2. Operations on Functions
Review
Given f (x) =
x3
x+1
and g(x) = x(x + 1) find each of the following:
1. ( f g)(x) =
2. ( f g)(−1) =
3. ( gf )(x) =
Simplify the following:
4.
5.
6.
7.
8.
9.
If f (x) = 2x + 4 and g(x) = 3x − 7, find ( f + g(x)).
If g(x) = 23 x + 12 and h(x) = 14 x + 7, find (g + h)(x)
If f (x) = −4x2 − 10 and g(x) = 5x2 − 2x − 3, find ( f + g)(x)
If f (x) = 6x2 − 3x + 5 and g(x) = 4x2 + 5x − 8, find (g − f )(x).
If g(x) = 6x − 8, find − 32 g(x).
If g(x) = 2x2 + 3 and h(x) = −3x − 6, find(g · h)(x).
Evaluate and graph:
10.
11.
12.
13.
14.
15.
634
If f (x) = 6x + 4 and g(x) = −7x − 8, find ( f + g)(3).
If f (x) − 41 x + 3 and h(x) = 23 x + 6, find (g + h)(12).
If g(x) = 5x2 − 4x + 3 and h(x) = 2x − 7, find (g − h)(2).
If g(x) = 4x3 − 3x find 5g(6).
If h(x) = |4x − 7| find 2h(−5).
If f (x) = x + 4 and g(x) = 3x − 6, find ( f · g)(1).
www.ck12.org
www.ck12.org
Chapter 4. Unit 4
16. If h(x) = x4 and g(x) = x − 12, find (h · g)(−2).
Try these more challenging problems.
Solve and graph.
17. If f (x) = 4x − 7, g(x) = 3x + 18, and h(x) = −5x + 2, find ( f + g − h)(x).
18. If f (x) = 6x − 8,y(x) − 21 x, and h(x) = x + 4, find ( f · g · h)(x)
19. If g(x) = 3x − 7 and (g · h)(x) = 15x2 − 47x + 28, find (h)(x).
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.15.
635
4.3. CK-12 Calculus Study Guide - Compositions of Functions
www.ck12.org
4.3 CK-12 Calculus Study Guide Compositions of Functions
Important Points - Compositions of Functions
Function compositions are functions that are created through chains of 2 or more intermediate functions where the
output of one intermediate function is taken as the input to another intermediate function.
Notation
Two ways exist to write function compositions. For example, the final result in the diagram above can be written as:
• f (g(x)) or
• ( f ◦ g)(x)
Notes
• Be careful not to confuse ( f ◦ g)(x) with ( f · g)(x). The former represents a function composition while
the latter represents a multiplication operation between functions. ( f ◦ g)(x) = f (g(x)) while ( f · g)(x) =
f (x) · g(x).
• It is important to be cautious about the domain of a function created by function compositions. Due to the
nested nature of the intermediate functions, it is important to check at every step whether or not an intermediate
function can accept a certain value as an input. Once all the constraints are taken into account, then the Domain
has to be determined such that it only contains satisfactory values of the independent variable for all of the
intermediate functions in the function composition. Example 10 goes over this in detail.
Example 1
Question:
Let g(x) = x3 and f (x) = x2 − 1.What is f (g(x))?
Analysis:
636
www.ck12.org
Chapter 4. Unit 4
f (g(x)) = (g(x))2 − 1
= (x3 )2 − 1
= x6 − 1
Answer:
f (g(x)) = x6 − 1
Below is the graph of the function for reference:
( f (g(x)) = x6 − 1
Example 2
Question:
Let g(x) = 4x and f (x) = x2 . What is f (g(x))?
Analysis:
f (g(x)) = (g(x))2
= (4x)2
= 16x2
Answer:
f (g(x)) = 16x2
Below is the graph of the function for reference:
f (g(x)) = 16x2
637
4.3. CK-12 Calculus Study Guide - Compositions of Functions
Example 3
Question:
Let g(x) = x and f (x) = x2 + 1. What is f (g(x))?
Analysis:
f (g(x)) = (g(x))2 + 1
= (x)2 + 1
= x2 + 1
Note that here, f (x) = f (g(x)).
Answer:
f (g(x)) = x2 + 1
Below is the graph of the function for reference:
f (g(x)) = x2 + 1
638
www.ck12.org
www.ck12.org
Chapter 4. Unit 4
Example 4
Question:
Let g(x) = 4 and f (x) = x2 . What is f (g(x))? Is there anything interesting about this function?
Analysis:
f (g(x)) = (g(x))2
= (4)2
= 16
It is interesting to note that f (g(x)) = 16 for every input value of x.
Answer:
f (g(x)) = 16
Below is the graph of the function for reference:
f (g(x)) = 16
639
4.3. CK-12 Calculus Study Guide - Compositions of Functions
www.ck12.org
Extension:
If g( f (x)) is analysed in a similar manner, it is seen that g( f (x)) = 4. This means that here, g( f (x)) = g(x).
Example 5
Question:
Let g(x) = x + 1 and f (x) = x2 . What is f (g(x))? What is g( f (x))? What can be said in general about the
commutativity of function compositions?
Analysis:
f (g(x)) = (g(x))2
= (x + 1)2
g( f (x)) = f (x) + 1
= x2 + 1
We note that f (g(x)) 6= g( f (x))
Answer:
f (g(x)) = (x + 1)2
g( f (x)) = x2 + 1
In general, function compositions are not commutative. This is another way of saying that order usually matters for
function compositions.
Below are the graphs of the functions for reference:
f (g(x)) = (x + 1)2
640
www.ck12.org
Chapter 4. Unit 4
g( f (x)) = x2 + 1
Example 6
Question:
Let f (g(x)) = x2 + 4x + 4 and f (x) = x2 , What is g(x)?
Analysis:
First the given information is re-written in a different form. It is seen that f (g(x)) = (g(x))2 = x2 +4x+4. This means
that x2 + 4x + 4 must be the square of the algebraic expression in g(x). Next it can be seen that x2 + 4x + 4 = (x + 2)2 .
This means that g(x) = (x + 2).
Answer:
g(x) = (x + 2)
641
4.3. CK-12 Calculus Study Guide - Compositions of Functions
www.ck12.org
Below is the graph of the function for reference:
g(x) = (x + 2)
Example 7
Question:
Let h(x) = x2 − 1, g(x) = x − 1, f (x) = x. What is
h( f (x))
g(x)
− f (x)?
Analysis:
First it can be noted that this problem involves both function composition and arithmetic operations on functions.
h( f (x))
g(x)
− f (x) =
( f (x))2 −1
g(x)
− f (x)
=
(x)2 −1
(x−1)
−x
=
(x)2 −1
(x−1)
−x
=
(x+1)(x−1)
(x−1)
−x
= (x + 1) − x
= 1 for Domain: {x ∈ R|x 6= 1}
Note that there is a constraint on the domain due to the (x − 1) in the denominator.
Answer:
h( f (x))
g(x)
− f (x) = 1 for Domain : {x ∈ R|x 6= 1}
Below is the graph of the function for reference:
642
www.ck12.org
Chapter 4. Unit 4
h( f (x))
− f (x) = 1 for Domain: {x ∈ R|x 6= 1}
g(x)
Example 8
Question:
1
Let g(x) = |x|, h(x) = x 4 , f (x) = x + 1 and l(x) = x2 . Express k(x) = x, {x ∈ R|x ≥ 0} as a composition of one or
more of g(x), h(x), f (x), and l(x).
Analysis:
k(x) = x needs to be constructed step by step:
1. l(x) = x2
1
2. (h ◦ l)(x) = x 2
3. (l ◦ h ◦ l)(x) = x, {x ∈ R|x ≥ 0}
Note that this answer is not unique. For example, the expression (h ◦ l ◦ l)(x) = x is also valid.
Answer:
k(x) = (l ◦ h ◦ l)(x) = x, {x ∈ R|x ≥ 0}
Below is the graph of the function for reference:
k(x) = (l ◦ h ◦ l)(x) = x, {x ∈ R|x ≥ 0}
643
4.3. CK-12 Calculus Study Guide - Compositions of Functions
www.ck12.org
Example 9
Question:
1
Let g(x) = |x|, h(x) = x 4 , f (x) = x + 1 and l(x) = x2 . Express k(x) =
g(x), h(x), f (x) and l(x).
p
|x|+4 as a composition of one or more of
Analysis:
p
k(x) = |x|+4 needs to be constructed step by step:
1.
2.
3.
4.
5.
6.
7.
g(x) = |x|
( f ◦ g)(x) = |x|+1
( f ◦ f ◦ g)(x) = |x|+2
( f ◦ f ◦ f ◦ g)(x) = |x|+3
( f ◦ f ◦ f ◦ f ◦ g)(x) = |x|+4
1
(h ◦ f ◦ f ◦ f ◦ f ◦ g)(x) = (|x|+4)
p 4
(l ◦ h ◦ f ◦ f ◦ f ◦ f ◦ g)(x) = (|x|+4)
Note that this answer is not unique. For example, the expression (h ◦ l ◦ f ◦ f ◦ f ◦ f ◦ g)(x) =
Answer:
k(x) = (l ◦ h ◦ f ◦ f ◦ f ◦ f ◦ g)(x) =
p
(|x|+4)
Below is the graph of the function for reference:
k(x) = (l ◦ h ◦ f ◦ f ◦ f ◦ f ◦ g)(x) =
644
p
(|x|+4)
p
(|x|+4) is also valid.
www.ck12.org
Chapter 4. Unit 4
Example 10
Question:
Let g(x) =
√
x and f (x) = 1x . What is g( f (x))? What is the domain of (g ◦ f )(x)?
Analysis:
(g ◦ f )(x) = g(
pf (x))
g( f (x)) = r f (x)
1
g( f (x)) =
x
The output of f (x) is given as the input to g(x). However g(x) can only accept input values ≥ 0 because of the square
root operator. f (x) = 1x will only produce outputs that are satisfactory if x > 0. f (x) > 0 for x > 0. Note that x = 0
is not defined for f (x). Therefore the domain of (g ◦ f )(x) is x > 0.
Answer:
The domain of (g ◦ f )(x) is x > 0.
Below is the graph of the function for reference:
r
g( f (x)) =
1
x
645
4.3. CK-12 Calculus Study Guide - Compositions of Functions
646
www.ck12.org
www.ck12.org
Chapter 4. Unit 4
4.4 Function Composition
Here you will learn a new type of transformation called composition. Composing functions means having one
function inside the argument of another function. This creates a brand new function that may not look like a regular
transformation of any of the basic functions.
Functions can be added, subtracted, multiplied and divided creating new functions and graphs that are complicated
combinations of the various original functions. One important way to transform functions is through function
composition. Function composition allows you to line up two or more functions that act on an input in tandem.
Is function composition essentially the same as multiplying the two functions together?
Composition of Functions
A common way to describe functions is a mapping from the domain space to the range space:
Function composition means that you have two or more functions and the range of the first function becomes the
domain of the second function.
There are two notations used to describe function composition. In each case the order of the functions matters
because arithmetically the outcomes will be different. Squaring a number and then doubling the result will be
647
4.4. Function Composition
www.ck12.org
different from doubling a number and then squaring the result. In the diagram above, f (x) occurs first and g(x) occurs
second. This can be written as:
g( f (x)) or (g ◦ f )(x)
You should read this “g of f of x.” In both cases notice that the f is closer to the x and operates on the x values first.
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/57958
Examples
Example 1
Earlier, you were asked if function composition is the same as multiplying two functions together. Function
composition is not the same as multiplying two functions together. With function composition there is an outside
function and an inside function. Suppose the two functions were doubling and squaring. It is clear just by looking at
the example input of the number 5 that 50 (squaring then doubling) is different from 100 (doubling then squaring).
Both 50 and 100 are examples of function composition, while 250 (five doubled multiplied by five squared) is an
example of the product of two separate functions happening simultaneously.
For the next two examples, use the functions below:
f (x) = x2 − 1
h(x) =
g(x) =
j(x) =
x−1
x+5
3ex − x
√
x+1
Example 3
Show f (h(x)) 6= h( f (x))
2
x−1
f (h(x)) = f x+5
= x−1
−1
x+5
h( f (x)) = h(x2 − 1) =
(x2 −1)−1
(x2 −1)+5
=
x2 −2
x2 +4
In order to truly show they are not equal it is best to find a specific counter example of a number where they
differ. Sometimes algebraic expressions may look different, but are actually the same. You should notice that
f (h(x))is undefined when x = −5 because then there would be zero in the denominator. h( f (x)) on the other hand
is defined at x = −5. Since the two function compositions differ, you can conclude:
f (h(x)) 6= h( f (x))
Example 4
What is f j(h(g(x))) ?
648
www.ck12.org
Chapter 4. Unit 4
These functions are nested within the arguments of the other functions. Sometimes functions simplify significantly
when composed together, as f and j do in this case. It makes sense to evaluate those two functions first together and
keep them on the outside of the argument.
√
x − x; j(x) =
f (x) = x2 − 1; h(x) = x−1
;
g(x)
=
3e
x+1
x+5
p
p
2
f ( j(y)) = f
y+1 =
y+1 −1 = y+1−1 = y
Notice how the composition of f and j produced just the argument itself?
Thus,
f j(h(g(x))) = h(g(x)) = h(3ex − x)
(3ex − x) − 1
(3ex − x) + 5
3ex − x − 1
= x
3e − x + 5
=
For the next two examples, use the graphs shown below:
f (x) = |x|
g(x) = ex
h(x) = −x
649
4.4. Function Composition
www.ck12.org
Example 4
Compose g( f (x)) and graph the result. Describe the transformation.
g( f (x)) = g(|x|) = e|x|
The positive portion of the exponential graph has been mirrored over the y axis and the negative portion of the
exponential graph has been entirely truncated.
Example 5
Compose h(g(x)) and graph the result. Describe the transformation.
h(g(x)) = h(ex ) = −ex
650
www.ck12.org
Chapter 4. Unit 4
The exponential graph has been reflected over the x-axis.
Review
For questions 1-9, use the following three functions: f (x) = |x|, h(x) = −x, g(x) = (x − 2)2 − 3.
1. Graph f (x), h(x) and g(x).
2. Find f (g(x)) algebraically.
3. Graph f (g(x)) and describe the transformation.
4. Find g( f (x)) algebraically.
5. Graph g( f (x)) and describe the transformation.
6. Find h(g(x)) algebraically.
7. Graph h(g(x)) and describe the transformation.
8. Find g(h(x)) algebraically.
9. Graph g(h(x)) and describe the transformation.
For 10-16, use the following three functions: j(x) = x2 , k(x) = |x|, m(x) =
√
x.
10. Graph j(x), k(x) and m(x).
11. Find j(k(x)) algebraically.
12. Graph j(k(x)) and describe the transformation.
13. Find k(m(x)) algebraically.
14. Graph k(m(x)) and describe the transformation.
15. Find m(k(x)) algebraically.
16. Graph m(k(x)) and describe the transformation.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.11.
651
4.5. Composition of Functions
www.ck12.org
4.5 Composition of Functions
Learning Objectives
Here you will explore the composition of functions. Function composition refers to the combination of two functions
by using the output of one function as the input of the other.
If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x))?
A function can be conceptualized as a ’black box’. The input, or x value is placed into the box, and the box performs
a specific set of operations on it. Once the operations are complete, the output (the "f(x)" or "y" value) is retrieved.
Once the output is retrieved, the box is ready to work on the next input.
Using this idea, function composition can be seen as a box inside of a box. The input x value goes into the inner
box, and then the output of the inner box is used as the input of the outer box.
Composition of Functions
Functions are often described in terms of “input” and “output.” For example, consider the function f (x) = 2x +
3. When we input an x value, we output a y value, or a function value. We find the output by taking the input x,
multiplying by 2, and adding 3. We can do this for any value of x. Now consider a second function g(x) = 5x. For
this function too, we can take an x value, input the x into g(x), and obtain an output. What happens if we take the
output of g and use it as the input of f ?
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/176068
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/187163
Examples
Example 1
Earlier, you were given a problem about finding a composite function.
652
www.ck12.org
Chapter 4. Unit 4
If f(x) = x + 2, and g(x) = 2x + 4, what is f(g(x))?
f(g(x)) = f(2x + 4) = (2x + 4) + 2 = 2x + 6
Example 2
Given the function definition above, g(x) = 5x. Therefore if x = 4, then we have g(4) = 5(4) = 20. What happens if
we then take the output of 20 and use it as the input of f ?
Substituting 20 in for x in f (x) = 2x + 3 gives: f (20) = 2(20) + 3 = 43.
The table below shows several examples of this same process:
TABLE 4.2:
Output from g
10
15
20
25
x
2
3
4
5
Output from f
23
33
43
53
Examining the values in the table, we can see a pattern: all of the final output values from f are 3 more than 10 times
the initial input. We have created a new function called h(x) out of f (x) = 2x + 3 in which g(x) = 5x is the input:
h(x) = f (5x) = 2(5x) + 3 = 10x + 3
When we input one function into another, we call this the composition of the two functions. Formally, we write the
composed function as f (g(x)) = 10x + 3 or write it as (f o g)x = 10x + 3
Example 3
Find f (g(x)) and g(f (x)):
a. f (x) = 3x + 1 and g(x) = x2
f (g(x)) = f (x2 ) = 3(x2 ) + 1 = 3x2 + 1
g(f (x)) = g(3x + 1) = (3x + 1)2 = 9x2 + 6x + 1
In both cases, the resulting function is quadratic.
b. f (x) = 2x + 4 and g(x) = (1/2) x - 2
f (g(x)) = 2((1/2)x - 2) + 4 = (2/2)x - 4 + 4 = (2/2)x = x
g(f (x)) = g(2x + 4) = (1/2)(2x + 4) - 2 = x+ 2 - 2 = x.
In this case, the composites were equal to each other, and they both equal x, the original input into the function. This
means that there is a special relationship between these two functions. We will examine this relationship in Chapter
3. It is important to note, however, that f (g(x) is not necessarily equal to g(f (x)).
Example 4
Decompose the function f (x) = (3x - 1)2 - 5 into a quadratic function g(x) and a linear function h(x).
653
4.5. Composition of Functions
www.ck12.org
When we compose functions, we are combining two (or more) functions by inputting the output of one function into
another. We can also decompose a function. Consider the function f (x) = (2x + 1)2 . We can decompose this function
into an “inside” and an “outside” function. For example, we can construct f (x) = (2x+ 1)2 with a linear function and
a quadratic function. If g(x) = x2 and h(x) = (2x + 1), then f (x) = g(h(x)). The linear function h(x) = (2x + 1) is the
inside function, and the quadratic function g(x) = x2 is the outside function.
Let h(x) = 3x - 1 and g(x) = x2 - 5. Then f (x) = g(h(x)) because g(h(x)) = g(3x - 1) = (3x - 1)2 - 5.
The decomposition of a function is not necessarily unique. For example, there are many ways that we could express
a linear function as the composition of other linear functions.
Example 5
Given:
f (x) = 5x + 3
g(x) = 3x2
Find: f (g(4))
To find f (g(4)), we need to know what g(4) is, so we know what to substitute into f (x):
Substitute 4 for x for the function g(x), giving: 3 · 42
Simplify: 3 · 16 = 48
∴ g(4) = 48
Substitute 48 for the x in the function f (x) giving: 5(48) + 3
Simplify: 240 + 3 = 243
∴ f (g(4)) = 243
Example 6
Given:
h(n) = 7n + 1 + 4(g(n))
g(t) = −t
f (x) = −2x + g(x)
Find: f (h(−5))
First, let’s solve for the value of the inner function, h(−5). Then we’ll know what to plug into the outer function.
h(−5) = (7)(−5) + 1 + 4(g(−5))
To solve for the value of h, we need to solve g(−5)
g(−5) = −(−5)
∴ g(−5) = 5
Now we have: h(−5) = (7)(−5) + 1 + (4)(5)
Simplify to get: h(−5) = −14
Now we know that h(−5) = −14. That tells us that f (h(−5)) is f (−14)
Find f (−14) = (−2)(−14) + g(−14)
So to solve for the value of f (−14), we need to solve for the value of g(−14)
654
www.ck12.org
Chapter 4. Unit 4
g(−14) = −(−14)
∴ g(−14) = 14
Now we can finish up!
f (−14) = (−2)(−14) + 14
∴ f (−14) = 42
Review
For problems 1-4:
f (x) = 2x − 1 g(x) = 3x h(x) = x2 + 1
1.
2.
3.
4.
Find:
Find:
Find:
Find:
f (g(−3))
f (h(7))
h(g(−4))
f (g(h(2)))
Evaluate each composition below:
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
Given: f (x) = −5x + 2 and g(x) = 21 x + 4. Find f (g(12)).
Given: g(x) = −3x + 6 and h(x) = 9x + 3. Find g(h( 31 )).
Given: f (x) = − 15 x + 4 and g(x) = 4x2 . Find f (g(10)).
3
Given g(x) = 3|x
√ − 4|+6 and h(x) = −x . Find h(g(4)).
Given f (x) = x + 2 and g(x) = |2x|. Find g( f (−7)).
Given f (x) = −3x + 2 and given g(x) √
= 2x2 and given h(x) = 4|7 − x|+6. Find f (g(h(1))).
Given f (x) = (−3) and given g(x) = 2x and given h(x) = |4x|−12. Find f (h(g(18))).
Are compositions commutative? In other words, does f (g(x)) = g( f (x)) ?
Given: f (x) = −22 − 5x and h(x) = 3x + 2. Find f (h(x)).
Two functions are inverses of each other if f (g(x)) = x and g( f (x)) = x If f (x) = x + 3, find its inverse: g(x)
A toy manufacturer has a new product to sell. The number of units to be sold, n, is a function of the price p
such that: n(p) = 30 − 25p. The revenue r earned from the sales is a function of the number of units sold n
such that: r(n) = 1000 − 14 x2 . Find the function for revenue in terms of price, p.
Review (Answers)
To see the Review answers, open this PDF file and look for section 1.16.
655
4.6. Composition of Functions and Invertible Functions
www.ck12.org
4.6 Composition of Functions and Invertible
Functions
Your mom mixed some ingredients like flour, butter, sugar, choco powder etc. and bake all the ingredients in a cake
bowl and after some time, cake is ready. We can convert this activity into mathematical terms as functions.
Consider f to be a function of mixing of ingredients and g to be a function of baking and as a result we get a new
function h as a cake. Here the new function h is a combination of two functions f and g and hence called composite
function and is denoted as g ◦ f = h, i.e, baking of ingredients makes cake.
Let f : A → B and g : B → C be two functions. Then the composition of functions g ◦ f will be a function g ◦ f : A →
C given by
g ◦ f (x) = g( f (x)) ∀ x ∈ A
Concept
If f (x) = x + 2, and g(x) = 2x + 4, what is f (g(x)) ?
A function can be conceptualized as a “black box”. The input, or x value is placed into the box, and the box performs
a specific set of operations on x. Once the operations are complete, the output (the “ f (x)” or “y” value) is provided.
Once the output is provided, the box is ready to work on the next input.
Using this idea, function composition can be seen as a box inside of a box. The input x value goes into the inner
box, and then the output of the inner box is used as the input of the outer box.
This lesson is all about boxes inside of boxes. See if you can use what you learn to answer the question above before
the review at the end.
656
www.ck12.org
Chapter 4. Unit 4
Watch This
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/57958
Watch the video at: http://www.youtube.com/watch?v=qxBmISCJSME
Guidance
Composition of Functions
Functions are often described in terms of “input” and “output”. For example, consider the function f (x) = 2x + 3.
When we input an x value, we output a y value, or a function value. We find the output by taking the input x,
multiplying by 2, and adding 3. We can do this for any value of x. Now consider a second function g(x) = 5x. For
this function too, we can take an x value, input the x into g(x), and obtain an output. What happens if we take the
output of g and use it as the input of f ?
Example 1
Given the function definition above, g(x) = 5x. Therefore if x = 4, then we have g(4) = 5(4) = 20. What happens if
we then take the output of 20 and use it as the input of f ?
Solution:
Substituting 20 in for x in f (x) = 2x + 3 gives: f (20) = 2(20) + 3 = 43.
The table below shows several examples of this same process:
TABLE 4.3:
x
2
3
4
5
Output from g
10
15
20
25
Output from f
23
33
43
53
Examining the values in the table, we can see a pattern: all of the final output values from f are 3 more than 10 times
the initial input. We have created a new function called h(x) out of f (x) = 2x + 3 in which g(x) = 5x is the input:
h(x) = f (5x) = 2(5x) + 3 = 10x + 3
When we input one function into another, we call this the composition of the two functions. Formally, we write the
composed function as f (g(x)) = 10x + 3 or write it as ( f ◦ g)x = 10x + 3.
657
4.6. Composition of Functions and Invertible Functions
www.ck12.org
Example 2
Find f (g(x)) and g( f (x)):
a) f (x) = 3x + 1 and g(x) = x2
b) f (x) = 2x + 4 and g(x) =
1
2
x−2
Solution:
a) f (x) = 3x + 1 and g(x) = x2
f (g(x)) = f (x2 ) = 3(x2 ) + 1 = 3x2 + 1
g( f (x)) = g(3x + 1) = (3x + 1)2 = 9x2 + 6x + 1
In both cases, the resulting function is quadratic.
b) f (x) = 2x + 4 and g(x) = 12 x − 2
2
2
1
x−2 +4 =
x−4+4 =
x=x
f (g(x)) = 2
2
2
2
1
g( f (x)) = g(2x + 4) =
(2x + 4) − 2 = x + 2 − 2 = x.
2
In this case, the composites were equal to each other, and they both equal x, the original input into the function. This
means that there is a special relationship between these two functions. We will examine this relationship in later
concepts. It is important to note, however, that f (g(x)) is not necessarily equal to g( f (x)).
Example 3
Decompose the function f (x) = (3x − 1)2 − 5 into a quadratic function g(x) and a linear function h(x).
Solution:
When we compose functions, we are combining two (or more) functions by inputting the output of one function into
another. We can also decompose a function. Consider the function f (x) = (2x+1)2 . We can decompose this function
into an “inside” and an “outside” function. For example, we can construct f (x) = (2x + 1)2 with a linear function
and a quadratic function. If g(x) = x2 and h(x) = (2x + 1), then f (x) = g(h(x)). The linear function h(x) = (2x + 1)
is the inside function, and the quadratic function g(x) = x2 is the outside function.
Let h(x) = 3x − 1 and g(x) = x2 − 5. Then f (x) = g(h(x)) because
g(h(x)) = g(3x − 1) = (3x − 1)2 − 5.
The decomposition of a function is not necessarily unique. For example, there are many ways that we could express
a linear function as the composition of other linear functions.
Concept Question Wrap-up
Can you answer the question at the beginning of the lesson now?
If f (x) = x + 2, and g(x) = 2x + 4, what is f (g(x)) ?
f (g(x)) = f (2x + 4) = (2x + 4) + 2 = 2x + 6
Once you get the idea, composite functions aren’t as difficult as they look!
658
www.ck12.org
Chapter 4. Unit 4
Guided Practice
1. Given:
f (x) = 5x + 3
g(x) = 3x2
Find: f (g(4))
2. Given:
h(n) = 7n + 1 + 4(g(n))
g(t) = −t
f (x) = −2x + g(x)
Find: f (h(−5))
3. Given:
g(x) = 5x2
h(x) = 5x2 − 2x − 4(g(x))
Find h(g(−1))
Solutions:
1. To find f (g(4)), we need to know what g(4) is, so we know what to substitute into f (x):
Substitute 4 for x for the function g(x), giving: 3 · 42
Simplify: 3 · 16 = 48
∴ g(4) = 48
Substitute 48 for the x in the function f (x) giving: 5(48) + 3
Simplify: 240 + 3 = 243
∴ f (g(4)) = 243
2. First, let’s solve for the value of the inner function, h(−5). Then we’ll know what to plug into the outer function.
h(−5) = (7)(−5) + 1 + 4(g(−5))
To solve for the value of h, we need to solve g(−5)
g(−5) = −(−5)
∴ g(−5) = 5
Now we have: h(−5) = (7)(−5) + 1 + (4)(5)
Simplify to get: h(−5) = −14
Now we know that h(−5) = −14. That tells us that f (h(−5)) is f (−14)
Find f (−14) = (−2)(−14) + g(−14)
So to solve for the value of f (−14), we need to solve for the value of g(−14)
g(−14) = −(−14)
∴ g(−14) = 14
659
4.6. Composition of Functions and Invertible Functions
www.ck12.org
Now we can finish up!
f (−14) = (−2)(−14) + 14
∴ f (−14) = 42
3. First, solve for the value of the inner function g(−1) to find what to plug into the outer function h(g(−1))
g(−1) = 5(−1)2
g(−1) = 5 · 1
∴ g(−1) = 5
Next, solve for h(g(−1)) which we now know is: h(5)
h(5) = 5(52 ) + (−2)(5) − 4(g(5))
To solve for the value of h, we need to solve for the value of g(5).
g(5) = 5(52 )
g(5) = 125
∴ h(5) = 5(52 ) + (−2)(5) + (−4)(125)
Finally: h(5) = −385
Inverse Functions
Concept
Most of the functions from previous concepts have been polynomial functions, or rational functions, or functions
generated using a finite number of terms, involving only the operations of addition, subtraction, multiplication,
division, and raising to integer or fractional powers. These functions are algebraic functions, and when there is an
inverse, the inverse is also an algebraic function. Functions that are not algebraic functions are called transcendental
functions, i.e., functions that “transcend” algebra in the sense that they cannot be expressed in terms of a finite set of
the algebraic operations of addition/subtraction, multiplication/division, and power/root. The three groups of transcendental functions that will be discussed in subsequent concepts, along with their inverses, are the “elementary”
transcendentals:
• Exponential/inverse exponential (logarithmic) functions,
• Trigonometric/inverse trigonometric functions, and
• Hyperbolic/inverse hyperbolic functions
Have you heard of hyperbolic functions, or any other transcendental functions? Because the functions in each of the
above groups has an inverse useful for solving certain types of problems, you should know why a function inverse is
important, and how to use it. See if you can state, before you read further, the basic property of an inverse, and the
condition that allows a function to have an inverse.
660
www.ck12.org
Chapter 4. Unit 4
Watch This
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/57953
Watch the video at: http://www.youtube.com/watch?v=qgezKpQYH2w
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/79163
Watch the video at: http://www.youtube.com/watch?v=UKhwZbgaT5M
Guidance
In this section we will look at the following topics relating to inverse functions:
•
•
•
•
What does it mean to have an inverse?
When does a function have an inverse?
Finding the inverse of a function
Graphs of Inverse Functions
1. What does it mean to have an inverse?
If a function f (x) can be reversed in such a way that the input of the function becomes the output and the output
becomes an input, and the resulting reversed relationship is itself a function h(x), then h(x) is the inverse of
f (x). This relationship can be written as:
f ◦ h = f (h(x)) = h ◦ f = x
The inverse function h is often written as f −1 , which is not to be confused as 1f . In general, f −1 6= 1f .
If a function has an inverse, it is said to be invertible.
661
4.6. Composition of Functions and Invertible Functions
www.ck12.org
Example 4
The two functions f (x) = 2x + 3 and h(x) = x−3
2 are inverses of each other since
x−3 =
f ◦ h = f (h(x)) = 2 2 + 3 = x − 3 + 3 = x, and h ◦ f = h( f (x)) = (2x+3)−3
2
2x
x
=x
Thus
f ◦ h = h ◦ f = x,
and f and h are inverses of each other. The function f (x) = 2x + 3 is invertible.
2. When does a function have an inverse?
The answer to the question is:
Existence of an Inverse Function
A function f (x) has an inverse if it is one-to-one in its domain and onto, hence it is a bijective.
Example 5
Determine whether these functions have inverses, i.e., are invertible:
1
1. (a) f (x) = |x| (b) h(x) = x 2
2. f (x) = 3x5 + 2x + 1
Solution:
1. It is best to graph both functions and draw on each a horizontal line. As you can see from the graphs, f (x) = |x|
1
is not one-to-one since the horizontal line intersects it at two points. The function h(x) = x 2 however, is indeed
one-to-one since only one point is intersected by the horizontal line.
662
www.ck12.org
Chapter 4. Unit 4
2. Finding the inverse of a function
To find the inverse of a one-to-one and onto function, perform the following steps if possible:
1. solve for the independent variable x in terms of the dependent variable y; and then
2. interchange x and y.
The resulting formula is the inverse y = f −1 (x).
Of course, it is not always easy or possible to perform the first step.
Note: It is also acceptable to perform Step 2 above first, then to solve for the dependent variable y.
Example 6
Find the inverse of f (x) =
√
4x + 1.
Solution:
First of all, we know the function is one-to-one over the domain x ≥ 14 , and therefore it has an inverse.
From the discussion above, we first solve for x:
y=
√
4x + 1
y2 = 4x + 1
x=
y2 − 1
4
Next, perform the variable interchange x and y to determine the inverse
y=
x2 −1
4 .
Replacing y = f −1 (x),
f −1 (x) =
x2 −1
4
which is the inverse of the original function f (x) =
√
4x + 1.
663
4.6. Composition of Functions and Invertible Functions
www.ck12.org
3. Graphs of Inverse Functions
What is the relationship between the graphs of f and f −1 ? If the point (a, b) is on the graph of f (x), then from the
definition of the inverse, the point (b, a) is on the graph of f −1 (x). In other words, when we reverse the coordinates
of a point on the graph of f (x) we automatically get a point on the graph of f −1 (x). Examination of the resulting
graphs will show that the function and its inverse are reflections of one another about the line y = x. That is, each is
a mirror image of the other about the √
line y = x. The figure below shows an example of y = x2 and, when the domain
is restricted to x ≥ 0, its inverse y = x and how they are reflected about y = x.
It is important to note that for the function f (x) = x2 to have an inverse, we must restrict its domain to 0 ≤ x < ∞,
since that is the domain in which the function is increasing.
Concept Question Wrap-up
Have you heard of hyperbolic functions, or any other transcendental functions?
See if you can state, before you read further, the basic property of an inverse, and the condition that allows a function
to have an inverse.
Other than the exponential, logarithmic, and trigonometric functions, you may not have heard of any other transcendental functions. But, there are quite a few that are used extensively in science and engineering applications,
including the hyperbolic functions which have properties similar to the trigonometric functions but are made up of
exponential functions. Other transcendental functions have names such as Bessel functions and Hankel functions,
and arise in specialized areas.
The basic property of the function inverse when it exists is: f −1 ◦ f = x = f ◦ f −1 .
For a function, f (x), to have an inverse , f −1 (x), the function must be one-to-one and onto.
Example 7
Using composition of functions, check is the function f : R → R defined as f (x) = 2x + 3 is
invertible and hence find its inverse.
Solution:
Let y be any element of Range f , then
664
www.ck12.org
Chapter 4. Unit 4
y = f (x)
⇒ y = 2x + 3
⇒ x = y−3
2 = g(y)
Consider ( f ◦ g)(y) = f (g(y)) = f
y−3
2
And (g ◦ f )(x) = g( f (x)) = g(2x + 3) =
=2
y−3
2
(2x+3)−3
2
+ 3 = y = IR
= x = IR
Hence, g ◦ f = f ◦ g = IR
Thus f is invertible and f −1 (x) =
x−3
2
Multimedia links
Composite Functions - Overview
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/142595
Watch the video at: http://www.ck12.org/flx/render/embeddedobject/142595
Composite Functions - Example 1
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/142596
Watch the video at: http://www.ck12.org/flx/render/embeddedobject/142596
Composite Functions - Example 2
Evaluate using a composition of functions
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/142597
Watch the video at: http://www.ck12.org/flx/render/embeddedobject/142597
665
4.6. Composition of Functions and Invertible Functions
www.ck12.org
Composite Functions - Example 3
Decompose a function into two functions
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/142599
Watch the video at: http://www.ck12.org/flx/render/embeddedobject/142599
How to find inverse of a function by using composition of functions? Let’s learn it by using following video link:
MEDIA
Click image to the left or use the URL below.
URL: http://www.ck12.org/flx/render/embeddedobject/142601
Watch the video at: https://www.youtube.com/watch?v=ipbMdjq3yjU
Practice
For problems 1-4: f (x) = 2x − 1 and g(x) = 3x and h(x) = x2 + 1.
1. Find: f (g(−3))
2. Find: f (h(7))
3. Find: h(g(−4))
4. Find: f (g(h(2)))
Evaluate each composition below:
5. Given: f (x) = −5x + 2 and g(x) = 21 x + 4 Find f (g(12))
6. Given: g(x) = −3x + 6 and h(x) = 9x + 3 Find g h 13
7. Given: f (x) = − 51 x + 4 and g(x) = 4x2 Find f (g(10))
8. Given: g(x) = 3|x − 4|+6 and h(x) = −x3 Find h(g(4))
√
9. Given: f (x) = x + 2 and g(x) = |2x| Find g( f (−7))
10. Given f (x) = −3x + 2 and given g(x) = 2x2 and given h(x) = 4|7 − x|+6 Find f (g(h(1)))
√
11. Given f (x) = (−3) and given g(x) = 2x and given h(x) = |4x|−12 Find f (h(g(18)))
12. Are compositions commutative? In other words, does f (g(x)) = g( f (x))?
13. Given: f (x) = −22 − 5x and h(x) = 3x + 2 Find f (h(x))
14. Two functions are inverses of each other if f (g(x)) = x and g( f (x)) = x. If f (x) = x + 3, find its inverse: g(x).
15. A toy manufacturer has a new product to sell. The number of units to be sold, n, is a function of the price p such
that: n(p) = 30 − 25p. The revenue r earned from the sales is a function of the number of units sold n such
that: r(n) = 1000 − 14 x2 Find the function for revenue in terms of price, p.
666
www.ck12.org
Chapter 4. Unit 4
In problems #16 - 19, find the inverse function of f and verify that f ◦ f −1 = f −1 ◦ f = x.
16. f (x) = 3x + 1
√
17. 3 x
18. f (x) =
19. h(x) =
x−1
3
4−x
6
In problems #20 - 21, use the functions f (x) = x + 4 and g(x) = 2x − 5 to find the specified functions.
20. g−1 ◦ f −1
21. ( f ◦ g)−1
Find the inverse, if it exists, of the following functions.
22. f (x) =
6x−1
3x+7
23. f (x) = x2 − 4x + 8 for x ≥ 2
24. f (x) =
13x
11x+5
25. f (x) = x7 − 2
√
1+ x
√
26. f (x) =
1− x
667
4.7. Function Operations
www.ck12.org
4.7 Function Operations
Learning Objectives
Here you’ll learn to add, subtract, multiply, divide and compose two or more functions.
√
The area of a rectangle is 2x2 . The length of the rectangle is x + 3. What is the width of the rectangle? What
restrictions, if any, are on this value?
Function Operations
We have already dealt with adding, subtracting, and multiplying functions. To add and subtract, you combine like
terms. When multiplying, you either FOIL or use the “box” method. When you add, subtract, or multiply functions,
it is exactly the same as what you would do with polynomials, except for the notation. Note that we don’t need to
write out the entire function f (x) − g(x), just f − g, for example. Let’s continue:
f (x) − g(x) = (x + 5) − (x2 − 4x + 8)
= x + 5 − x2 + 4x − 8
= −x2 + 5x − 3
Distribute the negative sign to the second function and combine like terms. Be careful! f (x) − g(x) 6= g(x) − f (x).
Also, this new function, f (x) − g(x) has a different domain and range that either f (x) or g(x).
√
If f (x) = x − 8 and g(x) = 21 x2 , let’s find f g and gf and determine any restrictions for gf .
First, even though the x is not written along with the f (x) and g(x), it can be implied that f and g represent f (x) and
g(x).
√
√
f g = x − 8 · 21 x2 = 21 x2 x − 8
To divide the two functions, we will place f over g in a fraction.
√
√
x−8 = 2 x−8
f
=
1 2
g
x2
x
2
To find the restriction(s) on this function, we need to determine what value(s) of x make the denominator zero
because we cannot divide by zero. In this case x 6= 0. Also, the domain of f (x) is only x ≥ 8, because we cannot take
the square root of a negative number. The portion of the domain where f (x) is not defined is also considered part of
the restriction. Whenever there is a restriction on a function, list it next to the function, separated by a semi-colon.
We will not write x 6= 0 separately because it is included in x
668
www.ck12.org
Chapter 5. Exam Review
C HAPTER
5
Exam Review
Chapter Outline
5.1
E XAM R EVIEW Q UESTIONS
5.2
E XAM R EVIEW A NSWERS
669
5.1. Exam Review Questions
5.1 Exam Review Questions
670
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
671
5.1. Exam Review Questions
672
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
673
5.1. Exam Review Questions
674
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
675
5.2. Exam Review Answers
5.2 Exam Review Answers
676
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
677
5.2. Exam Review Answers
678
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
679
5.2. Exam Review Answers
680
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
681
5.2. Exam Review Answers
682
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
683
5.2. Exam Review Answers
684
www.ck12.org
www.ck12.org
Chapter 5. Exam Review
685