Grade 10
Mathematics
Collectable
Marks
ISBN: 978-0-620-94859-3
Publisher: 7M Education (Pty) Ltd
Author: Kabelo Sedumedi
Copyright © 7M Education (Pty) Ltd 2021
Table of Contents
Examination Overview .............................................................................................................................................. 2
Chapter 1: Algebra ...................................................................................................................................................... 3
Chapter 2: Number Patterns ................................................................................................................................ 12
Chapter 3: Finance ................................................................................................................................................... 15
Chapter 4: Functions ............................................................................................................................................... 23
Chapter 5: Probability ............................................................................................................................................ 53
Chapter 6: Statistics ................................................................................................................................................ 66
Chapter 7: Analytical Geometry ......................................................................................................................... 72
Chapter 8: Trigonometry ...................................................................................................................................... 75
1
Copyright reserved
Examination overview
1. Breakdown of marks per section in examinations
Paper 1
A) Algebra
B) Number Patterns
C) Finance
D) Functions
E) Probability
Total
Marks
30±3
15±3
10±3
30±3
15±3
100
%
±30%
±15%
±10%
±30%
±15%
100%
Paper 2
A) Statistics
B) Analytical Geometry
C) Trigonometry
D) Euclidean Geometry
Total
Marks
15±3
15±3
40±3
30±3
%
±15%
±15%
±40%
±30%
100
100%
2. Breakdown of types of questions in examinations
Cognitive levels
Knowledge
Routine
Procedures
Complex
Procedures
Problem Solving
%
Description of skills to be demonstrated
20% Straight recall e.g. domain and range
35% Routine exercises e.g. proofs of theorem proofs, solve for
𝑥, general solutions
30% Routine exercises that require higher reasoning. There is
often not an obvious route to the solution e.g. complex
differentiation
15% Non-routine exercises that require the ability to break
down the question into smaller parts (or sections), may
or may not be difficult
3. How to prepare for each type of question
Cognitive levels
Knowledge
Routine
Procedures
Complex
Procedures
Problem Solving
How to prepare
It is important for learners to be aware that 55% of the
examination papers is based on knowledge and routine
questions. These are basic questions. Learners assume that the
paper is based on mostly complex questions and hence place
much effort on complex questions without mastering the basics.
Learners should practise complex procedure and problem
solving questions only after they have mastered the knowledge
and routine procedure questions. It is important for learners to
understand the Mathematical language as well as to know how
to apply the Mathematics tips in order to be able to attempt
complex procedure and problem solving questions. Learners
should break down these questions into smaller sections so as
to make them simpler to understand and to solve.
2
Copyright reserved
Chapter 1: Algebra
1 Factorise p4
a) highest common factors
b) difference of 2 squares
c) sum or difference of 2 cubes
d) trinomials
e) grouping
2 Simplify p4
a) multiplying or dividing fractions
b) adding or subtracting fractions
3 Solve for 𝑥 p6
a) solve by factorisation
b) fractions
c) making 𝑥 the subject
4 Inequalities p8
a) solve for 𝑥
b) number line
c) interval notation
5 Simultaneous equations p8
Steps:
a) make 𝑥 or 𝑦 the subject in the simpler
equation
b) substitute 𝑥 or 𝑦 in the other equation and
solve
6 Exponents (simplifying) p9
Two types:
a) when no addition or subtraction between
bases
b) addition or subtraction between bases
7 Exponents (solve for 𝑥) p10
when 𝑥 is an exponent
3
Copyright reserved
1 Factorise
a) 𝑥 2 − 2𝑥
d) 9𝑥 2 − 4
g) 𝑥 2 − 3𝑥 + 2
j) 𝑥𝑦 − 2𝑦 + 𝑥 2 − 4
b) 2𝑥 2 𝑦 − 4𝑥𝑦 3
e) 𝑥 3 − 8
h) 3𝑥 2 − 𝑥 − 2
k) 𝑥𝑦 − 2𝑦 + 𝑥 3 − 8
c) 𝑥 2 − 4
f) 27𝑥 3 + 8
i) 𝑥𝑦 − 2𝑦 + 𝑥 2 − 2𝑥
Answers:
a) 𝑥 2 − 2𝑥
= 𝑥 (𝑥 − 2)
b) 2𝑥 2 𝑦 − 4𝑥𝑦 3
= 2𝑥𝑦 (𝑥 − 2𝑦 2 )
c) 𝑥 2 − 4
= (𝑥 − 2)(𝑥 + 2)
d) 9𝑥 2 − 4
= (3𝑥 − 2)(3𝑥 + 2)
e) 𝑥 3 − 8
= (𝑥 − 2)(𝑥 2 + 2𝑥 + 4)
f) 27𝑥 3 + 8
= (3𝑥 + 2)(9𝑥 2 − 6𝑥 + 4)
g) 𝑥 2 − 3𝑥 + 2
= (𝑥 − 1)(𝑥 − 2)
h) 3𝑥 2 − 𝑥 − 2
= (3𝑥 + 2)(𝑥 − 1)
i) 𝑥𝑦 − 2𝑦 + 𝑥 2 − 2𝑥
= 𝑦 (𝑥 − 2) + 𝑥 (𝑥 − 2)
= (𝑥 − 2)(𝑦 + 𝑥)
j) 𝑥𝑦 − 2𝑦 + 𝑥 2 − 4
= 𝑦(𝑥 − 2) + (𝑥 − 2)(𝑥 + 2)
= (𝑥 − 2)(𝑦 + 𝑥 + 2)
k) 𝑥𝑦 − 2𝑦 + 𝑥 3 − 8
= 𝑦(𝑥 − 2) + (𝑥 − 2)(𝑥 2 + 2𝑥 + 4)
= (𝑥 − 2)(𝑦 + 𝑥 2 + 2𝑥 + 4)
2 Simplify
𝑥 2 − 2𝑥
a) 2
𝑥 −4
𝑥 2 + 3𝑥 + 2 𝑥 2 + 𝑥
c)
÷
𝑥−8
𝑥−1
2
𝑥
e)
+
𝑥−2 𝑥+2
2
𝑥
g) 2
+ 2
𝑥 − 2𝑥 𝑥 − 4𝑥 + 4
𝑥 2 − 2𝑥 𝑥 2 + 3𝑥 + 2
b) 2
×
𝑥 −4
𝑥−8
𝑥 2 − 2𝑥 𝑥 2 + 3𝑥 + 2 𝑥 2 + 𝑥
d) 2
×
÷
𝑥 −4
𝑥−8
𝑥−1
2
𝑥
f) 2
+
𝑥 − 2𝑥 𝑥 2 − 4
4
Copyright reserved
Answers:
a)
𝑥 2 − 2𝑥
𝑥2 − 4
𝑥 (𝑥 − 2)
=
(𝑥 − 2)(𝑥 + 2)
𝑥
=
(𝑥 + 2)
𝑥 2 + 3𝑥 + 2 𝑥 2 + 𝑥
c)
÷
𝑥−8
𝑥−1
𝑥 2 + 3𝑥 + 2 𝑥 − 1
=
× 2
𝑥−8
𝑥 +𝑥
(𝑥 + 1)(𝑥 + 2)
𝑥−1
=
×
𝑥−8
𝑥 (𝑥 + 1)
(𝑥 + 2)(𝑥 − 1)
=
𝑥 (𝑥 − 8)
2
𝑥 − 𝑥 + 2𝑥 − 2
=
𝑥 2 − 8𝑥
2
𝑥 +𝑥−2
= 2
𝑥 − 8𝑥
e)
2
𝑥
+
𝑥−2 𝑥+2
2( 𝑥 + 2) + 𝑥 ( 𝑥 − 2)
=
(𝑥 − 2)(𝑥 + 2)
2𝑥 + 4 + 𝑥 2 − 2𝑥
= 2
𝑥 + 2𝑥 − 2𝑥 − 4
𝑥2 + 4
= 2
𝑥 −4
b)
𝑥 2 − 2𝑥 𝑥 2 + 3𝑥 + 2
×
𝑥2 − 4
𝑥−8
(𝑥 + 1)(𝑥 + 2)
𝑥 (𝑥 − 2)
=
×
(𝑥 − 2)(𝑥 + 2)
𝑥−8
(
)
𝑥 𝑥+1
=
𝑥−8
2
𝑥 +𝑥
=
𝑥−8
𝑥 2 − 2𝑥 𝑥 2 + 3𝑥 + 2 𝑥 2 + 𝑥
d) 2
×
÷
𝑥 −4
𝑥−8
𝑥−1
𝑥 2 − 2𝑥 𝑥 2 + 3𝑥 + 2 𝑥 − 1
= 2
×
× 2
𝑥 −4
𝑥−8
𝑥 +𝑥
(𝑥 + 1)(𝑥 + 2)
𝑥 (𝑥 − 2)
𝑥−1
=
×
×
(𝑥 − 2)(𝑥 + 2)
𝑥−8
𝑥 (𝑥 + 1)
𝑥−1
=
𝑥−8
f)
2
𝑥
− 2
− 2𝑥 𝑥 − 4
2
𝑥
=
−
𝑥 (𝑥 − 2) (𝑥 − 2)(𝑥 + 2)
2(𝑥 + 2) − 𝑥 (𝑥 )
=
𝑥 (𝑥 − 2)(𝑥 + 2)
2𝑥 + 4 − 𝑥 2
=
𝑥 (𝑥 2 + 2𝑥 − 2𝑥 − 4)
−𝑥 2 + 2𝑥 + 4
=
𝑥 (𝑥 2 − 4)
−𝑥 2 + 2𝑥 + 4
=
𝑥 3 − 4𝑥
𝑥2
5
Copyright reserved
g)
2
𝑥
+
𝑥 2 − 2𝑥 𝑥 2 − 4𝑥 + 4
2
𝑥
=
−
𝑥 (𝑥 − 2) (𝑥 − 2)(𝑥 − 2)
2
𝑥
=
−
𝑥 (𝑥 − 2) (𝑥 − 2)2
2( 𝑥 − 2) − 𝑥 ( 𝑥 )
=
𝑥 (𝑥 − 2)2
2𝑥 − 4 − 𝑥 2
=
𝑥 (𝑥 − 2)(𝑥 − 2)
−𝑥 2 + 2𝑥 − 4
=
𝑥 (𝑥 2 − 2𝑥 − 2𝑥 + 4)
−𝑥 2 + 2𝑥 − 4
=
𝑥 (𝑥 2 − 4𝑥 + 4)
−𝑥 2 + 2𝑥 − 4
= 3
𝑥 − 4𝑥 2 + 4𝑥
3 Solve for 𝑥 by factorising
Solve for 𝑥:
a) 𝑥 2 − 3𝑥 = 4
𝑥2 𝑥 1
d)
− =
12 4 3
g) 𝑥 − 𝑎 = 𝑏
𝑥
j) = 𝑏
𝑑
m) 𝑐𝑥 − 𝑑𝑥 = 𝑏
p) 𝑐𝑥 2 + 𝑎 = 𝑏
b) 𝑥 2 − 3𝑥 = 0
𝑥 2 − 2𝑥 𝑥 2 5𝑥
e)
=
−
3
4 12
h) 𝑥 + 𝑎 = 𝑏
c) 𝑥 2 = 4
𝑥2 1
f)
=
12 3
i) 𝑐𝑥 = 𝑏
k) 𝑥 2 = 𝑏
l) √𝑥 = 𝑏
𝑥
o) − 𝑎 = 𝑏
𝑑
n) 𝑐𝑥 + 𝑎 = 𝑏
q)
√𝑥
−𝑎 =𝑏
𝑑
r) 𝑐𝑥 − 𝑑𝑥 − 𝑎𝑑 = 𝑎𝑏
Answers:
a)
𝑥 2 − 3𝑥 = 4
𝑥 2 − 3𝑥 − 4 = 0
(𝑥 + 1)(𝑥 − 4) = 0
b)
𝑥 2 − 3𝑥 = 0
𝑥(𝑥 − 3) = 0
c)
𝑥2 = 4
𝑥2 − 4 = 0
(𝑥 − 2)(𝑥 + 2) = 0
𝑥 = 0 or 𝑥 = 3
𝑥 = −1 or 𝑥 = 4
𝑥 = 2 or 𝑥 = −2
6
Copyright reserved
𝑥2 𝑥 1
d)
− =
12 4 3
𝑥 2 − 3(𝑥 ) 4 (1)
=
12
12
𝑥 2 − 3𝑥 = 4
𝑥 2 − 3𝑥 − 4 = 0
(𝑥 + 1)(𝑥 − 4) = 0
𝑥 = −1 or 𝑥 = 4
𝑥 =𝑏−𝑎
𝑥
= 𝑏 (× 𝑑 )
𝑑
𝑥 = 𝑏𝑑
k) 𝑥 2 = 𝑏
(÷ 𝑐 )
l) √𝑥 = 𝑏
(square)
𝑥 = 𝑏2
𝑐𝑥 = 𝑏 − 𝑎
𝑏
𝑐−𝑑
p) 𝑐𝑥 2 + 𝑎 = 𝑏
𝑥=
(−𝑎)
𝑐𝑥 2 = 𝑏 − 𝑎
𝑏−𝑎
𝑥2 =
𝑐
(÷ 𝑐 )
(root)
𝑏−𝑎
𝑐
(−𝑎)
n) 𝑐𝑥 + 𝑎 = 𝑏
𝑥(𝑐 − 𝑑 ) = 𝑏 (÷ (𝑐 − 𝑑 ))
𝑥 = ±√
(root)
i) 𝑐𝑥 = 𝑏
𝑏
𝑥=
𝑐
𝑥 = ±√𝑏
m) 𝑐𝑥 − 𝑑𝑥 = 𝑏 (HCF)
𝑥=
𝑥 = 2 or 𝑥 = −2
(−𝑎)
h) 𝑥 + 𝑎 = 𝑏
𝑥 =𝑏+𝑎
j)
𝑥2 1
f)
=
12 3
𝑥 2 4(1 )
=
12
12
𝑥2 = 4
𝑥2 − 4 = 0
(𝑥 − 2)(𝑥 + 2) = 0
𝑥 = 0 or 𝑥 = 3
(+𝑎)
g) 𝑥 − 𝑎 = 𝑏
𝑥 2 − 2𝑥 𝑥 2 5𝑥
e)
=
−
3
4 12
4(𝑥 2 − 2𝑥) 3(𝑥 2 ) − 5𝑥
=
12
12
4𝑥 2 − 8𝑥 = 3𝑥 2 − 5𝑥
𝑥 2 − 3𝑥 = 0
𝑥(𝑥 − 3) = 0
q)
(÷ 𝑐 )
𝑏−𝑎
𝑐
√𝑥
=𝑏
𝑑
√𝑥 = 𝑏𝑑
(× 𝑑 )
(square)
𝑥 = (𝑏𝑑 )2
𝑥 = 𝑏2 𝑑2
o)
𝑥
−𝑎 =𝑏
𝑑
𝑥
=𝑏+𝑎
𝑑
Copyright reserved
(× 𝑑 )
𝑥 = 𝑑 (𝑏 + 𝑎 )
𝑥 = 𝑏𝑑 + 𝑎𝑑
r) 𝑐𝑥 − 𝑑𝑥 + 𝑎𝑑 = 𝑎𝑐
𝑐𝑥 − 𝑑𝑥 = 𝑎𝑐 − 𝑎𝑑
𝑥 (𝑐 − 𝑑 ) = 𝑎 (𝑐 − 𝑑 )
𝑥=
𝑎 (𝑐 − 𝑑 )
(𝑐 − 𝑑 )
𝑥=𝑎
7
(+𝑎)
4 Inequalities
Solve for 𝑥 and represent on a number line and in interval notation:
a) 2𝑥 − 4 < 8
b) 2𝑥 − 4 ≤ 8
c) − 2𝑥 + 4 ≤ −8
d) − 8 < 2𝑥 − 4 < 4
e) − 8 ≤ 2𝑥 − 4 ≤ 4
f) − 4 < −2𝑥 + 4 ≤ 8
Answers:
a) 2𝑥 − 4 < 8
2𝑥 < 12
𝑥<6
(+4)
(÷ 2)
6
b) 2𝑥 − 4 ≤ 8
2𝑥 ≤ 12
𝑥≤6
(+4)
(÷ 2)
6
𝑥 ∈ (−∞; 6)
6
𝑥 ∈ (−∞; 6]
𝑥 ∈ [6; ∞)
d) − 8 < 2𝑥 − 4 < 4 (+4) e) − 8 ≤ 2𝑥 − 4 ≤ 4 (+4)
−4 < 2𝑥 < 8 (÷ 2)
− 4 ≤ 2𝑥 ≤ 8 (÷ 2)
−2 <
𝑥 <4
−2≤
𝑥 ≤4
−2
4
𝑥 ∈ (−2; 4)
−2
4
𝑥 ∈ [−2; 4]
5 Simultaneous equations
a) Solve for 𝑥 and 𝑦 if 𝑥 − 3𝑦 = 1 and 2𝑥 + 5𝑦 = 13 .
Answers:
𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟏:
𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞 𝐢𝐧𝐭𝐨 𝟐:
𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞 𝐢𝐧𝐭𝐨 𝟏:
𝑥 − 3𝑦 = 1
𝑥 = 1 + 3𝑦
2𝑥 + 5𝑦 = 13
2(1 + 3𝑦) + 5𝑦 = 13
2 + 6𝑦 + 5𝑦 = 13
2 + 11𝑦 = 13 (−2)
11𝑦 = 11 (÷ 11)
𝑦=1
𝑥 = 1 + 3𝑦
𝑥 = 1 + 3(1)
𝑥=4
8
Copyright reserved
c) − 2𝑥 + 4 ≤ −8 (−4)
− 2𝑥 ≤ −12 (÷ −2)
𝑥≥6
f) − 4 < −2𝑥 + 4 ≤ 8
− 8 < −2𝑥 ≤ 4
4>
𝑥 ≥ −2
−2
𝑥 ∈ [−2; 4)
4
b) Solve for 𝑥 and 𝑦 if 2𝑥−1 = 8𝑦 and 2𝑥 + 5𝑦 = 13 .
Answers:
𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧 𝟏:
𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞 𝐢𝐧𝐭𝐨 𝟐:
𝐒𝐮𝐛𝐬𝐭𝐢𝐭𝐮𝐭𝐞 𝐢𝐧𝐭𝐨 𝟏:
2𝑥−1 = 8𝑦
2𝑥−1 = 23𝑦
𝑥 − 1 = 3𝑦
𝑥 = 1 + 3𝑦
2𝑥 + 5𝑦 = 13
2(1 + 3𝑦) + 5𝑦 = 13
2 + 6𝑦 + 5𝑦 = 13
2 + 11𝑦 = 13 (−2)
11𝑦 = 11 (÷ 11)
𝑦=1
𝑥 = 1 + 3𝑦
𝑥 = 1 + 3(1)
𝑥=4
6 Exponents (simplifying)
a) Simplify the following:
4𝑥+2 . 8𝑥−1
12𝑥+2 . 72𝑥−1
i.
ii. 𝑥−1
16𝑥−1 . 2𝑥+2
16 . 54𝑥+2
Answers:
4𝑥+2 . 8𝑥−1
12𝑥+2 . 72𝑥−1
(prime the bases) ii. 𝑥−1
(prime the bases)
i.
16𝑥−1 . 2𝑥+2
16 . 54𝑥+2
(22 )𝑥+2 . (23 )𝑥−1
(22 . 3)𝑥+2 . (23 . 32 )𝑥−1
=
=
(24 )𝑥−1 . 2𝑥+2
(24 )𝑥−1 . (2. 33 )𝑥+2
2𝑥+4 3𝑥−3
2𝑥+4 𝑥+2 3𝑥−3 2𝑥−2
2
.2
2
.3 .2
.3
= 4𝑥−4 𝑥+2
=
4𝑥−4
𝑥+2
3𝑥+6
2
.2
2
.2 .3
= 22𝑥+4+3𝑥−3−4𝑥+4−𝑥−2
= 22𝑥+4+3𝑥−3−4𝑥+4−𝑥−2 . 3𝑥+2+2𝑥−2−3𝑥−6
= 23
= 23 . 3−6
23
=8
= 6
3
8
=
729
9
Copyright reserved
b) Simplify the following:
2𝑥+2 − 2𝑥
4𝑥+2 + 22𝑥−1
i.
ii.
2𝑥+3
11. 4𝑥+1
Answers:
2𝑥+2 − 2𝑥
i.
2𝑥+3
2 𝑥 . 22 − 2 𝑥
=
2 𝑥 . 23
𝑥( 2
2 2 − 1)
=
2 𝑥 . 23
4−1
=
8
3
=
8
4𝑥+2 + 22𝑥−1
ii.
11. 4𝑥+1
(22 )𝑥+2 + 22𝑥−1
=
11. (22 )𝑥+1
2𝑥+4
2
+ 22𝑥−1
=
11. 22𝑥+2
2𝑥 4
2 . 2 + 22𝑥 . 2−1
=
11. 22𝑥 . 22
2𝑥 ( 4
2 2 + 2−1 )
=
11. 22𝑥 . 22
1
16 +
2
=
11.4
33
= 2
44
33 1
=
×
2 44
3
=
8
(HCF is 2𝑥 )
(prime the bases)
(HCF is 22𝑥 )
7 Exponents (solve for 𝑥)
Solve for 𝑥:
a) 4
𝑥+1
=8
2𝑥+2
d) 2𝑥+2 = 1
b) 4
𝑥+1
=
1
1
2
c) ( 𝑥+1 ) = 82𝑥+2
2
𝑥+2
f) 2
− 3. 2𝑥 = 4
82𝑥+2
𝑥+2
e) 2
+ 2𝑥 = 20
Answers:
a)
4𝑥+1 = 82𝑥+2 (prime the bases)
b)
(22 )𝑥+1 = (23 )2𝑥+2
22𝑥+2 = 26𝑥+6
2𝑥 + 2 = 6𝑥 + 6
−4𝑥 = 4
𝑥 = −1
10
Copyright reserved
4𝑥+1 =
1
(prime the bases)
82𝑥+2
1
(22 )𝑥+1 = 3 2𝑥+2
(2 )
1
22𝑥+2 = 6𝑥+6
2
22𝑥+2 = 2−6𝑥−6
2𝑥 + 2 = −6𝑥 − 6
8𝑥 = −8
𝑥 = −1
1
2
(prime the 8)
) = 82𝑥+2
2𝑥+1
(2−𝑥−1 )2 = (23 )2𝑥+2
2−2𝑥−2 = 26𝑥+6
−2𝑥 − 2 = 6𝑥 + 6
−8𝑥 = 8
𝑥 = −1
c) (
e) 2𝑥+2 + 2𝑥 = 20
2𝑥 . 22 + 2𝑥 = 20
2𝑥 (22 + 1) = 20
2𝑥 (4 + 1) = 20
2𝑥 (5) = 20
2𝑥 = 4
2 𝑥 = 22
𝑥=2
d) 2𝑥+2 = 1
2𝑥+2 = 20
𝑥+2=0
𝑥 = −2
f) 2𝑥+2 − 3. 2𝑥 = 4
2𝑥 . 22 − 3. 2𝑥 = 4
2 𝑥 (22 − 3) = 4
2 𝑥 (4 − 3) = 4
2 𝑥 (1) = 4
2𝑥 = 4 (prime the 4)
2 𝑥 = 22
𝑥=2
(÷ 5)
(prime the 4)
11
Copyright reserved
(20 = 1)
Chapter 2: Number Patterns
Section 1: Definitions
 𝒏: 𝑛 is the position of each term. 𝑛 cannot be a negative number, zero, a decimal or a
fraction.
 𝑻𝒏 : 𝑇𝑛 is the term in the sequence. This is the actual number that you see in the
sequence.
Section 2: Sequence specific summary
Subsection
1 Term formula p13
2
3
4
5
Variables
Difference p13
Function
Function formulae
Linear
𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇𝑛 = 𝑑𝑛 + 𝑐
𝑎 = 𝑇1
𝑑 = 𝑇𝑛 − 𝑇𝑛−1
𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
Linear function
𝑇𝑛 = 𝒅𝑛 + 𝑐
𝑦 = 𝒎𝑥 + 𝑐
12
Copyright reserved
Examples
1 Term formula
𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
a) Given 1; 3; 5; 7 …
i. Determine the next 2 terms.
ii. Determine the nth term (or the general term or the 𝑇𝑛 formula).
iii. What is the 20th term?
iv. Which term is 57?
Answers:
ii. 𝑇𝑛 = 𝑎 + (𝑛 − 1)𝑑
𝑇𝑛 = 1 + (𝑛 − 1)(2)
𝑇𝑛 = 1 + 2𝑛 − 2
𝑇𝑛 = 2𝑛 − 1
i. 1 ; 3 ; 5 ; 7 ; 𝟗 ; 𝟏𝟏
2
2
2
𝟐
𝟐
iii. 𝑇𝑛 = 2𝑛 − 1
𝑇20 = 2(20) − 1
𝑇20 = 39
3
iv. 57 = 2𝑛 − 1
58 = 2𝑛
29 = 𝑛
Difference
𝑇2 − 𝑇1 = 𝑇3 − 𝑇2
a) Determine the value of 𝑥 if the following sequence is an arithmetic (linear) sequence:
𝑥 − 1; 5 − 𝑥; 3𝑥 − 1
Answers:
𝑥−1
;
5−𝑥
;
3𝑥 − 1
5 − 𝑥 − (𝑥 − 1) = 3𝑥 − 1 − (5 − 𝑥)
5 − 𝑥 − 𝑥 + 1 = 3𝑥 − 1 − 5 + 𝑥
6 − 2𝑥 = 4𝑥 − 6
−6𝑥 = −12
𝑥=2
𝐂𝐡𝐞𝐜𝐤: if 𝑥 = 2
1
;
2
3
;
5
(𝐂𝐨𝐦𝐦𝐨𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐜𝐞)
2
13
Copyright reserved
b) Determine the values of 𝑥 and 𝑦 if the following sequence is an arithmetic (linear)
sequence: 1; 𝑥; 𝑦; 7
Answers:
1
;
𝑥
𝑥−1
𝑦−𝑥
𝑥−1=𝑦−𝑥
(𝟏 )
2𝑥 − 𝑦 = 1
(𝟏 ):
𝐒𝐮𝐛𝐬 (𝟐):
;
2𝑥 − 𝑦 = 1
−𝑦 = 1 − 2𝑥
𝑦 = −1 + 2𝑥
or
𝑦
;
7
7−𝑦
𝑦−𝑥 =7−𝑦
(𝟐 )
2𝑦 − 𝑥 = 7
(÷ −1)
2𝑦 − 𝑥 = 7
2(−1 + 2𝑥) − 𝑥 = 7
−2 + 4𝑥 − 𝑥 = 7
−2 + 3𝑥 = 7
3𝑥 = 9
𝑥=3
𝐒𝐮𝐛𝐬 (𝟏):
𝑦 = −1 + 2𝑥
𝑦 = −1 + 2(3)
𝑦=5
𝐂𝐡𝐞𝐜𝐤: if 𝑥 = 3 and 𝑦 = 5
1
;
2
3
;
2
5
;
7
(𝐂𝐨𝐦𝐦𝐨𝐧 𝐝𝐢𝐟𝐟𝐞𝐫𝐞𝐧𝐜𝐞)
2
14
Copyright reserved
Chapter 3: Finance
Section 1: Definitions




𝑨: this the accumulated value.
𝒊: 𝑖 is the interest rate.
𝒏: this is the number of years.
𝑷: this is the principal value.
Section 2: Formula specific summary
1. Increasing interest
a) 𝐴 = 𝑃(1 + 𝑖𝑛) p16 Simple interest rate
b) 𝐴 = 𝑃(1 + 𝑖)𝑛 p16 Compound interest rate or Inflation
2. Hire purchase
𝐴 = 𝑃(1 + 𝑖𝑛) p18
3. Interest
Interest = 𝐴 − 𝑃 (or Interest = 𝑃𝑖𝑛 for simple interest) p21
4. Exchange rates
a) One pair of currencies p21
b) More than one pair of currencies p22
15
Copyright reserved
1
𝐴 = 𝑃(1 + 𝑖𝑛) or 𝐴 = 𝑃(1 + 𝑖)𝑛
a) Thabo deposits R10000 into an account. Determine how much Thabo will have in
his account 3 years after his deposit if the interest rate is 12% p.a.
Answers:
𝐴 = 𝑃(1 + 𝑖𝑛) (interest rate is not compounded hence this is a simple interest rate)
𝐴 = 10000(1 + 0,12 × 3)
𝐴 = R13600
b) Palesa borrows R12000 from the bank. Determine how much Palesa will have to
pay back to the bank after 5 years if the interest rate is 10% p.a. compounded
annually.
Answers:
𝐴 = 𝑃(1 + 𝑖)𝑛 (interest rate is compounded hence this is a compound interest rate)
𝐴 = 12000(1 + 0,10)5
𝐴 = R19326,12
c) Thabo deposits an amount into an account. The interest rate is 12% p.a. If Thabo
has R13600 in his account 3 years after his deposit, determine how much Thabo
deposited.
Answers:
𝐴 = 𝑃(1 + 𝑖𝑛)
13600 = 𝑃(1 + 0,12 × 3)
R10000 = 𝑃
(÷ (1 + 0,12 × 3))
d) Palesa borrows an amount from the bank. The interest rate is 10% p.a.
compounded annually. If Palesa pays back R19326,12 after 5 years, determine how
much Palesa borrowed.
Answers:
𝐴 = 𝑃(1 + 𝑖)𝑛
19326,12 = 𝑃(1 + 0,10)5
R12000 = 𝑃
(÷ (1 + 0,10)5 )
16
Copyright reserved
e) Thabo deposits R10000 into an account. If Thabo has R13600 in his account 3 years
after his deposit, determine the interest rate.
Answers:
𝐴 = 𝑃(1 + 𝑖𝑛)
13600 = 10000(1 + 𝑖 × 3)
1,36 = 1 + 3𝑖
0,36 = 3𝑖
0,12 = 𝑖
12% = 𝑖
(÷ 10000)
(−1)
(÷ 3)
(× 100%)
f) Palesa borrows R12000 from the bank. If Palesa pays back R19326,12 after 5 years,
determine the interest rate if the interest rate is compounded annually .
Answers:
𝐴 = 𝑃(1 + 𝑖)𝑛
19326,12 = 12000(1 + 𝑖)5
1,61051 = (1 + 𝑖)5
1,1 = 1 + 𝑖
0,1 = 𝑖
10% = 𝑖
(÷ 12000)
5
(√5th root)
(−1)
(× 100%)
g) Thabo deposits R10000 into an account. The interest rate is 12% p.a. Determine
how many years it will take Thabo to have R13600 in his account.
Answers:
𝐴 = 𝑃(1 + 𝑖𝑛)
13600 = 10000(1 + 0,12 × 𝑛)
1,36 = 1 + 0,12𝑛
0,36 = 0,12𝑛
3 years = 𝑛
(÷ 10000)
(−1)
(÷ 0,12)
17
Copyright reserved
h) Palesa borrows R12000 from the bank. The interest rate is 10% compounded
annually. Palesa pays back R19326,12 at the end of the term of the loan. How long
(in years) is the term of the loan?
Answers:
𝐴 = 𝑃(1 + 𝑖)𝑛
19326,12 = 12000(1 + 0,10)𝑛
11 𝑛
1,61051 = ( )
10
5 years = 𝑛
2
(÷ 12000)
(by substitution)
Hire purchase
a) Thabo purchases a television set for R10000 on a hire purchase agreement. The
term of the agreement is 3 years and the interest rate on the loan is 12% p.a.
i. Determine the total amount that will be paid by the end of the three years.
ii. Determine the monthly payment.
Answers:
i. 𝐴 = 𝑃(1 + 𝑖𝑛)
𝐴 = 10000(1 + 0,12 × 3)
𝐴 = R13600
𝐴
number of payments
13600
monthly payment =
36
monthly payment = 𝑅377,78
ii. monthly payment =
(Note: 12 × 3 = 36)
b) Thabo purchases a television set for R10000. He has to pay a 10% cash deposit on
the cost. The rest of the amount will be paid by a hire purchase agreement. The
term of the agreement is 3 years and the interest rate on the loan is 12% p.a.
i. Determine the deposit amount.
ii. Determine the total amount that will be paid by the end of the three years.
iii. Determine the monthly payment.
Answers:
i. Deposit = 10000 × 10%
Deposit = R1000
18
Copyright reserved
ii. Loan amount = 10000 − 1000
Loan amount = R9000
𝐴 = 𝑃(1 + 𝑖𝑛)
𝐴 = 9000(1 + 0,12 × 3)
𝐴 = R12240
Total = 1000 + 12240 = R13240
𝐴
number of payments
12240
monthly payment =
36
monthly payment = R340
iii. monthly payment =
(Note: 12 × 3 = 36)
c) Thabo purchases a television set for R10000. He has to pay a 10% cash deposit on
the cost. The rest of the amount will be paid by a hire purchase agreement. The
term of the agreement is 3 years and the interest rate on the loan is 12% p.a.
Thabo has to pay a monthly insurance premium of R15.
i. Determine the deposit amount.
ii. Determine the total amount that will be paid by the end of the three years.
iii. Determine the monthly payment.
Answers:
i. Deposit = 10000 × 10%
Deposit = R1000
ii. Loan amount = 10000 − 1000
Loan amount = R9000
𝐴 = 𝑃(1 + 𝑖𝑛)
𝐴 = 9000(1 + 0,12 × 3)
𝐴 = R12240
iii.
𝐴
number of payments
12240
=
36
= R340
Total = 1000 + 12240 = R13240
(Note: 12 × 3 = 36)
monthly payment = 340 + 15
monthly payment = R355
19
Copyright reserved
d) Thabo purchases a television set for a certain amount on a hire purchase
agreement. The term of the agreement is 3 years and the interest rate on the loan is
12% p.a. The monthly payment on the hire purchase agreement is R377,78.
i. Determine the total amount that will be paid by the end of the three years.
ii. Determine the loan amount.
Answers:
𝐴
number of payments
𝐴
(× 36)
377,78 =
36
R13600,08 = 𝐴
i. monthly payment =
ii.
𝐴 = 𝑃(1 + 𝑖𝑛)
13600,08 = 𝑃(1 + 0,12 × 3)
R10000,06 = 𝑃
e) Thabo purchases a television set for a certain amount on a hire purchase
agreement. He has to pay a 10% cash deposit on the cost. The rest of the amount
will be paid by a hire purchase agreement. The term of the agreement is 3 years
and the interest rate on the loan is 12% p.a. The monthly payment on the hire
purchase agreement is R340.
i. Determine the total amount that will be paid by the end of the three years
(excluding the deposit).
ii. Determine the loan amount.
iii. Determine the cost of the television.
Answers:
𝐴
number of payments
𝐴
340 =
36
R12240 = 𝐴
i. monthly payment =
ii.
𝐴 = 𝑃(1 + 𝑖𝑛)
12240 = 𝑃(1 + 0,12 × 3)
R9000 = 𝑃
9000
90%
Cost = R10000
iii. Cost =
(100% − 10% = 90%)
20
Copyright reserved
(× 36)
3
Interest = 𝐴 − 𝑃
i. Thabo deposits R10000 into an account. The interest rate is 12% p.a. Determine the
interest after 3 years.
Answers:
𝐴 = 𝑃(1 + 𝑖𝑛) (interest rate is not compounded hence this is a simple interest rate)
𝐴 = 10000(1 + 0,12 × 3)
𝐴 = R13600
Interest = 𝐴 − 𝑃
Interest = 13600 − 10000
Interest = R3600
OR
Interest = 𝑃𝑖𝑛
Interest = (10000)(0,12)(3)
Interest = R3600
ii. Palesa borrows R12000 from the bank. The interest rate is 10% p.a. compounded
monthly. Determine the interest after 5 years.
Answers:
𝐴 = 𝑃(1 + 𝑖)𝑛 (interest rate is compounded hence this is a compound interest rate)
𝐴 = 12000(1 + 0,1)5
𝐴 = R19326,12
Interest = 𝐴 − 𝑃
Interest = 19326,12 − 12000
Interest = R7326,12
4a) Exchange rates: One pair of currencies
The US Dollar to Rand exchange rate is $1 = R14.
i. If a phone costs $650, determine the cost of the phone in Rands.
ii. If a van costs R350 000, determine the cost of the van in US Dollars.
Answers:
i. 650 × 14 = R9100
ii. 350 000 ÷ 14 = R25000
21
Copyright reserved
4b) Exchange rates: More than one pair of currencies
The US Dollar to Rand exchange rate is $1 = R14.
The Euro to Rand exchange rate is €1 = R17.
The British Pound to Rand exchange rate is £0,05 = R1.
i. If a phone costs $650, determine the cost of the phone in Euros.
ii. If a van costs $25000, determine the cost of the van in British Pounds.
iii. If a laptop costs £430, determine the cost of the laptop in Euros.
Answers:
i. 650 × 14 = R9100
9100 ÷ 17 = €535,29
ii. 25000 × 14 = R350 000
350 000 × 0,05 = £17500
iii. 430 ÷ 0,05 = R8600
8600 ÷ 17 = €505,88
(convert US Dollars to 𝐑𝐚𝐧𝐝𝐬)
(convert 𝐑𝐚𝐧𝐝𝐬 to Euros)
(convert US Dollars to 𝐑𝐚𝐧𝐝𝐬)
(convert 𝐑𝐚𝐧𝐝𝐬 to British Pounds)
(convert British Pounds to 𝐑𝐚𝐧𝐝𝐬)
(convert 𝐑𝐚𝐧𝐝𝐬 to Euros)
22
Copyright reserved
Chapter 4: Functions
Section 1: Definitions
 Asymptotes: Asymptotes are lines that graphs continuously approach but never
touch. Asymptotes are vertical or horizontal straight line graphs and are therefore
written as equations.
 Axis of symmetry: An axis of symmetry is a line that divides a graph in half.
 Decreasing: Decreasing means that the 𝑦 values on a graph decrease as the 𝑥 values
increase.
 Domain: These are all the possible 𝑥 values.
 𝒇(𝒙): 𝑓(𝑥) is the 𝑦.
 Gradient: The gradient refers to the steepness and direction of the slope of a graph.
 Increasing: Increasing means that the 𝑦 values on a graph increase as the 𝑥 values
increase.
 Intercept: An intercept is a point where a graph meets an 𝑥 or 𝑦 axis.
 Intersection: An intersection is a point where two graphs meet. The 𝑥 and 𝑦 values of
the two graphs are equal at this point.
 Range: These are all the possible 𝑦 values.
 Transformation: Transformation refers to vertical shifts as well as reflections across
the 𝑥 axis. Transformations also include changes in steepness.
 Turning point: This is the point where the graph turns. The 𝑦 value at the turning
point is either the minimum or maximum value of the graph. Turning point only
applies to parabola (quadratic) functions.
23
Copyright reserved
Section 2: Formula specific summary
No. Subsection
1
Formulae
p28
2
𝑎 and 𝑚
Linear
𝑦 = 𝑎𝑥 + 𝑞
𝑦 = 𝑚𝑥 + 𝑐
Positive 𝒂
Parabola/quadratic
𝑦 = 𝑎𝑥 2 + 𝑞
Positive 𝒂
Hyperbola
𝑎
𝑦 = +𝑞
𝑥
Positive 𝒂
Exponential
𝑦 = 𝑎𝑏 𝑥 + 𝑞
Positive 𝒂
𝑏>1
0<𝑏<1
Negative 𝒂
Negative 𝒂
Negative 𝒂
Negative 𝒂
𝑏>1
0<𝑏<1
3
4
𝑐
𝑞
5
Turning
point p29
𝑥
asymptote
𝑦
asymptote
Axis of
symmetry
p29
6
7
8
𝑦 – intercept
𝑦 – intercept
𝑦 turning point
value
(0; 𝑞)
𝑦 asymptote
value
𝑦 asymptote
value
𝑥=0
𝑦=𝑞
𝑥=0
𝑦=𝑞
Increasing
function
𝑦 =𝑥+𝑞
Decreasing
function
𝑦 = −𝑥 + 𝑞
9
Steps for
drawing
graphs
p31
𝑥 – intercept
𝑦 – intercept
𝑥 – intercept
𝑦 – intercept
Turning point
24
Copyright reserved
𝑥 – intercept
𝑦 – intercept
𝑥 asymptote
𝑦 asymptote
𝑥 – intercept
𝑦 – intercept
𝑦 asymptote
No. Subsection
Linear
10 Finding the Find 𝑚 by
equation of using:
𝑦2 − 𝑦1
a graph
𝑚=
𝑥2 − 𝑥1
p33
𝑚1 = 𝑚2
Parabola/quadratic
𝑦 = 𝑎𝑥 2 + 𝑞
Hyperbola
𝑎
𝑦 = +𝑞
𝑥
Exponential
Equation will
be given
𝑥∈𝑅
𝑥 ∈ 𝑅; 𝑥 ≠ 0
𝑥∈𝑅
or
(−∞; 0) ∪ (0; ∞)
or
𝑥 < 0 or 𝑥 > 0
𝑦 ∈ 𝑅; 𝑦 ≠ 𝑞
𝑎 > 0; 𝑦 > 𝑞
or
or
(−∞; 𝑞) ∪ (𝑞; ∞) 𝑎 > 0;
or
𝑦 ∈ (𝑞; ∞)
𝑦 < 𝑞 or 𝑦 > 𝑞
𝑎 < 0; 𝑦 < 𝑞
or
𝑎 < 0;
𝑦 ∈ (−∞; 𝑞)
𝑚1 × 𝑚2 = −1
11
Domain
p38
12
Range
p39
Then use
𝑦 = 𝑚𝑥 + 𝑐
𝑥∈𝑅
𝑦∈𝑅
𝑎 > 0; 𝑦 ≥ 𝑞
or
𝑎 > 0; 𝑦 ∈ [𝑞; ∞)
𝑎 < 0; 𝑦 ≤ 𝑞
or
𝑎 < 0; 𝑦 ∈ (−∞; 𝑞]
25
Copyright reserved
Section 3: General function rules
1
a
b
c
d
e
f
g
h
i
j
k
l
m
n
o
p
2
a
Using graph to determine 𝒙 values
𝑓(0) p40
This represents the 𝑦 intercept(s).
𝑓 (𝑥) = 0 p40
This represents the 𝑥 intercept(s).
𝑓 (𝑥) = 𝑔(𝑥)
This is an intersection.
p40
𝑓 (𝑥) > 0 p41
Give the 𝑥 values where 𝑓(𝑥) is above the 𝑥 axis.
𝑓 (𝑥) < 0 p41
Give the 𝑥 values where 𝑓(𝑥) is below the 𝑥 axis.
(
)
𝑓 𝑥 > 𝑔(𝑥)
Give the 𝑥 values where 𝑓(𝑥) is above 𝑔(𝑥).
p42
𝑓 (𝑥) < 𝑔(𝑥)
Give the 𝑥 values where 𝑓(𝑥) is below 𝑔(𝑥).
p42
𝑓 (𝑥). 𝑔(𝑥) > 0 Give the 𝑥 values where both 𝑓(𝑥) and 𝑔(𝑥) are above the 𝑥 axis or
p43
both 𝑓(𝑥) and 𝑔(𝑥) are below the 𝑥 axis.
𝑓 (𝑥). 𝑔(𝑥) < 0 Give the 𝑥 values where one function is above the 𝑥 axis and the
p43
other function is below the 𝑥 axis.
𝑓 (𝑥 )
Give the 𝑥 values where both 𝑓(𝑥) and 𝑔(𝑥) are above the 𝑥 axis or
> 0 p44
both 𝑓(𝑥) and 𝑔(𝑥) are below the 𝑥 axis.
𝑔 (𝑥 )
𝑓 (𝑥 )
Give the 𝑥 values where one function is above the 𝑥 axis and the
< 0 p44
other function is below the 𝑥 axis.
𝑔 (𝑥 )
Increasing p45 Give the 𝑥 values where the 𝑦 values on a graph increase as the 𝑥
values increase.
Decreasing p45 Give the 𝑥 values where the 𝑦 values on a graph decrease as the 𝑥
values increase.
> or <
When we have these inequality signs we use round brackets ( or ).
When we have these inequality signs we use square brackets [ or ].
≥ or ≤
Asymptotes
𝑥 values that lie on 𝑥 asymptotes should have round brackets ( or )
p46
Transformations
𝑓(𝑥) → 𝑓(𝑥) + 𝑘
p47
b
c
𝑓(𝑥) → 𝑘. 𝑓(𝑥)
𝑓 (𝑥) → −𝑓(𝑥)
p48
This is a vertical shift. A positive 𝑘 means that we are shifting
upwards. A negative 𝑘 means that we are shifting downwards. We
write +𝑘 after the original equation.
This is a change in gradient.
This is a reflection across the 𝑥 axis. We write " − " in front of the
original equation.
26
Copyright reserved
3
a
b
c
d
e
f
g
Important points
Length p49
Vertical line
p49
Horizontal line
p49
(𝑥; 𝑓(𝑥))
Rational
hyperbolic
function p51
Average
gradient
Minimum or
maximum
Use 𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
Equation of a vertical line is 𝑥 = 𝑘 where 𝑘 is the 𝑥 value on the 𝑥
intercept.
Equation of a horizontal line is 𝑦 = 𝑘 where 𝑘 is the 𝑦 value on the 𝑦
intercept.
𝑓(𝑥) can be expressed in terms of 𝑥.
e.g. if 𝑓 (𝑥) = 2𝑥 2 we can express coordinates as (𝑥; 2𝑥 2 )
𝑎𝑥 + 𝑏
𝑎
Change format 𝑦 =
into format 𝑦 =
+ 𝑞 by using
𝑥−𝑝
𝑥−𝑝
long division or a simplifcation method.
𝑦2 − 𝑦1
𝑚=
𝑥2 − 𝑥1
This is the 𝑦 turning point value.
27
Copyright reserved
1
𝑦 = 𝑎𝑥 + 𝑞
a)
or 𝑦 = 𝑚𝑥 + 𝑐
𝑚>0
𝑚<0
𝑦
𝑦
𝑐
𝑦 = 𝑎𝑥 2 + 𝑞
b)
𝑐
𝑎>0
𝑎<0
(Note: 𝑥1 = −𝑥2 )
𝑦
𝑦
(0; 𝑞)
𝑥
𝑥1
𝑥
𝑥2
𝑥1
𝑥2
(0; 𝑞)
𝑦=
𝑎
+𝑞
𝑥
c)
𝑎>0
𝑎<0
𝑦
𝑦
𝑦=𝑞
𝑞
𝑦=𝑞
𝑞
𝑥
𝑥=0
𝑥=0
28
Copyright reserved
𝑥
𝑦 = 𝑎𝑏 𝑥 + 𝑞
d)
𝑎>0
𝑎<0
𝑦
𝑦
𝑦=𝑞
𝑞
𝑥
𝑞
5
𝑥
𝑦=𝑞
Turning point (0; 𝑞)
a) Determine the turning points of 𝑓 and 𝑔 if 𝑓(𝑥) = 2𝑥 2 − 8 and 𝑔(𝑥) = −2𝑥 2 + 8.
Answers:
𝑓(𝑥) = 2𝑥 2 − 8
𝑔(𝑥) = −2𝑥 2 + 8
Turning point is (0; −8)
Turning point is (0; 8)
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧:
𝑦
𝑦
−8
𝑥
𝑥
−8
8
Axis of symmetry
a) Determine the axis of symmetry of 𝑓 and 𝑔 if 𝑓(𝑥) = 2𝑥 2 − 8 and
𝑔(𝑥) = −2𝑥 2 + 8.
Answers:
𝑓(𝑥) = 2𝑥 2 − 8
𝑔(𝑥) = −2𝑥 2 + 8
𝑥=0
𝑥=0
29
Copyright reserved
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧:
𝑦
𝑦
𝑥
𝑥
𝑥=0
𝑥=0
b) Determine the axis of symmetry of 𝑓 if 𝑓(𝑥) =
4
+2
𝑥
Answers:
𝐈𝐧𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐚𝐱𝐢𝐬 𝐨𝐟 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐲
𝑦=𝑥+𝑞
𝑦=𝑥+2
𝐃𝐞𝐜𝐫𝐞𝐚𝐬𝐢𝐧𝐠 𝐚𝐱𝐢𝐬 𝐨𝐟 𝐬𝐲𝐦𝐦𝐞𝐭𝐫𝐲
𝑦 = −𝑥 + 𝑞
𝑦 = −𝑥 + 2
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧:
𝑦
𝑦=𝑥+2
2
𝑥
𝑦 = −𝑥 + 2
30
Copyright reserved
9
Steps for drawing graphs
a) Draw the graph of 𝑓(𝑥) = 2𝑥 − 4
Answers:
𝑥
𝑦
−2
−8
𝑦
−1
−6
0
−4
1
−2
2
0
0
−8
1
−6
2
0
𝑓
𝑥
2
(1; −2)
−4
(−1; −6)
(−2; −8)
b) Draw the graph of 𝑓(𝑥) = 2𝑥 2 − 8
Answers:
𝑥
𝑦
−2
0
−1
−6
𝑦
𝑓
−2
(−1; −6)
2
𝑥
(1; −6)
−8
31
Copyright reserved
c) Draw the graph of 𝑓(𝑥) =
4
+2
𝑥
Answers:
𝑥
𝑦
−2
0
−1
−2
0
error
𝒙 − 𝐚𝐬𝐲𝐦𝐩𝐭𝐨𝐭𝐞:
𝑥=0
1
6
𝑦
2
4
𝑓
(1; 6)
(2; 4)
𝒚 − 𝐚𝐬𝐲𝐩𝐦𝐭𝐨𝐭𝐞:
𝑦=2
𝑦=2
−2
(−1; −2)
𝑥=0
d) Draw the graph of 𝑓(𝑥) = 4. 2𝑥 − 16
Answers:
𝑥
𝑦
−2
−15
−1
−14
0
−12
𝒚 − 𝐢𝐧𝐭𝐞𝐫𝐜𝐞𝐩𝐭:
𝑦 = −16
1
−8
𝑦
𝑓
2
(−2; −15)
2
0
𝑥
(1; −8)
− 12
(−1; −14)
𝑦 = −12
32
Copyright reserved
10 Finding the equation of a graph
a) Determine the equation of 𝑓.
𝑦
𝑓
2
𝑥
−4
Answers:
𝑦2 − 𝑦1
𝑥2 − 𝑥1
−4 − 0
𝑚=
0−2
𝑚=2
𝑚=
𝑦 = 𝑚𝑥 + 𝑐
or
− 4 = (2)(0) + 𝑐
−4=𝑐
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − (−4) = 2(𝑥 − 0)
𝑦 + 4 = 2𝑥
𝑦 = 2𝑥 − 4
∴ 𝑦 = 2𝑥 − 4
b) Determine the equation of 𝑓 if 𝑓 is parallel to 𝑔 and 𝑔(𝑥) = 2𝑥 − 7 .
𝑦
𝑓
𝑔
𝑥
−4
Answers:
The gradient of 𝑔 is 2 therefore the gradient of 𝑓 will also be 2 since 𝑓 ∣∣ 𝑔.
𝑦 = 𝑚𝑥 + 𝑐
−4 = (2)(0) + 𝑐
−4 = 𝑐
or
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
𝑦 − (−4) = 2(𝑥 − 0)
𝑦 + 4 = 2𝑥
𝑦 = 2𝑥 − 4
∴ 𝑦 = 2𝑥 − 4
33
Copyright reserved
𝑥
c) Determine the equation of 𝑓 if 𝑓 is perpendicular to 𝑔 and 𝑔(𝑥) = − .
2
𝑦
𝑓
𝑥
−4
𝑔
Answers:
The gradient of 𝑔 is −
𝑚1 × 𝑚2 = −1
1
(− ) × 𝑚2 = −1
2
𝑚2 = 2
1
.
2
𝑦 = 𝑚𝑥 + 𝑐
or
− 4 = (2)(0) + 𝑐
−4=𝑐
d) Determine the equation of 𝑓 .
𝑓
𝑥
(−1; −6)
−8
Answers:
𝑦 = 𝑎𝑥 2 + 𝑞
−6 = 𝑎(−1)2 − 8
2 = 𝑎. 1
2=𝑎
[(−1)2 = 1 and − 6 + 8 = 2]
(𝑎. 1 = 𝑎)
𝑦 = 2𝑥 2 − 8
34
Copyright reserved
𝑦 − (−4) = 2(𝑥 − 0)
𝑦 + 4 = 2𝑥
𝑦 = 2𝑥 − 4
∴ 𝑦 = 2𝑥 − 4
𝑦
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
e) Determine the equation of 𝑓 .
𝑦
𝑓
2
𝑥
(−1; −6)
Answers:
𝑦 = 𝑎𝑥 2 + 𝑞
−6 = 𝑎(−1)2 + 𝑞
−6 = 𝑎. 1 + 𝑞
−6 = 𝑎 + 𝑞
(1)
𝑦 = 𝑎𝑥 2 + 𝑞
0 = 𝑎 (2)2 + 𝑞
0 = 4𝑎 + 𝑞
(𝟏) :
subs (𝟐):
subs (𝟏):
(2)
−6=𝑎+𝑞
−6 − 𝑞 = 𝑎
0 = 4𝑎 + 𝑞
0 = 4(−6 − 𝑞) + 𝑞
0 = −24 − 4𝑞 + 𝑞
0 = −24 − 3𝑞
3𝑞 = −24
𝑞 = −8
𝑎 = −6 − 𝑞
𝑎 = −6 − (−8)
𝑎=2
𝑦 = 2𝑥 2 − 8
35
Copyright reserved
f) Determine the equation of 𝑓 .
𝑓
𝑦=2
2
𝑥
(−1; −2)
Answers:
𝑎
+𝑞
𝑥
𝑎
−2 =
+2
−1
−4 = −𝑎
4=𝑎
𝑦=
𝑦=
(
𝑎
= −𝑎 and − 2 − 2 = −4)
−1
4
+2
𝑥
g) Determine the equation of 𝑓 .
𝑓
−2
(−1; −2)
𝑥
36
Copyright reserved
Answers:
𝑎
+𝑞
𝑥
𝑎
−2 =
+𝑞
−1
−2 = −𝑎 + 𝑞
𝑦=
𝑎
= −𝑎)
−1
(1)
(
𝑎
+𝑞
𝑥
𝑎
0=
+𝑞
−2
𝑎
(× −2)
−𝑞 =
−2
2𝑞 = 𝑎
(2)
𝑦=
(𝟏) :
− 2 = −𝑎 + 𝑞
𝑎 =𝑞+2
subs (𝟐):
2𝑞 = 𝑎
2𝑞 = 𝑞 + 2
𝑞=2
subs (𝟏):
𝑎 =𝑞+2
𝑎 =2+2
𝑎=4
𝑦=
4
+2
𝑥
h) Determine the equation of 𝑓 (𝑥) = 𝑏 𝑥 + 𝑞
𝑦
𝑓
2
𝑥
𝑦 = −4
−4
Answers:
𝑦 = 𝑏𝑥 + 𝑞
0 = 𝑏2 − 4
4 = 𝑏2
2=𝑏
𝑦 = 2𝑥 − 4
37
Copyright reserved
i) Determine the equation of 𝑓(𝑥) = 𝑎. 𝑏 𝑥 + 𝑞
𝑦
𝑓
2
𝑥
−9
−12
𝑦 = −12
Answers:
𝑦 = 𝑎. 𝑏 𝑥 + 𝑞
−9 = 𝑎. 𝑏0 − 12
3 = 𝑎. 1
3=𝑎
𝑦 = 𝑎. 𝑏 𝑥 + 𝑞
0 = 3. 𝑏2 − 12
12 = 3. 𝑏2
4 = 𝑏2
2=𝑏
[𝑏0 = 1 and − 9 + 12 = 3]
(÷ 3)
𝑦 = 3. 2𝑥 − 12
11 Domain
Determine the domain of 𝑓 .
𝑦
𝑓
2
𝑦=2
𝑥
𝑥=0
Answers:
𝑥 ∈ 𝑅; 𝑥 ≠ 0
𝐨𝐫
𝑥 ∈ (−∞; 0) ∪ (0; ∞) 𝐨𝐫
38
Copyright reserved
𝑥 < 0 or 𝑥 > 0
12 Range
a) Determine the range of 𝑓 and 𝑔 .
𝑦
𝑦
𝑓
8
𝑔
𝑥
𝑥
−8
Answers:
𝒇:
𝑦 ≥ −8
𝒈:
𝐨𝐫
𝑦 ∈ [−8; ∞)
𝑦≤8
𝐨𝐫
𝑦 ∈ (−∞; 8]
b) Determine the range of 𝑓 .
𝑦
𝑓
2
𝑦=2
𝑥
𝑥=0
Answers:
𝑦 ∈ 𝑅; 𝑦 ≠ 2
𝐨𝐫
𝑦 ∈ (−∞; 2) ∪ (2; ∞) 𝐨𝐫
39
Copyright reserved
𝑦 < 2 or 𝑦 > 2
c) Determine the range of 𝑓 and 𝑔 .
𝑦
𝑦
12
𝑦 = 12
𝑓
𝑥
𝑥
𝑦 = −12
𝑔
−12
Answers:
𝒇:
𝑦 > −12
𝐨𝐫
𝒈:
𝑦 < 12
𝑦 ∈ (−12; ∞)
𝐨𝐫
𝑦 ∈ (−∞; 12)
13 𝑓 (𝑥) = 0 , 𝑓(0) and 𝑓(𝑥) = 𝑔(𝑥)
Given 𝑓 and 𝑔 below:
𝑦
𝑓
E
D
A
B
𝑥
𝑔
C
Determine the coordinates at A , B , C , D and E if 𝑓(𝑥) = 2𝑥 2 − 8 and
𝑔(𝑥) = 2𝑥 + 4 .
Answers:
𝐀 𝐚𝐧𝐝 𝐁 (𝒇(𝒙) = 𝟎):
𝐂 𝐚𝐧𝐝 𝐃 (𝒇(𝟎)):
𝐄 (𝒇(𝒙) = 𝒈(𝒙)):
𝑓(𝑥) = 2𝑥 2 − 8
0 = 2𝑥 2 − 8 (÷ 2)
0 = 𝑥2 − 4
0 = (𝑥 + 2)(𝑥 − 2)
𝑥 = −2 or 𝑥 = 2
𝑓(𝑥) = 2𝑥 2 − 8
𝑓 (0) = 2( 0)2 − 8
𝑓(0) = −8
𝑓 (𝑥 ) = 𝑔 (𝑥 )
2𝑥 − 8 = 2𝑥 + 4
2
2𝑥 − 2𝑥 − 12 = 0 (÷ 2)
𝑥2 − 𝑥 − 6 = 0
(𝑥 + 2)(𝑥 − 3) = 0
𝑥 = −2 or 𝑥 = 3
𝑔(𝑥) = 2𝑥 + 4
𝑔 (0) = 2 (0) + 4
𝑔 (0) = 4
2
𝑓(𝑥) = 2𝑥 2 − 8
𝑓 (3) = 2(3)2 − 8
𝑓(3) = 10
A(−2; 0)
B(2; 0)
C(0; −8)
D(0; 4)
40
Copyright reserved
E(3; 10)
14 𝑓 (𝑥) > 0 and 𝑓(𝑥) < 0
a) Given 𝑓 below:
𝑦
−2
𝑓
2
𝑥
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥) > 0
ii. 𝑓(𝑥) ≥ 0
Answers:
i. 𝑥 ∈ (−∞; −2) ∪ (2; ∞)
ii. 𝑥 ∈ (−∞; −2] ∪ [2; ∞)
𝐨𝐫
𝐨𝐫
𝑥 < −2 or 𝑥 > 2
𝑥 ≤ −2 or 𝑥 ≥ 2
b) Given 𝑓 below:
𝑦
−2
𝑓
2
𝑥
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥) < 0
ii. 𝑓(𝑥) ≤ 0
Answers:
i. 𝑥 ∈ (−2; 2)
ii. 𝑥 ∈ [−2; 2]
𝐨𝐫
𝐨𝐫
−2<𝑥 <2
−2≤𝑥 ≤2
41
Copyright reserved
𝐨𝐫
𝐨𝐫
− 2 < 𝑥 and 𝑥 < 2
− 2 ≤ 𝑥 and 𝑥 ≤ 2
15 𝑓 (𝑥) > 𝑔(𝑥) and 𝑓(𝑥) < 𝑔(𝑥)
a) Given 𝑓 and 𝑔 below:
𝑓
𝑦
(3; 10)
−2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥) > 𝑔(𝑥)
ii. 𝑓(𝑥) ≥ 𝑔(𝑥)
Answers:
i. 𝑥 ∈ (−∞; −2) ∪ (3; ∞)
ii. 𝑥 ∈ (−∞; −2] ∪ [3; ∞)
𝐨𝐫
𝐨𝐫
𝑥 < −2 or 𝑥 > 3
𝑥 ≤ −2 or 𝑥 ≥ 3
b) Given 𝑓 and 𝑔 below:
𝑓
𝑦
(3; 10)
−2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥) < 𝑔(𝑥)
ii. 𝑓(𝑥) ≤ 𝑔(𝑥)
Answers:
i. 𝑥 ∈ (−2; 3)
ii. 𝑥 ∈ [−2; 3]
𝐨𝐫
𝐨𝐫
−2<𝑥 <3
−2≤𝑥 ≤3
42
Copyright reserved
𝐨𝐫
𝐨𝐫
− 2 < 𝑥 and 𝑥 < 3
− 2 ≤ 𝑥 and 𝑥 ≤ 3
16 𝑓 (𝑥). 𝑔(𝑥) > 0 and 𝑓 (𝑥). 𝑔(𝑥) < 0
a) Given 𝑓 and 𝑔 below:
𝑓
𝑦
−2
2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥). 𝑔(𝑥) > 0
ii. 𝑓(𝑥). 𝑔(𝑥) ≥ 0
Answers:
i. 𝑥 ∈ (2; ∞)
ii. 𝑥 ∈ [2; ∞)
𝐨𝐫
𝐨𝐫
𝑥>2
𝑥≥2
b) Given 𝑓 and 𝑔 below:
𝑓
𝑦
−2
2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
i. 𝑓(𝑥). 𝑔(𝑥) < 0
ii. 𝑓(𝑥). 𝑔(𝑥) ≤ 0
Answers:
i. 𝑥 ∈ (−∞; 2)
ii. 𝑥 ∈ (−∞; 2]
𝐨𝐫
𝐨𝐫
𝑥 < 2 ; 𝑥 ≠ −2
𝑥≤2
43
Copyright reserved
𝑓 (𝑥 )
17 𝑓 (𝑥)
> 0 and
<0
𝑔 (𝑥 )
𝑔 (𝑥 )
a) Given 𝑓 and 𝑔 below:
𝑓
𝑦
−2
2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
𝑓 (𝑥 )
i.
>0
𝑔 (𝑥 )
𝑓 (𝑥 )
ii.
≥0
𝑔 (𝑥 )
Answers:
i. 𝑥 ∈ (2; ∞)
ii. 𝑥 ∈ [2; ∞)
𝐨𝐫
𝐨𝐫
𝑥>2
𝑥≥2
b) Given 𝑓 and 𝑔 below:
𝑓
𝑦
−2
2
𝑥
𝑔
Determine the value(s) of 𝑥 for which:
𝑓 (𝑥 )
i.
<0
𝑔 (𝑥 )
𝑓 (𝑥 )
ii.
≤0
𝑔 (𝑥 )
Answers:
i. 𝑥 ∈ (−∞; 2) ; 𝑥 ≠ −2
𝐨𝐫
𝑥 < 2 ; 𝑥 ≠ −2
ii. 𝑥 ∈ (−∞; 2] ; 𝑥 ≠ −2
𝐨𝐫
𝑥 ≤ 2 ; 𝑥 ≠ −2
[𝑔 is a denominator and cannot equal 0 . ∴ we cannot include coordinate (−2; 0)]
44
Copyright reserved
18 Increasing and Decreasing
a) Given 𝑓 below:
𝑦
𝑓
𝑥
−8
i. Determine the value(s) of 𝑥 for which 𝑓 is increasing?
ii. Determine the value(s) of 𝑥 for which 𝑓 is strictly increasing?
Answers:
i. 𝑥 ∈ [0; ∞)
ii. 𝑥 ∈ (0; ∞)
𝐨𝐫
𝐨𝐫
𝑥≥0
𝑥>0
b) Given 𝑓 below:
𝑦
𝑓
𝑥
−8
i. Determine the value(s) of 𝑥 for which 𝑓 is decreasing?
ii. Determine the value(s) of 𝑥 for which 𝑓 is strictly decreasing?
Answers:
i. 𝑥 ∈ (∞; 0]
ii. 𝑥 ∈ (∞; 0)
𝐨𝐫
𝐨𝐫
𝑥≤0
𝑥<0
45
Copyright reserved
19 Asymptotes
a) Given 𝑓 below:
𝑦
𝑓
2
𝑦=2
−1
𝑥
𝑥=0
Determine the value(s) of 𝑥 for which 𝑓(𝑥) ≤ 0?
Answers:
𝑥 ∈ [−1; 0)
𝐨𝐫
−1≤𝑥 <0
𝐨𝐫
− 1 ≤ 𝑥 and 𝑥 < 0
b) Given 𝑓 below:
𝑦
𝑓
2
𝑦=2
−1
𝑥
𝑥=0
Determine the value(s) of 𝑥 for which 𝑓(𝑥) ≥ 0?
Answers:
𝑥 ∈ (−∞; −1] ∪ (0; ∞)
𝐨𝐫
𝑥 ≤ −1 or 𝑥 > 0
46
Copyright reserved
20 𝑓 (𝑥) → 𝑓(𝑥) + 𝑘
a) Given 𝑓 below:
𝑦
𝑓
𝑥
−8
Determine the coordinates at the turning point of:
i. 𝑗 if 𝑗(𝑥) = 𝑓(𝑥) − 3
ii. 𝑘 if 𝑘 (𝑥) = 𝑓(𝑥) + 3
Answers:
i. (0; −11)
ii. (0; −5)
(0 ; −𝟖 − 𝟑)
(0 ; −𝟖 + 𝟑)
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐬
𝑘
𝑦
𝑓
𝑥
𝑗
−5
−8
−11
b) If 𝑓(𝑥) = 2𝑥 2 − 8 determine the coordinates of the turning point of 𝑙 if
𝑙 (𝑥 ) = 𝑓 (𝑥 ) − 3
Answers:
𝑙 (𝑥 ) = 𝑓 (𝑥 ) − 3
𝑙(𝑥) = 2𝑥 2 − 8 − 3
𝑙(𝑥) = 2𝑥 2 − 11
The turning point coordinates are (0; −11) .
47
Copyright reserved
21 𝑓 (𝑥) → −𝑓(𝑥)
a) Given 𝑓 below:
𝑦
𝑓
𝑥
−8
Determine the coordinates at the turning point of:
i. ℎ if ℎ(𝑥) = −𝑓(𝑥)
ii. 𝑗 if 𝑗(𝑥) = −𝑓 (𝑥) − 3
Answers:
i. (1; 8)
ii. (1; 5)
(1 ; −(−𝟖))
(1 ; −(−𝟖) − 𝟑)
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧𝐬
𝑦
8
𝑦
𝑓
5
𝑓
(1; 5)
𝑥
𝑥
ℎ
𝑗
−8
−8
b) If 𝑓(𝑥) = 2𝑥 2 − 8 determine the coordinates of the turning point of 𝑗 if
𝑗(𝑥) = −𝑓 (𝑥) − 3 .
Answer:
𝑗(𝑥) = −𝑓 (𝑥) − 3
𝑗(𝑥) = −(2𝑥 2 − 8) − 3
𝑗(𝑥) = −2𝑥 2 + 8 − 3
𝑗(𝑥) = −2𝑥 2 + 5
The turning point coordinates are (0; 5) .
48
Copyright reserved
22 Vertical line, horizontal line and length
a) Given the graphs below:
𝑦
3
1
𝑥
Determine the equations of:
i. vertical line if it is parallel to the 𝑦 axis
ii. horizontal line if it is parallel to the 𝑥 axis
Answer:
i. 𝑥 = 1
ii. 𝑦 = 3
b) Given 𝑓 below:
𝑦
−1
𝑥
−3
Determine the equations of:
i. vertical line if it is parallel to the 𝑦 axis
ii. horizontal line if it is parallel to the 𝑥 axis
Answer:
i. 𝑥 = −1
ii. 𝑦 = −3
49
Copyright reserved
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧:
𝑥 is 1 throughout the line
𝑦
𝑦 is 3 throughout the line
𝑥=1
𝑦
(𝟏; 3)
(𝟏; 0)
(−1; 𝟑)
(0; 𝟑) (1; 𝟑)
𝑥
𝑦=3
𝑥
(𝟏; −3)
c) Given 𝑓 and 𝑔 below where 𝑓(𝑥) = 2𝑥 2 − 8 and 𝑔(𝑥) = 2𝑥 − 4:
𝑦
𝑓
A
−3
𝑥
C
B
D
If AB and CD are parallel to the 𝑦 axis (or perpendicular to the 𝑥 axis), determine
the length of:
i. AB
ii. CD
ii. BC
Answers:
i.
𝐀: 𝒙 = −𝟑 𝐚𝐭 𝐀
𝑓(𝑥) = 2𝑥 2 − 8
𝑓(−3) = 2(−3)2 − 8
𝑓(−3) = 10
𝐁: 𝒙 = −𝟑 𝐚𝐭 𝐁
𝑔(𝑥) = 2𝑥 − 4
𝑔(−3) = 2(−3) − 4
𝑔(−3) = −10
𝐀𝐁:
AB = 10 − (−10)
AB = 20 units
ii.
𝐂: 𝒙 = 𝟎 𝐚𝐭 𝐂
𝑔(𝑥) = 2𝑥 − 4
𝑔 (0) = 2(0 ) − 4
𝑔(0) = −4
𝐃: 𝒙 = 𝟎 𝐚𝐭 𝐃
𝑓(𝑥) = 2𝑥 2 − 8
𝑓 (0) = 2(0)2 − 8
𝑓(0) = −8
𝐂𝐃:
CD = −4 − (−8)
CD = 4 units
50
Copyright reserved
iii.
23
𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
BC = √((−3) − (0))2 + ((−10) − (−4))2
BC = 3√5 units
𝑎𝑥 + 𝑏
𝑎
into 𝑦 = + 𝑞
𝑥
𝑥
𝑥+3
𝑎
a) Change 𝑦 =
into format 𝑦 = + 𝑞 .
𝑥
𝑥
Change 𝑦 =
Answer:
(𝑥 ÷ 𝑥 = 1)
1
𝑥 𝑥+3
− 𝑥
3
𝑦=
(1 × 𝑥 = 𝑥 )
(𝑥 + 3 − 𝑥 = 3)
3
+1
𝑥
OR
𝑥+3
𝑥
𝑥 3
𝑦= +
𝑥 𝑥
3
𝑦 =1+
𝑥
𝑦=
𝑦=
3
+1
𝑥
b) Change 𝑦 =
2𝑥 + 3
𝑎
into format 𝑦 =
+𝑞.
𝑥
𝑥−𝑝
Answer:
2
𝑥 2𝑥 + 3
− 2𝑥
3
𝑦=
(2𝑥 ÷ 𝑥 = 2)
(2 × 𝑥 = 2𝑥)
(2𝑥 + 3 − 2𝑥 = 3)
3
+2
𝑥
51
Copyright reserved
OR
2𝑥 + 3
𝑥
2𝑥 3
𝑦=
+
𝑥
𝑥
3
𝑦 =2+
𝑥
𝑦=
𝑦=
3
+2
𝑥
52
Copyright reserved
Chapter 5: Probability
Section 1: Venn diagrams p55
not (A or B)
Set
A
B
𝐀𝐝𝐝𝐢𝐭𝐢𝐨𝐧𝐚𝐥 𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧
A only
A and B
B only
A or B
not (A or B)
(outside the circles)
not (A or B or C)
A and B
Set
A
B
(A and B) only
A only
B only
not (A or B or C)
(outside the circles)
(B and C) only
A and C
B and C
C only
C
53
Copyright reserved
𝐀𝐝𝐝𝐢𝐭𝐢𝐨𝐧𝐚𝐥 𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧
A or B or C
A and B and C
(A and C) only
𝐍𝐨𝐭𝐞:
𝐀𝐧𝐝 can be written as ∩
𝐎𝐫 can be written as ∪
𝐍𝐨𝐭 can be written as ′
Section 2: Contingency tables p64
𝐁
𝐃
𝐓𝐨𝐭𝐚𝐥
𝐀
A and B
A and D
A
𝐂
C and B
C and D
C
𝐓𝐨𝐭𝐚𝐥
Total B
Total D
Total
n(A)
n(Total)
To determine P(A):
P(A) =
To determine P(A and B):
P(A and B) =
To determine P(A or B):
P(A or B) = P(A) + P(B) − P(A and B)
n(A and B)
n(Total)
A and B are 𝐢𝐧𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭 if: P(A and B) = P(A) × P(B)
54
Copyright reserved
1 Venn Diagrams
a) A group of 200 learners where asked to state if they do Mathematics and History.
Below are the results:
(Learners doing Mathematics)
n(M) = 130
(Learners doing History)
n(H) = 80
(Learners doing Mathematics and History or n(M ∩ H))
n(M and H) = 20
n(not M or H) = 10 (Learners not doing Mathematics or History or n(M ∪ H)′ )
i. Draw a Venn diagram.
ii. Determine the probability that a learner does Mathematics (or P(M)).
iii. Determine the probability that a learner does Mathematics only (P(M only)).
iv. Determine the probability that a learner does Mathematics and History(P(M and H))
v. Determine the number of learners who do Mathematics or History (or n(M or H))
vi. Determine the probability that a learner does Mathematics or History (P(M or H))
vii. Determine the probability that a learner does not do Mathematics or History.
Answer:
i.
10
M
H
110
20
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧
10
200
130 − 20 20 80 − 20
55
Copyright reserved
200
60
130 13
=
200 20
110 11
iii. P(M only) =
=
200 20
20
1
iv. P(M and H) =
=
200 10
v. n(M or H) = 190
190 19
vi. P(M or H) =
=
200 20
10
1
vii. P(M or P)′ =
=
200 20
ii. P(M) =
(can be written as 0,65 or 65%)
(can be written as 0,55 or 55%)
(can be written as 0,1 or 10%)
(110 + 20 + 60
or
200 − 10)
(can be written as 0,95 or 95%)
(can be written as 0,05 or 5%)
b) A group of 200 learners where asked to state if they do Mathematics and History.
Below are the results:
(Learners doing Mathematics)
n(M) = 130
(Learners doing History)
n(H) = 80
n(not M or H) = 10 (Learners not doing Mathematics or History or n(M ∪ H)′ )
Draw a Venn diagram.
Answer:
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 [𝐥𝐞𝐭 𝐧(𝐌 𝐚𝐧𝐝 𝐇) = 𝒙]
10
M
130 − 𝑥
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧
130 − 𝑥 + 𝑥 + 80 − 𝑥 + 10 = 200
220 − 𝑥 = 200
−𝑥 = −20
𝑥 = 20
H
𝑥
80 − 𝑥
(÷ −1)
56
Copyright reserved
200
𝐅𝐢𝐧𝐚𝐥 𝐀𝐧𝐬𝐰𝐞𝐫
10
M
110
(130 − 20)
H
20
200
60
(80 − 20)
c) A group of 200 learners where asked to state if they do Mathematics and History.
Below are the results:
(Learners doing Mathematics)
P(M) = 0,65
(Learners doing History)
P(H) = 0,4
(
)
(
P not M and H = 0,05 Learners not doing Mathematics or History or n(M ∪ H)′ )
Draw a Venn diagram AND determine P(M or H).
Answer:
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 [𝐥𝐞𝐭 𝐏(𝐌 𝐚𝐧𝐝 𝐇) = 𝒙]
0,05
M
0,65 − 𝑥
H
𝑥
57
Copyright reserved
0,4 − 𝑥
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧
0,65 − 𝑥 + 𝑥 + 0,4 − 𝑥 + 0,05 = 1
1,1 − 𝑥 = 1
−𝑥 = −0,1
𝑥 = 0,1
𝐅𝐢𝐧𝐚𝐥 𝐀𝐧𝐬𝐰𝐞𝐫
0,05
(÷ −1)
M
0,55
(0,65 − 0,1)
H
0,1
0,3
(0,4 − 0,1)
P(M or H) = 0,95
d) A group of 250 learners where asked to state if they do Mathematics, History and
Afrikaans. Below are the results:
(Learners doing Mathematics)
n(M) = 130
(
)
(
n H = 80
Learners doing History)
(Learners doing Afrikaans)
n(A) = 110
(Learners doing Mathematics and History)
n(M and H) = 20
(Learners doing Mathematics and Afrikaans)
n(M and A) = 40
(Learners doing History and Afrikaans)
n(H and A) = 25
(
)
(
n M and H and A = 20
Learners doing Mathematics, History and Afrikaans)
(Learners not doing Mathematics or History)
n(not M or H or A) = 10
i. Draw a Venn diagram.
ii. Determine the probability that a learner does Mathematics.
iii. Determine the probability that a learner does Mathematics only (P(M only)).
iv. Determine the probability that a learner does Mathematics and History
v. Determine the probability that a learner does Mathematics and History only.
vi. Determine the probability that a learner does Mathematics, History and Afrikaans.
vii. Determine the probability that a learner does Mathematics or History.
viii. Determine the probability that a learner does Mathematics, History or Afrikaans.
58
Copyright reserved
ix. Determine the probability that a learner does not do Mathematics, History or
Afrikaans.
Answer:
i.
10
M
75
H
15
250
40
5
35
20
50
A
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧𝐬
15 = 20 − 5
35 = 40 − 5
20 = 25 − 5
75 = 130 − 15 − 5 − 35
40 = 80 − 15 − 5 − 20
50 = 110 − 35 − 5 − 20
130 13
ii. P(M) =
=
250 25
75
3
iii. P(M only) =
=
250 10
20
2
iv. P(M and H) =
=
250 25
15
3
v. P(M and H only) =
=
250 50
5
1
vi. P(M and H and A) =
=
250 50
(can be written as 0,52 or 52%)
(can be written as 0,3 or 30%)
(can be written as 0,08 or 8%)
(can be written as 0,06 or 6%)
(can be written as 0,02 or 2%)
59
Copyright reserved
190 19
=
250 25
240 24
viii. P(M or H or A) =
=
250 25
10
1
ix. P(M or H or A)′ =
=
250 25
vii. P(M or H) =
(190 = 75 + 35 + 15 + 5 + 40 + 20)
(240 = 75 + 35 + 15 + 5 + 40 + 20 + 50)
e) A group of 250 learners where asked to state if they do Mathematics, History and
Afrikaans. Below are the results:
(Learners doing Mathematics)
n(M) = 130
(
)
(
n H = 80
Learners doing History)
(Learners doing Afrikaans)
n(A) = 110
(Learners doing Mathematics and Afrikaans)
n(M and A) = 40
(Learners doing History and Afrikaans)
n(H and A) = 25
(Learners doing Mathematics, History and Afrikaans)
n(M and H and A) = 20
(
)
(Learners not doing Mathematics, History or Afrikaans)
n not M or H or A = 10
Draw a Venn diagram.
Answer:
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 [𝐥𝐞𝐭 𝐧(𝐌 𝐚𝐧𝐝 𝐇) = 𝒙]
10
M
90 − 𝑥
H
𝑥
55 − 𝑥
5
35
20
50
A
60
Copyright reserved
250
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧𝐬
35 = 40 − 5
20 = 25 − 5
90 − 𝑥 = 130 − 𝑥 − 5 − 35
55 − 𝑥 = 80 − 𝑥 − 5 − 20
50 = 110 − 15 − 5 − 20
90 − 𝑥 + 𝑥 + 55 − 𝑥 + 35 + 5 + 20 + 50 + 10 = 250
265 − 𝑥 = 250
−𝑥 = −15
𝑥 = 15
(÷ −1)
𝐅𝐢𝐧𝐚𝐥 𝐀𝐧𝐬𝐰𝐞𝐫
10
M
75
(90 − 15)
H
15
250
40
(55 − 15)
5
35
20
50
A
61
Copyright reserved
f) A group of 250 learners where asked to state if they do Mathematics, History and
Afrikaans. Below are the results:
(Learners doing Mathematics)
P(M) = 0,52
(Learners doing History)
P(H) = 0,32
(
)
(
P A = 0,44
Learners doing Afrikaans)
(Learners doing Mathematics and Afrikaans)
P(M and A) = 0,16
(Learners doing History and Afrikaans)
P(H and A) = 0,1
(Learners doing Mathematics, History and Afrikaans)
P(M and H and A) = 0,02
(Learners not doing Mathematics or History)
P(not M and H) = 0,04
Draw a Venn diagram AND determine P(M or H).
Answer:
𝐈𝐥𝐥𝐮𝐬𝐭𝐫𝐚𝐭𝐢𝐨𝐧 [𝐥𝐞𝐭 𝐏(𝐌 𝐚𝐧𝐝 𝐇) = 𝒙]
0,04
M
0,36 − 𝑥
H
𝑥
0,22 − 𝑥
0,02
0,14
0,08
0,2
A
𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐢𝐨𝐧𝐬
0,14 = 0,16 − 0,02
0,08 = 0,1 − 0,02
0,36 − 𝑥 = 0,52 − 𝑥 − 0,02 − 0,14
0,22 − 𝑥 = 0,32 − 𝑥 − 0,02 − 0,08
0,2 = 0,44 − 0,14 − 0,02 − 0,08
62
Copyright reserved
0,36 − 𝑥 + 𝑥 + 0,22 − 𝑥 + 0,14 + 0,02 + 0,08 + 0,2 + 0,04 = 1
1,06 − 𝑥 = 1
−𝑥 = −0,06
𝑥 = 0,06
𝐅𝐢𝐧𝐚𝐥 𝐀𝐧𝐬𝐰𝐞𝐫
0,04
M
0,3
(0,36 − 0,06)
H
0,06
0,16
(0,22 − 0,06)
0,02
0,14
0,08
0,2
A
P(M or H) = 0,76
63
Copyright reserved
(÷ −1)
2 Contingency Tables
a) A group of 200 learners where asked to state if they do Mathematics and History.
Below are the results:
𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬
78
42
120
𝐁𝐨𝐲𝐬
𝐆𝐢𝐫𝐥𝐬
𝐓𝐨𝐭𝐚𝐥
i.
ii.
iii.
iv.
v.
𝐇𝐢𝐬𝐭𝐨𝐫𝐲
52
28
80
𝐓𝐨𝐭𝐚𝐥
130
70
200
Determine the probability that a learner does Mathematics.
Determine the probability that the learner is a boy.
Determine the probability that a boy does Mathematics.
Determine the probability that a learner is a boy 𝐨𝐫 does Mathematics.
Determine whether being a boy and doing Mathematics are 𝐢𝐧𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭 or not.
Answer:
i.
P(M) =
ii. P(B) =
120 3
=
200 5
130 13
=
200 20
iii. P(B and M) =
78
39
=
200 100
iv. P(B or M) = P(B) + P(M) − P(B and M)
130 120 78
P(B or M) =
+
−
200 200 200
43
P(B or M) =
50
39
(From iii. )
100
130 120
39
P(B) × P(M) =
×
=
200 200 100
v. P(B and M) =
P(B and M) = P(B) × P(M)
Therefore being a boy and doing Mathematics are independent.
64
Copyright reserved
b) A group of 200 learners where asked to state if they do Mathematics and History.
Below are the results:
𝐁𝐨𝐲𝐬
𝐆𝐢𝐫𝐥𝐬
𝐓𝐨𝐭𝐚𝐥
𝐌𝐚𝐭𝐡𝐞𝐦𝐚𝐭𝐢𝐜𝐬
78
𝑎
120
𝐇𝐢𝐬𝐭𝐨𝐫𝐲
52
28
80
Determine the values of 𝑎, 𝑏 and 𝑐.
Answer:
𝑎 = 120 − 78 = 42
or
𝑎 = 70 − 28 = 42
or
𝑐 = 130 + 70 = 200
𝑏 = 78 + 52 = 130
𝑐 = 120 + 80 = 200
65
Copyright reserved
𝐓𝐨𝐭𝐚𝐥
𝑏
70
𝑐
Chapter 6: Statistics
Section 1: Definitions
 Decile: Decile is the value below which a tenth (1/10) of items in a set fall.
 Frequency: Frequency is the number of items in a given interval.
 Interquartile range (𝑰𝑸𝑹): This is the difference between the upper quartile and the
lower quartile.
 Lower quartile (𝑸𝟏 ): This is the value below which twenty five percent (25%) of
items in a set fall.
 Maximum (𝑸𝟒 ): Maximum is the highest number.
̅ ): Mean is the average number.
 Mean ( 𝒙
 Median (𝑸𝟐 ): This is the middle number or the value below which fifty percent (50%)
of items in a set fall.
 Minimum: Minimum is the lowest number.
 Mode: Mode is the number that appears the most.
 Modal class: Modal class is the interval that has the highest frequency.
 𝒏: 𝑛 is the total frequency.
 Percentile: Percentile is the value below which a given percentage of items in a set fall
 Range: Range is the difference between the maximum and the minimum.
 Upper quartile (𝑸𝟑 ): This is the value below which seventy five percent (75%) of
items in a set fall.
66
Copyright reserved
Section 2: Formulae
No. Subsection
1 Median
Formula
Position of median =
2
Lower quartile
3
Upper quartile
4
Decile
5
Percentile
6
Mean p68
7
8
Range
Interquartile range
𝑛+1
2
𝑛+1
4
3(𝑛 + 1)
Position of upper quartile =
4
𝑥(𝑛
+
1)
Position of 𝑥 𝑡ℎ decile =
10
𝑥(𝑛 + 1)
Position of 𝑥 𝑡ℎ percentile =
100
∑𝑛𝑖 𝑥𝑖
∑𝑛𝑖 𝑓𝑖 𝑥̅𝑖
𝑥=
or 𝑥 =
(frequency table)
𝑛
𝑛
range = maximum − minimum
𝐼𝑄𝑅 = 𝑄3 − 𝑄1
Position of lower quartile =
Section 3: Box and Whisker p68
Note: 25% of the data lies between each quartile see below.
Skewness: symmetrical
25%
25%
25%
25%
Skewness: skewed left
Min
Q1
Q2
Q3
Max
Interquartile range (IQR)
Skewness: skewed right
Range
Section 4: Frequency table p70
No. Detail
1 Frequency column relates to histogram or frequency polygon.
2 Modal class is the interval that has the highest frequency, the highest bar on the
histogram and the steepest gradient on the ogive.
∑𝑛𝑖 𝑓𝑖 𝑥̅𝑖
3
To calculate the mean we use the formula 𝜇 =
𝑛
67
Copyright reserved
1 Mean and box and whisker
a) The maximum temparature each day is recorded for 9 consective days. Below are the
temperatures for the nine days.
Day
Temperature
(ᵒC)
1
2
3
4
5
6
7
8
9
20
21
25
26
28
30
31
33
36
i. Determine the mean.
ii. Draw the box and whisker.
Answer:
i.
20 + 21 + 25 + 26 + 28 + 30 + 31 + 33 + 36
= 27,8 ℃
9
𝑛+1
4
9+1
Position =
4
Position = 2,5
Q1 = 21 + 0,5(25 − 21) = 23 ℃
𝑛+1
2
9+1
Position =
2
Position = 5
Q 2 = 28℃
3(𝑛 + 1)
4
3(9 + 1)
Position =
4
Position = 7,5
Q 3 = 31 + 0,5(33 − 31) = 32 ℃
ii. Position =
Position =
Position =
20
23
28
32
68
Copyright reserved
36
b) The maximum temperature of each day is recorded for 20 consective days. Below is a
box and whisker diagram representing the maximum temperatures (in ℃) of the
20 days.
20
23
28
32
36
i. What was the highest (maximum) temperature experienced during the 20 days?
ii. What was the lowest (minimum) temperature experienced during the 20 days?
iii. Determine the range of the temperatures.
iv. Determine the interquartile range (IQR) of the temperatures.
v. Comment on the skewness of the data.
vi. What percentage of days had a maximum temperature that is less than 23 ℃?
vii. What percentage of days had a maximum temperature that is more than 23 ℃?
viii. What percentage of days had a maximum temperature that is less than 28 ℃?
ix. What percentage of days had a maximum temperature that is less than 32 ℃?
x. What percentage of days had a maximum temperature that is more than 32 ℃?
xi. How many days had a maximum temperature that is less than 23 ℃?
xii. How many days had a maximum temperature that is less than 28 ℃?
Answer:
i.
36 ℃
ii. 20 ℃
iii. range = maximum − minimum
range = 36 − 20
range = 16 ℃
iv. IQR = Q 3 − Q1
IQR = 32 − 23
IQR = 9 ℃
v. The data is symmetric (or has a normal distribution).
vi. 25%
vii. 75%
(25% + 25% + 25% or 3 × 25%)
viii. 50%
(25% + 25% or 2 × 25%)
ix. 75%
(25% + 25% + 25% or 3 × 25%)
x.
25%
xi. 25% × 20 = 5 days
69
Copyright reserved
xii. 50% × 20 = 10 days
2 Frequency table
a) The maximum temperature of each day is recorded for 165 consective days. Below is a
frequency table representing the maximum temperatures (in ℃) of the 165 days.
Temperature
(ᵒC)
20 ≤ 𝑥 ≤ 25
25 < 𝑥 ≤ 30
30 < 𝑥 ≤ 35
35 < 𝑥 ≤ 40
Total
Frequency
(days)
34
50
58
23
165
Determine the following:
i.
ii.
iii.
vi.
v.
modal class
mean temperature
minimum temperature
maximum temperature
range
Answer:
i.
30 < 𝑥 ≤ 35
ii. 29,62 ℃
(the modal class has the highest frequency which is 58 in this case)
(see below two methods to determine the mean)
𝐌𝐞𝐭𝐡𝐨𝐝 𝟏: 𝐂𝐚𝐥𝐜𝐮𝐥𝐚𝐭𝐨𝐫
Press Mode → STATS → 1 − VAR → enter values as indicated below → AC → Shift → 1
→ VAR → 𝑥̅ → =
X
(20 + 25)
2
(25 + 30)
2
(30 + 35)
2
(35 + 40)
2
Total
70
Copyright reserved
FREQ
34
50
58
23
165
𝐌𝐞𝐭𝐡𝐨𝐝 𝟐: 𝐄𝐪𝐮𝐚𝐭𝐢𝐨𝐧
Temperature (ᵒC)
𝒇
20 ≤ 𝑥 ≤ 25
34
25 < 𝑥 ≤ 30
50
30 < 𝑥 ≤ 35
58
35 < 𝑥 ≤ 40
23
Total
165
̅
𝒙
(20 + 25)
= 22,5
2
(25 + 30)
= 27,5
2
(30 + 35)
= 32,5
2
(35 + 40)
= 37,5
2
(34 × 22,5) + (50 × 27,5) + (58 × 32,5) + (23 × 37,5)
165
Estimated mean = 29,62 ℃
Estimated mean =
iii. 20 ℃
vi. 40 ℃
v. 40 − 20 = 20 ℃
71
Copyright reserved
Chapter 7: Analytical Geometry
Section 1: Definitions
 Bisect: To bisect means to cut in half.
 Collinear: Points are collinear if they have the same gradient or lie on the same line.
 Decreasing: Decreasing means that the 𝑦 values on a graph decrease as the 𝑥 values
increase.
 Gradient: Gradient refers to the steepness and direction of a graph.
 𝒇(𝒙): 𝑓(𝑥) is the 𝑦.
 Increasing: Increasing means that the 𝑦 values on a graph increase as the 𝑥 values
increase.
 Intercept: An intercept is a point where a graph meets an 𝑥 or 𝑦 axis.
 Intersection: An intersection is a point where two graphs meet. The 𝑥 and 𝑦 values of
the two graphs are equal at this point.
 Midpoint: The middle point between two points.
 Perpendicular: This is an angle of 90°.
 Transformation: Transformation refers to horizontal or vertical shifts as well as
reflections across the 𝑥 or 𝑦 axis or across other lines.
 Vertex: The corner of a shape where two lines intersect.
Important Note: General function rules from Chapter 4 apply in Analytical Geometry.
Section 2: Formulae
No. Formula
1 𝑚 = 𝑦2 − 𝑦1
𝑥2 − 𝑥1
2 𝑚1 = 𝑚2 p73
3 𝑚1 × 𝑚2 = −1 p73
4
5
6
𝑑 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2
𝑥1 + 𝑥2
𝑦1 + 𝑦2
(𝑥 =
)
;𝑦 =
2
2
p74
𝑦 = 𝑚𝑥 + 𝑐
or 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )
Use when
given 2 points
given parallel lines or a shape that has parallel lines
given perpendicular lines or a shape that has
perpendicular lines
given or asked to calculate length
asked to calculate midpoint or given that a line
bisects another line
asked to find the equation of a straight line
72
Copyright reserved
2 𝑚1 = 𝑚2
a) Show that 𝑓 is parallel to 𝑔 .
𝑦
𝑓
𝑔
8
(−1; 4)
(1; 4)
𝑥
Answer:
𝑦2 − 𝑦1
𝑚=
𝑥2 − 𝑥1
8−4
𝑚𝑓 =
0 − (−1)
𝑚𝑓 = 4
𝑚𝑓 = 𝑚𝑔
𝑦2 − 𝑦1
𝑥2 − 𝑥1
4−0
𝑚𝑔 =
1−0
𝑚𝑔 = 4
𝑚=
∴ 𝑓 is parallel to 𝑔 .
3 𝑚1 × 𝑚2 = −1
a) Show that 𝑓 is perpendicuar to 𝑔 .
𝑦
𝑓
8
𝑔
(−1; 4)
15
Answer:
𝑦2 − 𝑦1
𝑚=
𝑥2 − 𝑥1
8−4
𝑚𝑓 =
0 − (−1)
𝑚𝑓 = 4
1
= −1
4
𝑚𝑓 × 𝑚𝑔 = −1
𝑥
𝑦2 − 𝑦1
𝑥2 − 𝑥1
4−0
𝑚𝑔 =
−1 − 15
1
𝑚𝑔 = −
4
𝑚=
4×−
∴ 𝑓 is perpendicuar to 𝑔 .
73
Copyright reserved
5
𝑥1 + 𝑥2
𝑦1 + 𝑦2
)
;𝑦 =
2
2
a) Determine the coordinates at B, the midpoint of A and C.
(𝑥 =
𝑦
C(1; 6)
B
𝑥
A(−3; −2)
Answer:
𝑥1 + 𝑥2
2
(−3) + (1)
𝑥=
2
𝑥 = −1
𝑦1 + 𝑦2
2
(−2) + (6)
𝑦=
2
𝑦=2
𝑥=
𝑦=
B(−1; 2)
b) Determine the coordinates at C if B is the midpoint of A and C.
𝑦
C
B(−1; 2)
𝑥
A(−3; −2)
Answer:
𝑥1 + 𝑥2
2
(−3) + 𝑥
−1 =
2
−2 = −3 + 𝑥
1=𝑥
𝑥=
𝑦1 + 𝑦2
2
(−2) + 𝑦
2=
2
4 = −2 + 𝑦
6=𝑦
𝑦=
(× 2)
C(1; 6)
74
Copyright reserved
(× 2)
Chapter 8: Trigonometry
Section 1: Definitions
 Amplitude: Amplitude is the height from the middle value of a trigonometric function
(the amplitude value is always positive).
 Asymptote: These are lines that graphs continuously approach but never touch.
 Intersection: An intersection is a point where two graphs meet. The 𝑥 and 𝑦 values of
the two graphs are equal at this point.
 Period: Period is a full horizontal cycle of a function.
Section 2: Basics of Trigonometry
1
a
Trigonometric ratios
Ratios
𝐬𝐢𝐧 𝜽
𝑜
ℎ
b
Inverses
2
Quadrants and triangle
Quadrant 2
Quadrant 1
cosec 𝜃 =
sin 𝜃 > 0
ℎ
𝑜
𝐜𝐨𝐬 𝜽
𝑎
ℎ
sec 𝜃 =
𝐭𝐚𝐧 𝜽
𝑜
𝑎
𝑎
cot 𝜃 =
𝑜
ℎ
𝑎
all ratios > 0
hypotenuse (ℎ)
opposite (𝑜)
0°
360°
180°
𝜃
adjacent (𝑎)
tan 𝜃 > 0
cos 𝜃 > 0
Quadrant 3
3
Quadrant 4
Special angles
30°
2
45°
√3
2
60°
45°
1
√2
75
Copyright reserved
√2
Section 3: Trigonometric diagrams p77
No. Step
1 Step 1: Identify the
quadrant
(if the quadrant is not
given)
2
3
Detail
i. Use a restriction (e.g. 90° < 𝜃 < 270°)
−1
ii. Use signs (e. g. sin 𝜃 =
or sin θ < 0 which both
2
mean that sin 𝜃 is negative therefore we use the third or
forth quadrant)
Step 2: draw the triangle i. Opposite = 𝑥
and label the sides of the ii. Adjacent = 𝑦
triangle.
iii. Hypotenuse = 𝑟
iv. Use Pythagoras to find the third side
Step 3: Use the diagram Use the sides from the diagram to answer the questions.
to answer the questions
Section 4: Calculating angles p78
No.
1
2
3
4
5
6
7
Step
sin 𝜃 + 𝑎 = 𝑏
sin 𝜃 − 𝑎 = 𝑏
𝑎 sin 𝜃 = 𝑏
sin 𝜃
=𝑏
𝑎
cosec 𝜃 = 𝑏
sin(𝜃 + 𝑎) = 𝑏
sin 𝑎𝜃 = 𝑏
𝑎, 𝑏 ∈ 𝑍
𝑎, 𝑏 ∈ 𝑍
𝑎, 𝑏 ∈ 𝑍
𝑎, 𝑏 ∈ 𝑍
𝑏∈𝑍
𝑎, 𝑏 ∈ 𝑍
𝑎, 𝑏 ∈ 𝑍
Detail
Subtract 𝑎
Add 𝑎
Divide by 𝑎
Multiply by 𝑎
Substitute inverse
sin−1 𝑏 and subtract 𝑎
sin−1 𝑏 and divide by 𝑎
76
Copyright reserved
3
Trigonometric diagrams
a) Use the following diagram to answer the questions below.
𝑦
P(−3; 4)
𝜃
O
𝑥
Determine:
i. the length of OP.
ii. cos 𝜃
Answer:
i.
𝑥2 + 𝑦2 = 𝑟2
(−3)2 + (4)2 = 𝑟 2
25 = 𝑟 2
5=𝑟
𝑦
4
5
𝜃
ii.
cos 𝜃 =
−3
5
b) Given cos 𝜃 = −
a calculator.
i.
ii.
−3
𝑥
3
(0° ≤ 𝜃 ≤ 180°), determine the following without the use of
5
Draw a sketch.
tan 𝜃
Answer:
i.
𝑥2 + 𝑦2 = 𝑟2
(−3)2 + 𝑦 2 = (5)2
9 + 𝑦 2 = 25
𝑦 2 = 16
𝑦=4
4
𝑦
• cos 𝜃 is negative in
quadrants 2 and 3
5
• 0° ≤ 𝜃 ≤ 180° refers
to quadrants 1 and 2
𝑥
• ∴ we use quadrant 2
−3 𝑥
( )
• cos 𝜃 =
5
𝑟
𝑟 is always positive
𝜃
−3
• ∴ determine the value
of 𝑦
77
Copyright reserved
ii.
tan 𝜃
4
=
−3
c) Given sin 𝜃 =
calculator.
i.
ii.
4
(tan 𝜃 < 0), determine the following without the use of a
5
Draw a sketch.
tan 𝜃
Answer:
i.
𝑥2 + 𝑦2 = 𝑟2
𝑥 2 + (4)2 = (5)2
𝑥 2 + 16 = 25
𝑥2 = 9
𝑥 = −3
4
𝑦
• sin 𝜃 is positive in
quadrants 1 and 2
5
• tan 𝜃 is negative in
quadrants 2 and 4
𝑥
• ∴ we use quadrant 2
𝜃
−3
• sin 𝜃 =
4 𝑦
( )
5 𝑟
• ∴ determine the value
of 𝑥
ii.
4
tan 𝜃
4
=
−3
Calculating angles
a) Simplify the following:
5
i. sin 𝑥 + 2 =
2
sin 𝑥 1
iv.
=
2
4
1
vii. sin 3𝑥 =
2
ii.
sin 𝑥 − 2 = −
v.
cosec 𝑥 = 2
viii. 2 sin 𝑥 + 3 = 4
78
Copyright reserved
3
2
iii. 2 sin 𝑥 = 1
vi. sin(𝑥 + 20°) =
ix. 2 sin 3𝑥 = 1
1
2
Answer:
i.
5
2
1
sin 𝑥 =
2
(−2)
sin 𝑥 + 2 =
ii.
1
𝑥 = sin−1 ( )
2
𝑥 = 30°
iv.
1
𝑥 = sin−1 ( )
2
𝑥 = 30°
v.
cosec 𝑥 = 2
1
=2
sin 𝑥
1
sin 𝑥 =
2
𝑥 = 30°
vii. sin 3𝑥 =
(invert
2
)
1
(−20°)
𝑥 + 20° = 30°
𝑥 = 10°
viii. 2 sin 𝑥 + 3 = 4
(−3)
2 sin 𝑥 = 1
(÷ 2)
1
2
𝑥 = 30°
(÷ 3)
sin 𝑥 =
𝑥 = 10°
1
3𝑥 = sin−1 ( )
2
3𝑥 = 30°
𝑥 = 10°
1
2
1
𝑥 + 20° = sin−1 ( )
2
1
3𝑥 = sin−1 ( )
2
(÷ 2)
(÷ 3)
79
Copyright reserved
(× 2)
1
𝑥 = sin−1 ( )
2
𝑥 = 30°
1
2
ix. 2 sin 3𝑥 = 1
1
sin 3𝑥 =
2
sin 𝑥 1
=
2
4
1
sin 𝑥 =
2
vi. sin(𝑥 + 20°) =
3𝑥 = 30°
(+2)
1
𝑥 = sin−1 ( )
2
𝑥 = 30°
(÷ 2)
iii. 2 sin 𝑥 = 1
1
sin 𝑥 =
2
3
sin 𝑥 − 2 = −
2
1
sin 𝑥 =
2