Focus TG4 KSSM (Addmaths) Terbitan Penerbitan Pelangi Sdn Bhd

CONTENTS

Mathematical Formulae iv

1Chapter

Functions 1

1.1 Functions 2
1.2 Composite Functions 14
1.3 Inverse Functions 19
SPM Practice 1 27

2Chapter Quadratic Functions 29
30
2.1 Quadratic Equations and Inequalities 37
2.2 Types of Roots of Quadratic Equations 39
2.3 Quadratic Functions 52
SPM Practice 2

3Chapter Systems of Equations 55

3.1 Systems of Linear Equations in Three Variables 56
3.2 Simultaneous Equations involving One Linear Equation and One Non-Linear
Equation 60
SPM Practice 3 65

4Chapter Indices, Surds and Logarithms 67

4.1 Laws of Indices 68
4.2 Laws of Surds 70
4.3 Laws of Logarithms 75
4.4 Applications of Indices, Surds and Logarithms 80
SPM Practice 4 85

5Chapter Progressions 87
88
5.1 Arithmetic Progressions 96
5.2 Geometric Progressions 104
SPM Practice 5

ii

6Chapter Linear Law 108
6.1 Linear and Non-Linear Relations 109
6.2 Linear Law and Non-Linear Relations 114
6.3 Applications of Linear Law 121
SPM Practice 6 124

7Chapter Coordinate Geometry 128
8Chapter
7.1 Divisor of a Line Segment 129
7.2 Parallel Lines and Perpendicular Lines 132
7.3 Areas of Polygons 137
7.4 Equations of Loci 142
SPM Practice 7 145

Vectors 148

8.1 Vectors 149
8.2 Addition and Subtraction of Vectors 155
8.3 Vectors in a Cartesian Plane 166
SPM Practice 8 173

9Chapter Solution of Triangles 179
1 0Chapter
9.1 Sine Rule 180
9.2 Cosine Rule 184
9.3 Area of a Triangle 188
9.4 Application of Sine Rule, Cosine Rule and Area of a Triangle 191
SPM Practice 9 194

Index Numbers 196

10.1 Index Numbers 197
10.2 Composite Index 200
SPM Practice 10 211

Pre-SPM Model Paper 216
Answers 227



iii

2Chapter Learning Area : Algebra

Quadratic Functions

KEYWORDS
• Axis of symmetry – Paksi simetri
• Completing the square – Penyempurnaan kuasa dua Concept
• Discriminant – Pembezalayan Map
• Factorisation – Pemfaktoran
• General form – Bentuk am
• Imaginary root – Punca khayalan
• Intersection – Persilangan
• Maximum point –Titik maksimum
• Minimum point – Titik minimum
• Parabola – Parabola
• Quadratic equation – Persamaan kuadratik
• Quadratic function – Fungsi kuadratik
• Quadratic inequality – Ketaksamaan kuadratik
• Real root – Punca nyata
• Root – Punca
• Vertex form – Bentuk verteks

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29

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

2.1 Quadratic Equations and Solution Express the
Inequalities (a) x2 + 6x – 2 = 0 equation in the
x2 + 6x = 2 form of ax2 + bx + c
with a = 1.
6 2 = 2 + 62
A Solving quadratic equations using the     x2 + 6x + 2 2
method of completing the square and
using the quadratic formula x2 + 6x + 32 = 2 + 9  Addb 2 on
(x + 3)2 = 11 2a
Chapter 2 x + 3 = ± 11
1. Certain quadratic equations can be solved by both sides of the
factorisation.
equation.

x = –3 ± 11
= –3 + 11 or –3 – 11
For example: x2 + x – 2 = 0
(x – 1)(x + 2) = 0 = 0.3166 or –6.3166
x – 1 = 0  or  x + 2= 0
x = 1 x = –2 SPM Tips

2. Quadratic equations such as x2 = 16, (x – 2)2 = 6 Use the calculator to check your answer by
can be solved easily because the left hand side of substituting the answer into the quadratic equation
the equation is a complete square. involved.

For example: (x – 2)2 = 6 6 (b) For the case when the coefficient of x2 is not 1,
x – 2 = ± ± 6 change the coefficient to 1 by dividing each term in
= 2 the equation by that coefficient before completing
x the square.
= 2 + 6 or  2 – 6
= 4.449 or  –0.4495
[to 4 significant figures]
2x2 + 4x – 1 = 0
2 4 1 Divide each term
3. For quadratic equations that are difficult to 2 x2 + 2 x – 2 = 0 by the coefficient
factorise or cannot be factorised easily such as of x2.
1
x2 + 6x – 2 = 0 and 2x2 + 4x – 1 = 0, these equations x2 + 2x = 2
can be solved using the
(a) completing the square method,     x2 + 2x + 2 2 = 1 + 22
2 2 2
–b ± b2 – 4ac 1
(b) quadratic formula method, 2a . x2 + 2x + 1 = 2 +1

(x + 1)2 = 3
2
Derivation of the 3
Quadratic Formula x + 1 = ± 2

INFO x = –1 ± 1.2247

4. It is easier to use the quadratic formula method = 0.2247 or –2.2247
to solve the quadratic equation compared to
completing the square method. Try Question 1 in ‘Try This! 2.1’

1 2

Determine the roots for each of the following quadratic Solve each of the following quadratic equations by
equations by using the completing the square. using the quadratic formula method.
(a) x2 – 2 + 6x = 0 (a) 3x2 = 2x + 5
(b) 2x2 + 4x – 1 = 0 (b) 6 – 2x = 5x2

30

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

Solution Solution
(a) Rearrange 3x2 = 2x + 5 to the general form of Applying the Pythagoras theorem,
ax2 + bx + c = 0. AC2 = AB2 + BC2
3x2 – 2x – 5 = 0 (p + 4)2 = (p – 3)2 + (p + 1)2
p2 + 8p + 16 = p2 – 6p + 9 + p2 + 2p + 1
Therefore, a = 3, b = –2 and c = –5. p2 – 12p – 6 = 0
p = –(–12)
Substitute the values of a, b and c into the quadratic ±  (–12)2 – 4(1)(–6)
formula, 2(1) Chapter 2
–(–2) ± (–2)2 – 4(3)(–5)
x = 2(3) = 12 ± 168
2
= 2 ± 4+ 60 x –b ± b2 – 4ac + 168 12 – 168
6 = 2a 12 2 2
=   or  
2 ± 64
= 6 = 12.48  or  –0.4807 (not acceptable)

= 2 ± 8
6
2 + 8 2–8 Therefore, p = 12.48
6 6
x =   or x =  Try Questions 3 – 7 in ‘Try This! 2.1’

= 5   or =  –1 C A L C U L A T O R Corner
3
Solve the quadratic equation x2 – 14x – 39 = 0.
(b) Rearrange 6 – 2x = 5x2 to the general form of
ax2 + bx + c = 0. 1: Press the MODE key until EQN is displayed.
5x2 + 2x – 6 = 0 2: Press 1
3: ‘Unknowns?’ is displayed, press 
Therefore, a = 5, b = 2 and c = –6. 4: ‘Degree?’ is displayed, press 2
5: ‘a?’ is displayed, press 1 =
Substitute the values of a, b and c into the quadratic 6: ‘b?’ is displayed, press – 1 4 =
formula, 7: ‘c?’ is displayed, press – 3 9 =
–2 ± (2)2 – 4(5)(–6) 8: One of the roots is, ‘x1 = 16.38’ is displayed
x = 2(5) 8: Press the  key. Another root is, ‘x2 = –2.381’ is

= –2 ± 124 x = –b ± b2 – 4ac displayed
10 2a

x = –2 + 124   or   x = –2 – 124
10 10

= 0.9136  or = –1.314

Try Question 2 in ‘Try This! 2.1’ B Forming quadratic equations from
given roots
3 (p + 4) cm
1. If α and β are the roots of a quadratic equation,
A then
x = α and x = b
(p – 3) cm (x – α) = 0  and  (x – b) = 0
(x – α)(x – b) = 0
B (p + 1) cm C x2 – αx – bx + αb = 0
x2 – (α + b)x + αb = 0
The diagram above shows a right-angled triangle α + b is called the sum of the roots (SOR) and ab
ABC with AB = (p – 3) cm, BC = (p + 1) cm and is called the product of the roots (POR).
AC = (p + 4) cm. Find the value of p. [Give your answer
correct to 4 significant figures.] x2 – (SOR)x + (POR) = 0

31

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

2. One of the methods to form a quadratic equation Alternative Method
from the given roots of α and b is by expanding 2
the product of (x – α)(x – b) = 0. x = 23 and x = 5
3x = and x = 5
3. The second method is by finding the sum of 3x – 2 = 0 and x – 5 = 0
roots, (α + b) and the product of roots, (αb)
Chapter 2 to be substituted into the quadratic equation (3x – 2)(x – 5) = 0 One root only means
x2 – (SOR)x + (POR) = 0. 3x2 – 17x + 10 = 0 there are two equal
(c) SOR = –4 + (–4) = –8 roots.
4. To determine the sum of roots and product of POR = –4 × (–4) = 16
roots of a quadratic equation The quadratic equation is
x2 – (SOR)x + (POR) = 0
Quadratic equation x2 – (–8)x + (16) = 0
ax2 + bx + c = 0 x2 + 8x + 16 = 0
where a ≠ 0

Divide each term of the Try Question 8 in ‘Try This! 2.1’
equation by a so that the
coefficient of x2 becomes 1.

x2 + b x + c =0 5
a a

Sum of roots =– b Given α and b are the roots of a quadratic equation
a 5x2 – 2x – 4 = 0. Form a quadratic equation with
roots

c (a) (5a + 1) and (5b + 1)
a 2 2
Product of roots = (b) a and b



4 Solution
Divide each term of the equation by 5, so that the
coefficient of x2 becomes 1.

Form a quadratic equation with the given roots. x2 – 2 x – 4 =0
2 5 5
(a) 2 and –3 (b) 3 and 5 (c)  –4 only 2
a+b= 5 Given a and b are the
Solution roots of the quadratic
(a) SOR = 2 + (–3) = –1 4 equation.
5
POR = 2 × (–3) = –6 ab = –

The quadratic equation is (a) The new roots are (5a + 1) and (5b + 1).
x2 – (SOR)x + (POR) = 0
x2 – (–1)x + (–6) = 0 New SOR = (5a + 1) + (5b + 1)
= 5(a + b) + 2
x2 + x – 6 = 0
= 5 2  + 2
2 17 5
(b) SOR = 3 +5= 3 =4

POR =  2 (5) = 10 New POR = (5a + 1)(5b + 1)
3 3 = 25ab + 5a + 5b + 1
The quadratic equation is
= 25ab + 5(a + b) + 1
  x2 – 17 x+ 10 = 0 Multiply each term of
3 3 the equation by 3. = 25– 4  + 5 2  + 1
5 5
3x2 – 17x + 10 = 0 = –17

32

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

The new quadratic equation is: POR = (a)(2a) = 25
x2 – (SOR)x + (POR) = 0 2
x2 – (4)x + (–17) = 0 25
Therefore, x2 – 4x – 17 = 0 2a2 = 2

a2 = 25
4
(b) The new roots are 2 and 2 . 5
2a b a = ± 2 ............... b Chapter 2
2 b
New SOR = a + Substitute b into a:

= 2b + 2a When  a= 5 , k = 6 5 
ab 2 2
= 15
= 2(b + a)  When  5 5
ab a=– 2 , k = 6 – 2

= 2 2  = –15
5
– 4 Therefore, the possible values of k are 15 and –15.
5
= –1
SPM Tips
 2  2 
New POR = a b This example can also be solved by using a and 1  a
2
= 4 as the roots of the equation.
ab
4
= 4
= – 5 Try Questions 10 – 17 in ‘Try This! 2.1’
–5

The new quadratic equation is: SPM Highlights Paper 1 SPM
x2 – (SOR)x + (POR) = 0 2016
x2 – (–1)x + (–5) = 0
The quadratic equation 4x2 + 5x – 6 = 0 has roots h and
h k
Therefore, x2 + x – 5 = 0 k. Form a quadratic equation with roots 2 and 2 .

Try Question 9 in ‘Try This! 2.1’ Solution: 4x2 + 5x – 6 = 0

6 x2 + 5 x – 3 = 0
One root of the quadratic equation 2x2 – kx + 25 = 0 4 2
is half of the other root of the equation. Find the
possible values of k. SOR =h + k =– 5
4
3
POR = hk = – 2

Solution h + k h+k – 5 5
2 2 2 4 8
New SOR = = = =–
2
2x2 – kx + 25 = 0 Divide each term of the
k 25 equation by 2. 3
x2 – 2 + 2 = 0 h k hk – 2
x   New POR = 2 2 4 3
= = 4 =– 8

Assume one of the roots is a. The new equation is
Then, the other root is 2a.
–– k  x2 – – 5 x + – 3  = 0
SOR = a + 2a = 2 8 8

3a = k 8x2 + 5x – 3 = 0
2
k = 6a..............

33

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

SPM Highlights Paper 1 SPM 7
2017 Determine the range of values of x that satisfy each
of the following quadratic inequalities by using graph
One of the roots of the quadratic equation sketching method, number line method or table
x2 + (n – 5)x – 2n2 = 0, where n is a constant, is the method.
negative of the other root of the equation. Find the (a) x2 + x – 6  0
product of the roots. (b) 6 + x – x2 > 0
(c) 2x2 – 5x + 2 . 0
Chapter 2 Solution:
Assume that the roots for the equation Solution
x2 + (n – 5)x – 2n2 = 0 are a and – a. (a) x2 + x – 6  0

SOR = a + (– a) = –(n – 5) Graph sketching method
0 = –n + 5 When f(x) = 0, x2 + x – 6 = 0
n = 5............. (x – 2)(x + 3) = 0

POR = –2n2
= –2(5)2
= –50

SPM Highlights Paper 1 SPM x – 2 = 0 or x + 3 = 0
2018 x = 2 x = –3

The quadratic equation px2 – 4x + q = 0, where p and q The coefficient of x2, a = 1 . 0 ⇒ the shape of the
are constants, has roots a and 3a. Express p in terms graph is .
of q.
The graph sketch:
Solution: f(x)
px2 – 4x + q = 0
q
x2 – 4 x + p = 0 –3 2 x
p

 SOR = a 4
+ 3a = – – p

4a = 4 Therefore, the range of values of x that satisfy the
p quadratic inequality x2 + x – 6  0 is –3  x  2.
1
a = p . .............

POR = (a)(3a) = q (b) 6 + x – x2 > 0
p
q
3a2 = p ............. Number line method
Factorise the quadratic equation,
Substitute  into . 6 + x – x2 = 0

  3 1 2 q (2 + x)(3 – x) = 0
p p (2 + x) = 0  or (3 – x) = 0
=

p32 = q x = –2 x = 3
p
3p = p2q
3p – p2q = 0 (2 + x) ≥ 0 when x ≥ –2

p(3 – pq) = 0 –+ +

p = 0  or 3 – pq = 0 –2 3
pq = 3 + +
(unacceptable) (3 – x) ≥ 0 when x ≤ 3 + –

p = 3
q

C Solving quadratic inequalities (+) × (+) = (+)

1. The range of values of x that satisfy a quadratic Therefore, the range of values of x that satisfy
inequality can be determined by using the quadratic inequality 6 + x – x2 > 0 is
• graph sketching method –2 < x < 3.
• number line method
• table method SPM Tips
INFO
Use the graph sketching method to check the
answer.

34

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

(c) 2x2 – 5x + 2 . 0 Example of HOTS Question
Table method
2x2 – 5x + 2 = 0 6.6 m
(2x – 1)(x – 2) = 0
xm
Factorise the 1 1 1 xm 4.8 m
2 2 2
quadratic x x= x 2 x=2 x.2 Chapter 2
equation

(2x – 1) – 0 + ++ The diagram above shows a rectangular piece of
– – 0+ land with a length of 6.6 m and a width of 4.8 m.
(x – 2) – 0 – 0+ Amin wants to lay square tiles with side x m around
the land to build a walkway. If the area of the
(2x – 1)(x – 2) + region where the tiles are laid is 12.24 m2, find the
value of x.
From the table, it shows that (2x – 1)(x – 2) . 0
1
when x  2 or x . 2.

Try Question 18 in ‘Try This! 2.1’ Solution

8 Area of the region where the tiles are laid
= 2 × 6.6 × x + 2 × (4.8 – 2x) × x
Find the range of values of x for (2x – 1)(x + 4) < 4 + x. = 13.2x + 9.6x – 4x2
= 22.8x – 4x2
Solution
(2x – 1)(x + 4) < 4 + x Hence, 22.8x – 4x2 = 12.24
2x2 + 8x – x – 4 < 4 + x 4x2 – 22.8x + 12.24 = 0
2x2 + 7x – 4 – 4 – x < 0 400x2 – 2 280x + 1 224 = 0
2x2 + 6x – 8 < 0 50x2 – 285x + 153 = 0
x2 + 3x – 4 < 0
a = 1 . 0 ⇒ the shape of the graph is x = –(–285) ± (–285)2 – 4(50)(153)
2(50)
285 ± 225
. = 100

When x2 + 3x – 4 = 0 x = 510   or x = 60
(x – 1)(x + 4) = 0 100 100

x – 1 = 0 or x + 4 = 0 x = 5.1  or  x = 0.6
x = 1 or x = –4

The graph intersects the x-axis at x = 1 and x = –4 5.1 m is longer than the width of the land which is
4.8 m. So, x = 5.1 is unacceptable.

Therefore, x = 0.6

x
–4 1

Therefore, the range of values of x is –4 < x < 1. Try this HOTS question

SPM Tips The profit made by a factory that is producing
nuts in packets is given by
Eliminating the common expression of (x + 4) in the P(t) = 40t2 – 7t – 5 676, where t is the time, in
inequality will cause the range of values of x obtained hours, the production process is running. Find
the time of production needed for the factory to
incomplete. (2x – 1)(x + 4) < (4 + x) get back its capital.
2x – 1 < 1
2x < 1 + 1 Answer: The time of production must be at least
x < 1 12 hours.


Try Questions 20 – 21 in ‘Try This! 2.1’

Try Question 19 in ‘Try This! 2.1’

35

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

Try This! 2.1 10. Given one of the roots of the quadratic equation
8x2 + 2x – m = 0 is twice the other root. Find the
1. Solve each of the following quadratic equations by possible values of m.

using the completing the square method. 11. One of the roots of the quadratic equation
(a) x2 – x = 3 (b) x2 + x – 4 = 0 3x2 + hx = 6x – 25 is one third the other root. Find
(c) x2 + 5x + 2 = 0 (d) x(x – 2) = 6 the possible values of h.
(e) x(x + 2) – 3 = 0 (f) 2x2 + 3x – 2 = 0
Chapter 2 (g) 3x2 + 6x – 1 = 0 (h) 4x2 – 3x – 2 = 0 12. Given the quadratic equation qx2 – 12x + 8 = 0,
q ≠ 0, has two equal roots. Find the possible values
2. Solve each of the following quadratic equations by of q.

using the formula. (b) 2x2 = 3x + 4 13. Given a and b are the roots of the quadratic equation
(a) x2 + 4x – 3 = 0 (d) 4x2 – 5x = 2
(c) 3x2 + 7x – 5 = 0 5x2 – 6x + h = 0, whereas a and b are the roots
(e) 3x(x + 3) = 4 k k
of the quadratic equation 2x2 + 3x – 5 = 0. Find the
values of h and k.
3. The sum of the squares for two consecutive positive
integers is 2 521. Find the values of the two integers. 14. Given the quadratic equation x2 – (p – 10)x + 5q = 0
has roots 6 and 10. Find the values of p and q.
4. Given two squares with side lengths of x cm and
(x – 2) cm respectively. The total area of the two 15. Given q and 4 are the roots of the quadratic equation
squares is 340 cm2. Find the total perimeter for both (2x – h)2 = 4x. Find the possible values of h and q.
squares.
16. The roots of the quadratic equation (x + 1)(x – 5) =
5. Nesa’s age is four times her son’s age in this year. p(q – x) – 9 are 2 and – 6. Find the possible values
The product of Nesa’s age and her son’s age two of p and q.
years later will be 270. Find Nesa’s age and her
son’s age in this year. 17. Given –3 and (k – 2) are the roots of the quadratic
equation x2 + (h – 5)x – 12 = 0, where h and k are
6. The product of two consecutive odd integers is constants. Find the values of h and k.
1 763. Find the values of these two integers.

7. 18. Find the range of values of x that satisfy the following
quadratic inequalities by using graph sketching
(2x – 1) cm
method, number line method or table method.
x cm (a) x2 + x – 6 . 0
(b) x2 – 3x – 10 < 0
(c) 4 + 3x – x2 > 0
(d) 9 + 5x – 4x2 , 0

The diagram above shows a triangle with an area of 19. Find the range of values of x that satisfy the following
quadratic inequalities.
60 cm2. Find (a) 3x2 + 13x < 10
(a) the value of x, (b) 7 – 2x , (x + 4)2
(c) 3x2 – x – 21 > x(2x + 3)
(b) the perimeter of the triangle. HOTS (d) (2x + 1)(x – 5) > 3(2x + 1)

Applying

8. Form quadratic equations which have the following

roots. 20. 300 packages can be produced when a packaging
machine operates at a rate of x packages per
(a) 2 and 3 (b) – 4 and 5

(c) –2 and –5 (d) 2 only minute. A study found out that when the rate of
3 operation of the machine is increased to (x + 3)
1 (f) 3 + 2 and 3 – 2
(e) –3 and 2 packages per minute, the time saved is 5 minutes

9. Given a and b are the roots of a quadratic equation for 300 packages. Determine the new operating rate
2x2 – 8x – 5 = 0. Form quadratic equations with the
following roots. of the machine. HOTS
Evaluating

(a) 2a and 2b 21. In an experiment, a stone was thrown upwards
at a speed of 15 m s–1 from a platform 5 m
(b) (3a + 1) and (3b + 1) above the ground. The position of the stone from
the ground can be represented by the function
(c) a and b f(t) = 5 + 15t – 4.9t 2, where t represents the time, in
2 2 seconds, after the stone was thrown. HOTS

(d) 4 and 4 Analysing
a b

36

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

5 m Ground (b) 9x2 = 6x – 1 Chapter 2
9x2 – 6x + 1 = 0

Find the time of the stone to reach the ground. [Give Compare with ax2 + bx + c = 0.
a = 9, b = –6, c = 1
your answer correct to two significant figures.] b2 – 4ac = (–6)2 – 4(9)(1)
=0
2.2 Types of Roots of Quadratic Therefore, equation 9x2 = 6x – 1 has two equal real
Equations roots.

A Relating the types of roots (c) x2 + 4x + 5 = 0
of quadratic equations to the Compare with ax2 + bx + c = 0.
discriminant value a = 1, b = 4, c = 5
b2 – 4ac = 42 – 4(1)(5)
1. The types of roots of a quadratic equation = –4 , 0
ax2 + bx + c = 0 depends on the value of the Therefore, equation x2 + 4x + 5 = 0 has no real
roots.
expression b2 – 4ac.
2. The expression b2 – 4ac is called the discriminant SPM Tips

of the quadratic equation. For the equation x2 + 4x + 5 = 0 in (c), the imaginary root
3. The relation between the types of roots and the
can be written as
values of the discriminant are shown in the table b2 – 4ac –4 ± –4
below. x  = –b ± 2a = 2(1)

= –2 ± –14
2
= –2 ± –1
= –2 ± i
Discriminant, Types of roots
b2 – 4ac The imaginary part, that is –1, is represented by the
Two different real symbol i.
(a) . 0 roots
Imaginary roots are widely used in the fields of
(b) = 0 Two equal real electric and electronics.
roots
(c)  0 Try Question 1 in ‘Try This! 2.2’
No roots (or
imaginary roots )

9 B Solving problems involving types of
Determine the types of roots for each of the following roots of quadratic equations
quadratic equations.
(a) 3x2 – 5x – 6 = 0 1. The relation between the types of roots and
(b) 9x2 = 6x – 1 the values of the discriminants for quadratic
(c) x2 + 4x + 5 = 0 equations can be used to
(a) find the value or values of unknown
Solution variables,
(a) 3x2 – 5x – 6 = 0 (b) find a relation between two variables in the
quadratic equation.
Compare with ax2 + bx + c = 0.
a = 3, b = –5, c = –6 10
b2 – 4ac = (–5)2 – 4(3)(–6) Find the range of values of p if the quadratic equation
= 97 . 0 3x2 + 4 = 8x + 2px2, where p is a constant, has two
Therefore, equation 3x2 – 5x – 6 = 0 has two different real roots.
different real roots.

37

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

Solution The condition for no real roots is
3x2 + 4 = 8x + 2px2 b2 – 4ac , 0
3x2 – 2px2 – 8x + 4 = 0 42 – 4(2 – m)(–3) , 0
(3 – 2p)x2 – 8x + 4 = 0 16 + 24 – 12m , 0
a = 3 – 2p, b = –8, c = 4 12m 4100
m . 3
For two different real roots, .

Chapter 2 b2 – 4ac . 0 Try Question 4 in ‘Try This! 2.2’
(–8)2 – 4(3 – 2p)(4) . 0
64 – 48 + 32p . 0
16 + 32p . 0 13
32p . –16 Given the quadratic equation (n – 2m)x2 – 4mx + m = 0,
16 where m ≠ 0, has two equal roots. Find the relation
p . – 32 between m and n.

p . – 1 Solution
2 (n – 2m)x2 – 4mx + m = 0
a = n – 2m, b = –4m, c = m
Try Question 2 in ‘Try This! 2.2’ For two equal real roots,
b2 – 4ac = 0
11 (–4m)2 – 4(n – 2m)(m) = 0
16m2 – 4mn + 8m2 = 0
The quadratic equation x2 – 4kx + (k + 3)2 = 0, where k 24m2 – 4mn = 0
is a constant and k . 0, has two equal roots. Find the 4m(6m – n) = 0
value of k. m = 0  or  6m – n = 0

Solution Since m ≠ 0, therefore 6m = n
x2 – 4kx + (k + 3)2 = 0 n
a = 1, b = –4k, c = (k + 3)2 m = 6

The condition for two equal roots is Try Questions 5 – 8 in ‘Try This! 2.2’
b2 – 4ac = 0
(–4k)2 – 4(1)(k + 3)2 = 0
16k2 – 4(k2 + 6k + 9) = 0 SPM Highlights Paper 1 SPM
2011

16k2 – 4k2 – 24k – 36 = 0 The quadratic equation (1 + 2q)x2 – 4qx + 2q – 4 = 0 has
12k2 – 24k – 36 = 0 two equal roots. Find the value of q.
k2 – 2k – 3 = 0
(k + 1)(k – 3) = 0 Solution:
k + 1 = 0  or k – 3 = 0 (1 + 2q)x2 – 4qx + 2q – 4 = 0
k = –1 k = 3 a = 1 + 2q, b = –4q, c = 2q – 4

Since k . 0, therefore k = 3. For two equal real roots,

Try Question 3 in ‘Try This! 2.2’ b2 – 4ac = 0
(–4q)2 – 4(1 + 2q)(2q – 4) = 0
16q2 – 8q + 16 – 16q2 + 32q = 0
24q = –16

q = – 2
3

12 Try This! 2.2
The quadratic equation (2 – m)x2 = 3 – 4x has no real
roots. Find the range of values of m. 1. Find the discriminant for each of the following

Solution quadratic equations. Hence, determine the type of
(2 – m)x2 = 3 – 4x
(2 – m)x2 + 4x – 3 = 0 roots for each of the quadratic equation.
(a) 9x2 – 12x + 4 = 0 (b) 5x2 – 3x + 6 = 0
a = 2 – m, b = 4, c = –3 (c) 2x2 + x – 3 = 0 (d) x2 = 4x + 7

38

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

(e) 2x(4x – 3) = 5 (f) 7x = 5x2 + 3 5. Given the quadratic equation mx2 – 5nx + m = 0
(g) 2(5x – 9) = 3x2 (h) (x + 2)(3x + 1) = x – 1 has two equal roots. Find the relation between m
and n.
2. Find the range of values of p if each of the following
6. Given the quadratic equation p(x2 + 4) = 8qx has two
quadratic equations has two different real roots. equal roots. Find the ratio of p : q. Hence, find the
(a) x2 – 6x + 4p = 0 (b) 2px2 + 5(x – 1) = 0 roots.
(c) x(3x + 7) = – 6 – p (d) x(9x + 1) = p(6x – p)
7. If the ratio of h : k = 3 : 2, determine the types of
3. Find the value of k if each of the following quadratic roots of the quadratic equation 9kx2 + k = 4hx. Chapter 2
equations has two equal real roots.
(a) 3x2 – 2kx + k = 0 8. Given that 5 and – 4 are the roots of the quadratic
(b) kx2 – 4x + 3k = 4 equation px2 – 2x + q = 0. Find
(c) 4kx2 = x(3k – x) – 1 (a) the value of p and the value of q,
(d) x(4x – k) = k – 2(x + 1) (b) the value of m where the quadratic equation
px2 – 2x + q = m has two equal roots.
4. Find the range of values of h if each of the following

quadratic equations has no roots.
(a) x2 – 2x + 5h = 0 (b) x2 + 2hx + (2 – h)2 = 0
(c) 5 – 2x = (3 – h)x2 (d) 4x2 – 4hx + h2 = 5x

2.3 Quadratic Functions

A Analysing the effects of the changes of a, b and c towards the shape and position of the
graph of f (x) = ax2 + bx + c

a.0 a0

f(x) a f(x)
1
a > = 1 x
0
0<a<1
0< a <1
Only the 0x a >1
value
of a a = –1

changes When the value of a changes, y-intercept is unchanged but the shape and width of the graph will
Only the change.

value For the case of a . 0: For the case of a , 0:
of b • the shape of the graph is  . • the shape of the graph is  .
changes • a . 1 and the larger the value of a, the width • a . 1 and the larger the value of a, the width

of the graph decreases. of the graph decreases.
• 0 , a , 1 and the smaller the value of a, the • 0 , a , 1 and the smaller the value of a, the

width of the graph increases. width of the graph increases.

Consider c = 0 in this case

f(x) f(x)

b>0 b=0 b<0 x
0 Move to
Move to 0 Move to x Move to
the left when the left the right
b>0 the right when b<0 b>0
b<0
b<0 b=0 b>0

When the value of b changes, the shape of the graph and the y-intercept are unchanged but the
position of vertex will be shifted to the left or to the right of the y-axis.

39

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

For the case of a . 0: For the case of a , 0:
When the value of b changes, the position of When the value of b changes, the position of
vertex vertex
• will be shifted to the left when b . 0. • will be shifted to the right when b . 0.
• will be shifted to the right when b , 0. • will be shifted to the left when b , 0.

Chapter 2 SPM Tips

• The larger the value of |b |, the further the position of vertex is displaced from the y-axis.
• When b = 0, the vertex is on the y-axis.
• When the same value of b changes sign from positive to negative or from negative to

positive, the position of the graph is reflected on the y-axis.

Moves f(x) f(x) c>0
upwards c>0
when Moves upwards
c>0 c=0 when
c>0
x
0

Moves 0 x Moves c=0
downwards c<0 downwards c<0
Only the when when
value of c c<0 c<0
changes
When the value of c changes, the shape of the graph is unchanged but the position of the graph
• will be shifted vertically upwards when c . 0.
• will be shifted vertically downwards when c  0.

SPM Tips The effects of the
changes in a, b

c is the value of the y-intercept. and c on the graph
INFO

14 The y-intercept does not change.

f(x) = x2 + 2x + 2 f(x)

The diagram below shows the sketch of the graph of
f(x) = x2 + 2x + 2. Make generalisations on the shape
and position of the graph of the given functions when
compared to the values of a, b and c for the following f(x) = —12 x2 + 2x + 2 2
0x
functions. Hence, sketch the graphs of the functions.
1
(a) f(x) = 2 x2 + 2x + 2 (b) f(x) = x2 – 2x + 2
When the value of b changes from +2 to –2, the
(b) f(x) = x2 – 2x + 2 position of the graph reflects on the y-axis.
(c) f(x) = x2 + 2x – 1 The shape and y-intercept do not change.
f(x)
f(x)

f(x) = x2 + 2x + 2 f(x) = x2 + 2x + 2 f(x) = x2 – 2x + 2

2

0x 2
0x
Solution 1
2
(a) f (x) = x2 + 2x + 2 1 (c) f(x) = x2 + 2x – 1
2
When the value of a changes from 1 to , the When the value of c changes from +2 to –1, the
width of the graph increases. position of the graph is shifted vertically. The
y-intercept is –1. The shape of the graph does not
change.

40

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

f(x) = x2 + 2x + 2 f(x) (c) If b2 – 4ac , 0, the equation f (x) = 0 has no
2 roots. Therefore, the graph of the quadratic
function f (x) does not intersect or touch
0 x the x-axis.
–1
f(x) = x2 + 2x – 1 f(x) f(x) Chapter 2
Ox
SPM Tips
Ox
When the value of c changes from +2 to –1, the
difference in the value of c is 3, so the graph a . 0 a,0
moves 3 units downwards.

Try Questions 1 – 2 in ‘Try This! 2.3’ 3. To determine whether a straight line,
f(x) = mx + n intersects the curve of a quadratic
B Relating the position of the graph of function, f(x) = px2 + qx + r :
a quadratic function with the types of • Equalise both the equations and express
roots it in the form of f (x) = ax2 + bx + c.
• Hence, calculate the value of b2 – 4ac.
1. The position of the graph of a quadratic function (i) If b2 – 4ac . 0, there are two intersection
f (x) = ax2 + bx + c depends on the types of the points.
roots of the quadratic equation f (x) = 0. Roots of (ii) If b2 – 4ac = 0, there is only one intersection
a quadratic equation are the values of x when the point.
graph intersects or touches the x-axis. (iii) If b2 – 4ac , 0, there is no intersection
point.
2. The discriminant for a quadratic function is the
value of b2 – 4ac. The relationships between the 15
discriminants and the positions of the graph are
as follows. Determine the types of roots for each of the following
(a) If b2 – 4ac . 0, the equation f (x) = 0 has equations f (x) = 0.
two different roots. Therefore, the graph of
the quadratic function f (x) intersects the (a)  f(x) (b)  f(x)
x-axis at two different points.

f(x) f(x)
f(x) = ax2 + bx + c
4 9
x O1 x –3 O x
4 x
O xO
(c)  f(x) (d)  f(x)
f(x) = ax2 + bx + c
10 O
a . 0 a , 0 –8
O x
(b) If b2 – 4ac = 0, the equation f (x) = 0 has
two equal roots. Therefore, the graph of the
quadratic function f (x) touches the x-axis at
only one point.

f(x) f(x) Solution
(a) The graph of f (x) intersects the x-axis at two
Ox
different points, therefore f (x) = 0 has two different
O x roots.
(b) The graph of f (x) touches the x-axis at only one
a . 0 a,0 point, therefore f (x) = 0 has two equal roots.

41

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

(c) The graph of f (x) intersects the x-axis at two This function has two equal roots and the graph
different points, therefore f (x) = 0 has two different intersects the x-axis at one point.
roots.

(d) The graph of f (x) does not intersect or touches the
x-axis, therefore f (x) = 0 has no roots.

Chapter 2 Try Question 3 in ‘Try This! 2.3’

16 Try Question 4 in ‘Try This! 2.3’
Determine the types of roots for the following
quadratic functions. Hence, sketch the position of the 17
graph of the function with respect to the x-axis. Given that the quadratic function f (x) = x2 + 2x + 1 – k
(a) f(x) = 3x2 – 4x – 2 intersects the x-axis at two different points. Find the
(b) f(x) = –5x2 + 3x – 1 range of values of k.
(c) f(x) = x2 – 8x + 16 Solution
f (x) = x2 + 2x + 1 – k
Solution a = 1, b = 2 and c = 1 – k
(a) f(x) = 3x2 – 4x – 2 f (x) = 0 has two different roots when
b2 – 4ac . 0
a = 3, b = –4, c = –2 (2)2 – 4(1)(1 – k) . 0
b2 – 4ac = (–4)2 – 4(3)(–2) 4 – (4 – 4k) . 0
= 40 . 0 4 – 4 + 4k . 0
4k . 0
a = 3 . 0, therefore the graph of the function is in k . 0
the shape of .
This function has two different roots and the Try Questions 5 – 6 in ‘Try This! 2.3’
graph intersects the x-axis at two points.

18

Given that the quadratic function
(b) f(x) = –5x2 + 3x – 1 f (x) = p2x2 + (p – 1)x + —14 does not intersect the x-axis.
Find the range of values of p.
a = –5, b = 3, c = –1
b2 – 4ac = 32 – 4(–5)(–1) Solution 1)x + —41 = 0 —14
= –11 , 0 p2x2 + (p – p– 1 and c =
a = p2, b =
a = –5 , 0, therefore the graph of the function is
in the shape of  . f (x) = 0 does not have real roots when
This function has no real roots and the graph does b2 – 4ac , 0
not intersect the x-axis.
1 2 (p – 1)2 – 4(p2) —14 , 0
p2 – 2p + 1 – p2 , 0
(c) f(x) = x2 – 8x + 16 –2p + 1 , 0
1 , 2p
a = 1, b = –8, c = 16 —21
b2 – 4ac = (–8)2 – 4(1)(16) p , p
= 0 . —12

a = 1 . 0, therefore the graph of the function is in Try Questions 7 – 10 in ‘Try This! 2.3’
the shape of .

42

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

19 • For the case kes a , 0, the vertex (h, k) is a
maximum point and k is the maximum value
Show that the straight line y = 2x – 1 intersects the of f(x).
curve of the graph of f (x) = 3x2 – 4x + q at one point
for q = 2. 3. The method of completing the squares is used
to change f(x) = ax2 + bx + c to the vertex
Solution
y = 2x – 1................................................1 form f(x) = a(x – h)2 + k where h = – b and Chapter 2
f (x) = 3x2 – 4x + q......................................2 2a
4ac – b2 .
2 = 1 at the point of intersection, k = 4a

3x2 – 4x + q = 2x – 1 4. By using the factorisation method or the
3x2 – 6x + q + 1 = 0 quadratic formula, functions in the general form
a = 3, b = –6 and c = q + 1
f(x) = ax2 + bx + c can be changed to the intercept
There is only one point of intersection: form f(x) = a(x – p)(x – q) where p and q are the
b2 – 4ac = 0 roots of f(x).
(–6)2 – 4(3)(q + 1) = 0
36 – 12q – 12 = 0 INFO Steps of deriving
24 – 12q = 0 Formula of Vertex
12q = 24 Form
q = 2

Try Question 11 in ‘Try This! 2.3’

C Making relation between the vertex 20
form of a quadratic function,
 Express the quadratic function f(x) = 2 x+ 5 2– 9
f(x) = a(x – h)2 + k with other forms of 4 8
the quadratic function in the intercept form, f(x) = a(x – p)(x – q) where a, p
and q are constants and p . q. Hence, state the values
1. A quadratic function can be expressed in either of a, p and q.
the vertex form, general form or intercept form.
Solution
Vertex form Change the vertex form of the quadratic function into
f(x) = a(x – h)2 + k general form first.

Completing the Expansion  f(x) = 2x+ 5 2– 9
squares 4 8
  5 25 9
= 2 x2 + 2 x + 16 – 8

General form = 2x2 + 5x + 25 – 9
f(x) = ax2 + bx + c 8 8

Expansion Factorisation or = 2x2 + 5x + 2
using formula
Hence, change the general form of the quadratic
function into intercept form.
Intercept form f(x) = (2x + 1)(x + 2)
f(x) = a(x – p)(x – q)   1
= 2 x+ 2 (x + 2)

2. The quadratic function f(x) = a(x – h)2 + k has Compare to f(x) = a(x – p)(x – q),
vertex (h, k) and it is symmetrical about the line
x = h. therefore a = 2, p = – 1 and q = –2.
• For the case a . 0, the vertex (h, k) is a 2
minimum point and k is the minimum value
of f(x). Try Questions 12 – 14 in ‘Try This! 2.3’

43

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

21 = x2 – 8x + (–4)2 – (–4)2 + 7
Express the quadratic function f(x) = 2x2 – 4x + 5 in = (x – 4)2 – 16 + 7
the vertex form by using the completing the squares = (x – 4)2 – 9.....................................1
method. Hence, determine the coordinates of the Compare 1 to y = (x – p)2 + q,
minimum point. p = 4, q = –9.

Chapter 2 Solution (b) The coordinates of the turning point are (p, q)
= (4, –9). Since a = 1 . 0, the turning point is a
f(x) = 2x2 – 4x + 5 minimum point.
  4 5
= 2 x2 – 2 x + 2 Try Questions 16 – 18 in ‘Try This! 2.3’

= 2 x2 – 2x +  –2 2 –  –2 2 + 5 
2 2 2
  5
= 2 (x – 1)2 – 1 + 2

  3 23
= 2 (x – 1)2 + 2
The minimum value of the quadratic function
= 2(x – 1)2 + 3 y = x2 + 2kx + 5k – 3 is –17, where is k a constant. Find
the possible values of k.
The coordinates of the minimum point are (1, 3).

Try Question 15 in ‘Try This! 2.3’ Solution
y = x2 + 2kx + 5k – 3
22
1 2 1 2 = x2 + 2kx + —22k– 2 – —22k– 2 + 5k – 3
Express the quadratic function y = x2 – 8x + 7 in the
form of y = (x – p)2 + q , where p and q are constants. = x2 + 2kx + k2 – k2 + 5k – 3
Find = (x + k)2 – k2 + 5k – 3
(a) the values of p and q,
(b) the coordinates of the turning point and hence, The minimum value = –k2 + 5k – 3

determine whether that point is a maximum or –k2 + 5k – 3 = –17 or k – 7 = 0
minimum point. k2 – 5k – 14 = 0 or k = 7
(k + 2)(k – 7) = 0
Solution k + 2 = 0
(a) y = x2 – 8x + 7 k = –2

1 2 1 2 = x2 – 8x + —–28– 2 – —–28– 2 + 7 Try Questions 19 – 22 in ‘Try This! 2.3’

D Analysing the effects of the changes of a, h and k on the shape and position of the
graph of f(x) = a(x – h)2 + k

a.0 a0

Only f(x) a f(x)
the 1
value a > = 1 0
of a
changes x

0<a<1

0x 0< a <1
a >1

a = –1

44

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

When the value of a changes, the shape and width of the graph will REMEMBER!
change but the axis of symmetry is unchanged.
The minimum or maximum
value remain unchanged.

For the case of a . 0: For the case of a  0: Chapter 2
• the shape of the graph is and it has a • the shape of the graph is   and it has a
minimum point. maximum point
• a . 1 and the larger the value of a, the width of • a . 1 and the larger the value of a, the width of
the graph decreases. the graph decreases.
• 0  a  1 and the smaller the value of a, the • 0  a  1 and the smaller the value of a, the
width of the graph increases. width of the graph increases.

The vertical line which is passing through the vertex of the graph divides the graph into two
symmetrical parts. Therefore, the equation of the axis of symmetry of the graph for the quadratic
function is given by x = h.

f(x) f(x)

x=h x=h

Only Shift to the 0 Shift to the x Shift to the 0 x
the left as the left as the left as the x=h
value value of h value of h value of h Shift to the
of h decreases increases decrease right as the
changes value of h
increases

x=h

When h changes, the shape of the graph is unchanged but the REMEMBER!
vertex of the graph or the axis of symmetry
• is shifted to the right when the value of h increases. The minimum or maximum
• is shifted to the left when the value of h decreases. value remain unchanged.

Shift upwards f(x) Shift upwards f(x)
as the as the value
value of k k>0 of k k>0
increases k=0 increases

0 x

Only 0 x Shift downwards k=0
the k<0 as the value
value Shift downwards of k k<0
of k as the value of decreases
changes k decreases

When k changes, the shape of the graph is unchanged REMEMBER!
but the vertex of the graph
• is shifted upwards when the value of k increases. • The axis of symmetry is unchanged.
• is shifted downwards when the value of k decreases. • The minimum or maximum value

changes.

45

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

24 f(x) = 3(x + 5)2 + 2 f(x)
The diagram below shows the graph for f(x) = 3(x + 1)2 + 2
f(x) = 3(x + 1)2 + 2, with a = 3, h = –1 and k = 2.
2
f(x) x

Chapter 2 f(x) = 3(x + 1)2 + 2 –5 –1 0
(c) When the value of k changes from 2 to –3, the
5
2 shape of the graph is unchanged but the position of
the graph is shifted vertically 4 units downwards.
–1 0 x The minimum value becomes –3 and the equation
of the axis of symmetry remains the same, that is x
Make generalisations on the shape and position of the = –1.
graph of the given function when compared to the
values of a, h and k of the following functions. Hence, f(x)
s(ak)e tc(hi)t hef(xg)ra=ph13s(oxf+th1e)2fu+n2ctions.
(ii) f(x) = 4(x + 1)2 + 2 f(x) = 3(x + 1)2 + 2
(b) f(x) = 3(x + 5)2 + 2
(c) f(x) = 3(x + 1)2 – 3 f(x) = 3(x + 1)2 – 3 5
2

–1 0 x

Solution –3
(a) (i) When the
voaflutheeofgraapchhainngcerseafsreosm. T3hetoax31is , Try Questions 23 – 25 in ‘Try This! 2.3’
the width
of symmetry and the minimum value of the

graph are unchanged. 25

f(x) = —31 (x + 1)2 + 2 f(x) f(x) = 3(x + 1)2 + 2 f(x)

5 5
0x
2
The diagram above shows the graph for the quadratic
–10 x function f(x) = 2(x – 2)2 + 3p – 1, where p is a constant.
(a) Given that the minimum value for the function is
(ii) When the value of a changes from 3 to 4,
the width of the graph decreases. The axis 5, find the value of p.
of symmetry and the minimum value of the (b) State the equation of the axis of symmetry for the
graph are unchanged.
f(x) f(x) = 3(x + 1)2 + 2 curve.
f(x) = 4(x + 1)2 + 2 Solution
(a) f(x) = 2(x – 2)2 + 3p – 1
5 When f(x) = 3p – 1 the value of x – 2 = 0.
3p – 1 = 5
2 p = 2

–1 0 x (b) x – 2 = 0
x = 2
Therefore, the equation of the axis of symmetry is
(b) When the value of h changes from –1 ke –5, the
shape of the graph is unchanged but the position x = 2.
of the graph is shifted horizontally 4 units to the
left. The equation of the axis of symmetry becomes Try Questions 26 – 28 in ‘Try This! 2.3’
x = –5 and the minimum value of the graph is
unchanged.

46

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

E Sketching the graph of quadratic Therefore, the graph of f(x) intersects the x-axis at two
functions different points.
f(x) = 2x2 + 3x – 2
Sketching the graph of a quadratic function   3
f(x) = ax2 + bx + c = 2 x2 + 2 x – 1

Determine the shape of the graph of the = 2 x2 + 3 x +  3 2 –  3 2 – 1 Chapter 2
quadratic function by identifying the value of a. 2 4 4
3 9
Determine the position of the graph of the = 2x + 4 2 – 16 – 1
quadratic function by calculating the value of
  =2 x+ 3 2– 25
the discriminant b2 – 4ac. 4 8
3 25
Determine the maximum or minimum point  Minimum point is – 4 , – 8  .
by expressing the quadratic function in the
vertex form f(x) = a(x – h)2 + k where the When f(x) = 0, 2x2 + 3x – 2 = 0
vertex is (h, k) and the equation of the axis of (2x – 1)(x + 2) = 0

symmetry is x = h. 2x – 1 = 01  or x + 2 = 0
x = 2 x = –2

f(x)
When x = 0, f(0) = –2.

x = – —34

Solve the quadratic equation f(x) = 0 to –2 0 —21 x
determine the points of intersection on –2
the x-axis, which are the real roots of the
quadratic equation, if the roots exist. ( )– —43 , –2—89
Determine the point of intersection of the
graph with the y-axis by finding the value of The equation of the axis of symmetry is x = – 3 .
4
f(x) when x = 0.
Try Question 29 in ‘Try This! 2.3’

F Solving problems involving quadratic
functions

Draw a smooth parabola which is symmetrical 27
at the line x = h and passes through all the The curve of the quadratic function f(x) = –2(x – p)2 + 2q
points that have been determined. cuts the x-axis at points (1, 0) and (5, 0). The line y = 8
touches the curve at its maximum point.
26 (a) Find the values of p and q.
(b) Sketch the graph of f(x) for 0 < x < 6.
Sketch the graph for the quadratic function (c) If the graph is reflected on the x-axis, write the
f(x) = 2x2 + 3x – 2. Hence, state the axis of symmetry
for the graph of the quadratic function. equation of the curve.

Solution Solution
f(x) = 2x2 + 3x – 2 (a) f(x) = –2(x – p)2 + 2q
a=2.0
The graph is in the shape of The coordinates of vertex of f(x) are (p, 2q).
p = 1 + 5
with a minimum point. 2   2q = 8
q = 4
b2 – 4ac = 32 – 4(2)(–2) =3
= 25 . 0

47

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

(b) The maximum point is (3, 8) Try this HOTS question
When x = 0, f(0) = –2(0 – 3)2 + 8 = –10 x
When x = 6, f(6) = f(0) = –10
y
f(x)
(3, 8)

Chapter 2 01 x The diagram above shows a rectangular barn
56 with the length of x m and width of y m. A wire
fence of 60 m length will be used to fence up the
–10 barn into two parts of the same size as shown.
(c) When the graph is reflected on the x-axis, (a) Find the area, A m2, in terms of x.
(b) Find
f(x) = –[–2(x – 3)2 + 8] (i) the value of x, such that the area of the
= 2(x – 3)2 – 8
barn is maximum.
Try Questions 30 – 31 in ‘Try This! 2.3’ (ii) the maximum area, in m2.

Example of HOTS Question Answer: 3
2
The daily profit, k, in RM, for a company that sells (a) A = 30x – x2

children’s toys can be estimated by using a quadratic (b) (i) 10
1
function, k(x) = – 2 x2 + 50x + 400, where x is the (ii) 150 m2

number of units of the toys sold each day.

(a) Express k(x) in the form of a(x + p)2 + q, where a, SPM Highlights Paper 1 SPM
p and q are constants. 2017

(b) Find
(i) the number of units of the toys that must be

sold each day to obtain the maximum profit,
(ii) the maximum profit.

Solution: Find the range of values of x such that the quadratic function
f(x) = 10 + 3x – x2 is negative.
1
(a) k(x) = – 2 x2 + 50x + 400

= – 1 (x2 – 100x – 800) Solution
2 f(x) , 0
10 + 3x – x2 , 0
= – 1 x2 – 100x + (–50)2 – (–50)2 – 800) x2 – 3x – 10 . 0
2 (x – 5)(x + 2) . 0

= – 21(x – 50)2 – 2 500 – 800

= – 12(x – 50)2 + 1 650 x
–2 5
where a = – 1 , p = –50 and q = 1 650.
2 The range of values of is x , –2 or x . 5.

(b) (i) For the profit, k(x), to be maximum, the value
of x = 50.

Therefore, the number of units of the toys that
must be sold is 50.

(ii) When x = 50, k(x) = 1 650
Therefore, the maximum profit = RM1 650

48

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

SPM Highlights Paper 2 SPM 2. The diagram below shows the sketch of the graph
2016 for f(x) = 5 + 3x – 2x2.

The curve of the quadratic function f (x) = 3(x – p)2 + 4q f(x)
intersects the x-axis at points (0, 0) and (4, 0). The line
y = –12 touches the minimum point of the curve. 5 f(x) = 5 + 3x – 2x2
(a) Find the values of p and q.
(b) Hence, sketch the graph of f(x) for –1 < x < 5. 0x
(c) If the graph is reflected on the x-axis, write the
Chapter 2
equation of the curve. Analyse and make generalisations on the shape and

Solution position of the graph of the given functions when
(a) f (x) = 3(x – p)2 + 4q. compared to the values of a, b and c of the following
The minimum point is (p, 4q), such that p is the
functions. Hence, sketch the graphs of the functions.
midpoint of 0 and 4. (a) f(x) = 5 – 3x – 2x2
 p= 4+0 =2 (b) f(x) = –1 + 3x – 2x2
(c) f(x) = 2x2 – 3x – 5
2
4q = –12 3. Based on the graph for each of the following
  q = –3
quadratic functions, state the types of roots of the
(b) f(x)
functions.

(a) f(x) (b) f(x)

f (x) = 3(x – p)2 + 4q x
x

x (c) f(x) (d) f(x)
x
–1 0 1 2 3 45

–12 y = –12

x
(c) f(x) = –3(x – 2)2 + 12

(e) f(x) (f) f(x)

xx

Try This! 2.3

1. The diagram below shows the sketch of the graph 4. Determine the types of roots for each of the following
f(x) = x2 – 3x – 4. quadratic functions. Hence, sketch the position of
the graph of the function with respect to x-axis.
f(x) (a) f(x) = 4x2 – 2x – 5 (b) f(x) = –2x2 + 4x + 1
f(x) = x2 – 3x – 4 (c) f(x) = 3x2 – 4x + 2 (d) f(x) = 2 + 6x – 3x2
(e) f(x) = 4x2 – 12x + 9 (f) f(x) = –1 + 4x – 4x2
0x
5. The graph of the quadratic function
–4 f (x) = 2x2 – 5x + p + 3 intersects the x-axis at two
different points. Find the range of values of p.

6. The graph of the quadratic function
Analyse and make generalisations on the shape and f (x) = qx2 + 2qx – 6 + q intersects the x-axis at two
position of the graph of the given functions when different points. Find the range of values of q.
compared to the values of a, b and c of the following
7. Show that the quadratic function f (x) = 3x2 – 4x + k + 1
functions. Hence, sketch the graphs of the functions. does not intersect the x-axis for k . 31.
(a) f(x) = x2 + 3x – 4 8. The graph of the quadratic function
(b) f(x) = x2 + 3x + 2
(c) f(x) = 3x2 – 3x – 4 f (x) = x2 – 6x + 2p – 3 does not intersect the x-axis.
Find the range of values of p.

49

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

9. The graph of the quadratic function f (x) = px2 – 2x + p (a) Express f (x) in the form of (x + m)2 + n, where
touches the x-axis at only one point. Find the m and n are constants.

possible values of p. (b) State the equation of the axis of symmetry for
10. The graph of the quadratic function f (x) = 2hx2 – 6hx + 9 the curve.
touches the x-axis at only one point. Find the value
Chapter 2 19. Given that the minimum value of f (x) = x2 – 5x + p
of h. is – 94. Find the value of p.
11. The line y = 1 – 5x intersects the curve y = t – x – 2x2
at two different points. Find the range of values of t. 20. Given that the maximum value of f (x) = k – 6x – x2
12. Given that f(x) = 2(x – 4)2 – 8 = a(x – p)(x – q) for is 3. Find the value of k.

all values of x. Find the values of a, p and q where 21. The quadratic function, f (x) = 1 – 2p2 + 3px – x2
p  q. has a maximum value of (q2 – p) where p and q are

13. Express each of the following the vertex form of constants. By using completing the squares method,
quadratic functions into general form and intercept p
show that q = + 2 .
form. 2
(a) f(x) = (x + 2)2 – 9
22. The quadratic function f (x) = x2 + 4px + 5p2 – 1 has
1 2(b) x 9 2 1 a minimum value of (q2 + 2p – 2), where p and q are
f(x) = – 4 – 8
constants. By using completing the squares method,
(c) f(x) = 4 – (2x – 1)2 express p in terms of q.

14. Find the vertex of the quadratic function 23. The diagram below shows the graph of the function
f(x) = – 31(x + 3)2 – 6 and change it into general f (x) = –2(x – 3)2 + 5, where a = –2, h = 3 and k = 5.

form. f(x)

5

15. Express each of the following quadratic functions 03 x
in the form of f(x) = a(x – h)2 + k where a, h and k
are constants, by using the completing the squares f(x) = –2(x – 3)2 + 5
method. Hence, state the maximum or minimum
value for f(x) and the corresponding value of x. (a) Determine the maximum point and the equation
(a) f(x) = x2 + 4x – 2 (b) f(x) = 2x2 – 4x + 1 of the axis of symmetry.
(c) f(x) = 4x2 + 10x – 3 (d) f(x) = –2x2 – 4x + 5
(e) f(x) = 2 – 5x – 3x2 (f) f(x) = 4x2 – 8x + 1 (b) Make generalisations on the shapes and
position of the graph of the given functions
16. Express the quadratic function y = x2 – 6x + 11 when compared to the values of a, h and k
in the form of y = (x + p)2 + q, where p and q are of the following functions. Hence, sketch the
constants. Find graphs of the functions.
(a) the values of p and q,
(b) the coordinates of the vertex. Hence, determine (i) f (x) = – 4(x – 3)2 + 5
whether the vertex is a maximum or minimum (ii) f (x) = –2(x – 1)2 + 5
point. (iii) f (x) = –2(x – 3)2 – 1

17. Express the quadratic function y = 4 + 12x – 3x2 in 24. The diagram below shows the graph of the function
the form of y = a(x – h)2 + k, where a, h and k are f (x) = (x + 4)2 + 2q, where q is a constant. The
constants. Find minimum point of the graph is (p, –3).
(a) the values of a, h and k,
(b) the maximum point. f(x)

18. f(x) f(x) = (x + 4)2 + 2q
n

y=4 0 x
4x (p, –3)
y = f(x)
–1 0 (a) State the values of p, q and n.
(b) If the graph is shifted 3 units to the right,
The diagram above shows the graph of the quadratic
function y = f (x). Given that the line y = 4 is a determine the equation of the axis of symmetry
tangent to the curve of y = f (x). for the graph.
(c) If the graph is shifted 2 units upwards, determine
the minimum value.

50

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

25. Compare the graph of the quadratic function 31. The curve of the quadratic function f (x) = (x – h)2 + k Chapter 2
f (x) = 12(x – 2)2 – 3 to the graph of the function intersects the x-axis at points (–2, 0) and (3, 0). The
line y = –614 touches the curve at its minimum point.
f (x) = x2. Sketch and comment on the differences (a) Find the values of h and k.
between these two graphs. (b) Sketch the graph of f (x) for –3 < x < 4.
(c) If the graph is reflected at the x-axis, write the
26. equation of the curve.

f(x)

32. The weekly profit, K(x), in RM, for a company that

k0 x sells calculators is given by the quadratic function,
K(x) = –0.5x2 + 40x – 200, where x is the number of

The diagram above shows the graph of the quadratic calculators sold every week.
(a) Express K(x) in the vertex form.

1 2function f (x) = (b) Find

x – 7 2 – 2, where k is the vertex of (i) the number of calculators that must be
2
sold per week to obtain maximum profit,

the graph. State (ii) the maximum profit. HOTS
(a) the value of k, Applying

(b) the equation of the axis of symmetry of the 33. The daily cost, H, in RM, for a factory that produces

curve. decorative lamps is given by the quadratic function,

27. f(x) H(x) = 500 – 8x + 1 x2, where x is the number of
4

(p, –6) 0 x units of decorative lamps produced per day.
(a) Express H(x) in the vertex form.

(b) Find

(i) the number of decorative lamps that must

be produced per day to achieve minimum

cost,

The diagram above shows the graph of the quadratic (ii) the minimum cost. HOTS
function f (x) = –(x + 4)2 – 3q, where q is a constant. Applying
Given that (p, – 6) is the maximum point of the
graph. State 34. xm
(a) the values of p and q,
(b) the equation of the axis of symmetry of the ym
graph.
Rahim suggests to fence up two adjacent pieces of
28. The quadratic function f (x) = –(x + h)2 + 3k – 2 has rectangular land as shown above to plant flowers.
a maximum point at (4, 16). Find He has a wire fence of 24 m length.
(a) the values of h and k, (a) Find the total area of the land, A m2, in terms of x.
(b) the equation of the axis of symmetry for the (b) Find
graph of function. (i) the value of x and the value of y, such that
the area A is a maximum,
29. Sketch the graph of each of the following quadratic (ii) the maximum area, in m2. HOTS
functions.
(a) f(x) = x2 – 2x + 1 Evaluating
(b) f(x) = x2 + 2x – 8
(c) f(x) = 2x2 – 10x + 8 35. x
(d) f(x) = x2 + 6x + 9
(e) f(x) = 4x2 – 6x + 5 y
(f) f(x) = –4x2 + 12x – 9
(g) f(x) = 3x – x2 – 2 The diagram above shows a piece of land that
(h) f(x) = –x2 + 5x – 4 already fenced up by using a wire fence of 120 m.
Both of the rectangular lands are placed next to
30. The quadratic function f (x) = –x2 + 3x + 4 is each other with both lands have equal length, x m,
expressed in the form of f (x) = –(x + p)2 + 2q and width, y m, respectively.
where p and q are constants. (a) Determine the values of x and y such that the
(a) Find the value of p and of q. whole area of the land is maximum.
(b) (i) Find the value of x when f (x) = 0. (b) Find the maximum area of the land.
(ii) Hence, sketch the graph for f (x) > 0.
HOTS
Evaluating

51

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 2

SPM Practice

Chapter 2 PAPER 1 11. Find the range of values of x for 3x2 + 11x > 4.

1. Given that the quadratic equation 2x2 + px – 18 = 0, 12. Given that the quadratic function f(x) = 14x – 2x2.
Find
where p is a constant, find the value of p if (a) the coordinates of the vertex of the quadratic
(a) one of the roots of the equation is 2, function,
(b) the sum of the equation roots is –1. (b) the range of values of x when f(x) is positive.

2. Given that a and b are the roots of the quadratic 13. f(x)
equation 2x2 – 6x – 3 = 0. Form a quadratic equation
with roots 2a and 2b.

3. Given that a and b are the roots of the quadratic
SPM equation 6x2 + 4x – 3 = 0. Form a quadratic equation
2016 a b
2 2
with roots and . 02 7 x

4. Find the range of values of x such that the quadratic The diagram above shows the graph of the quadratic
SPM function f(x) = 7 + 6x – x2 is negative.
2017 1 2function f(x) =9 2 25
x– 2 4
– . State

5. Given that the quadratic function (a) the coordinates of the minimum point of the
2S0P1M6 f(x) = x2 + 2px + p – 6, where p is a constant, is
always positive when m , p , n. Find the value curve,

of m and the value of n. (b) the equation of the axis of symmetry for the

curve,
(c) the range of values of x when f(x) is negative.

6. The graph of the quadratic function 14. The quadratic function f(x) = x2 – 2kx + 6 + k, where
2S0P1M5 f(x) = hx2 – 6x – 2k, where h and k are constants, SPM k is a constant, is always positive m k n.
2018 Find the values of m and n. for  
has a minimum value.
(a) Given that h is an integer such that –2 , h , 2, 15. (a) State the minimum value of f(x) = 2(x – 1)2 + 7
and the corresponding value of x.
state the value of h.
(b) Based on the value of h in (a), find the value of (b) Hence, sketch the graph of f(x) = 2(x – 1)2 + 7.

k if the graph touches the x-axis at one point.

7. The quadratic equation 4px2 – 6qx + 2p = 0, where 16. f(x)
SPM p q
2017 p and are constants, has two equal roots. Find
: q.
k

8. Given that the curve y = (k – 3)x2 – x + 6 where k is x
SPM a constant, intersects the straight line y = 5x 0 (4, 0)
2018 two points. Find the range of values of k. – 1 at

9. The quadratic equation x2 + 4(3x + p) = 0, where p The diagram above shows the graph of the quadratic
is a constant, has roots m and 2m, m ≠ 0. function f(x) = (x – p)2 + q, where p and q are
(a) Find the value of m and the value of p. constants. State
(b) Hence, form a quadratic equation with roots (a) the values of p and k,
m + 6 and m – 1 . (b) the equation of the axis of symmetry for the
curve.
10. Given that the quadratic equation 2x2 + mx + n = 0
has roots 4 and – 23, find 17. The graph of the quadratic function f(x) = mx2 – 8x + 8n,
(a) the value of m and the value of n, 2 S0P1M5 where m and n are constants, has one minimum
(b) the range of values of k such that the quadratic
equation 2x2 + mx + n = k has imaginary roots. point.
(a) State the value of m if –2  m  2 where m is

an integer.
(b) Based on the answer in (a), find the value of n

if the graph touches the x-axis at one point.

52

Additional Mathematics  Form 4  Chapter 2  Quadratic Functions 

18. The diagram below shows a rectangular piece of 3. A piece of wire with length of 76 cm is cut into two
SPM
2014 land with length of 6.6 m and width of 4.6 m. A parts with different lengths. Each part is bent to form
path of width with x m around the land has been
a square such that the total area of both squares is

cemented. The area of the path is 12 m2. Find the 185 cm2. Find the lengths of both parts of the cut
value of x.
wire. HOTS
Analysing

xm 4. Initially, a group of n students share the cost of
xm
RM150.00 to buy a present for Emily’s birthday. Chapter 2

4.6 m When 4 more students join the initial group of

students to share the cost of the present, it is found

that each student pays RM6.25 less than the original

payment. Find the amount to be paid by each

6.6 m student of the new group. HOTS
Evaluating

19. A rectangular wall of Amin’s bedroom is white 5. (a) Express f(x) = – 4x2 + 4x – 1 in the form of
SPM coloured with length of 4x m and width of 3x m. He f(x) = a(x – h)2 + k, where a, h and k are
2018 draws similar squares with sides x m at every corner constants.

of the wall and paints the squares with blue colour. (b) Sketch the graph of f(x) = – 4x2 + 4x – 1 and
Find the range of values of x if the part of the wall state the coordinates of the maximum point.
that remains white coloured is at least (x2 + 7) m2.
6. The quadratic function f(x) = x2 – hx – 5 has a
PAPER 2 minimum point at (2, k).
(a) Find the values of h and k.
1. The quadratic equation x(x – 4) = 2p – 3, where p is (b) Find the intersection point of the graph of
function f(x) with the x-axis. Hence, sketch the
graph of f(x).

a constant, has roots a and b. 7. (a) Express f(x) = –3x2 + 8x – 11 in the form of
(a) Find the range of values of p if a ≠ b. f(x) = a(x – m)2 + n, where a, m and n are
(b) Another quadratic equation 3x2 + qx – 4 = 0, constants.

where q is a constant, has roots a and b . (b) Sketch the graph of f(x).
Find the values of p and q. 3 3 (c) State the axis of symmetry of the graph of the

function.

2. 8. (a) Find the value of p if the graph of a quadratic
function f(x) = p – 2 + 2px – x2 touches the
SPM x-axis at one point.
2017
(b) Hence, sketch the graph f(x) for negative values
Chocolate bar of p.

C 9. (a) Sketch f(x) = 2x2 – 8x + 5. State the coordinates
of the vertex and the x-intercept of the graph of
14.8 cm the function.

The diagram above shows the cross-section of a (b) Based on the graph in (a), find
cylindrical-shaped chocolate bar stored inside a box (i) the values of x that satisfy the inequality
in the shape of cuboid. The height of the box is
enough to store the entire chocolate bar such that 2x2 + 5 < 8x,
the chocolate bar touches the side surface and the (ii) the range of values of t if 2x2 – 8x + 5 + t = 0
base of the box. It is found that point C of the has no roots.
chocolate bar is at 2 cm from the side of the box
and 1 cm from the base of the box. The width of 10. The curve of the quadratic function
the box is 14.8 cm. Determine whether three similar SPM
chocolate bars can be stored into the box. Justify f(x) = –  32 (x + m)2 + 2n intersects the x-axis at
your answer. 2016

points (–2, 0) and (4, 0). The straight line y = 6

touches the maximum point of the curve.
(a) Find the value of m and of n.
(b) Hence, sketch the graph f(x) for –2 < x < 4.
(c) If the graph is reflected on the x-axis, write the

equation of the curve.

53

  Additional Mathematics  Form 4  Chapter 2  Quadratic Functions

11. Given that the straight line 2x – y + 6 = 0 is a 15. E F
tangent to the curve y = x2 + kx + 7.
(a) Find the possible value of k if k . 0.
(b) Hence, sketch the graph of y = x2 + kx + 7.

12. The weekly income, P(x), in RM, for a factory that M
x cm
produces certain electronic components is given by

Chapter 2 the quadratic function, P(x) = 800x – 1 x2, where 2x cm
10
x is the number of electronic components sold per H N G

week. The diagram above shows a rectangle EFGH with
(a) Express P(x) in the vertex form. length EF = 64 cm and FG = 40 cm. A triangle FMN
is drawn inside rectangle EFGH where MH = x cm
(b) Find the and HN = 2x cm.
(a) Express the area of triangle FMN, A cm2, in
(i) number of electronic components that terms of x.
(b) Find the minimum area of triangle FMN and the
must be produced per week to obtain corresponding value of x.

maximum income, HOTS
Analysing
(ii) maximum income. HOTS
Analysing

13. A projectile is launched vertically upwards from the 16.
surface of the ground. The height, h, in m, at t
seconds after the launch is h(t) = 30t – 5t2. Wall
(a) Express h(t) in the form of a(t – p)2 + q, where
a, p and q are constants. xm
(b) Determine the
(i) maximum height achieved by the projectile, The diagram above shows a rectangular piece of
(ii) time taken to achieve the maximum height. land fenced right next to a wall. The total length of
the wire used to fence the land is 80 m.
HOTS (a) Show that the area of the fenced land, A m2 is
Applying A = 80x – 2x2 .
(b) Hence, find the maximum area of the fenced
14. A stone is thrown vertically upwards from a platform land and the corresponding value of x.

that is 4 m from the surface of the ground. The HOTS
height of the stone, h, in m, at time t seconds after Applying
the throw is h(t) = 4 + 30t – 5t2.

(a) Find the

(i) time taken by the stone to reach the

maximum height,

(ii) maximum height.

(b) Find the range of time when the stone is at a

height more than 4 m from the surface of the

ground. HOTS
Analysing

54

Matematik Tambahan  Tingkatan 4  Jawapan 

ANSWERS

1Chapter (b) y
5
Functions
2
Try This! 1.1 1 Range: 0 ≤ f(x) ≤ 5
–3 –2 0 x
1. (a) 3 (b) 3 (c) 7
(c) y 3
2. (a) Function. Each object only has one image, even
though the element 6 does not have an object. 5 Range: 0 ≤ f(x) ≤ 5
3
(b) Not a function. Object 2 has three images, which are
2, 4, and 6. 0 x
58
(c) Not a function. Element s does not have an image. (d)
y
3. (a) This graph is not a graph of y as a function of x. There 7
are vertical lines which intersect the graph at two 5
different points. 3

(b) This graph is a graph of y as a function of x. Each
vertical line intersects the graph at one point only.

(c) This graph is a graph of y as a function of x. Each
vertical line intersects the graph at one point only.

(d) This graph is not a graph of y as a function of x. All
vertical lines intersect the graph at two different points,
except at the y-axis where the vertical line intersects
at one point only.

4. (a) (i) f : x → x3 or f(x) = x3
(ii) h : x → x + 1 or h(x) = x + 1
(b) A : r → πr 2 or A(r) = πr 2
(c) (i) f : x → 2x2 + 3x – 1 or f(x) = 2x2 + 3x – 1
(ii) g : x → sin x or g(x) = sin x

5. (a) 0 (b) –3 (c) —41 (d) – —32

6. (a) False (b) True (c) True

7. (a) Domain = {x, y, z}, range = {p, q} –4 – 2–3 0 x
(b) Domain = {–2, –1, 1, 2}, range = {1, 4} 2
(c) Domain = {1, 2, 3, 4, 5}, range = {12, 24, 36, 48, 50}
(d) Domain is x ∈ , range is y ∈ . 11. (a) –1 (b) 5 – t (c) 3 – 2t
(e) Domain is x ∈ , range is y > 3.
(f) Domain is x ∈ , x ≠ –5, range is y ∈ , y ≠ 0. 12. (a) 1 (b) 9x + 1 (c) 6z – 2
(g) Domain is x ∈ , range is y > 0.
(h) Domain is x . 0, range is L . 0.

8. (a) Domain 0 < x < 8, range 3 < y < 15. 13. (a) 2 (b) 4.5 (c) 5 – 2x
(b) Domain –6 < x < 1, range 0 < y < 9.
14. (a) 10 (b) –5, 3 (c) 5
9. (a) When the domain is , all values of y are possible.
The range is y ∈ . 15. 5 (b) – —32 , 3 (c) 1, —35

(b) When x is limited to negative values, all values of 16. (a) 7, 5
y will be at least 2. The range is y > 2.
17. (a) 3
(c) The range is {2, 3, 4, 5, 6}. (b) 1, 6

10. (a) y 18. (a) h = 3, k = 1
(b) 0 < f(x) < 9

–2 0 Range: 0 ≤ f(x) ≤ 2 19. (a) RM77 000
x (b) RM10 500
2 (c) 3
(d) V(n) is a function, Each input n will give rise to one
and only one output value of V(n).

227