Module & More Add Math Tg 4

MoMduOlRe E&

Pembelajaran BERPANDU dan SISTEMATIK

MATEMATIK 4TINGKATAN
TAMBAHAN
KSSM
Additional Mathematics

Tee Hock Tian

PENERBITAN PELANGI SDN. BHD. (89120-H)
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ISBN: 978-967-2930-01-3

Cetakan Pertama 2021

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KANDUNGAN

BAB Fungsi 1 BAB Indeks, Surd dan Logaritma 60

1 Functions 4 Indices, Surds and Logarithms

1.1 Fungsi................................................................................... 1 4.1 Hukum Indeks...................................................................60
Functions Laws of Indices
1.2 Fungsi Gubahan.................................................................. 7 4.2 Hukum Surd......................................................................64
Composite Functions Laws of Surds
1.3 Fungsi Songsang...............................................................13 4.3 Hukum Logaritma............................................................71
Inverse Functions Laws of Logarithms
4.4 Aplikasi Indeks, Surd dan Logaritma.............................78
Praktis SPM 1.............................................................................19 Applications of Indices, Surds and Logarithms
Sudut KBAT...............................................................................21
Online Quick Quiz QR code ........................................................21 Praktis SPM 4.............................................................................80
Sudut KBAT...............................................................................82
BAB Online Quick Quiz QR code ........................................................82

2 Fungsi Kuadratik 22

Quadratic Functions BAB

2.1 Persamaan dan Ketaksamaan Kuadratik.......................22 5 Janjang 83
Quadratic Equations and Inequalities
2.2 Jenis-jenis Punca Persamaan Kuadratik........................28 Progressions
Types of Roots of Quadratic Equations
2.3 Fungsi Kuadratik...............................................................31 5.1 Janjang Aritmetik..............................................................83
Quadratic Functions Arithmetic Progressions
5.2 Janjang Geometri..............................................................91
Praktis SPM 2.............................................................................38 Geometric Progressions
Sudut KBAT...............................................................................43
Online Quick Quiz QR code ........................................................43 Praktis SPM 5...........................................................................101
Sudut KBAT.............................................................................105
Online Quick Quiz QR code ......................................................105

BAB Sistem Persamaan 44 BAB Hukum Linear 106

3 Systems of Equations 6 Linear Law

3.1 Sistem Persamaan Linear dalam Tiga 6.1 Hubungan Linear dan Tak Linear................................106
Pemboleh Ubah ................................................................44 Linear and Non-Linear Relations
6.2 Hukum Linear dan Hubungan Tak Linear..................112
Systems of Linear Equations in Three Variables Linear Law and Non-Linear Relations
3.2 Persamaan Serentak yang melibatkan Satu Persamaan 6.3 Aplikasi Hukum Linear..................................................118
Application of Linear Law
Linear dan Satu Persamaan Tak Linear.........................50
Simultaneous Equations involving One Linear Equation and Praktis SPM 6...........................................................................122
Sudut KBAT.............................................................................125
One Non-Linear Equation Online Quick Quiz QR code ......................................................125

Praktis SPM 3.............................................................................57
Sudut KBAT...............................................................................59
Online Quick Quiz QR code ........................................................59

iii

BAB Geometri Koordinat 126 9.3 Luas Segi Tiga..................................................................172
Area of a Triangle
7 Coordinate Geometry 9.4 Aplikasi Petua Sinus, Petua Kosinus dan Luas

7.1 Pembahagi Tembereng Garis........................................126 Segi Tiga...........................................................................174
Divisor of a Line Segment Application of Sine Rule, Cosine Rule and Area of a Triangle
7.2 Garis Lurus Selari dan Garis Lurus Serenjang............130
Parallel Lines and Perpendicular Lines Praktis SPM 9...........................................................................175
7.3 Luas Poligon....................................................................135 Sudut KBAT.............................................................................178
Areas of Polygons Online Quick Quiz QR code ......................................................178
7.4 Persamaan Lokus............................................................137
Equations of Loci BAB Nombor Indeks 179

Praktis SPM 7...........................................................................140 10 Index Numbers
Sudut KBAT.............................................................................143
Online Quick Quiz QR code ......................................................143 10.1 Nombor Indeks...............................................................179
Index Numbers
BAB Vektor 144 10.2 Indeks Gubahan..............................................................185
Composite Index
8 Vectors
Praktis SPM 10........................................................................190
8.1 Vektor...............................................................................144 Sudut KBAT.............................................................................193
Vectors Online Quick Quiz QR code ......................................................194
8.2 Penambahan dan Penolakan Vektor............................148
Addition and Subtraction of Vectors Kertas Pra-SPM............................................................... 195
8.3 Vektor dalam Satah Cartes............................................152
Vectors in a Cartesian Plane Jawapan

Praktis SPM 8...........................................................................159 http://www.epelangi.com/Module&More2021/
Sudut KBAT.............................................................................164 MatematikTambahan/T4/JawapanKeseluruhan.pdf
Online Quick Quiz QR code ......................................................164

BAB Penyelesaian Segi Tiga 165 Lembaran PBD

9 Solution of Triangles http://www.epelangi.com/Module&More2021/
MatematikTambahan/T4/LPBD.pdf
9.1 Petua Sinus.......................................................................165
Jawapan Lembaran PBD
Sine Rule
9.2 Petua Kosinus..................................................................169 http://www.epelangi.com/Module&More2021/
MatematikTambahan/T4/JawapanLPBD.pdf
Cosine Rule

BONUS Lembaran PBD dengan Jawapan

untuk Guru http://www.epelangi.com/Module&More2021/MatematikTambahan/T4/
BonusLPBD.

iv

Rekod Pencapaian Pentaksiran Murid

Matematik Tambahan Tingkatan 4 / Additional Mathematics Form 4

Nama murid: Kelas:
Student’s name Class

Bab Tahap Penguasaan
Chapter penguasaan Achievement
Performance
1 Deskriptor (3) (7)
FUNGSI level Descriptor Menguasai Belum
FUNCTIONS Menguasai
Achieve Not yet achieve
2
FUNGSI 1 Mempamerkan pengetahuan asas tentang fungsi.
KUADRATIK Demonstrate the basic knowledge of functions.
QUADRATIC
FUNCTIONS 2 Mempamerkan kefahaman tentang fungsi.
Demonstrate the understanding of functions.

Mengaplikasikan kefahaman tentang fungsi untuk
3 melaksanakan tugasan mudah.

Apply the understanding of functions to perform simple tasks.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi dalam konteks penyelesaian masalah rutin
4 yang mudah.
Apply appropriate knowledge and skills of functions in the context of
simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi dalam konteks penyelesaian masalah rutin
5 yang kompleks.
Apply appropriate knowledge and skills of functions in the context of
complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi dalam konteks penyelesaian masalah bukan
6 rutin secara kreatif.
Apply appropriate knowledge and skills of functions in the context of
non-routine problem solving in a creative manner.

1 Mempamerkan pengetahuan asas tentang fungsi kuadratik.
Demonstrate the basic knowledge of quadratic functions.

2 Mempamerkan kefahaman tentang fungsi kuadratik.
Demonstrate the understanding of quadratic functions.

Mengaplikasikan kefahaman tentang fungsi kuadratik untuk
3 melaksanakan tugasan mudah.

Apply the understanding of quadratic functions to perform simple tasks.

v © Penerbitan Pelangi Sdn. Bhd.

  Matematik Tambahan  Tingkatan 4  Rekod Pencapaian Pentaksiran Murid

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi kuadratik dalam konteks penyelesaian
4 masalah rutin yang mudah.
Apply appropriate knowledge and skills of quadratic functions in the
context of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi kuadratik dalam konteks penyelesaian
5 masalah rutin yang kompleks.
Apply appropriate knowledge and skills of quadratic functions in the
context of complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang fungsi kuadratik dalam konteks penyelesaian
6 masalah bukan rutin secara kreatif.
Apply appropriate knowledge and skills of quadratic functions in the
context of non-routine problem solving in a creative manner.

1 Mempamerkan pengetahuan asas tentang sistem persamaan.
Demonstrate the basic knowledge of systems of equations.

Mempamerkan kefahaman tentang penyelesaian sistem
2 persamaan.

Demonstrate the understanding of systems of equations.

Mengaplikasikan kefahaman tentang sistem persamaan
untuk melaksanakan tugasan mudah.
3 Apply the understanding of systems of equations to perform simple

tasks.

3 Mengaplikasikan pengetahuan dan kemahiran yang sesuai
SISTEM tentang sistem persamaan dalam konteks penyelesaian
PERSAMAAN 4 masalah rutin yang mudah.
SYSTEMS OF Apply appropriate knowledge and skills of systems of equations in the
EQUATIONS context of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang sistem persamaan dalam konteks penyelesaian
5 masalah rutin yang kompleks.
Apply appropriate knowledge and skills of systems of equations in the
context of complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang sistem persamaan dalam konteks penyelesaian
6 masalah bukan rutin secara kreatif.
Apply appropriate knowledge and skills of systems of equations in the
context of non-routine problem solving in a creative manner.

Mempamerkan pengetahuan asas tentang indeks, surd dan
4 1 logaritma.
INDEKS, SURD Demonstrate the basic knowledge of indices, surds and logarithms.

DAN LOGARITMA
INDICES, SURDS
AND LOGARITHMS Mempamerkan kefahaman tentang indeks, surd dan
2 logaritma.
Demonstrate the understanding of indices, surds and logarithms.

© Penerbitan Pelangi Sdn. Bhd. vi

  Matematik Tambahan  Tingkatan 4  Rekod Pencapaian Pentaksiran Murid 

Mengaplikasikan kefahaman tentang indeks, surd dan
logaritma untuk melaksanakan tugasan mudah.
3 Apply the understanding of indices, surds and logarithms to perform

simple tasks.

Mengaplikasikan pengetahuan dan kemahiran yang
sesuai tentang indeks, surd dan logaritma dalam konteks
4 penyelesaian masalah rutin yang mudah.
Apply appropriate knowledge and skills of indices, surds and logarithms
in the context of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang indeks dan logaritma dalam konteks penyelesaian
5 masalah rutin yang kompleks.
Apply appropriate knowledge and skills of indices and logarithms in the
context of complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang indeks dan logaritma dalam konteks penyelesaian
6 masalah bukan rutin secara kreatif.
Apply appropriate knowledge and skills of indices and logarithms in the
context of non-routine problem solving in a creative manner.

1 Mempamerkan pengetahuan asas tentang janjang.
Demonstrate the basic knowledge of progressions.

Mempamerkan kefahaman tentang janjang aritmetik dan
janjang geometri.
2 Demonstrate the understanding of arithmetic progressions and

geometric progressions.

Mengaplikasikan kefahaman tentang janjang aritmetik dan
janjang geometri untuk melaksanakan tugasan mudah.
3 Apply the understanding of arithmetic progressions and geometric

progressions to perform simple tasks.

5 Mengaplikasikan pengetahuan dan kemahiran yang sesuai
JANJANG tentang janjang aritmetik dan janjang geometri dalam
PROGRESSIONS 4 konteks penyelesaian masalah rutin yang mudah.
Apply appropriate knowledge and skills of arithmetic progressions and
geometric progressions in the context of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang janjang aritmetik dan janjang geometri dalam
konteks penyelesaian masalah rutin yang kompleks.
5 Apply appropriate knowledge and skills of arithmetic progressions

and geometric progressions in the context of complex routine problem
solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang janjang aritmetik dan janjang geometri dalam
konteks penyelesaian masalah bukan rutin secara kreatif.
6 Apply appropriate knowledge and skills of arithmetic progressions and

geometric progressions in the context of non-routine problem solving in
a creative manner.

vii © Penerbitan Pelangi Sdn. Bhd.

  Matematik Tambahan  Tingkatan 4  Rekod Pencapaian Pentaksiran Murid

Mempamerkan pengetahuan asas tentang garis lurus
1 penyuaian terbaik.

Demonstrate the basic knowledge of lines of best fit.

Mempamerkan kefahaman tentang garis lurus penyuaian
2 terbaik.

Demonstrate the understanding of lines of best fit.

Mengaplikasikan kefahaman tentang hukum linear untuk
3 melaksanakan tugasan mudah.

Apply the understanding of linear law to perform simple tasks.

6 Mengaplikasikan pengetahuan dan kemahiran yang sesuai
HUKUM LINEAR tentang hukum linear dalam konteks penyelesaian masalah
4 rutin yang mudah.
LINEAR LAW Apply appropriate knowledge and skills of linear law in the context of
7 simple routine problem solving.

GEOMETRI Mengaplikasikan pengetahuan dan kemahiran yang sesuai
KOORDINAT tentang hukum linear dalam konteks penyelesaian masalah
COORDINATE 5 rutin yang kompleks.
Apply appropriate knowledge and skills of linear law in the context of
GEOMETRY complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang hukum linear dalam konteks penyelesaian masalah
6 bukan rutin secara kreatif.
Apply appropriate knowledge and skills of linear law in the context of
non-routine problem solving in a creative manner.

Mempamerkan pengetahuan asas tentang pembahagi
1 tembereng garis.

Demonstrate the basic knowledge of divisor of line segments.

Mempamerkan kefahaman tentang pembahagi tembereng
2 garis.

Demonstrate the understanding of divisor of line segments.

Mengaplikasikan kefahaman tentang geometri koordinat
untuk melaksanakan tugasan mudah.
3 Apply the understanding of coordinate geometry to perform simple

tasks.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang geometri koordinat dalam konteks penyelesaian
4 masalah rutin yang mudah.
Apply appropriate knowledge and skills of coordinate geometry in the
context of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang geometri koordinat dalam konteks penyelesaian
5 masalah rutin yang kompleks.
Apply appropriate knowledge and skills of coordinate geometry in the
context of complex routine problem solving.

© Penerbitan Pelangi Sdn. Bhd. viii

  Matematik Tambahan  Tingkatan 4  Rekod Pencapaian Pentaksiran Murid 

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang geometri koordinat dalam konteks penyelesaian
6 masalah bukan rutin secara kreatif.
Apply appropriate knowledge and skills of coordinate geometry in the
context of non-routine problem solving in a creative manner.

1 Mempamerkan pengetahuan asas tentang vektor.
Demonstrate the basic knowledge of vectors.

2 Mempamerkan kefahaman tentang vektor.
Demonstrate the understanding of vectors.

Mengaplikasikan kefahaman tentang vektor untuk
3 melaksanakan tugasan mudah.

Apply the understanding of vectors to perform simple tasks

8 Mengaplikasikan pengetahuan dan kemahiran yang sesuai
VEKTOR tentang vektor dalam konteks penyelesaian masalah rutin
VECTORS 4 yang mudah.
Apply appropriate knowledge and skills of vectors in the context of
9 simple routine problem solving.
PENYELESAIAN
Mengaplikasikan pengetahuan dan kemahiran yang sesuai
SEGI TIGA tentang vektor dalam konteks penyelesaian masalah rutin
SOLUTION OF 5 yang kompleks.
TRIANGLES Apply appropriate knowledge and skills of vectors in the context of
complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang vektor dalam konteks penyelesaian masalah bukan
6 rutin secara kreatif.
Apply appropriate knowledge and skills of vectors in the context of non-
routine problem solving in a creative manner.

Mempamerkan pengetahuan asas tentang petua sinus dan
1 petua kosinus.

Demonstrate the basic knowledge of sine rule and cosine rule.

Mempamerkan kefahaman tentang petua sinus dan petua
2 kosinus.

Demonstrate the understanding of sine rule and cosine rule.

Mengaplikasikan kefahaman tentang petua sinus, petua
kosinus dan luas segi tiga untuk melaksanakan tugasan
3 mudah.
Apply the understanding of sine rule, cosine rule and area of a triangle
to perform simple tasks.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang penyelesaian segi tiga dalam konteks penyelesaian
4 masalah rutin yang mudah.
Apply appropriate knowledge and skills of solution of triangles in the
context of simple routine problem solving.

ix © Penerbitan Pelangi Sdn. Bhd.

  Matematik Tambahan  Tingkatan 4  Rekod Pencapaian Pentaksiran Murid

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang penyelesaian segi tiga dalam konteks penyelesaian
5 masalah rutin yang kompleks.
Apply appropriate knowledge and skills of solution of triangles in the
context of complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang penyelesaian segi tiga dalam konteks penyelesaian
6 masalah bukan rutin secara kreatif.
Apply appropriate knowledge and skills of solution of triangles in the
context of non-routine problem solving in a creative manner.

1 Mempamerkan pengetahuan asas tentang nombor indeks.
Demonstrate the basic knowledge of index numbers.

2 Mempamerkan kefahaman tentang nombor indeks.
Demonstrate the understanding of index numbers.

Mengaplikasikan kefahaman tentang nombor indeks untuk
3 melaksanakan tugasan mudah.

Apply the understanding of index numbers to perform simple tasks.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang nombor indeks dalam konteks penyelesaian masalah
10 4 rutin yang mudah.
NOMBOR INDEKS Apply appropriate knowledge and skills of index numbers in the context
INDEX NUMBERS
of simple routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang nombor indeks dalam konteks penyelesaian masalah
5 rutin yang kompleks.
Apply appropriate knowledge and skills of index numbers in the context
of complex routine problem solving.

Mengaplikasikan pengetahuan dan kemahiran yang sesuai
tentang nombor indeks dalam konteks penyelesaian masalah
6 bukan rutin secara kreatif.
Apply appropriate knowledge and skills of index numbers in the context
of non-routine problem solving in a creative manner.

© Penerbitan Pelangi Sdn. Bhd. x

BAB Fungsi   Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

1 Functions Functions / Fungsi

1.1 Fungsi

Functions

NOTA IMBASAN

Fungsi / Function Bukan fungsi / Not a function

x f(x) y x f(x) y
2
28
3 125 x
5 27 4

Tatatanda fungsi / Function notation y
6
NOTA IMBASAN f : x → x3
z
f(x) = x3 7

f(x) f(x)

x 0x
0
Garis mencancang memotong pada dua titik
Ujian garis mencancang / Vertical line test Vertical line cuts at 2 points

Fungsi mutlak, f(x) = |x|

3 9 Absolute function, f(x) = |x|
| x| =  –x,x  , jikjiaka/ if x > 0
16 / if x > 0

4 49 f(x)

7 64

Domain = {3, 4, 7} 0 x
Kodomain / Codomain = {9, 16, 49, 64}
Objek / Objects = 3, 4, 7 1 NOTA
Julat / Range = {9,16,49}

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

1. Tentukan sama ada setiap hubungan yang berikut ialah satu fungsi atau bukan. Beri sebab untuk jawapan

anda.  TP 1

Determine whether each of the following relations is a function. Give reason for your answer. 

BAB 1 CONTOH (a)

p p4
a q8

q

b r 12
r

c

s

Suatu fungsi. Setiap objek hanya mempunyai satu Suatu fungsi. Setiap objek mempunyai hanya

imej. satu imej.

A function. Each object has only one image. A function. Each object has only one image.

(b) (c)

3m 6x

5n 9y

7p 13 z

11
Bukan fungsi. 6 mempunyai lebih daripada satu

Bukan fungsi. 11 tidak mempunyai imej. imej.
Not a function. 6 has more than 1 image.
Not a function. 11 does not have any image.

2. Menggunakan ujian garis mencancang untuk menentukan sama ada graf berikut ialah satu fungsi atau bukan.
Using the vertical line test to determine whether each of the following graph is a function or not.  TP 2

CONTOH (a)

y

y

3

0x 02 x



Satu fungsi. Garis mencancang memotong graf

Satu fungsi. Garis mencancang memotong graf itu itu pada satu titik sahaja.
pada satu titik sahaja.
A function. The vertical line cuts the graph at only one
A function. The vertical line cuts the graph at only one
point.

point.

2

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

(b) y (c) y

9 –25
–2 0
–4 0 4 x x BAB 1

–9

Bukan fungsi. Garis mencancang memotong graf Satu fungsi. Garis mencancang memotong graf
itu pada lebih daripada satu titik.
itu pada satu titik sahaja.
Not a function. The vertical line cuts the graph at more
A function. The vertical line cuts the graph at only one
than one point.
point.

3. Tulis setiap fungsi yang berikut menggunakan tatatanda fungsi.   TP 2

Write each of the following by using the function notation.

CONTOH (a)

g

f

–3 –27 –5 5
4 64 7
5 125 –2 11

f : x → x3 4 g : x → |2x – 1|
f(x) = x3 g(x) = |2x – 1|




(b) h (c) k

3 –19 1 5
5 2–15 3 11
6 3–16 4 14

1
x2
h:x→ k : x → 3x + 2
k(x) = 3x + 2
h(x) = 1
x2

3

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

4. Lengkapkan jadual yang berikut.   TP 1

Complete the following table. 

BAB 1 Fungsi Domain Kodomain Objek Imej Julat

Function Domain Codomain Object Image Range

CONTOH {0 < x < 6} {−3 < f(x) < 10} 0 < x < 6 –3 < f(x) < 13 {0 < f(x) < 13}

f(x)

13

1 6 x

02 {−2 < x < 4} {0 <f(x) < 10} −2 < x < 4 0 < f(x) < 10 {0 < f(x) < 10}
–3

(a)

f(x)

10

6

2

–2 0 34 x

(b) {−1, 0, 1, 2, 3} {−2, 0, 2, 4, 6} −1, 0, 1, 2, 3 −2, 0, 2, 4, 6 {−2, 0, 2, 4, 6}

f(x)

6 123 x
5
4
3
2
1

–1–10
–2

(c) {4, 10, 18} {2, 5, 9, 11} 4, 10, 18 2, 5, 9 {2, 5, 9}

42
10 5
18 9

11

(d) {(3, 10), (5, 26), {3, 5, 6, 9} {10, 26, 37, 3, 5, 6, 9 10, 26, 37, 82 {10, 26, 37,
(6, 37),(9, 82)} 82} 82}

4

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

5. Lakar graf bagi setiap fungsi berikut. Seterusnya, nyatakan julat sepadan bagi domain yang diberikan.
Sketch the graph of the following functions. Hence, state the corresponding range for the given domain.   TP 4

CONTOH (a) y = |x2 − 1|, −3 < x < 3

f(x) = |1 − 2x|, −1 < x < 3 f(x) BAB 1

x –1 0 1 1 2 3 (–3, 8) 8 (3, 8)
­2
7
f(x) 3 1 0 1 3
56

5

f(x) 4
5
4 3
3
2
1

–3 –2 –1 0 1 2 3 x

2 Julat/ Range: 0 < f(x) < 8
1

–1 0 123 x

Julat/ Range: 0 < f(x) < 5

6. Selesaikan setiap yang berikut.   TP 5

Solve each of the following.

CONTOH 1

Fungsi f ditakrifkan sebagai f : x → 4x + 3 , x ≠ 0. Cari
x
Function f is defined as f : x → 4x + 3 , x≠ 0. Find
x
(i) f(−2)
1
(ii) imej bagi 4 di bawah f.

the image of 1 under f.
4

(iii) nilai-nilai f apabila imej ialah 13.
the values of f when the image is 13.

Penyelesaian: 3 (iii) f(x) = 13
x
(i)   f(x) = 4x + 4x + 3 = 13
x
3 1
f(–2) = –8 – 2 = –9 2 4x2 + 3 = 13x

4x2 – 13x + 3 = 0

(ii)  f(x) = 4x + 3 (4x – 1)(x – 3) = 0
x
 f1 41 2 1 3 x = 1 ,  x = 3
= 41 4 2 + 1 4

4

= 1 + 12 = 13

5

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

CONTOH 2

Fungsi g ditakrifkan sebagai g : x → |2 – 8x|, cari

BAB 1 Function g is defined as g : x → |2 – 8x|, find

(i) g1 1 2
2

(ii) nilai-nilai x yang memetakan hubungan dirinya.

the values of x which maps to itself.

Penyelesaian: (ii) g(x) = x
|2 – 8x| = x
(i) g(x) = |2 – 8x|

g1 12 2 = |2 − 4|
= |−2|
2 – 8x = x 2 – 8x = –x
= 2
2 = x + 8x 2 = 8x – x

9x = 2 7x = 2

x = 2 x = 2
9 7

(a) Fungsi f ditakrifkan sebagai f : x → 12 , x ≠ 2. (b) Diberi g(x) = u3x – 7u, cari
Cari x–2
Given g(x) = u3x – 7u,
The function f is defined as f : x → 12 , x ≠ 2. Find
x–2 (i) nilai bagi g(−2) dan g(5).
(i) nilai bagi f(−4) dan f(10).
the value of f(−4) and f(10). the value of g(−2) and g(5).
­3x
(ii) nilai-nilai x dengan keadaan f(x) = . (ii) objek dengan keadaan imej ialah 14.

the values of the objects for which the image is 14.

x for which f(x)= 3 . (i) g(x) = u3x – 7u
x g(–2) = u–6 – 7u = 13
g(5) = u15 – 7u = 8
(i) f(x) = 12
x–2

f(–4) = 12 = –2
–6
(ii) g(x) = 14
f(10) = 12 = 3 u3x – 7u = 14
8 2 3x – 7 = 14
3x = 21
3x – 7 = –14
x = 7
(ii) f(x) = 3 3x = –7 7
4 3
x = –

x1–22 = 3
4
3(x – 2) = 48

x – 2 = 16

x = 18

6

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

(c) Diberi ℎ(x) = 2x2 + mx − 30 dengan keadaan m (d) Fungsi f ditakrifkan sebagai f(x) = 7 + |3x|. Cari BAB 1
ialah satu pemalar dan ℎ(2) = −34. Cari
The function f is defined as f(x) = 7 + |3x|. Find
Given that ℎ(x) = 2x2 + mx − 30 for which m is a constant
and ℎ(2) = −34. Find (i) nilai bagi f(−1) dan f(6).

(i) nilai m. the values of f(−1) and f(6).

the value of m. (ii) objek-objek dengan keadaan imejnya ialah
19.
(ii) nilai-nilai x yang memetakan kepada dirinya.
N OTAtheIMvaBluAeSs AofNx which map to itself. the objects for which the image is 19.

(i) ℎ(x) = 2x2 + mx − 30 (i) f(x) = 7 + |3x|
ℎ(2) = 8 + 2m − 30 = −34 f(−1) = 7 + |−3| = 10
2m = −34 + 22 f(6) = 7 + |18| = 25
m = −6
(ii) f(x) = 19
(ii) ℎ(x) = x
7 + |3x| = 19
2x2 − 6x − 30 = x
|3x| = 12
2x2 − 7x − 30 = 0
3x = 12 3x = –12
(x + 5)(x − 6) = 0
5  x = 4   x = –4
2
x = − , x = 6

1.2 Fungsi Gubahan

Composite Functions

NOTA IMBASAN

Af B gC Dalam rajah, fungsi f memetakan set A kepada set B dan fungsi g
memetakan set B kepada set C. Fungsi yang memetakan set A kepada set
x f(x) gf(x) C ialah fungsi gubahan, gf.
In the diagram, the function f maps set A to set B and the function g maps set B to set C.
The function which maps set A to set C is the composite function, gf.

gf

7. Bagi setiap pasangan fungsi yang berikut, cari (a) f(x) = 3x + 5
For each pair of the following function, find   TP 3 g(x) = 1 – 2x
(i) fg(x), (ii) gf(x).
(i) fg(x) = f(1 – 2x)
CONTOH = 3(1 – 2x) + 5
= 3 – 6x + 5
f(x) = 2x + 3 = 8 – 6x
g(x) = 1 – x
(ii) gf(x) = g(3x + 5)
Penyelesaian: = 1 – 2(3x + 5)
(i) fg(x) = f(1 – x) fg(x) = f(g(x)) = 1 – 6x – 10
= 2(1 – x) + 3 = –9 – 6x
= 2 – 2x + 3
= 5 – 2x

(ii) gf(x) = g(2x + 3) gf(x) = g(f(x))
= 1 – (2x + 3)
= 1 – 2x – 3
= –2 – 2x

7

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi (c) f(x) = 1 – x

(b) f(x) = x + 3 g(x) = 4
g(x) = 5x2 – 2 x

BAB 1 (i) fg(x) = f(5x2 – 2) (i) fg(x) = f 1 4 2
= 5x2 – 2 + 3 x
= 5x2 + 1
= 1– 4 , x≠ 0
(ii) gf(x) = g(x + 3) x
= 5(x + 3)2 – 2
= 5(x2 + 6x + 9) – 2 (ii) gf(x) = g(1 – x)
= 5x2 + 30x + 43
= 1 4 x , x ≠ 1
–

8. Bagi setiap pasangan fungsi yang berikut, cari

For each pair of the following functions, find   TP 3
(i) fg(3) (ii) gf(4)

CONTOH (a) f : x → 3x + 2,   g : x → 2 – x2

f : x → 5x,   g : x → 4 – 2x (i) g(3) = 2 – (3)2 = –7
fg(3) = f(–7)
Penyelesaian: = 3(–7) + 2
(i) g(3) = 4 – 2(3) = –2 = –19

fg(3) = f(–2) (ii)  f(4) = 3(4) + 2 = 14
gf(4) = g(14)
= 5(–2) = 2 – (14)2
= –194
= –10 1 Gantikan 4 ke dalam f, anda
akan dapat nilai 20.
(ii) f(4) = 5(4) = 20
gf(4) = g(20) Insert 4 into f, you will get 20.
= 4 – 2(20)
= –36 2 Gantikan 20 ke dalam g.
Insert 20 into g.

(b) f : x→ x– 3,  g:x → 3  , x≠ 0 (c) f : x → 2 – 8x,  g : x → x x 5 , x ≠ –5
x +

(i) g(3) = 3 =1 (i) g(3) = 3 3 = 3
3 +5 8

fg(3) = f(1) fg(3) = f1 3 2
8
= 1 – 3 = –2
= 2 – 81 3 2 = –1
8

(ii) f(4) = 4 – 3 = 1 (ii) f(4) = 2 – 8(4) = –30
gf(4) = g(1)
gf(4) = g(–30)

= 3 = –30
1 –30 +5

= 3 = 6
5

8

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

9. Bagi setiap pasangan fungsi yang berikut, cari
For each pair of the following functions, find   TP 3
(i) fg(x) (ii) gf(x) (iii) fg(2) (iv) gf(3)

CONTOH (a) f(x) = 3x + 1, g(x) = 2 – x BAB 1

f(x) = 2x,  g(x) = x + 5 (i) fg(x) = f(2 – x)
= 3(2 – x) + 1
Penyelesaian: = 7 – 3x
(i) fg(x) = f(x + 5)
= 2(x + 5) (ii) gf(x) = g(3x + 1)
= 2x + 10 = 2 – (3x + 1)
= 1 – 3x
(ii) gf(x) = g(2x)
= 2x + 5 (iii) fg(2) = 7 – 3(2)
= 1
(iii) fg(2) = 2(2) + 10
= 14 (iv) gf(3) = 1 – 3(3)
= –8
(iv) gf(3) = 2(3) + 5
= 11

(b) f(x) = x – 6,  g(x) = x2 + 1 (c) f : x→ 2+ 5x,  g: x → x , x≠ 1
x–1
(i) fg(x) = f(x2 + 1)
= x2 + 1 – 6 (i) fg(x) = f 1 x x 1 2
= x2 – 5 –

(ii) gf(x) = g(x – 6) = 2 + 51 x x 1 2
= (x – 6)2 + 1 –
= x2 – 12x + 37
= 2(x – 1) + 5x
(iii) fg(2) = 22 – 5 x–1
= 4 – 5
= –1 = 7x – 2 ,  x ≠ 1
x–1
(iv) gf(3) = (3)2 – 12(3) + 37
= 9 –36 + 37 (ii) gf(x) = g(2 + 5x)
= 10 2 + 5x
= 2 + 5x – 1

= 2 + 5x ,  x ≠ – 1
1 + 5x 5

(iii) fg(2) = 7(2) – 2 = 12
2–1

(iv) gf(3) = 2 + 5(3) = 17
1 + 5(3) 16

9

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

BAB 1 10. Selesaikan setiap yang berikut. (a) Diberi f : x → x + 3, g : x → 9 – 2x dan fg(x) = 6.
Solve each of the following.   TP 4
Given f : x → x + 3, g : x → 9 – 2x and fg(x) = 6.
CONTOH

Diberi f : x → 2x + 4, g : x → x – 2 dan fg(x) = 2. fg(x) = f(9 – 2x)
= 9 – 2x + 3
Given f : x → 2x + 4, g : x → x – 2 and fg(x) = 2. = 12 – 2x

Penyelesaian:

fg(x) = f(x – 2) 1 Cari fungsi gubahan fg(x). fg(x) = 12 – 2x = 6
= 2(x – 2) + 4 Find the composite function fg(x).

= 2x 2x = 6

fg(x) = 2x = 2 2 Samakan fg(x) dengan 2. x = 3
x = 1 Equate fg(x) with 2.

(b) Diberi f : x → x – 3, g : x → 3 – 5x dan gf(x) = –2. (c) Diberi f : x→ 3 , g: x → 2x + 1 dan gf(x) = –1.
x
Given f : x → x – 3, g : x → 3 – 5x and gf(x) = –2. Given f : x 3 :x → 2x + 1 and gf(x) = –1.
x ,g
→

gf(x) = g(x – 3) gf(x) = g1 3 2
= 3 – 5(x – 3) x
= 18 – 5x
21 3 2
gf(x) = 18 – 5x = –2 = x + 1

5x = 20 = 6 + x ,  x ≠ 0
x
x = 4

gf(x) = 6 + x = –1
x
6 + x = –x

2x = –6

x = –3

11. Cari fungsi g. (a) f(x) → x + 2,   fg(x) = 7 – 3x
Find the function g.   TP 4
f(x) = x + 2
CONTOH 1 fg(x) = g(x) + 2
7 – 3x = g(x) + 2
f(x) = x + 1,   fg(x) = x + 7 g(x) = 7 – 3x – 2
= 5 – 3x
Penyelesaian: 1 Gantikan g(x) ke dalam f(x). g : x → 5 – 3x
f(x) = x + 1 Insert g(x) into f(x).
2 Samakan kedua-dua fg(x).
fg(x) = g(x) + 1 Equate both fg(x).

x + 7 = g(x) + 1 3 Selesaikan untuk g(x).
Solve for g(x).
g(x) = x + 7 – 1
= x + 6
g : x → x + 6

10

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

(b) f(x) → 2x – 1,   fg(x) = 9 – 4x (c) f(x) = 4 – x,   fg(x) = 16 – x
8

f(x) = 2x – 1 f(x) = 4 – x

fg(x) = 2g(x) – 1 fg(x) = 4 – g(x) BAB 1

9 – 4x = 2g(x) – 1 16 – x = 4 – g(x)
8 = 4 –
2g(x) = 9 – 4x + 1 1 2 16 – x
8
2g(x) = 10 – 4x g(x)

g(x) = 10 – 4x = 32 – (16 – x)
2 8

= 5 – 2x 16 + x
8
g : x → 5 – 2x =

g : x → 16 + x
8

CONTOH 2 (d) f(x) = 5 – x,   gf(x) = 18 – x

f(x) = x + 2,   gf(x) = 7 – 3x gf(x) = 18 – x

Penyelesaian: g(5 – x) = 18 – x

gf(x) = 7 – 3x Katakan y = 5 – x
g(x + 2) = 7 – 3x
Katakan/ Let y = x + 2 1 Gantikan f(x) = x + 2 ke x = 5 – y
dalam gf(x).
g(y) = 18 – (5 – y)
Insert f(x) = x + 2 into gf(x).
= 13 + y
x = y – 2 2 Gantikan y = x + 2 dan
g(y) = 7 – 3(y – 2) x = y – 2 ke dalam g(x + 2). g(x) = 13 + x
= 7 – 3y + 6
= 13 – 3y Insert y = x + 2 and g : x → 13 + x
g(x) = 13 – 3x x = y – 2 into g(x + 2).
g : x → 13 – 3x
3 Selesaikan untuk g(y) ≈ g(x).
Solve for g(y) ≈ g(x).

(e) f : x → 6 , x ≠ 0,  gf : x→ 2x + 3 (f) f(x) = x + 2,   gf(x) = 8 – 3x2
x
gf(x) = 8 – 3x2
gf(x) = 2x + 3
g(x + 2) = 8 – 3x2
6
g1 x 2 = 2x + 3 Katakan y = x + 2

6 x = y – 2
x
Katakan y = g(y) = 8 – 3(y – 2)2

6 = 8 – 3(y2 – 4y + 4)
y
x = = 8 – 3y2 + 12y – 12

21 6 2 = –3y2 + 12y – 4
y
g(y) = + 3 g(x) = –3x2 + 12x – 4

12 g : x → –3x2 + 12x – 4
y
= + 3

g(x) = 12 + 3
x

g : x → 12 + 3, x ≠ 0
x

11

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

12. Selesaikan setiap yang berikut.
Solve each of the following.   TP 5
CONTOH 1

BAB 1 Fungsi g ditakrifkan sebagai g : x → 4x – 1, cari

Function g is defined as g : x → 4x – 1, find

(a) g2 (b) g21 12 2

Penyelesaian:

(a) g2(x) = gg(x) (b) g21 1 2 = 161 1 2 – 5
= g(4x – 1) 2 2
= 4(4x – 1) – 1
= 16x – 5 = 3
g2 : x → 16x – 5

Fungsi f ditakrifkan sebagai f :x → 4  , x ≠ k.
x
4
The function f is defined as f : x → x , x ≠ k.

(a) Tentukan nilai k. (b) Cari f 2 dan seterusnya cari nilai f 2(3).
Determine the value of k.
Find f 2 and hence find the value of f 2(3).

f(x) = 4 (b) f 2(x) = ff(x)
x
= f1 4 2
x ≠ 0 x

\ k = 0 = 4

1 4 2
x
= x

f 2 : x → x

f 2(3) = 3

CONTOH 2

Diberi fungsi f : x → 9 – 2x, g : x → ax + b dan fg : x → 1 – 6x. Cari nilai bagi a dan b.

Given the function f : x → 9 – 2x, g : x → ax + b and fg : x → 1 – 6x. Find the value of a and of b.

Penyelesaian: 1 Cari fungsi gubahan fg(x).
Find composite function fg(x).
fg(x) = f(ax + b) 2 Samakan kedua-dua fg(x).
= 9 – 2(ax + b) Equate both fg(x).
= 9 – 2b – 2ax

9 – 2b – 2ax = 1 – 6x

3 Bandingkan pemalar.
Compare the constant.



Bandingkan pemalar:  9 – 2b = 1, –2a = –6

Compare the constants: 2b = 8 a = 3  

b = 4

12

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

(c) Diberi fungsi f : x → 4x + k, g : x → x – 3 dan fg : x → mx – 8. Cari nilai bagi k dan m.

Given the function f : x → 4x + k, g : x → x – 3 and fg : x → mx – 8. Find the value of k and of m.

fg(x) = f(x – 3) BAB 1

= 4(x – 3) + k

= 4x – 12 + k

4x – 12 + k = mx – 8

Bandingkan pemalar: 

Compare the constants:

4x = mx, –12 + k = –8
k = 4
m = 4

(d) Diberi fungsi f : x → 5x – 1, g : x → 2x dan gf : x → ax + b. Cari nilai bagi a dan b.

Given the function f : x → 5x – 1, g : x → 2x and gf : x → ax + b. Find the value of a and of b.

gf(x) = g(5x – 1)
= 2(5x – 1)
= 10x – 2
10x – 2 = ax + b

Bandingkan pemalar:

Compare the constants:

10x = ax, b = –2

a = 10



1.3 Fungsi Songsang

Inverse Functions

NOTA IMBASAN

1. f B A f –1 B

A 23
56
23 89
56
89

Fungsi f memetakan unsur dalam Fungsi f –1 memetakan unsur dalam
set A kepada unsur dalam set B. set B kepada unsur dalam set A.
The function f maps the elements in The function f –1 maps the elements
set A onto the elements in set B. in set B onto the elements in set A.

2. f –1 ialah fungsi songsang bagi f. Tip
f –1 is called the inverse function of f.
f –1(x) ≠ 1
3. Hanya fungsi dengan hubungan satu kepada satu mempunyai fungsi songsang. f(x)
Only a function with one-to-one relation has an inverse function.

13

BAB 1 NOTA IMBASAN

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

4. f dan g ialah fungsi songsang antara satu sama lain jika dan hanya jika
f and g are inverse functions of one another if and only if
(i) fg(x) = x, x dalam domain g, dan
fg(x) = x, x in domain g, and
(ii) gf(x) = x, x dalam domain f.
gf(x) = x, x in domain f.

5. Jika f dan g ialah fungsi songsang antara satu sama lain, maka
If f and g are inverse function of one another then
(i) domain f = julat / range g, dan / and
(ii) domain g = julat / range f.
(iii) Graf g adalah pantulan graf f pada garis y = x.
Graph g is a reflection of graph f in the line y = x.

6. Jika f dan g ialah fungsi songsang antara satu sama lain maka titik (a, b) berada pada graf f dan titik (b, a) berada pada
graf g.
If f and g are inverse function of one another then point (a, b) lies on the graph f and point (b, a) lies on the graph g.

7. Kaedah untuk mendapatkan fungsi songsang:
Method to find inverse funtion:
(i) Katakan/ Let:  f –1(x) = y → f(y) = x.
(ii) Jadikan y sebagai perkara rumus bagi f(y) = x.
Make y into subject for f(y) = x.
(iii) Gantikan y ke dalam f –1(x) = y.
Insert y into f –1(x) = y.

13. Cari nilai bagi setiap yang berikut. TP 3 (a) Dalam gambar rajah anak panah yang berikut,
fungsi f memetakan x kepada y. Cari
Find the values for each of the following.
In the following arrow diagram, the function f maps x
CONTOH to y. Find

Dalam gambar rajah anak panah yang berikut, f
fungsi f memetakan x kepada y. Cari xy
4
In the following arrow diagram, the function f maps x to
y. Find 2

f 1
xy –34

4 –6 –1
5

2 –2

–1 –8 (i) f(1)

(i) f(2) (ii) f −1(−1)
(ii) f −1(−8)
(iii) f −1(−2) (iii) f −1(2)
(iv) f(4)
1(iv) f −1 3 2
Penyelesaian: 4
(i) f(2) = 4
(ii) f(−1) = −8, f −1(−8) = −1 (i) f(1) = −1
(iii) f(5) = −2, f −1(−2) = 5
(iv) f(2) = 4, f −1(4) = 2 (ii) f(1) = −1, f −1(−1) = 1

(iii) f(4) = 2, f −1(2) = 4

(iv) f(−6) = 3 , 1f −1 3 2 = −6
4 4

14

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

(b) Dalam gambar rajah anak panah yang berikut, (c) Dalam gambar rajah anak panah yang berikut, BAB 1

fungsi f memetakan x kepada y. Cari fungsi f memetakan x kepada y. Cari

In the following arrow diagram, the function f maps x In the following arrow diagram, the function f maps x
to y. Find to y. Find

f f
xy xy
5 –32 81

34 3 –5

–2 6 –2 –7

(i) f(3) (i) f(–2)
(ii) f −1(−7)
(ii) f −1(4) (iii) f −1(1)
(iv) f −1(−5)
(iii) f −1(6)
(i) f(–2) = 1
1(iv) f −1 3 2 (ii) f(3) = −7, f −1(−7) = 3
2 (iii) f(–2) = 1, f −1(1) = –2
(iv) f(8) = –5, f −1(−5) = 8
(i) f(3) = 3
2

(ii) f(–2) = 4, f −1(4) = –2

(iii) f(5) = 6, f −1(6) = 5

(iv) f(3) = 3 , 1f −1 3 2 =3
2 2

14. Tentukan sama ada setiap fungsi f berikut mempunyai fungsi songsang atau tidak. Beri satu sebab untuk

jawapan anda. TP 3

Determine whether each of the following function f has an inverse function or not. Give a reason for your answer.

CONTOH (a)

f(x) f
xy

x p –3
0 q5
r7

f bukan fungsi satu dengan satu kerana garis f ialah fungsi satu dengan satu. f mempunyai

mengufuk memotong graf itu pada dua titik. f tidak fungsi songsang.

ada fungsi songsang. f is a one-to-one function. f has an inverse function.

f is not a one-to-one function because the horizontal line
cuts the graph at two points. f has no inverse function.

15

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi (c)

(b) f(x)

f(x)

BAB 1 x x
0 0

f ialah fungsi satu dengan satu. f mempunyai f bukan fungsi satu dengan satu kerana garis

fungsi songsang. mengufuk memotong graf itu pada dua titik. f

f is a one-to-one function. f has an inverse function. tidak ada fungsi songsang.

f is not a one-to-one function because the horizontal line

cuts the graph at two points.. f has no inverse function.

15. Rajah berikut menunjukkan graf bagi fungsi satu dengan satu, f. Dalam setiap kes, lakar graf bagi f −1. TP 4

The following diagrams show the graph of one-to-one function, f. In each case, sketch the graph of f −1. 

CONTOH (a)

f(x) f(x)
(3, 5) y = x y=x

5

f

(5, 3) Tip –3 0 x
5
f (1, 0) → (0, 1)
(0, 1) x (5, 3) → (3, 5) f –1
0 (1, 0)
–3

(b) (c)

f(x) f(x) y=x
y=x 8 x

4 f –1 f –1 8
f
3
04 x f

03

16

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

16. Cari fungsi songsang bagi setiap fungsi yang berikut.   TP 3

Find the inverse function for each of the following functions.  

CONTOH (a) f : x → 3 – 4x

f:x→x–8 1 Katakan f –1(x) = y Katakan f –1(x) = y BAB 1
Penyelesaian: Let f –1(x) = y
Let f(y) = x

Katakan/ Let f –1(x) = y 3 – 4y = x
f(y) = x
y – 8 = x 2 Jadikan y sebagai perkara y = 3–x
y = x + 8 rumus. 4

Make y into subject of the f –1(x) = 3–x
equation. 4

\  f –1(x) = x + 8 3 Gantikan y ke dalam
f –1 : x → x + 8 f –1(x) = y

Insert y into f  –1(x) = y

(b) g : x → 5x (c) h:x →– x 2 1  , x ≠1
–

Katakan g –1(x) = y Katakan h –1(x) = y

Let g(y) = x Let h(y) = x

5y = x 2 = x
–
y = x y 1 2
5 x
y – 1 =
g –1(x) = x
5 2
y = x +1

h –1(x) = 2 + 1, x ≠ 0
x

17. Selesaikan setiap yang berikut.
Solve each of the following.   TP 4

CONTOH 1 (a) Fungsi f ditakrifkan sebagai f : x → 2x – 5  ,
x ≠ –2. Cari x+2
Fungsi f ditakrifkan sebagai f : x → 2x – 7. Cari
The function f is defined as f : x → 2x – 5  , x ≠ –2. Find
The function f is defined as f : x → 2x – 7. Find x+2
(i)   f –1(x) (ii) f –1(5)
(i) f –1(x)
(ii) f –1(3) (i) Katakan f –1(x) = y

Penyelesaian: Let f(y) = x

(i) Katakan/ Let f –1(x) = y 2y – 5 = x
y+2
f(y) = x

2y – 7 = x 2y – 5 = x(y + 2)

y = x+7 2y – 5 = xy + 2x
2
2y – xy = 2x + 5
x+7
\   f –1(x) = 2 y(2 – x) = 2x + 5

y = 2x +5
2 –x

(ii) f –1(3) = 3+7 f –1(x) = 2x +5  , x ≠ 2
2 2 –x

=5 (ii) f –1(5) = 2(5) + 5
2–5

= 15
–3

= –5

17

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

CONTOH 2

Diberi dua fungsi f(x) → x – 3 dan g(x) = x 2 1, x ≠ –1. Cari
+
BAB 1 2
Given two functions f(x) → x – 3 and g(x) = x + 1 , x ≠ 1. Find

(i)  f –1g (ii) gf –1

Penyelesaian:

Katakan/ Let f –1(x) = y 1 2(i) 2 (ii) gf –1 = g(x + 3)
f(y) = x f –1g(x) = f –1 x+1
y – 3 = x 2
y = x + 3 2 = x + 3+ 1
\   f –1(x) = x + 3 +
= x 1 + 3 2
+
2 + 3(x + 1) = x 4 ,  x ≠ –4
x+1
=

= 5 + 3x , x ≠ –1
x + 1

(b) Diberi dua fungsi f(x) = 6 – 3x dan g(x) = 1 1 , x ≠ 1 . Cari
2x – 2
1 1
Given two function f(x) = 6 – 3x and g(x) = 2x – 1 , x ≠ 2 . Find

(i) f –1g (ii) gf –1

Katakan f –1(x) = y 1 2(i) 1 1 2(ii) gf –1(x) = g6–x
f(y) = x f –1g(x) = f –1  2x – 1 3
6 – 3y = x
6 – 1 1 2 = 1
2x –
y = 6– x = 1 21 6 – x 2 – 1
3 3
3
1
f –1(x) = 6– x = 6(2x – 1) – 1 =
3 3(2x – 1) 1 12 – 2x – 32
3
= 12x – 7 ,  x ≠ 1
6x – 3 2 = 3 ,  x ≠ 9
9 – 2x 2

(c) Diberi fungsi f : x → 7 – 4x. Cari

Given the function f : x → 7 – 4x. Find

(i) f f –1 (ii) f –1f

Katakan f –1(x) = y 1 2(i) 7–x (ii) f –1f(x) = f –1(7 – 4x)
ff –1(x) = f  4
f(y) = x 7 – (7 – 4x)
= 4
7 – 4y = x 1 7 – x 2
= 7 – 4 4
(7 – x) 7 – 7 + 4x
y = 4 = 4
= 7 – 7 + x

f –1(x) = 7– x  = x = 4x
4 4

=x

18

PRAKTIS   Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

SPM 1

Kertas 1 3. Diberi fungsi f : x → 4 – 3x, cari BAB 1

1. Rajah menunjukkan fungsi gubahan fg yang 2 01 6 Given the function f : x → 4 – 3x, find

2015 memetakan set A kepada set C. (a) nilai x apabila f(x) memeta kepada diri sendiri,
the value of x when f(x) maps onto itself,
Diagram shows the composite function fg that maps set
(b) nilai h dengan keadaan f(3 – h) = 2h.
A to set C.
the value of h such that f(3 – h) = 2h.
ABC
(a) f(x) = x
gf 4 – 3x = x
x 2x 2x + 1 4 = x + 3x
x = 1
fg
(b) f(3 – h) = 2h
Nyatakan 4 – 3(3 – h) = 2h
4 – 9 + 3h = 2h
State –5 + 3h = 2h
3h – 2h = 5
(a) fungsi yang memetakan set A kepada set C. h = 5

the function that maps set A to set C.

(b) f –1(x)

(a) fg(x) = 2x + 1

(b) f –1(x) = x – 1

2. Rajah menunjukkan graf bagi fungsi f : x → |2x – 3|

2017 untuk domain –1 < x < 6.

Diagram shows the graph of the function f : x → |2x – 3|

for the domain –1 < x < 6.

f(x)

4. Diberi fungsi f : x → 2x – 7, g : x → px + 3 dan

9 2 01 6 gf : x → 2px + 5q. Ungkapkan q dalam sebutan p.

(–1, 5) Given the functions f : x → 2x – 7, g : x → px + 3 and

06 gf : x → 2px + 5q. Express q in terms of p.

Nyatakan x

State gf(x) = g(2x – 7)

(a) objek bagi 9, = p(2x – 7) + 3

the object of 9, 2px + 5q = 2px – 7p + 3

(b) imej bagi 4, 5q = –7p + 3

the image of 4, q = 3 – 7p
5
(c) domain bagi 0 < f(x) < 5.

the domain of 0 < f(x) < 5.

(a) 6
(b) f(4) = |2(4) – 3| = 5
(c) bila x = –1,  f(x) = 5
bila x = 4,  f(x) = 5
domain: –1 < x < 4

19

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi

Kertas 2 2. Diberi fungsi f : x → 4 – x dan fungsi g : x → ax2 +
bx, dengan keadaan a dan b ialah pemalar. Fungsi
BAB 1 1. Diberi bahawa f : x → 4 – 5x dan g : x → 3x – 5.
2018 gubahan gf –1 diberi oleh
It is given that f : x → 4 – 5x and g : x → 3x – 5.
Given the function f : x → 4 – x and function g : x → ax2
2018 (a) Cari + bx, such that a and b are constants. The composite
function gf –1 is given by
Find
gf –1 → x2 – 6x + 8
(i) f(4), 1 (a) Cari nilai a dan nilai b.
8
(ii) nilai p jika g(p + 3) =  f(4). Find the value of a and of b.
=
the value of p if g(p + 3) 1  f(4). (b) Lakar graf y = |gf −1(x)| untuk domain 0 < x < 7
8 dan nyatakan julat yang sepadan dengannya.
(iii) gf(x)
Sketch the graph of y = |gf −1(x)| for the domain
(b) Seterusnya lakarkan graf y = ugf(x)u untuk 0 < x < 7 and state the corresponding range.

–1 < x < 2. Nyatakan julat bagi y. (a) f(x) = 4 – x
Hence, sketch the graph of y = ugf(x)u for –1 < x < 2. 4 – x = y
4 – y = x
State the range of y. f –1(x) = 4 – x
gf –1(x) = x2 – 6x + 8
(a) (i) f(4) = 4 – 5(4) = –16 g(4 – x) = x2 – 6x + 8
4 – x = u
(ii) g(p + 3) = 1  f(4) x = 4 – u
8 g(u) = (4 – u)2 – 6(4 – u) + 8
1 = 16 – 4u – 4u + u2 – 24 + 6u + 8
3 (p + 3) – 5 = 8  (–16) = u2 – 2u
g(x) = x2 – 2x
3p + 9 – 5 = –2 g(x) = ax2 + bx
\  a = 1,  b = –2
3p = –6
(b) y = |gf  –1(x)|
p = –2 = |x2 – 6x + 8|
= |(x – 2)(x – 4)|
(iii) gf(x) = g(4 – 5x) \  y = 0  →  x = 2,  x = 4
= 3(4 – 5x) – 5 \  x = 0  →  y = 8
= 12 – 15x – 5 \  x = 7  →  y = |(7 – 2)(7 – 4)| = 15
= 7 – 15x
f(x)
(b) –1 0 7 2
x 15 15

y 22 7 0 23



y

23 8
22
(3, 1)
7 x 0 24 7 x
–1 0 1–75– 2
Julat/Range : 0 < f(x) < 15
Julat/Range y :  0 < y < 23

Praktis
SPM
Ekstra

20

  Matematik Tambahan  Tingkatan 4  Bab 1 Fungsi 

Sudut KBAT KBAT

Ekstra

BAB 1

xyz

1 –2
2



Rajah di atas menunjukkan pemetaan y kepada x di bawah fungsi f(y) = 5y – 3 dan pemetaan y kepada z di
m 1
bawah fungsi g(y) = 4y – 1  , y ≠ 4 .

Diagram above shows the mapping of y onto x by the function f(y) = 5y – 3 and the mapping of y onto z by
m 1
the function of g(y) = 4y – 1  , y ≠ 4  .

(a) Cari nilai m.

Find the value of m.

(b) Cari fungsi yang memetakan x kepada y.

Find the function which maps x onto y.

(c) Cari fungsi yang memetakan x kepada z.

Find the function which maps x onto z.

(a) g(y) = m  2(c) gf –1(x) = gx+3
4y – 1 5
–6
g(1) = m 1 = –2 =
4(1) – 4 x + 3 2
5 – 1
m = –6

= –6

(b) f(y) = 5y – 3  4x + 12 – 5 2
5
Katakan f –1(y) = x

Let f(x) = y 30 7
4x + 4
5x – 3 = y gf –1(x) = – ,  x ≠ –
y+3 7
x = 5

f –1(y) = y+3
5
x+3
\  f –1(x) = 5

Kuiz 1

21

BAB

2 Fungsi Kuadratik
Quadratic Functions

NOTA IMBASAN

2.1 Persamaan dan Ketaksamaan Kuadratik

Quadratic Equations and Inequalities

NOTA IMBASAN

Persamaan dan ketaksamaan kuadratik
Quadratic equations and inequalities

Penyelesaian persamaan Hasil tambah punca (HTP) dan hasil Penyelesaian ketaksamaan
Solving equation darab punca (HDP). kuadratik
Sumofroots(SOR)andproductofroots(POR) Solving quadratic inequalities
(a) Penyempurnaan kuasa dua
Completing the square (a) Punca-punca ialah α dan β (a) Lakaran graf
Graph sketching
(b) Rumus kuadratik Roots are α and β b (b) Garis nombor
Quadratic formula a Number line
(b) HTP/SOR = a + b = – (c) Jadual / Table
–b ±   b2 – 4ac (c) c
x= 2a HDP/POR = ab = a

(d) Persamaan / Equation

x2 – (HTP)x + HDP = 0

x2 – (SOR)x + POR = 0

1. Selesaikan setiap persamaan berikut menggunakan kaedah penyempurnaan kuasa dua. TP 3

Solve each of the following equations using completing the square method. 

CONTOH (a) x2 – 8x + 10 = 0

−2x2 + 8x + 13 = 0 x 2 – 8x = –10
Penyelesaian:
−2x2 + 8x = −13 Pindah sebutan pemalar ke kanan. x2 – 8x + 1– 8 22 = –10 + 1– 8 22
x2 – 4x = ­123 Move the constant to the right. 2 2
Pekali x2 = +1
Coefficient of x2 = +1 (x – 4)2 = –10 + 16

√(x – 4)2 = ±√6

x2 – 4x + 1– 4 22 = 13 + 1– 4 22 1 2Tambah sebutanpb2ers2admi aan. x – 4 = √6    atau x – 4 = –√6
2 2 2 = 6.449   or = 1.551
kedua-dua belah

(x – 2)2 = 13 + 4 1 2Add the term b 2 on both
2
sides of equatio2n.

√(x – 2)2 = ±√10.5 Masukkan ± apabila
mengambil punca kuasa
x – 2 = √10.5 x – 2 = –√10.5 dua.
Put in ± when taking
x = 5.240 x = –1.240 square roots.

22

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(b) –2x2 + 13x – 8 = 0 (c) –3x2 – 18x + 7 = 0

2x2 – 13x + 8 = 0 –3x2 – 18x = –7

x2 – 123x = – 4 x2 + 6x = 7
3

1 2 1 2 x2 –13x + – 13 2 = –4 + – 13 2 x2 + 6x + (3)2 = 7 + (3)2 BAB 2
4 4 4 3

1 2x– 13 2 = – 4 + 169 (x + 3)2 = 34
4 16 3

1 2x –13 2 ± 105 √(x + 3)2 = ± 34
4 16 3
=

x – 13 = 105    x – 143 = – 105 x+3= 34   x + 3 = – 34
4 16 16 3 3

x = 5.812 = 0.6883 x = 0.3665 x = –6.3665

2. Selesaikan setiap persamaan berikut menggunakan persamaan kuadratik.  TP 3

Solve each of the following equations using quadratic equation. 

CONTOH (a) –3x2 + 16x – 9 = 0

4x2 − 17x + 5 = 0 a = –3,  b = 16,  c = –9

Penyelesaian: Bandingkan dengan bentuk am. x = –16 ± √(16)2 – 4(–3)(–9)
Compare with the general form. 2(–3)
  ax2 + bx + c = 0
a = 4,  b = –17,  c = 5 = –16 + √148
–6
x = –b ± √b2 – 4ac Guna rumus
2a Use the formula –16 + √148 –16 – √148
x = –6   atau x= –6
–(–17) ± √(–17)2 – 4(4)(5)
= 2(4) or
= 0.6391 x = 4.6943

= 17 ± √209
8

x = 17 + √209   x= 17 – √209
8 8

= 3.9321 = 0.3179

(b) 5x2 – 19x + 8 = 0 (c) –7x2 + 22x – 13 = 0

a = 5,  b = –19,  c = 8 a = –7,  b = 22,  c = –13

x = –b ± √b2 – 4ac x = –b ± √b2 – 4ac
2a 2a

x = –(–19) ± √(–19)2 – 4(5)(8) x = –(22) ± √(22)2 – 4(–7)(–13)
2(5) 2(–7)

= 19 ± √201 = –22 ± √120
10 –14

x = 19 + √201   x = 19 – √201 x = –22 + √120   x = –22 – √120  
10 10 –14 –14

= 3.3177 = 0.4823 = 0.7890 = 2.3539

23

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

3. Tentukan hasil tambah dan hasil darab punca bagi setiap persamaan kuadratik yang berikut.  TP 2

Determine the sum and the product of roots for each of the following quadratic equations.

CONTOH (a) 6x2 + 14x – 7 = 0

5x2 – 12x – 9 = 0 a = 6,  b = 14,  c = –7

BAB 2 Penyelesaian: Bandingkan dengan bentuk am. HTP/SOR = – b =– 14 =– 7
ax2 + bx + c = 0 Compare with the general form. a 6 3
a = 5,  b = –12,  c = –9
HDP/POR = c =– 7
a 6
HTP/SOR = – b =– –12 = 12
a 5 5

HDP/POR = c = – 9
a 5

(b) –4x2 + 12x – 3 = 0 (c) 9x2 – 6x – 11 = 0

a = –4,  b = 12,  c = –3 a = 9,  b = –6,  c = –11

HTP/SOR = – b = – 12 = 3 HTP/SOR = – b =– –6 = 2
a –4 a 9 3

HDP/POR = c = –3 = 3 HDP/POR = c = –11
a –4 4 a 9

4. Bentukkan satu persamaan kuadratik 3x2 – 6x – 8 = 0 dengan punca-punca yang diberi. TP 4

Form a quadratic equation 3x2 – 6x – 8 = 0 with the given roots.

CONTOH (a) a + 5, b + 5

a – 4,  b – 4

HTP/SOR:

Penyelesaian: Tentukan HTP dan HDP bagi (a + 5) + (b + 5) = (a + b) + 10
persamaan yang diberi.
a = 3,  b = –6,  c = –8 Determine SOR and POR for = 2 + 10 = 12
the equation given.
b –6 HDP/POR:
a 3
a + b = – =– =2 (a + 5)(b + 5) = ab + 5a + 5b + 25

ab = c =– 8 = 1– 8 2 + 5(2) + 25
a 3 3

HTP/SOR: (a – 4) + (b – 4) = a + b – 8 = 97
= 2 – 8 = –6 3

HDP/POR: (a – 4)(b – 4) = ab – 4a – 4b + 16 Persamaan kuadratik / Quadratic equation
x2 – (HTP)x + HDP = 0
= 1– 8 2 – 4(a + b) + 16
3 x2 – (12)x + 19372 = 0

= 1– 8 2 – 4(2) + 16 = 16 3x2 – 36x + 97 = 0
3 3

Persamaan kuadratik / Quadratic equation

x2 – (HTP)x + (HDP) = 0

x2 – (–6)x + 136 = 0
3x2 + 18x + 16 = 0

24

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(b) 3a, 3b (c) a2, b2



HTP/SOR: HTP/SOR:

3a + 3b = 3(a + b)  a2 + b2 = (a + b)2 – 2ab

= 3(2) = 6 = 22 – 21– 8 2 = 28
3 3
HDP/POR: BAB 2

(3a)(3b) = 9ab HDP/POR:

= 91– 8 2 (a2)(b2) = (ab)2
3
1– 8 22 64
= –24 = 3 = 9

Persamaan kuadratik / Quadratic equation Persamaan kuadratik / Quadratic equation
x2 – (HTP)x + HDP = 0
x2 – (HTP)x + HDP = 0
x2 – 12382x + 16942 = 0
x2 – 6x + (–24) = 0
9x2 – 84x + 64 = 0
x2 – 6x – 24 = 0

5. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

Jika a dan b ialah punca-punca persamaan x2 – 4x + 7 = 0, bentukkan persamaan kuadratik dengan
punca-punca 3a dan 3b.

If α and β are roots of the quadratic equation x2 – 4x + 7 = 0, form a quadratic equation with roots 3α and 3b.

Penyelesaian:

a = 1,  b = –4,  c = 7

HTP/SOR =– b HDP/POR = c
a a

a + b = – 1 –4 2 = 4 ab = 7 = 7
1 1

HTP baharu/ New SOR = 3a + 3b HDP baharu/ New POR = 3a × 3b
= 3(a + b) = 3(4) = 12 = 9ab = 9(7) = 63

Persamaan baharu ialah/ New equation is
x2 – (HTP)x + HDP = 0
x2 – 12x + 63 = 0

Jika a dan b ialah punca-punca persamaan 3x2 – 9x + 5 = 0, bentukkan persamaan kuadratik dengan
punca-punca 2a dan 2b.

If α and β are roots of quadratic equation 3x2 – 9x + 5 = 0, form a quadratic equation with roots 2α and 2b.

a = 3,  b = –9,  c = 5

HTP = – b HDP = c Persamaan baharu ialah
a a
New equation is
a + b = – 1 –9 2 = 3 ab = 5
3 3 x2 – (HTP)x + HDP = 0

HDP baharu/ New POR = 2a × 2b x2 – 6x + 20 = 0
3
HTP baharu/ New SOR = 2a + 2b
= 2(a + b) = 2(3) = 6 = 4ab = 41 5 2 = 20 3x2 – 18x + 20 = 0
3 3

25

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

6. Selesaikan setiap yang berikut.   TP 5

Solve each of the following.

CONTOH

BAB 2 Salah satu daripada punca bagi persamaan 2x2 – 8x + p = 0 ialah tiga kali punca yang satu lagi. Cari
punca-punca itu dan nilai p.

One of the roots of quadratic equation 2x2 – 8x + p = 0 is three times the other. Find the roots and the value of p.

Penyelesaian:

a = 2,  b = –8,  c = p ;  Katakan punca-punca ialah a dan 3a./ Let roots are a and 3a.

HTP/ SOR =– b HDP/ POR = c
a a

a + 3a = – 1 –8 2 = 4 a× 3a = p
2 2
p
4a = 4   →   a = 1 3a2 = 2

3a = 3(1) = 3 p = 6a2 = 6(1)2 = 6

\  Punca-punca ialah 1 dan 3. Manakala, nilai p ialah 6.
Roots are 1 and 3. While, value of p is 6.

Salah satu punca bagi persamaan x2 – 15x + m = 0 ialah dua kali punca yang satu lagi. Cari punca-punca
itu dan nilai m.

One of the roots of quadratic equation x2 – 15x + m = 0 is two times the other. Find the roots and the value of m.

a = 1,  b = –15,  c = m ;  Katakan punca-punca ialah a dan 2a./ Let roots are a and 2a

HTP = – b HDP = c
a a

a + 2a = – 1 –15 2 a × 2a = m
1 1

3a = 15 m = 2a2
a = 5 = 2(5)2
2a = 2(5) = 10
= 50

Punca-punca ialah 5 dan 10. Nilai m ialah 50.

Roots are 5 and 10. The value of m is 50.

7. Tentukan ketaksamaan berikut.  TP 5 CONTOH 2 

Solve the following inequalities.

CONTOH 1 

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

Penyelesaian: Penyelesaian:

(i) Kaedah 1:  lakaran graf/ Method 1: graph sketching (i) Kaedah 1:  lakaran graf/ Method 1: graph sketching

a . 0 bentuk graf ialah/ the shape of the graph is a , 0 bentuk graf ialah/ the shape of the graph is

apabila/ when x2 – 7x + 10 < 0 apabila/ when –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

x – 2 = 0   atau/ or x – 5 = 0 –x + 4 = 0 atau/ or   x – 3 = 0

x = 2 x=5 x = 4 x=3

2 5 34

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4
26

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(ii) Kaedah 2:  garis nombor/ Method 2: number line (ii) Kaedah 2:  garis nombor/ Method 2: number line

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

Pertimbangkan x – 2 > 0 dan/ and   x – 5 > 0 Pertimbangkan –x + 4 > 0 dan/ and  x – 3 > 0

Consider x > 2 x>5 Consider x < 4 x>3 BAB 2

–– + –+ +

–+ + ++ –

+2 – 5+ –3 + 4 –

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4

(iii) Kaedah 3:  jadual / Method 3: table (iii) Kaedah 3:  jadual/ Method 3: table

x2 – 7x + 10 < 0 –x2 + 7x – 12 < 0

(x – 2)(x – 5) < 0 (–x + 4)(x – 3) < 0

Pertimbangkan x – 2 = 0 dan/ and  x – 5 = 0 Pertimbangkan –x + 4 = 0 dan/ and  x – 3 = 0

Consider x = 2 x=5 Consider x = 4 x=3

(x – 2): – + + (x – 3) : – + +

(x – 5): – – + (–x + 4) : + + –

(x – 2)(x – 5): + – + (x – 3)(–x + 4) : – + –
25 34

Maka/ Thus, 2 < x < 5 Maka/ Thus, x < 3  atau/ or  x > 4

(a) x2 – 6x + 8 > 0 (b) 4x2 + 8x – 45 , 0

Apabila x2 – 6x + 8 = 0 Apabila 4x2 + 8x – 45 = 0

When (x – 2)(x – 4) = 0 When (2x – 5)(2x + 9) = 0 9
5 2
x – 2 = 0 atau/or   x – 4 = 0 x = 2 atau/or   x= –

x = 2 x=4

24 – –29 –52

Maka/Thus, x < 2  atau  x > 4 Maka/Thus, – 9 , x , 5
2 2

(c) –3x2 + 17x – 10 . 0 (d) 7x2 – 24x – 16 > 0

Apabila –3x2 + 17x – 10 = 0 Apabila 7x2 – 24x – 16 = 0

When (–x + 5)(3x – 2) = 0 When (7x + 4)(x – 4) = 0
2 4
x=5  atau/or  x = 3 x = – 7   atau/or   x=4

–23 5 – –47 4

Maka/Thus, 2 , x , 5 Maka/Thus, x < – 4  atau/or  x > 4
3 7

27

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

2.2 Jenis-jenis Punca Persamaan Kuadratik

Types of Roots of Quadratic Equations

NOTA IMBASAN

BAB 2 1. Jenis punca persamaan kuadratik ditentukan oleh nilai pembeza layan, b2 – 4ac.
The types of roots of a quadratic equation are determined by the value of discriminant, b2 – 4ac.

2. b2 – 4ac . 0 b2 – 4ac = 0 b2 – 4ac , 0

Dua punca yang berbeza. Dua punca yang sama. Tiada punca.
Two different roots. Two equal roots. No roots.

8. Tentukan jenis punca bagi setiap persamaan kuadratik yang berikut. TP 4

Determine the type of roots for each of the following quadratic equations.

CONTOH

(i) 9x2 – 12x + 4 = 0 (ii) 4x2 – 13x + 3 = 0 (iii) 6x2 + 7x + 5 = 0

Penyelesaian:

(i) a = 9, b = –12, c = 4 (ii) a = 4, b = –13, c = 3 (ii) a = 6, b = 7, c = 5

b2 – 4ac = (–12)2 – 4(9)(4) b2 – 4ac = (–13)2 – 4(4)(3) b2 – 4ac = 72 – 4(6)(5)

= 144 – 144 = 169 – 48 = 49 – 120

=0 = 121 . 0 = –71 , 0

Persamaan itu mempunyai Persamaan itu mempunyai Persamaan itu tidak
mempunyai punca.
dua punca yang sama. dua punca yang berbeza.
The equation has no roots.
The equation has two equal roots. The equation has two distinct

roots.

(a) 2x2 – 5x – 4 = 0 (b) 3x2 + 7x + 8 = 0

a = 2,  b = –5,  c = –4 a = 3,  b = 7,  c = 8

b2 – 4ac = (–5)2 – 4(2)(–4) b2 – 4ac = 72 – 4(3)(8)

= 25 + 32 = 49 – 96

= 57 . 0 = –47 , 0

Persamaan itu mempunyai dua punca yang Persamaan itu tidak mempunyai punca.

berbeza. The equation has no roots.

The equation has two distanct roots.

(c) 4x2 – 28x + 49 = 0 (d) 6x2 – 9x + 2 = 0

a = 4,  b = –28,  c = 49 a = 6,  b = –9,  c = 2

b2 – 4ac = (–28)2 – 4(4)(49) b2 – 4ac = (–9)2 – 4(6)(2)

= 784 – 784 = 81 – 48

=0 = 33 . 0

Persamaan itu mempunyai dua punca yang Persamaan itu mempunyai dua punca yang

sama. berbeza.

The equation has two equal roots. The equation has two distanct roots.

28

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

9. Cari nilai-nilai p jika setiap persamaan kuadratik berikut mempunyai dua punca yang sama. TP 4

Find the values of p if each of the following quadratic equations has two equal roots.

CONTOH (a) x2 + 2px + 3p + 4 = 0

x2 – 2px + 6p + 16 = 0 a = 1, b = 2p, c = 3p + 4 BAB 2

Penyelesaian: b2 – 4ac = 0
a = 1,  b = –2p,  c = 6p + 16
b2 – 4ac = 0 Syarat untuk dua (2p)2 – 4(1)(3p + 4) = 0
(–2p)2 – 4(1)(6p + 16) = 0 punca yang sama.
4p2 – 24p – 64 = 0 Condition for two 4p2 – 12p – 16 = 0
p2 – 6p – 16 = 0 equal roots.
(p + 2)(p – 8) = 0 p2 – 3p – 4 = 0

(p + 1)(p – 4) = 0

p + 1 = 0   atau/or p – 4 = 0

p = –1 p=4

p + 2 = 0   atau/or p – 8 = 0

p = –2 p=8

(b) 3x2 + px + 12 = 0 (c) 4x2 – 4px + 8p + 9 = 0

a = 3, b = p, c = 12 a = 4, b = –4p, c = 8p + 9

b2 – 4ac = 0 b2 – 4ac = 0

p2 – 4(3)(12) = 0 (–4p)2 – 4(4)(8p + 9) = 0

p2 – 144 = 0 16p2 – 128p – 144 = 0

p2 = 144 p2 – 8p – 9 = 0

p = ±12 (p + 1)(p – 9) = 0

p + 1 = 0   atau/or p – 9 = 0

p = –1 p=9

10. Cari julat nilai p jika setiap persamaan kuadratik berikut mempunyai dua punca yang berbeza. TP 4

Find the range of values of p if each of the following quadratic equations has two different roots.

CONTOH (a) x2 + 2x + p – 3 = 0

x2 – 4x – 3 + p = 0 b2 – 4ac . 0

Penyelesaian: 22 – 4(1)(p – 3) . 0
b2 – 4ac . 0
(–4)2 – 4(1)(–3 + p) . 0 Syarat untuk dua 4 – 4p + 12 . 0
16 + 12 – 4p . 0 punca yang berbeza.
4p , 28 Condition for two 16 – 4p . 0
p , 7 distinct roots.
4p , 16

p , 4

(b) 2x2 – 7x + p = 0 (c) (p + 1)x2 + 4x – 9 = 0

b2 – 4ac . 0 b2 – 4ac . 0

(–7)2 – 4(2)(p) . 0 42 – 4(p + 1)(–9) . 0

49 – 8p . 0 16 + 36p + 36 . 0

8p , 49 36p . –52

p , 49 p . – 13
8 9

29

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

11. Cari julat nilai p jika setiap persamaan kuadratik berikut tidak mempunyai punca. TP 4

Find the range of values of p if each of the following quadratic equations has no roots.

CONTOH (a) x2 – 2x + p – 6 = 0

BAB 2 x2 + 6x + p – 4 = 0 b2 – 4ac , 0

Penyelesaian: Syarat untuk tiada punca. (–2)2 – 4(1)(p – 6) , 0
b2 – 4ac , 0 Condition for no roots.
62 – 4(1)(p – 4) , 0 4 – 4p + 24 , 0
36 – 4p + 16 , 0 Kesalahan Lazim
52 – 4p , 0 28 – 4p , 0
4p . 52 Tidak songsangkan simbol
p . 13 ketaksamaan apabila 4p . 28
darabkan ketaksamaan
dengan satu nombor negatif. p . 7
Does not change the
inequality sign when multiply
with negative number.

(b) 4x2 + 3x + p = 0 (c) (2p – 1)x2 – 6x + 8 = 0

b2 – 4ac , 0 b2 – 4ac , 0

32 – 4(4)(p) , 0 (–6)2 – 4(2p – 1)(8) , 0

9 – 16p , 0 36 – 64p + 32 , 0

–16p , –9 68 , 64p

p . 9 64p . 68
16
p . 17
16

12. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

Cari julat nilai p jika persamaan kuadratik x2 – (p + 5)x + 4 = 0 mempunyai dua punca yang berbeza.

Find the range of values of p if the quadratic equation x2 – (p + 5)x + 4 = 0 has two different roots.

Penyelesaian:

b2 – 4ac . 0 p + 1 = 0 p
[–(p + 5)]2 – 4(1)(4) . 0 p = –1 –9 –1
p2 + 10p + 25 – 16 . 0
p2 + 10p + 9 . 0 p + 9 = 0
(p + 1)(p + 9) . 0 p = –9

Maka/ Thus, p , –9  atau  p . –1

Cari julat nilai p jika persamaan kuadratik x2 – 2px + 4p – 3 = 0 tidak mempunyai punca.

Find the range of values of p if the quadratic equation x2 – 2px + 4p – 3 = 0 has no roots.

b2 – 4ac , 0 p – 1 = 0 p
(–2p)2 – 4(1)(4p – 3) , 0 p = 1 13
4p2 – 16p + 12 , 0
p2 – 4p + 3 , 0 p – 3 = 0 Maka/ Thus, 1 , p , 3
(p – 1)(p – 3) , 0 p = 3

30

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

2.3 Fungsi Kuadratik

Quadratic Functions

NOTA IMBASAN BAB 2

1. Bentuk am bagi fungsi kuadratik ialah f(x) = ax2 + bx + c, dengan keadaan a, b dan c adalah pemalar dan a ≠ 0.
The general form of a quadratic function is f(x) = ax2 + bx + c, where a, b and c are constants and a ≠ 0.

2. (a) Jika a . 0, bentuk graf ialah . (b) Jika a , 0, bentuk graf ialah .
If a . 0, the shape of the graph is . If a , 0, the shape of the graph is .

3. Kedudukan graf / Position of the graph

Nilai a b2 – 4ac . 0 b2 – 4ac = 0 b2 – 4ac , 0
Value of a

a.0 x

xx

xx

a,0 x

Persamaan f(x) = 0 mempunyai Persamaan f(x) = 0 mempunyai Persamaan f(x) = 0 tidak

dua punca yang berbeza. dua punca yang sama. mempunyai punca.

The equation f(x) = 0 has two different The equation f(x) = 0 has two equal The equation f(x) = 0 has no roots.
roots. roots.
NOTA IMBASAN
4. Dengan kaedah penyempurnaan kuasa dua f(x) = ax2 + bx + c boleh diungkapkan dalam bentuk f(x) = a(x – h)2 + k di mana

a, h dan k adalah pemalar.

By completing the square, f(x) = ax2 + bx + c can be expressed in the form f(x) = a(x – h)2 + k where a, h and k are constants.

(a) Jika a . 0, fungsi kuadratik mempunyai nilai minimum k apabila x = h dan titik minimum (h, k).

If a . 0, the quadratic function has a minimum value k when x = h and minimum point (h, k).

(b) Jika a , 0, fungsi kuadratik mempunyai nilai maksimum k apabila x = h dan titik maksimum (h, k).

If a , 0, the quadratic function has maximum value k when x = h and maximum point (h, k).

(c) Paksi simetri ialah satu garis menegak yang melalui titik maksimum atau titik minimum.

x = h adalah persamaan paksi simetri.

Axis of symmetry is a vertical line passing through the maximum point or minimum point.

x = h is the equation of axis of symmetry. b
2a
(d) Paksi simetri boleh ditentukan dengan menggunakan x = – .
Axis of symmetry can be determined
by using x = – b .
2a

5. Langkah-langkah untuk melakar graf fungsi kuadratik:

Steps for sketching the graph of quadratic function:

(a) Kenal pasti nilai a dan lakarkan bentuk graf itu. NOTA
Identify the value of a and sketch the shape of the graph.

(b) Cari nilai b2 – 4ac untuk menentukan kedudukan graf.

Find the value of b2 – 4ac to determine the position of the graph.

(c) Ungkapkan f(x) = ax2 + bx + c dalam bentuk f(x) = a(x – h)2 + k dengan kaedah penyempurnaan kuasa dua untuk

menentukan titik minimum atau titik maksimum (h, k).

Expressed f(x) = ax2 + bx + c in the form of f(x) = a(x – h)2 + k by completing the square to determine the minimum or maximum point (h,k).

(d) Cari titik persilangan antara graf dengan paksi-y dengan menggantikan x = 0.

Find the point of intersection of the graph with the y-axis by substituting x = 0.

(e) Cari titik persilangan antara graf dengan paksi-x dengan menyelesaikan f(x) = 0.

Find the point of intersection of the graph with the x-axis by solving f(x) = 0.

(f ) Lakarkan graf dengan menyambungkan semua titik diperoleh daripada langkah di atas.

Sketch the graph by joining all the points obtained in the steps above.

31

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

13. Tentukan sama ada setiap fungsi yang berikut ialah fungsi kuadratik atau bukan. TP 1

Determine whether each of the following functions is a quadratic function.

CONTOH (a) f(x) = 3 – 4x – 2x2

BAB 2 f(x) = 5x2 – x + 9 Kuasa tertinggi bagi x ialah 2

Penyelesaian: Maka f(x) ialah fungsi kuadratik.
Kuasa tertinggi bagi x ialah 2/Highest power of x is 2
Maka f(x) ialah fungsi kuadratik. Highest power of x is 2.
Thus, f(x) is a quadratic function.
Thus, f(x) is a quadratic function.

(b) f(x) = 7x3 – x2 + 8 (c) f(x) = –x – 3x2

Kuasa tertinggi bagi x ialah 3 Kuasa tertinggi bagi x ialah 2
Maka f(x) bukan fungsi kuadratik. Maka f(x) ialah fungsi kuadratik.

Highest power of x is 3. Highest power of x is 2.
Thus, f(x) is not a quadratic function. Thus, f(x) is a quadratic function.

14. Bagi setiap fungsi kuadratik berikut, tentukan bentuk graf itu dan tentukan juga jenis punca apabila
f(x) = 0. TP 3

For each of the following quadratic functions, determine the shape of the graph and determine also the type of roots when
f(x) = 0.

CONTOH

(i) f(x) = x2 – 6x + 3 (ii) f(x) = –4x2 + 8x – 4 (iii) f(x) = 3x2 – 9x + 7

a = 1 . 0, bentuk graf ialah/ a = –4 , 0, bentuk graf ialah/ a = 3 . 0, bentuk graf ialah/

shape of graph is shape of graph is shape of graph is

b2 – 4ac = (–6)2 – 4(1)(3) b2 – 4ac = (8)2 – 4(–4)(–4) b2 – 4ac = (–9)2 – 4(3)(7)

= 36 – 12 = 64 – 64 = 81 – 84

= 24 . 0 =0 = –3 , 0

Maka, f(x) = 0 mempunyai Maka, f(x) = 0 mempunyai Maka, f(x) = 0 tidak

dua punca yang berbeza. dua punca yang sama. mempunyai punca.
Thus, f(x) = 0 has two distinct roots. Thus, f(x) = 0 has two equal roots.
Thus, f(x) = 0 has no roots.


(a) f(x) = 2x2 – 7x + 5 (b) f(x) = –4 – 6x – 3x2

a = 2 . 0, bentuk graf ialah/shape of graph is a = –3 , 0, bentuk graf ialah/shape of graph is

b2 – 4ac = (–7)2 – 4(2)(5) b2 – 4ac = (–6)2 – 4(–3)(–4)

= 49 – 40 = 36 – 48

=9.0 = –12 , 0

Maka, f(x) = 0 mempunyai dua punca yang Maka, f(x) = 0 tidak mempunyai punca.
Thus, f(x) = 0 has no roots.
berbeza.
Thus, f(x) = 0 has two distinct roots.

(c) f(x) = 9x2 – 12x + 4 (d) f(x) = 5x2 + 3x + 1

a = 9 . 0, bentuk graf ialah/shape of graph is a = 5 . 0, bentuk graf ialah/shape of graph is

b2 – 4ac = (–12)2 – 4(9)(4) b2 – 4ac = (3)2 – 4(5)(1)

= 144 – 144 = 9 – 20

=0 = –11 , 0

Maka, f(x) = 0 mempunyai dua punca yang Maka, f(x) = 0 tidak mempunyai punca.
Thus, f(x) = 0 has no roots.
sama.
Thus, f(x) = 0 has two equal roots.

32

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

15. Cari julat nilai bagi k jika setiap graf bagi fungsi kuadratik berikut menyilangi paksi-x pada dua titik berlainan.

Find the range of values of k if each of the following graphs of quadratic function intersects the x-axis at two different
points.  TP 4

CONTOH (a) f(x) = (2k – 3)x2 – 4x – 8

f(x) = 3x2 – 8x + k – 6 a = 2k – 3,  b = –4,  c = –8 BAB 2

Penyelesaian: f(x) mempunyai dua punca yang berlainan apabila

a = 3,  b = –8,  c = k – 6 f(x) has two distinct roots when

f(x) mempunyai dua punca yang berlainan apabila b2 – 4ac . 0

f(x) has two distinct roots when (–4)2 – 4(2k – 3)(–8) . 0

b2 – 4ac . 0  16 + 64k – 96 . 0

(–8)2 – 4(3)(k – 6) . 0 64k . 80

64 – 12k + 72 . 0 k . 5
4
12k , 136

k , 34
3

16. Cari nilai-nilai m jika setiap graf bagi fungsi kuadratik berikut menyilangi paksi-x pada satu titik. TP 4

Find the values of m if each of the following graphs of quadratic function intersects the x-axis at one point.

CONTOH (a) f(x) = x2 + 2mx + m + 6

f(x) = mx2 – 6x + 9 a = 1,  b = 2m,  c = m + 6

Penyelesaian: f(x) mempunyai dua punca yang sama apabila
a = m,  b = –6,  c = 9 f(x) has two equal roots when
f(x) mempunyai dua punca yang sama apabila
f(x) has two equal roots when b2 – 4ac = 0

b2 – 4ac = 0  (2m)2 – 4(1)(m + 6) = 0

(–6)2 – 4(m)(9) = 0 4m2 – 4m – 24 = 0
36 – 36m = 0
36m = 36 m2 – m – 6 = 0
m = 1
(m – 3)(m + 2) = 0

m – 3 = 0 atau   m + 2 = 0

m = 3 m = –2

17. Cari julat nilai p jika setiap graf fungsi kuadratik berikut tidak menyilangi paksi-x. TP 4

Find the range of values of p if each of the following graphs of quadratic function does not intersects the x-axis.

CONTOH (a) f(x) = x2 + 2(p + 1)x + p2 – 1

f(x) = (2p + 5)x2 – 6x + 9 a = 1,  b = 2(p + 1),  c = p2 – 1

Penyelesaian: f(x) tidak mempunyai punca apabila
a = 2p + 5,  b = –6,  c = 9 f(x) has no roots when
f(x) tidak mempunyai punca apabila
f(x) has no roots when b2 – 4ac , 0

b2 – 4ac , 0  [2(p + 1)]2 – 4(1)(p2 – 1) , 0

(–6)2 – 4(2p + 5)(9) , 0 4(p + 1)2 – 4p2 + 4 , 0
36 – 72p – 180 , 0
–72p , 144 4p2 + 8p + 4 – 4p2 + 4 , 0
p . –2
8p , –8

p , –1

33

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

18. Ungkapkan setiap fungsi kuadratik yang berikut dalam bentuk a(x – h)2 + k. Nyatakan nilai maksimum atau
minimum dan nilai sepadan bagi x. TP 4

Express each of the following quadratic functions in the form a(x – h)2 + k. State the maximum or minimum value and the
corresponding value of x.

CONTOH

BAB 2 (i) f(x) = 2x2 + 8x – 1 (ii) f(x) = 2 + 6x – 3x2

Penyelesaian: Penyelesaian:

f(x) = 2x2 + 8x – 1 f(x) = –3x2 + 6x + 2

= 2(x2 + 4x) – 1 = –3(x2 – 2x) + 2

= 32 x2 + 4x + 1 4 22 – 1 4 224 – 1 = 3–3 x2 – 2x + 1 –2 22 – 1 –2 224 + 2
2 2 2 2

= 2(x + 2)2 – 8 – 1 = –3(x – 1)2 + 3 + 2

= 2(x + 2)2 – 9 = –3(x – 1)2 + 5

Oleh sebab a . 0, f(x) mempunyai nilai minimum Oleh sebab a , 0, f(x) mempunyai nilai
–9 apabila (x + 2) = 0 iaitu x = –2. maksimum 5 apabila (x – 1) = 0 iaitu x = 1.

Since a . 0, f(x) has minimum value of –9 when Since a , 0, f(x) has maximum value of 5 when
(x + 2) = 0 which is x = –2. (x – 1) = 0 which is x = 1.

(a) f(x) = x2 + 4x – 3 (b) f(x) = –x2 + 5x – 11

f(x) = –(x2 – 5x) – 11

f(x) = x2 + 4x + 1 4 22 – 1 4 22 – 3 = 3– x2 – 5x + 1 –5 22 – 1 –5 224 – 11
2 2 2 2

1 2 = 5 2+ 25
= (x + 2)2 – 4 – 3 – x – 2 4 – 11

= (x + 2)2 – 7 = – 1x – 5 22 – 19
2 4
Oleh sebab a . 0, f(x) mempunyai nilai minimum –7
apabila (x + 2) = 0 iaitu x = –2. Oleh sebab a , 0, f(x) mempunyai nilai maksimum

Since a . 0, f(x) has minimum value of –7 when (x + 2) = 0 1 2 19 x– 5 5
which is x = –2. 1 2 4 2 2
– apabila   has = 0 iaitu x = .
–
wSihncicehais,x0=, f(52x). maximum value of 19 when   x – 5 =0
4 2

(c) f(x) = 2x2 – 8x + 15 (d) f(x) = –2x2 – 12x + 9

f(x) = 2(x2 – 4x) + 15 f(x) = 3–2 x2 + 6x + 1 6 22 – 1 6 224 + 9
2 2
= 32 x2 – 4x + 1 –4 22 – 1 –4 224 + 15 = –2(x + 3)2 + 18 + 9
2 2
= –2(x + 3)2 + 27
= 2(x – 2)2 – 8 + 15

= 2(x – 2)2 + 7 Oleh sebab a , 0, f(x) mempunyai nilai maksimum
27 apabila (x + 3) = 0 iaitu x = –3.
Oleh sebab a . 0, f(x) mempunyai nilai minimum 7
apabila (x – 2) = 0 iaitu x = 2. Since a , 0, f(x) has maximum value of 27 when
(x + 3) = 0 which is x = –3.
Since a . 0, f(x) has minimum value of 7 when (x – 2) = 0
which is x = 2.

(e) f(x) = 4x2 – 8x + 17 (f) f(x) = –5 – 21x – 3x2
f(x) = 4(x2 – 2x) + 17
f(x) = –3x2 – 21x – 5

= 34 x2 – 2x + 1 –2 22 – 1 –2 224 + 17 = –3(x2 + 7x) – 5
2 2
= 3–3 x2 + 7x + 1 7 22 – 1 7 224 – 5
= 4(x – 1)2 – 4 + 17 2 2
= 4(x – 1)2 + 13 7 147
1 2 = 2 2+ 4
–3 x + –5

Oleh sebab a . 0, f(x) mempunyai nilai minimum 13 = –31x + 27 2 + 127
apabila (x – 1) = 0 iaitu x = 1. 4
2
Since a . 0, f(x) has minimum value of 13 when (x – 1) = 0
which is x = 1. Oleh sebab a , 0, f(x) mempunyai nilai maksimum

1 2127 x+ 7 = 0 iaitu x = –  72 .
4 2
apabila

1 2 127 7
4 2

Since a , 0, f(x) has maximum value of when   x + =0

which is x = – 7 .
2

34

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

19. Selesaikan setiap yang berikut. TP 5

Solve each of the following.

CONTOH

y x Rajah menunjukkan graf bagi fungsi f(x) = –(x – k)2 – 9 dengan keadaan BAB 2
O
–11 y = f(x) k ialah pemalar. Cari
(6, –11)
The diagram shows the graph of the function f(x) = –(x – k)2 – 9, where k is a
constant. Find

(a) nilai k.

the value of k.

(b) persamaan paksi simetri.

the equation of axis of symmetry.

(c) koordinat titik maksimum.

the coordinates of the maximum point.

Penyelesaian:

(a) Titik tengah bagi (0, –11) dan (6, –11)/ Midpoint of (0, –11) and (6, –1)

0 + 6 –11 – 11
2 2
1 2 = , = (3, –11)

Pada titik maksimum/ At the maximum point, x = 3

3–k=0
k = 3

(b) Persamaan paksi simetri ialah/ Equation of axis of symmetry is x = 3

(c) f(x) = –(x – 3)2 – 9
Maka, titik maksimum ialah/ Thus, maximum point is (3, –9).

Rajah menunjukkan bentuk bagi graf fungsi kuadratik f(x) = a(x + m)2 + n. Tentukan nilai-nilai a, m dan n.

The diagram shows the shapes of the graph of quadratic function f(x) = a(x + m)2 + n. Determine the values of a, m and n.

(a) f(x) (b) f(x)
–4 3
6 x

O4 x –22
–2

x + m = 0 x + m = 0

4 + m = 0 3 + m = 0

m = –4 m = –3

n = nilai minimum/minimum value n = nilai maksimum/maximum value

= –2 = –4

f(x) = a(x – 4)2 – 2 f(x) = a(x – 3)2 – 4

Pada titik (0, 6), 6 = a(0 – 4)2 – 2 Pada titik (0, –22), –22 = a(0 – 3)2 – 4

At point 16a = 8 At point 9a = –18
8
a = 16 a = –2

= 1
2

35

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

20. Lakarkan graf bagi setiap fungsi kuadratik yang berikut. Nyatakan persamaan paksi simetri bagi setiap
graf. TP 5

Sketch the graph of each of the following quadratic functions. State the equation of the axis of symmetry for each graph.

CONTOH 1

BAB 2 f(x) = x2 + 8x + 12

Penyelesaian: 1 Tentukan bentuk graf. the graph. Apabila/When x = 0 4 Tentukan pintasan-y.
a=1.0 Determine the shape of Determine y-intercept.
b2 – 4ac = (8)2 – 4(1)(12) f(x) = (0)2 + 8(0) + 12 5 Lakar graf.
Sketch the graph.
= 16 . 0 = 12

Maka, graf f(x) berbentuk dengan titik minimum

dan menyilangi paksi-x pada dua titik yang berbeza. f(x)

Thus, graph f(x) has shape with minimum point and

intersect the x-axis at two distinct points.

f(x) = x2 + 8x + 12 12

1 2 1 2 = x2 + 8x + 8 2 8 2 2 Tentukan titik minimum
2 2 atau maksimum.
– + 12
Determine the minimum
= (x + 4)2 – 16 + 12 or maximum point. –6(–4, –4)–2 0 x

= (x + 4)2 – 4

Titik minimum ialah/ Minimum point is (–4, –4).

Apabila/When f(x) = 0, Persamaan paksi simetri ialah/Equation of axis of
x2 + 8x + 12 = 0 symmetry is x = –4.
(x + 2)(x + 6) = 0 3 Tentukan pintasan-x jika ada.
x = –2   atau/or   x = –6 Determine x-intercept if exist.

CONTOH 2

f(x) = –2x2 + 6x – 5 Apabila/When x = 0
f(x) = –2(0)2 + 6(0) – 5
Penyelesaian: = –5
a = –2 , 0
b2 – 4ac = (6)2 – 4(–2)(–5) f(x) x
= –4 , 0
Maka, graf f(x) berbentuk dengan titik maksimum 0
dan tidak menyilang paksi-x.
΂–32 , – –12 ΃
Thus, graph f(x) has shape with maximum point and
does not intersect the x-axis.

f(x) = –2x2 + 6x – 5

= 3–2 x2 – 3x + 1 –3 22 – 1 –3 224 – 5
2 2

1 2 = –2 x– 3 2 9 –5 –5
2 2
+

1 2 = –2 x– 3 2 1
2 2
–

Persamaan paksi simetri ialah/ Equation of axis of
Titik maksimum ialah / Maximum point is 3
symmetry is x = 2 .
1 3 1 2.
2 , – 2

36

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

(a) f(x) = x2 – 6x – 7

a=1.0 Apabila/When x = 0
f(x) = (0)2 – 6(0) – 7
b2 – 4ac = (–6)2 – 4(1)(–7) = –7

= 64 . 0 f(x)

Maka, graf f(x) berbentuk dengan titik –10 7 x BAB 2
minimum dan menyilangi paksi-x pada dua –7
titik yang berbeza.

Thus, graph f(x) has shape with minimum point and

intersect the x-axis at two distinct points.

f(x) = x2 – 6x – 7 (3, –16)

1 2 1 2 = x2 – 6x + –6 2 –6 2 – 7
2 2
– Persamaan paksi simetri ialah x = 3.

= (x – 3)2 – 9 – 7 Equation of axis of symmetry is x = 3.

= (x – 3)2 – 16

Titik minimum ialah/Minimum point is (3, –16).

Apabila f(x) = 0,
x2 – 6x – 7 = 0
(x + 1)(x – 7) = 0
x = –1   atau/or   x = 7

(b) f(x) = –x2 + 6x – 5

a = –1 , 0 Apabila x = 0
f(x) = –(0)2 + 6(0) – 5
b2 – 4ac = (6)2 – 4(–1)(–5) = –5

= 16 . 0

Maka, graf f(x) berbentuk dengan titik f(x)
maksimum dan menyilang paksi-x pada dua titik (3, 4)
yang berbeza.
x
Thus, graph f(x) has shape with maximum point

and intersect the x-axis at two distinct points.

f(x) = –x2 + 6x – 5 01 5
–5
= 3– x2 – 6x + 1 –6 22 – 1 –6 224 – 5
2 2

= –(x – 3)2 + 9 – 5

= –(x – 3)2 + 4

Titik maksimum ialah/Maximum point is (3, 4). Persamaan paksi simetri ialah x = 3.

Apabila f(x) = 0, Equation of axis of symmetry is x = 3.
–x2 + 6x – 5 = 0
x2 – 6x + 5 = 0
(x – 1)(x – 5) = 0
x = 1   atau   x = 5

37

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik

(c) f(x) = 2x2 – 5x + 4

a=2.0 Apabila x = 0
f(x) = 2(0)2 – 5(0) + 4
b2 – 4ac = (–5)2 – 4(2)(4) = 4

= –7 , 0

BAB 2 Maka, graf f(x) berbentuk dengan titik f(x)
minimum dan tidak menyilang paksi-x. 4

Thus, graph f(x) has shape with minimum point and

intersect the x-axis.

f(x) = 2x2 – 5x + 4

= 32 x2 – 5 x + 1 –5 22 – 1 –5 224 + 4 ΂–45 , –78΃
2 4 4
x
1 2 = 2 5 2 25 0
x– 4 8 +4
–

1 2 = 2 x– 5 2 7 Persamaan paksi simetri ialah x= 5 .
4 8 4
+

5 7 Equation of axis of symmetry is x = 5 .
4 8 4
1 2 .
Titik minimum ialah ,

Minimum point is

PRAKTIS SPM 2

Kertas 1 2. Cari julat nilai x dengan keadaan fungsi kuadratik
SPM f(x) = 4 + 3x – x2 ialah negatif.
1. Diberi −3 ialah salah satu punca persamaan
SPM kuadratik (x − p)2 = 25, dengan keadaan p ialah 2017 Find the range of value of x such that the quadratic
2015 pemalar. Cari nilai-nilai p. function f(x) = 4 + 3x – x2 is negative.

Given −3 is one of the roots of the quadratic equation f(x) = 4 + 3x – x2
(x − p)2 = 25, where p is a constant. Find the values of p.
4 + 3x – x2 , 0

x2 – 3x – 4 . 0

(x − p)2 = 25 (x – 4)(x + 1) . 0
x − p = ±5

Apabila/When x = −3,

−3 − p = 5 , −3 − p = −5 –1 4

p = −8 , p = 2 x , –1  atau/or  x . 4


38

  Matematik Tambahan  Tingkatan 4  Bab 2 Fungsi Kuadratik 

3. Diberi fungsi kuadratik f(x) = (2m + 1)x2 – 3mx + (a) Nilai x bagi xtiotifkthteenmgidapho=int–22+ 8 =3

SPM 2(m – 2), dengan keadaan m ialah pemalar, adalah The value of
2016
sentiasa positif apabila m > p atau m < q. Cari nilai
h = 3,  m = –5
p dan nilai q. \  Q(3, –5)

Given the quadratic function f(x) = (2m + 1)x2 – 3mx

+ 2(m – 2), where m is a constant, is always positive (b) y = a(x – 3)2 – 5 BAB 2

when m . p or m , q. Find the value of p and of q.

Gantikan/Replace x = 8, y = 0

a = 2m + 1, b = –3m, c = 2m – 4 0 = a(8 – 3)2 – 5

b2 – 4ac < 0 0 = 25a – 5

(–3m)2 – 4(2m + 1)(2m – 4) < 0 25a = 5

9m2 – 4(4m2 – 8m + 2m – 4) < 0 a = 1
5
9m2 – 16m2 + 24m + 16 < 0

–7m2 + 24m + 16 < 0

7m2 – 24m – 16 > 0

(7m + 4)(m – 4) > 0 5. Diberi bahawa lengkung y = (k – 3)x2 – 6x + 1,
SPM dengan keadaan k ialah pemalar, bersilang dengan
7 m + 4 = 0 atau/or m – 4 = 0 2018 garis lurus y = 2x + 5 pada dua titik. Cari julat

(7m + 4) : – + + nilai k.

(m – 4) : – – + It is given that the curve y = (k – 3)x2 – 6x + 1, where k
is a constant, intersects with the straight line y = 2x + 5
+–+ at two points. Find the range of values of k.
4 4
– 7 y = (k – 3)x2 – 6x + 1 …… 1
y = 2x + 5 …… 2
m < – 4 m > 4 Gantikan 1 ke dalam 2,
7
Replace 1 into 2,
\ p = 4 dan/and q = – 4
7 2x + 5 = (k – 3)x2 – 6x + 1
(k – 3)x2 – 6x – 2x – 5 + 1 = 0
4. Rajah menunjukkan graf y = a(x – h)2 + m, dengan (k – 3)x2 – 8x – 4 = 0
b2 – 4ac . 0
SPM keadaan a, h dan m ialah pemalar. Garis lurus y = –5 (–8)2 – 4(k – 3)(–4) . 0
2018 ialah tangen kepada lengkung pada titik Q. 64 + 16k – 48 . 0
16k . –16
Diagram shows the graph y = a(x – h)2 + m, where a, k . –1
h and m are constants. The straight line y = – 5 is the

tangent to the curve at point Q.

y

6. Graf fungsi kuadratik g(x) = hx2 + (k − 1)x + 4,

x SPM dengan keadaan h dan k ialah pemalar, mempunyai
8 2015 satu titik minimum.

–2 0 The graph of a quadratic function g(x) = hx2 + (k − 1)x + 4,

where h and k are constants, has a minimum point.

(a) Nyatakan nilai h jika h ialah suatu integer

Q dengan keadaan −1 < h < 1.
State the value of h if h is an integer such that
(a) Nyatakan koordinat Q. −1 < h < 1.

State the coordinates of Q. (b) Dengan menggunakan jawapan di (a), cari

(b) Cari nilai a. julat nilai k jika graf itu tidak menyilang

Find the value of a. paksi-x.
Using the answer from (a), find the range of values

of k if the graph does not intersect the x-axis.

39