TOP ONE ADDMATHS TG4

Contents




CHAPTER Functions 1 CHAPTER Linear Law 82
1
6
Fungsi Hukum Linear
1.1 Functions 1 6.1 Linear and Non-Linear Relations 82
Fungsi Hubungan Linear dan Tak Linear
1.2 Composite Functions 6 6.2 Linear Law and Non-Linear Relations 88
Fungsi Gubahan Hukum Linear dan Hubungan Tak Linear
1.3 Inverse Functions 10 6.3 Application of Linear Law 90
Fungsi Songsang Aplikasi Hukum Linear
SPM Practice 1 15 SPM Practice 6 94
HOTS Challenge 17 HOTS Challenge 97
QQ 17 QQ 97

7
CHAPTER Quadratic Functions 18 CHAPTER Coordinate Geometry 98
Geometri Koordinat
2
Fungsi Kuadratik
7.1 Divisor of a Line Segment 98
2.1 Quadratic Equations and Inequalities 18 Pembahagi Tembereng Garis
Persamaan dan Ketaksamaan Kuadratik 7.2 Parallel Lines and Perpendicular Lines 101
2.2 Types of Roots of Quadratic Equations 23 Garis Lurus Selari dan Garis Lurus Serenjang
Jenis-jenis Punca Persamaan Kuadratik 7.3 Areas of Polygons 106
2.3 Quadratic Functions 25 Luas Poligon
Fungsi Kuadratik 7.4 Equations of Loci 110
SPM Practice 2 32 Persamaan Lokus
HOTS Challenge 34 SPM Practice 7 112
QQ 34 HOTS Challenge 114
QQ 114
CHAPTER Systems of Equations 35 CHAPTER Vectors 115
8
3
Sistem Persamaan Vektor
8.1 Vectors 115
3.1 Systems of Linear Equations in Three Variables 35 Vektor
Sistem Persamaan Linear dalam Tiga Pemboleh Ubah 8.2 Addition and Subtraction of Vectors 119
3.2 Simultaneous Equations involving One Linear Penambahan dan Penolakan Vektor
Equation and One Non-Linear Equation 40 8.3 Vectors in a Cartesian Plane 122
Persamaan Serentak yang melibatkan Satu Persamaan Vektor dalam Satah Cartes
Linear dan Satu Persamaan Tak Linear SPM Practice 8 128
SPM Practice 3 47 HOTS Challenge 131
HOTS Challenge 48 QQ 131
QQ 48
CHAPTER Solution of Triangles 132
9
CHAPTER Indices, Surds and Logarithms 49 Penyelesaian Segi Tiga
4
Indeks, Surd dan Logaritma 9.1 Sine Rule 132
Petua Sinus
4.1 Laws of Indices 49 9.2 Cosine Rule 137
Hukum Indeks Petua Kosinus
4.2 Laws of Surds 51 9.3 Area of a Triangle 139
Hukum Surd Luas Segi Tiga
4.3 Laws of Logarithms 56 9.4 Application of Sine Rule, Cosine Rule and Area
Hukum Logaritma of a Triangle 142
4.4 Applications of Indices, Surds and Logarithms 62 Aplikasi Petua Sinus, Petua Kosinus dan Luas Segi Tiga
Aplikasi Indeks, Surd dan Logaritma SPM Practice 9 144
SPM Practice 4 63 HOTS Challenge 146
HOTS Challenge 64 QQ 146
CHAPTER Index Numbers 147
QQ 64
10
Nombor Indeks
CHAPTER Progressions 65 10.1 Index Numbers 147
5
Janjang Nombor Indeks
5.1 Arithmetic Progressions 65 10.2 Composite Index 152
Janjang Aritmetik Indeks Gubahan
5.2 Geometric Progressions 71 SPM Practice 10 158
Janjang Geometri HOTS Challenge 161
SPM Practice 5 79 QQ 161
HOTS Challenge 81 Year-End Assessment 162
QQ 81
Answers in QR codes 177


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2 Quadratic Functions

Fungsi Kuadratik





2.1 Quadratic Equations and Inequalities Textbook
Persamaan dan Ketaksamaan Kuadratik
pg. 36 – 44

SMART Notes

1. A quadratic equation in general form can be written as 3. If a and b are the roots of a quadratic equation, then
ax + bx + c = 0 where a, b and c are constants and (x – a)(x – b) = 0 or x – (a + b)x + ab = 0.
2
2
a ≠ 0. Jika a dan b ialah punca-punca bagi suatu persamaan
Suatu persamaan kuadratik dalam bentuk am boleh ditulis kuadratik, maka (x – a)(x – b) = 0 atau x – (a + b)x + ab = 0.
2
sebagai ax + bx + c = 0 dengan keadaan a, b dan c ialah 4. For x – (a + b)x + ab = 0, a + b is the sum of roots
2
2
pemalar dan a ≠ 0.
and ab is the product of roots.
2. Methods used to solve quadratic equations: Bagi x – (a + b)x + ab = 0, a + b ialah hasil tambah punca
2
Kaedah-kaedah yang digunakan untuk menyelesaikan dan ab ialah hasil darab punca.
persamaan kuadratik: 5. For a quadratic equation in the form
(a) Completing the square
Kaedah penyempurnaan kuasa dua (x – a)(x – b) = 0, where a , b,
(b) Formula Bagi suatu persamaan kuadratik dalam bentuk
Kaedah rumus (x – a)(x – b) = 0, dengan keadaan a , b,
(a) if (x – a)(x – b) > 0, then x , a or x . b,
b – 4ac
2
x = –b ±  jika (x – a)(x – b) > 0, maka x , a atau x . b,
2a (b) if (x – a)(x – b) < 0, then a , x , b.
jika (x – a)(x – b) < 0, maka a , x , b.
1. Solve the following quadratic equations by using ‘completing the square’ method. Give your answers correct
to four decimal places. PL 3
Selesaikan persamaan kuadratik berikut dengan menggunakan kaedah ‘penyempurnaan kuasa dua’. Berikan jawapan anda betul
kepada empat tempat perpuluhan.
Example (a) x – 4x – 3 = 0
2
x + 3x – 6 = 0
2
2
Add the term x – 4x = 3
2
x + 3x = 6  coefficient of x  x – 4x + – 4 2 = 3 + – 4 2
2
2
3
3
2
x + 3x +   2 = 6 +   2 on the left and  2   2 
2
2
2
2
the right of the x – 4x + (–2) = 3 + (–2)
2
2
 x + 3  2 = 33 equation. (x – 2) = 7
2
4
x – 2 = ±√7
3
x + = ± 33 x = 4.6458 or x = –0.6458
2 4
x = 1.3723 or x = –4.3723
(b) –x + 6x + 1 = 0 (c) 3x + 9x – 2 = 0
2
2
2
x – 6x – 1 = 0 x + 3x – = 0
2
2
x – 6x = 1 3 2
2
2


x – 6x + – 6  2 = 1 + – 6  2 x + 3x = 3
2
2
2
2
3
3
2
x – 6x + (–3) = 1 + (–3) x + 3x +   2 = +   2
2
2
2
2
3
2
(x – 3) = 10
2

x – 3 = ± 10  x + 3  2 = 35
12
2
x = 6.1623 or x = –0.1623
3
x + = ± 35
2 12
x = 0.2078 or x = –3.2078
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2. Solve the following quadratic equations by using the formula method. Give your answers correct to four
decimal places. PL 3
Selesaikan persamaan kuadratik berikut dengan menggunakan kaedah rumus. Berikan jawapan anda betul kepada empat tempat
perpuluhan.
Example (a) 2x – 5x – 9 = 0
2
2
x – 6x + 7 = 0
a = 2, b = –5, c = –9
a = 1, b = –6, c = 7 –(–5) ± 
(–5) – 4(2)(–9)
2
(–6) ±  x = 2(2)
(–6) – 4(1)(7)
2
x = –  B
2(1) = 5 ± 97

6 ± 8 4
=


2 x = 5 + 97 or x = 5 – 97


6 + 8 6 – 8 4 4
x = or x = = 3.7122 = –1.2122
2 2
= 4.4142 = 1.5858
2
2
(b) 3x – 5x + 1 = 0 (c) –x + 8x – 9 = 0
a = 3, b = –5, c = 1 a = –1, b = 8, c = –9
–(–5) ±  –8 ± 
(–5) – 4(3)(1)
2
2
x = x = (8) – 4(–1)(–9)
2(3) 2(–1)


= 5 ± 13 = –8 ± 28
6 –2


5 + 13 5 – 13  
x = or x = x = –8 + 28 or x = –8 – 28
6 6 –2 –2
= 1.4343 = 0.2324 = 1.3542 = 6.6458
3. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut. (2x + 3) cm
(a) The product of 3x and (x + 2) is (–x + 12). Find (b)
the values of x. Give your answers correct to
three decimal places. A x cm B (2x + 3) cm
Hasil tambah 3x dan (x + 2) ialah (–x + 12). Cari nilai-
nilai x. Berikan jawapan anda betul kepada tiga tempat The diagram shows two squares, A and B. If the
perpuluhan. area of the square A is equal to the perimeter
3x(x + 2) = –x + 12 of the square B, find the value of x. Give your
2
3x + 6x = –x + 12 answer correct to three significant figures.
3x + 7x – 12 = 0 Rajah di atas menunjukkan dua buah segi empat sama,
2
A dan B. Jika luas segi empat sama A bersamaan dengan
7
x + x – 4 = 0 perimeter segi empat sama B, cari nilai x. Berikan jawapan
2
3
7
x + x = 4 anda betul kepada tiga angka bererti.
2
3 2
x + x +   2 = 4 +   2 x = 4(2x + 3)
7
7
x = 8x + 12
2
3
3
7
2
3 2 2 x – 8x – 12 = 0
2
2
7
7
7
x + x +   2 = 4 +   2 x = –(–8) ± √(–8) – 4(1)(–12)
2
3 6 6 2(1)
8 ± √112
2
 x + 7  = 193 = 2
36
6
7
x + = ± 193 x = 8 + √112 or x = 8 – √112
6 36 2 2
x = 1.149 or x = –3.482 = 9.29 = –1.29 (rejected)
∴ x = 9.29
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

4. Form quadratic equations by using the given roots. PL 3
Bentukkan persamaan kuadratik dengan menggunakan punca-punca yang diberikan.

Example (a) 1 and / dan 6 (b) –1 and / dan 1
2 and / dan –5 4
Let a = 1 and b = 6. 1
Let a = 2 and b = –5. Let a = –1 and b = .
4
Sum of roots, a + b Sum of roots, a + b = 1 + 6 1
= 2 + (–5) = 7 Sum of roots, a + b = –1 + 4
= –3 Product of roots, ab = 1 × 6 = – 3
Product of roots, ab = 6 4
= 2 × (–5) 1
= –10 x – (a + b)x + (ab) = 0 Product of roots, ab = –1 × 4
2
2
2
x – (a + b)x + (ab) = 0 x – 7x + 6 = 0 = – 1
4
2
x – (–3)x + (–10) = 0 2
x + 3x – 10 = 0 x – (a + b)x + (ab) = 0
2

1
x – – 3  x – = 0
2
Alternative Method 4 4
3
1
(x – 2)(x + 5) = 0 x + x – = 0
2
x + 5x – 2x – 10 = 0 4 4
2
2
x + 3x – 10 = 0 4x + 3x – 1 = 0
2
5. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.
Example
If a and b are the roots of the quadratic equation 2x – x – 15 = 0, form new quadratic equations using the
2
roots
Jika a dan b ialah punca-punca bagi persamaan kuadratik 2x – x – 15 = 0, bentukkan persamaan kuadratik yang baru
2
menggunakan punca-punca
2
(i) 2 and / dan (ii) (a + 3) and / dan (b + 3)
a b
2
2x – x – 15 = 0 (i) Sum of roots: (ii) Sum of roots:
2
2 + = 2b + 2a (a + 3) + (b + 3) = a + b + 6
b a b ab
a + b = – = 1 + 6
a 2(a + b) 2
= 13
= 1 ab = 2
2 1
2  
c = 2 Product of roots:
ab =
a – 15 (a + 3)(b + 3) = ab + 3a + 3b + 9
= – 15 2 = ab + 3(a + b) + 9
2 = – 2 15 1
15 = – + 3   + 9
2 2
Product of roots: = 3
4
2
2
   = ab x – x + 3 = 0
2 13
b
a
2
4 2x – 13x + 6 = 0
2
=
– 15
2
= – 8
15
2
8


x – – 15  x + – 15  = 0
2
2
x + 2 x – 8 = 0
15 15
15x + 2x – 8 = 0
2
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
(a) If the roots of the quadratic equation x – 4x – 5 = 0 are a and b, form new quadratic equations using
the roots
Jika punca-punca bagi persamaan kuadratik x – 4x – 5 = 0 ialah a dan b, bentukkan persamaan kuadratik baru menggunakan
2
punca-punca
1
(i) 1 and / dan (ii) (a + b) only / sahaja
2
a b
x – 4x – 5 = 0 (i) Sum of roots: (ii) Sum of roots:
2
2
1
1 + = b + a (a + b) + (a + b) 2
b a b ab = 4 + 4 2
2
a + b = –
a = – 4 = 32
= 4 5
c Product of roots: Product of roots:
ab = 1 (a + b) × (a + b)
2
2
1
1
a    =
2
= –5 a b ab = 4 × 4 2
= – 1 = 256
5
x – 32x + 256 = 0
2
1

2
x – – 4  x – = 0
5
5
1
4
2
x + x – = 0
5 5
5x + 4x – 1 = 0
2
1
(b) It is given that the roots of the quadratic equation 4x – px – 4 = 0 are q and – . Find the values of
2
4
p and q.
1
Diberi punca-punca bagi persamaan kuadratik 4x – px – 4 = 0 ialah q dan – . Cari nilai-nilai p dan q.
2
4
4x – px – 4 = 0 Product of roots = –1 Sum of roots = p
2
p 1 4

x – x – 1 = 0 ∴ q –  = –1
2
4 4 1 p

q = 4 ∴ q + – 4  = 4
1
4 – = p
4 4
p
15 =
4 4
p = 15
(c) Find the possible values of k if one of the roots of the quadratic equation x – kx + 8 = 0 is twice the
2
other.
2
Cari nilai-nilai k yang mungkin jika salah satu punca bagi persamaan kuadratik x – kx + 8 = 0 adalah dua kali ganda
punca yang satu lagi.
Let the roots of the quadratic equation Sum of roots = k
x – kx + 8 = 0 be a and 2a. ∴ a + 2a = k
2
k = 3a
Product of roots = 8 k = 3(2) or k = 3(–2)
∴ 2a = 8 = 6 = –6
2
a = 4
2

a = ± 4
= 2 or –2


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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

6. Find the range of the values of x which satisfy the following quadratic inequalities. PL 3
Cari julat nilai x yang memenuhi ketaksamaan kuadratik berikut.

Example
2x + 3x – 5 > 0
2
(x – 1)(2x + 5) > 0 Write in the form of (x – a)(x – b), where
a and b are the roots
Method 1: Graph sketching Method 2: Number line
When (x – 1)(2x + 5) = 0, x = 1 or x = – 5 Point testing:
2
x = –3 x = 0 x = 2
(–3 – 1)(–6 + 5) 0 (0 – 1)(0 + 5) 0 (2 – 1)(4 + 5) 0
x
x 5 5 5 1 x 1
x
5 1 x – – – 1
– 2 2 2
2
For (x – 1)(2x + 5) > 0, the range of the values of For (x – 1)(2x + 5) > 0, the range of the values of
5
x is x < – or x > 1. 5
2 x is x < – or x > 1.
2

Methohd 3: Table
The range of the values of x
5
5
x < – – < x < 1 x > 1
2 2
(x – 1) – – +
(2x + 5) – + +
(x – 1)(2x + 5) + – +

5
For (x – 1)(2x + 5) > 0, the range of the values of x is x < – or x > 1.
2
(a) x + 4x + 3 > 0 (b) (x + 3)(x – 5) < 0
2

2
x + 4x + 3 > 0 (x + 3)(x – 5) < 0
(x + 1)(x + 3) > 0
When (x + 3)(x – 5) < 0, x = –3 or x = 5
When (x + 1)(x + 3) = 0, x = –1 or x = –3

x
x –3 5
–3 –1
For (x + 3)(x – 5) < 0, the range of the values
For (x + 1)(x + 3) > 0, the range of the values of x is –3 < x < 5.
of x is x < –3 or x > –1.

(c) –x + 5x + 6 , 0 (d) –3x + 17x – 10 . 0
2
2
–x + 5x + 6 , 0 –3x + 17x – 10 . 0
2
2
(x + 1)(–x + 6) , 0 (3x – 2)(–x + 5) . 0
When (x + 1)(–x + 6) = 0, x = –1 or x = 6 2
When (3x – 2)(–x + 5) = 0, x = or x = 5
3
x
–1 6 x
2 5
3
For (x + 1)(–x + 6) , 0, the range of the values For (3x – 2)(–x + 5) . 0, the range of the values
2
of x is x , –1 or x . 6. of x is , x , 5.
3


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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2.2 Types of Roots of Quadratic Equations Textbook
Jenis-jenis Punca Persamaan Kuadratik
pg. 45 – 48


SMART Notes

1. The type of roots of a quadratic equation 2. Two different real roots
2
ax + bx + c = 0 can be determined by finding the b – 4ac . 0 Dua punca nyata yang berbeza
2
value of discriminant, b – 4ac.
2
Jenis-jenis punca bagi suatu persamaan kuadratik b – 4ac = 0 Two equal real roots
2
ax + bx + c = 0 boleh ditentukan dengan mencari nilai Dua punca nyata yang sama
2
pembezalayan, b – 4ac. No real roots
2
b – 4ac , 0
2
Tiada punca nyata
7. Find the value of discriminant for each of the following quadratic equations. Then, determine the type of
roots of the quadratic equation. PL 3
Cari nilai pembezalayan bagi setiap persamaan kuadratik berikut. Kemudian, tentukan jenis punca bagi persamaan kuadratik tersebut.
Example (a) x – 8x = –16
2
(x + 2) = 12x – 11
2
x – 8x + 16 = 0
2
x + 4x + 4 = 12x – 11
2
x – 8x + 15 = 0 Discriminant, b – 4ac
2
2
= (–8) – 4(1)(16)
2
Discriminant, b – 4ac = 0
2
2
= (–8) – 4(1)(15)
= 4 . 0
∴ two equal real roots
∴ The equation has two different real roots.
1
(b) x(6x – 5) = 1 (c) x + 2 = – x
2
2
6x – 5x – 1 = 0 2(x + 2) = –x
2
2
2x + 4 = –x
2
Discriminant, b – 4ac 2x + x + 4 = 0
2
2
2
= (–5) – 4(6)(–1)
2
= 49 . 0 Discriminant, b – 4ac
= (1) – 4(2)(4)
2
∴ two different real roots = –31 , 0
∴ no real roots
8. Find the range of values of p if each of the following quadratic equations has two different real roots. PL 3
Cari julat nilai p jika setiap persamaan kuadratik berikut mempunyai dua punca nyata yang berbeza.
Example (a) 4x – 5x + 3p – 1 = 0 (b) x + px + 4 = 0
2
2
3x – 12x + p = 0 For two different real roots:
2
For two different real roots: b – 4ac . 0
2
b – 4ac . 0
2
For two different real roots: 2 (p) – 4(1)(4) . 0
2
b – 4ac . 0 (–5) – 4(4)(3p – 1) . 0 p – 16 . 0
2
2
(–12) – 4(3)(p) . 0 25 – 16(3p – 1) . 0 (p – 4)(p + 4) . 0
2
144 – 12p . 0 25 – 48p + 16 . 0
–12p . –144 41 – 48p . 0 When (p – 4)(p + 4) = 0,
12p , 144 –48p . –41 p = 4 or p = –4
p , 12 48p , 41
p , 41
48 p
–4 4
∴ p , –4 or p . 4
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

9. Find the value of q if each of the following quadratic equations has two equal real roots. PL 4
Cari nilai q jika setiap persamaan kuadratik berikut mempunyai dua punca nyata yang sama.

Example (a) qx + 2x – 3 = 0 (b) x + (q – 1)x + 1 = 0
2
2
x – 6x + q = 0
2
For two equal real roots: For two equal real roots:
2
2
b – 4ac = 0 b – 4ac = 0
For two equal real roots: 2 2
2
b – 4ac = 0 (2) – 4(q)(–3) = 0 (q – 1) – 4(1)(1) = 0
2
2
(–6) – 4(1)(q) = 0 4 + 12q = 0 q – 2q + 1 – 4 = 0
2
36 – 4q = 0 12q = –4 1 q – 2q – 3 = 0
–4q = –36 q = – 3 (q – 3)(q + 1) = 0
q = 9 q = 3 or q = –1
10. Find the range of values of m if each of the following quadratic equations has no real roots. PL 4
Cari julat nilai m jika setiap persamaan kuadratik berikut tidak mempunyai punca nyata.
Example (a) 2x + 6x – (3 – m) = 0 (b) x – (3m + 1)x + 12m + 4 = 0
2
2
2
x – 3x + 7m = 0
2
2x + 6x + (–3 + m) = 0 No real roots:
2
No real roots: b – 4ac , 0
No real roots: 2 2
2
b – 4ac , 0 2 b – 4ac , 0 (–3m – 1) – 4(1)(12m + 4) , 0
2
2
(–3) – 4(1)(7m) , 0 (6) – 4(2)(–3 + m) , 0 9m + 6m + 1 – 48m – 16 , 0
2
2
9 – 28m , 0 (6) – 8(–3 + m) , 0 9m – 42m – 15 , 0
2
–28m , –9 36 + 24 – 8m , 0 3m – 14m – 5 , 0
28m . 9 60 – 8m , 0 (3m + 1)(m – 5) , 0
m . 9 –8m , –60 When (3m + 1)(m – 5) = 0,
1
28 8m . 60 m = – or m = 5
3
m . 15
2
m
1 5
–
3
1
∴ – , m , 5
3
11. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.
(a) It is given the quadratic equation x(2x – 5) = p – 4, where p is a constant, has two different real roots.
Diberi persamaan kuadratik x(2x – 5) = p – 4, dengan keadaan p ialah pemalar, mempunyai dua punca nyata yang berbeza.
(i) Find the range of values of p.
Cari julat nilai p.
(ii) State the value of p if the quadratic equation has two equal real roots.
Nyatakan nilai p jika persamaan kuadratik itu mempunyai dua punca nyata yang sama.

x(2x – 5) = p – 4
2x – 5x – p + 4 = 0
2
(i) For two different real roots, (ii) For two equal real roots,
b – 4ac = 0
2
b – 4ac . 0
2
(–5) – 4(2)(–p + 4) . 0 Hence, p = 7
2
25 – 8(–p + 4) . 0 8
25 + 8p – 32 . 0
8p – 7 . 0
8p . 7
p . 7
8



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02 TOP 1 + MATH F4.indd 24 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
(b) It is given the quadratic equation (q + 1)x – 8x + p = 0, where p and q are constants, has two equal
real roots. Express p in terms of q.
Diberi persamaan kuadratik (q + 1)x – 8x + p = 0, dengan keadaan p dan q ialah pemalar, mempunyai dua punca nyata
2
yang sama. Ungkapkan p dalam sebutan q.

(q + 1)x – 8x + p = 0 (–8) – 4(q + 1)(p) = 0
2
2
64 – 4p(q + 1) = 0
For two equal real roots, 64 – 4pq – 4p = 0
b – 4ac = 0 16 – pq – p = 0
2
pq + p = 16
p(q + 1) = 16
p = 16
q + 1

(c) The quadratic equation 3x – 5x + 2p + 3 = 0 where p is a constant, has no real roots. Find the range
2
of values of p.
2
Persamaan kuadratik 3x – 5x + 2p + 3 = 0 dengan keadaan p ialah pemalar, tidak mempunyai punca nyata. Cari julat
nilai p.
3x – 5x + 2p + 3 = 0 (–5) – 4(3)(2p + 3) , 0
2
2
25 – 12(2p + 3) , 0
No real roots, 25 – 24p – 36 , 0
b – 4ac , 0 –24p – 11 , 0
2
–24p , 11
p . – 11
24




2.3 Quadratic Functions Textbook
Fungsi Kuadratik
pg. 49 – 64


SMART Notes

1. A quadratic function can be expressed in the form b – 4ac = 0
2
f(x) = ax + bx + c, where a, b and c are constants and Two equal real roots
2
a ≠ 0. Dua punca nyata yang sama
Fungsi kuadratik boleh diungkapkan dalam bentuk
f(x) = ax + bx + c, dengan keadaan a, b dan c ialah pemalar a 0 a 0
2
dan a ≠ 0. x
2. (a) If a . 0, graph has the shape which passes
through a minimum point. x
Jika a . 0, graf berbentuk yang melalui titik minimum.
2
b – 4ac , 0
(b) If a , 0, graph has the shape which passes No real roots
through a maximum point.
Jika a , 0, graf berbentuk yang melalui titik Tiada punca nyata
maksimum. a 0 a 0
x
3. The relationship between the position of the graph
f(x) = ax + bx + c on the x-axis and its type of roots:
2
2
Hubungan antara kedudukan graf f(x) = ax + bx + c pada paksi-x x
dan jenis puncanya:
4. A quadratic function can be expressed in the form
b – 4ac . 0 f(x) = a(x – h) + k where a, h and k are constants.
2
2
Two different real roots Fungsi kuadratik boleh diungkapkan dalam bentuk
Dua punca nyata yang berbeza f(x) = a(x – h) + k dengan keadaan a, h dan k ialah pemalar.
2
a 0 a 0 5. In f(x) = a(x – h) + k, x = h is an axis of symmetry and
2
(h, k) is the coordinates of the minimum or maximum
x x point.
Dalam f(x) = a(x – h) + k, x = h ialah paksi simetri dan (h, k)
2
ialah koordinat titik minimum atau maksimum.
25 © Penerbitan Pelangi Sdn. Bhd.





02 TOP 1 + MATH F4.indd 25 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

12. Find the range of values of k if the graphs of the following quadratic functions intersect the x-axis at two
points. PL 3
Cari julat nilai k jika graf bagi fungsi kuadratik berikut menyilang paksi-x pada dua titik.
Example (a) f(x) = kx + 8x + 6 (b) f(x) = x + (k – 3)x + 1
2
2
f(x) = x – 3x – k
2
has two different real roots has two different real roots
b – 4ac . 0 b – 4ac . 0
2
2
The graph has (8) – 4(k)(6) . 0 (k – 3) – 4(1)(1) . 0
2
2
two different real roots. 64 – 24k . 0 k – 6k + 9 – 4 . 0
2
2
b – 4ac . 0 2
k – 6k + 5 . 0
(–3) – 4(1)(–k) . 0 –24k . –64 (k – 5)(k – 1) . 0
2
24k , 64
9 + 4k . 0 8
4k . –9 k , 3 When (k – 5)(k – 1) = 0,
k . – 9 k = 5 or k = 1
4
k
1 5
∴ k , 1 or k . 5
13. Find the value of p if the graphs of the following quadratic functions touch the x-axis at one point only. PL 3
Cari nilai p jika graf bagi fungsi kuadratik berikut menyentuh paksi-x pada satu titik sahaja.
Example (a) f(x) = x – 2x + (p – 5) (b) f(x) = (p – 3)x + 4x – 5
2
2
f(x) = x – 4x – p
2
has two equal real roots has two equal real roots
2
2
The graph has b – 4ac = 0 b – 4ac = 0
2
2
two equal real roots. (–2) – 4(1)(p – 5) = 0 (4) – 4(p –3)(–5) = 0
2
b – 4ac = 0 4 – 4(p – 5) = 0 16 + 20(p – 3) = 0
2
(–4) – 4(1)(–p) = 0 4 – 4p + 20 = 0 16 + 20p – 60 = 0
16 + 4p = 0 –4p = –24 20p = 44
4p = –16 p = 6 p = 11
5
p = –4
14. Find the range of values of h if the following quadratic functions do not intersect the x-axis. PL 3
Cari julat nilai h jika graf bagi fungsi kuadratik berikut tidak menyilang paksi-x.
Example (a) f(x) = 8x + 4x + h (b) f(x) = (2h – 3)x + 2hx – 1
2
2
2
f(x) = x + 5x – h
does not have real roots does not have real roots
b – 4ac , 0 b – 4ac , 0
2
2
The graph does not 2 2
have real roots. (4) – 4(8)(h) , 0 (2h) – 4(2h – 3)(–1) , 0
2
b – 4ac , 0 16 – 32h , 0 4h + 4(2h – 3) , 0
2
2
(5) – 4(1)(–h) , 0 –32h , –16 4h + 8h – 12 , 0
2
2
25 + 4h , 0 32h . 16 h + 2h – 3 , 0
4h , –25 h . 1 (h – 1)(h + 3) , 0
2
h , – 25 When (h – 1)(h + 3) = 0,
4 h = 1 or h = –3
h
–3 1

∴ –3 , h , 1





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02 TOP 1 + MATH F4.indd 26 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

15. Express each of the following quadratic functions in the form of f(x) = a(x – h) + k. Then, state the minimum
2
or maximum value, axis of symmetry and minimum or maximum point. PL 3
2
Ungkapkan setiap fungsi kuadratik berikut dalam bentuk f(x) = a(x – h) + k. Kemudian, nyatakan nilai minimum atau maksimum,
paksi simetri dan titik minimum atau maksimum.
Minimum / Minimum /
maximum Axis of maximum
Quadratic function f(x) = a(x – h) + k value symmetry point
2
Fungsi kuadratik
Nilai minimum / Paksi simetri Titik minimum /
maksimum maksimum
Example
2
f(x) = x – 2x + 5 f(x) = x – 2x + 5 Minimum Minimum
2


= x – 2x + – 2  2 + 5 – – 2  2 value point
2
= (1, 4)
a . 0, shape 2 2 = 4 x = 1
2
2
which passes = x – 2x + (–1) + 5 – (–1) 2
through minimum = (x – 1) + 5 – 1
2
point Add and subtract
= (x – 1) + 4 2
2
 coefficient of x 
2
(a) f(x) = x – 6x – 3 f(x) = x – 6x – 3 Minimum x = 3 Minimum
2
2


2
= x – 6x + – 6  2 – 3 – – 6  2 value point
2
2
= –12
= (3, −12)
= x – 6x + (–3) – 3 – (–3) 2
2
2
2
= (x – 3) – 3 – 9
2
= (x – 3) – 12
2
2
(b) f(x) = –x + 2x – 8 f(x) = –x + 2x – 8 Maximum x = 1 Maximum
= –[x – 2x + 8] value point
2

2
2



= – x – 2x + – 2  2 + 8 – – 2  = –7 = (1, −7)
2
2
2
= –[x – 2x + (–1) + 8 – (–1) ]
2
2
= –[(x – 1) + 8 – 1]
2
2
= –[(x – 1) + 7]
= –(x – 1) – 7
2
2
2
(c) f(x) = 2x + 10x + 7 f(x) = 2x + 10x + 7 Minimum x = – 5 Minimum

7
= 2 x + 5x + 2 value 11 2 point 5 11 
2

= –
= – , –
2 

5
5
7
= 2 x + 5x +   2 + –   2 2 2 2
2
2
2

4 
7

= 2 x + 5  2 + – 25
2
2

4 

= 2 x + 5  2 – 11
2

= 2 x + 5  2 – 11
2
2
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02 TOP 1 + MATH F4.indd 27 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

16. Solve the following problems. PL 4
Selesaikan masalah-masalah berikut.

Example (a) The diagram shows the graph of quadratic
2
2
Given the quadratic function f(x) = x + 6x – 5 has function f(x) = (x – p) – 8.
a minimum point (h, k), find Rajah di bawah menunjukkan graf bagi fungsi kuadratik
2
Diberi fungsi kuadratik f(x) = x + 6x – 5 mempunyai titik f(x) = (x – p) – 8.
2
minimum (h, k), cari f(x)
(i) the values of h and k,
nilai-nilai h dan k,
(ii) the equation of the axis of symmetry,
persamaan paksi simetri, –3 0 1 x
(iii) the minimum value of f(x).
nilai minimum f(x).
Find / Cari
2
(i) f(x) = x + 6x – 5 (i) the equation of the axis of symmetry,
6
6
persamaan paksi simetri,
2
= x + 6x +   2 – 5 –   2 (ii) the value of p,
2
2
2
= x + 6x + (3) – 5 – (3) 2 nilai p,
2
= (x + 3) – 5 – 9 (iii) the coordinates of the minimum point.
2
= (x + 3) – 14 koordinat bagi titik minimum.
2
∴ h = –3 and k = –14
(i) x = –3 + 1 (ii) p = –1
2
(ii) x = –3 2
= – (iii) (–1, –8)
2
(iii) –14 = –1
2
2
(b) A quadratic function is given by f(x) = –x + 8x (c) The quadratic function f(x) = ax + bx + c has the
+ k , where k is a constant. Find minimum point (–2, –9) and f(–1) = –7. Find
2
Suatu fungsi kuadratik diberi oleh f(x) = –x + 8x + k , Fungsi kuadratik f(x) = ax + bx + c mempunyai titik
2
2
2
dengan keadaan k ialah pemalar. Cari minimum (–2, –9) dan f(–1)= –7. Cari
(i) the equation of the axis of symmetry, (i) the values of a, b and c,
persamaan paksi simetri, nilai-nilai a, b dan c,
(ii) the possible values of k if the quadratic (ii) the equation of the axis of symmetry.
function f(x) has a maximum value of 25. persamaan paksi simetri.
nilai-nilai k yang mungkin jika fungsi kuadratik f(x)
mempunyai nilai maksimum 25. (i) In the form f(x) = a(x – h) + k
2
h = –2, k = –9
2
(i) f(x) = –x + 8x + k 2 ∴ f(x) = a(x + 2) – 9
2
2
2
2
= –[x – 8x – k ] = a(x + 4x + 4) – 9
= ax + 4ax + 4a – 9
2

2
8



2
= – x – 8x + – 8  2 – k – – 2  Comparing: 2
2
2
2
2
= –[x – 8x + (–4) – k – (–4) ] ax + bx + c = ax + 4ax + 4a – 9
2
2
2
∴ b = 4a and c = 4a – 9
2
= –[(x – 4) – k – 16]
2
2
2
= –(x – 4) + k + 16 f(–1) = a(–1) + 4a(–1) + 4a – 9
2
–7 = a – 4a + 4a – 9
Therefore, the equation of axis of symmetry –7 = a – 9
is x = 4. a = 2
(ii) k + 16 = 25 b = 4(2)
2
k = 9 = 8
2
k = ±√9
k = 3 or k = –3 c = 4(2) – 9
= –1
(ii) x = –2
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

17. Sketch the graph of each of the following quadratic functions. PL 4
Lakarkan graf bagi setiap fungsi kuadratik berikut.

Example
2
f(x) = x – x – 6 for / untuk –3 < x < 4 When x = 0, f(0) = (0) – 0 – 6
2
= –6
a . 0, then the shape of the graph is .
∴ The y-intercept is (0, –6).
2
2
Discriminant, b – 4ac = (–1) – 4(1)(–6)
2
= 1 + 24 When x = –3, f(–3) = (–3) – (–3) – 6
= 25 . 0 = 9 + 3 – 6
2
∴ f(x) = x – x – 6 has two different real roots. = 6
∴ (−3, 6) Range: –3 < x < 4
f(x) = x – x – 6 When x = 4, f(4) = (4) – 4 – 6
2
2


2
= x – x + – 1  2 – 6 – – 1  2 = 16 – 4 – 6
2
2
= 6

= x – 1  2 – 6 – 1 ∴ (4, 6) f(x)
2
4

= x – 1  2 – 25 (–3, 6) x = 1 2 (4, 6)
2 4
∴ The minimum point is  1 , –6 1  .
2
4
1
The axis of symmetry is x = .
2 –2 0 3 x
When f(x) = 0, x – x – 6 = 0 2
2
(x – 3)(x + 2) = 0 f(x) = x – x – 6
= 3 or x = –2
–6 1 1
( , –6 )
∴ The x-intercepts are (3, 0) and (–2, 0). 2 4
(a) f(x) = x – 4x – 5 for / untuk –2 < x < 6
2
a . 0, then the shape of the graph is . When x = 0, f(0) = (0) – 4(0) – 5
2
= –5
2
2
Discriminant, b – 4ac = (–4) – 4(1)(–5)
= 16 + 20 ∴ The y-intercept is (0, –5).
= 36 . 0
When x = –2, f(–2) = (–2) – 4(–2) – 5
2
2
∴ f(x) = x – 4x – 5 has two different real roots. = 4 + 8 – 5
= 7
∴ (–2, 7)
2
f(x) = x – 4x – 5


2
= x – 4x + – 4  2 – 5 – – 4  2 When x = 6, f(6) = (6) – 4(6) – 5
2
2
2
= 36 – 24 – 5

2
2
= x – 4x + (–2) – 5 – (–2) 2 = 7
2
= (x – 2) – 5 – 4 ∴ (6, 7)
2
= (x – 2) – 9
f(x)
∴ The minimum point is (2, –9). (–2, 7) x = 2 (6, 7)
The axis of symmetry is x = 2.
x
–1 0 5
2
When f(x) = 0, x – 4x – 5 = 0
(x – 5)(x + 1) = 0 –5 f(x) = x – 4x – 5
2
x = 5 or x = –1

∴ The x-intercepts are (5, 0) and (–1, 0). (2, –9)
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02 TOP 1 + MATH F4.indd 29 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
(b) f(x) = –2x + 7x + 4 for / untuk –1 < x < 5
2
a , 0, then the shape of the graph is . When f(x) = 0, –2x + 7x + 4 = 0
(2x + 1)(4 – x) = 0
1
Discriminant, b – 4ac = (7) – 4(–2)(4) x = – or x = 4
2
2
2
= 49 + 32
1
= 81 . 0 ∴ The x-intercepts are (– , 0) and (4, 0).
2
2
∴ f(x) = –2x + 7x + 4 has two different real roots. When x = 0, f(0) = –2(0) + 7(0) + 4
2
= 4

f(x) = –2x + 7x + 4 ∴ The y-intercept is (0, 4).
2

7
2
= –2 x – x – 2  When x = 5, f(5) = –2(5) + 7(5) + 4
2
2
2
= –2x – x +   2 – 2 –   = –50 + 35 + 4
7
7
–
–
= –11
2
2
7
2
2 2 2 ∴ (5, –11)
= –2 x – x + – 7 2 – 2 – – 4  When x = –1, f(–1) = –2(–1) + 7(–1) + 4

2
7
2
7


 
2
4
2

= –2 – 7 + 4
= –5

49

= –2 x – 7  2 – 2 – 16 ∴ (–1, –5)
4
f(x)

81

= –2 x – 7  2 – 16 (1 , 10 )
3
1
4
8
4

2
= –2 x – 7  2 + 81 4 f(x) = –2x + 7x + 4
8
4

3

∴ The maximum point is 1 , 10 1  . – 1 2 0 4 x
4
8
3
The axis of symmetry is x = 1 . (–1, –5)
4
x = 1 3
4 (5, –11)
18. Solve the following problems. PL 5
Selesaikan masalah-masalah berikut.
Example
2
(i) Given f(x) = 3x – 27, find the range of values of x such that f(x) is always positive.
Diberi f(x) = 3x – 27, cari julat nilai x dengan keadaan f(x) sentiasa positif.
2
(ii) Given f(x) = 3x – 9, find the range of values of x such that f(x) < –6x.
2
2
Diberi f(x) = 3x – 9, cari julat nilai x dengan keadaan f(x) < –6x.
(i) f(x) . 0 (ii) f(x) < –6x
3(x – 9) . 0 3x – 9 < –6x
2
2
2
3(x – 3)(x + 3) . 0 3x + 6x – 9 < 0
x + 2x – 3 < 0
2
When 3(x – 3)(x + 3) = 0, (x – 1)(x + 3) < 0
x = 3 or x = –3
When (x – 1)(x + 3) = 0,
x = 1 or x = –3
x
–3 3
x
–3 1
∴ For 3x – 27 . 0, x , –3 or x . 3.
2
2
∴ For 3x – 9 < –6x, –3 < x < 1.
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Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
(a) Given f(x) = 5x – 9x – 2, find the range of values (b) (i) Find the range of values of x for
2
of x such that f(x) is always (x + 2) , 11x + 12.
2
Diberi bahawa f(x) = 5x – 9x – 2, cari julat nilai x dengan Cari julat nilai x bagi (x + 2) , 11x + 12.
2
keadaan f(x) sentiasa (ii) Given the quadratic equation x(3k – x) = 4
(i) positive / positif, does not have any real roots, find the range
(ii) negative / negatif. of values of p.
Diberi persamaan kuadratik x(3k – x) = 4 tidak
2
(i) 5x – 9x – 2 . 0 mempunyai punca nyata, cari julat nilai p.
(x – 2)(5x + 1) . 0
2
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1 (i) x + 4x + 4 , 11x + 12
5 x – 7x – 8 , 0
2
(x – 8)(x + 1) , 0
x When (x – 8)(x + 1) = 0, x = 8 or x = –1
–1 1
4
1
∴ For 5x – 9x – 2 . 0, x , – or x . 2. x
2
5 –1 8
2
(ii) 5x – 9x – 2 , 0 2
(x – 2)(5x + 1) , 0 ∴ For x – 7x – 8 , 0, –1 , x , 8.
When (x – 2)(5x + 1) = 0, x = 2 or x = – 1 (ii) 3kx – x = 4
2
5
x – 3kx + 4 = 0
2
No real roots, b – 4ac , 0
2
x (–3k) – 4(1)(4) , 0
2
1 2
– 9k – 16 , 0
2
5
(3k – 4)(3k + 4) , 0
1
2
∴ For 5x – 9x – 2 , 0, – , x , 2.
5 4 4
When (3k – 4)(3k + 4) = 0, k = or k = –
3 3
x
4 4
–
3 3
4
4
∴ For 9k – 16 , 0, – , k , .
2
3 3
2
(c) Hashim threw a ball. The height of the ball is given by the function h(x) = –x + 6x + 1, where h is the
height of the ball from the ground, in m, and x is the horizontal distance of the ball from Hashim’s
position, in m. Find Daily Application
2
Hashim membaling sebiji bola. Tinggi bola itu diberi oleh fungsi h(x) = –x + 6x + 1, dengan keadaan h ialah tinggi bola
dari permukaan tanah, dalam m, dan x ialah jarak mengufuk bola dari kedudukan Hashim, dalam m. Cari
(i) the maximum height, in m, of the ball thrown by Hashim,
tinggi maksimum, dalam m, bola yang dibaling oleh Hashim,
(ii) the horizontal distance between Hashim and the ball, in m, when the ball touches the ground.
jarak mengufuk di antara Hashim dan bola, dalam m, apabila bola itu menyentuh permukaan tanah.
2
(i) h(x) = –x + 6x + 1 (ii) h(x) = 0
2
= –(x – 6x – 1) –x + 6x + 1 = 0 2
2
2 

–6
–6
= – x – 6x +   2 – 1 –   2 x = –b ± √b – 4ac
2
2a
2

2
= –[x – 6x + (–3) – 1 – (–3) ] = –6 ± √6 – 4(–1)(1)
2
2
2
= –[(x – 3) – 10] 2(–1)
2
2
= –(x – 3) + 10 = –6 ± √40
–2
–6 + √40 –6 – √40
a = –1 , 0, so the graph has a maximum point. x = or x =
–2 –2
The maximum value is 10 when x = 3. = –0.16 (ignore) = 6.16
Hence, the maximum height is 10 m.
Hence, the horizontal distance is 6.16 m.
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02 TOP 1 + MATH F4.indd 31 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

SPM Practice 2




2
Paper 1 7. A quadratic function is defined as f(x) = x + 6x + h,
SPM where h is a constant.
2017
Suatu fungsi kuadratik ditakrifkan sebagai f(x) = x + 6x + h,
2
1. It is given the graph of a quadratic function dengan keadaan h ialah pemalar.
2
SPM f(x) = px + 6x + q, where p and q are constants, has a
2
2015 (a) Express f(x) in the form (x + m) + n, where m and n
minimum point. are constants.
Diberi graf bagi fungsi kuadratik f(x) = px + 6x + q, dengan Ungkapkan f(x) dalam bentuk (x + m) + n, dengan
2
2
keadaan p dan q ialah pemalar, mempunyai satu titik minimum. keadaan m dan n ialah pemalar.
(a) If p is an integer such that –2 , p , 2, state the value (b) If the minimum value of f(x) is 7, find the value of h.
of p. Jika nilai minimum f(x) ialah 7, cari nilai h.
Jika p ialah integer dengan keadaan –2 , p , 2, nyatakan [4]
nilai p. Ans: (a) (x + 3) + h – 9
2
(b) Using the answer in 1(a), find the value of q when the (b) h = 16
graph touches the x-axis at one point only.
Menggunakan jawapan di 1(a), cari nilai q apabila graf
tersebut menyentuh paksi-x pada satu titik sahaja. 8. Find the range of values of x such that the quadratic
2
[3] SPM function f(x) = 10 + 3x – x is negative.
2017
Ans: (a) p = 1 Cari julat nilai x dengan keadaan fungsi kuadratik
2
(b) q = 9 f(x) = 10 + 3x – x adalah negatif.
[3]
2
2. Find the range of values of x for 4x + 7x < 2. Ans: x , –2 or x . 5
2
SPM Cari julat nilai x untuk 4x + 7x < 2.
2015
[2] 9. (a) Given one of the roots of the quadratic equation
2
Ans: –2 < x < 1 SPM x + (p + 4)x – p = 0, where p is a constant, is the
2
4 2017
negative of the other, find the value of product of
3. Given the quadratic equation (x + p) = 49, where p is roots.
2
SPM a constant, find the values of p if one of the roots of the Diberi satu daripada punca-punca persamaan kuadratik
2015
equation is 4. x + (p + 4)x – p = 0, dengan keadaan p ialah pemalar,
2
2
2
Diberi persamaan kuadratik (x + p) = 49, dengan keadaan adalah negatif kepada yang satu lagi, cari nilai hasil darab
p ialah pemalar, cari nilai-nilai p jika satu daripada punca punca.
persamaan tersebut ialah 4. [2]
2
[2] (b) Given the quadratic equation mx – 9nx + m = 0,
Ans: p = 3 or p = –11 where m and n are constants, has two equal roots,
find m : n.
2
4. The quadratic equation 16x – 24x + 9 = 0 has roots a and Diberi persamaan kuadratik mx – 9nx + m = 0, dengan
2
SPM b. Form a quadratic equation with the roots 3a and 3b. keadaan m dan n ialah pemalar, mempunyai dua punca
2016
Persamaan kuadratik 16x – 24x + 9 = 0 mempunyai punca a yang sama, cari m : n.
2
dan b. Bentukkan persamaan kuadratik dengan punca-punca [2]
3a dan 3b. Ans: (a) –16
[3] (b) 9 : 2
2
Ans: 16x – 72x + 81= 0
10. Encik Samad has a rectangular land with a dimension 4x m
5. The quadratic function f(x) = –x – 2kx + 4k – 5, where SPM in length and 3x m in width. Three square regions of the
2
2018
SPM k is a constant, is always negative when m , k , n. Find land are planted with different types of vegetables. Each
2016 region has a side length of x m. Find the range of values
the values of m and n.
2
Fungsi kuadratik f(x) = –x – 2kx + 4k – 5, dengan keadaan of x if the region that is not planted with vegetables is at
2
least (x + 32) m .
2
k ialah pemalar, adalah sentiasa negatif apabila m , k , n. Encik Samad mempunyai sebidang tanah berbentuk
Cari nilai-nilai m dan n. segi empat tepat yang berukuran 4x m panjang dan
[3]
Ans: m = –5, n = 1 3x m lebar. Tiga kawasan berbentuk segi empat sama ditanam
dengan pelbagai jenis sayur. Setiap kawasan itu mempunyai
panjang sisi x m. Cari julat nilai x jika kawasan yang tidak ditanam
6. Find the range of values of x for x + 30 , 11x. dengan sayuran adalah sekurang-kurangnya (x + 32) m .
2
2
2
2
Cari julat nilai x untuk x + 30 , 11x. [3]
[3] Ans: x > 2
Ans: 5 , x , 6
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02 TOP 1 + MATH F4.indd 32 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
11. Given the curve y = (p – 3)x – x + 8, where p is a constant, Paper 2
SPM intersects the straight line y = 4x + 6 at two points, find
2018
the range of values of p.
2
Diberi lengkung y = (p – 3)x – x + 8, dengan keadaan p ialah 1. It is given a and b are the roots of the quadratic equation
pemalar, menyilang garis lurus y = 4x + 6 pada dua titik, cari SPM x(x – 8) = 3h – 15, where h is a constant.
2015
julat nilai p. Diberi a dan b ialah punca-punca bagi persamaan kuadratik
[3] x(x – 8) = 3h – 15, dengan keadaan h ialah pemalar.
49 (a) Find the range of values of h if a ≠ b.
Ans: p ,
8 Cari julat nilai h jika a ≠ b.
[3]
2
12. Given the quadratic equation hx – 5x + k = 0, where h and a b
SPM k are constants, has roots b and 3b, express h in terms of k. (b) Given 2 and 2 are the roots of another quadratic
2018
Diberi persamaan kuadratik hx – 5x + k = 0, dengan keadaan equation x + kx + 3 = 0, where k is a constant, find
2
2
h dan k ialah pemalar, mempunyai punca-punca b dan 3b. the values of h and k.
Ungkapkan h dalam sebutan k. Diberi a dan b ialah punca-punca suatu persamaan
[3] 2 2
2
75 kuadratik yang lain, x + kx + 3 = 0, dengan keadaan k
Ans: h =
16k ialah pemalar, cari nilai-nilai h dan k.
[4]
13. It is given 5 and h + 1 are the roots of the quadratic Ans: (a) h . – 1
equation x + (k – 1)x – 5 = 0, where h and k are constants. 3
2
Find the values of h and k. (b) h = 1, k = –4
Diberi 5 dan h + 1 ialah punca-punca bagi persamaan kuadratik
2
x + (k – 1)x – 5 = 0 dengan keadaan h dan k ialah pemalar. 2. The curve of the quadratic function f(x) = 2(x – p) + 3q
2
Cari nilai-nilai h dan k. SPM intersects the x-axis at points (3, 0) and (7, 0). The straight
2016
[3] line y = –6 touches the minimum point of the curve.
Ans: h = –2, k = –3 Lengkung bagi fungsi kuadratik f(x) = 2(x – p) + 3q menyilang
2
paksi-x pada titik (3, 0) dan (7, 0). Garis lurus y = –6
14. The diagram shows the graph of a quadratic function menyentuh titik minimum lengkung tersebut.
y = f(x). (a) Find the values of p and q.
Rajah di bawah menunjukkan graf bagi suatu fungsi kuadratik Cari nilai-nilai p dan q.
y = f(x). [2]
y (b) Hence, sketch the graph of f(x) for 0 < x < 8.
3 Seterusnya, lakar graf f(x) untuk 0 < x < 8 [3]
y = f(x)
(c) If the curve is reflected about the x-axis, write the
x equation of the curve.
–4 0 6
Jika lengkung tersebut dipantulkan pada paksi-x, tulis
State / Nyatakan persamaan bagi lengkung tersebut. [1]
(a) the roots of the equation when f(x) = 0, Ans: (a) p = 5, q = –2
punca-punca persamaan tersebut apabila f(x) = 0, (b) Refer to Answer Section
(b) the equation of the axis of symmetry of the curve. (c) f(x) = –2(x – 5) + 6
2
persamaan paksi simetri lengkung itu.
[2] 3. The quadratic equation x – 7x + 12 = 0 has two roots p
2
Ans: (a) –4 and 6 and q, where p . q.
(b) x = 1 Persamaan kuadratik x – 7x + 12 = 0 mempunyai dua punca
2
p dan q, dengan keadaan p . q.
(a) Find / Cari
2
15. It is given the quadratic function f(x) = x – 4x + 3 can (i) the values of p and q,
be expressed in the form f(x) = (x – 2) + m, where m is a nilai-nilai p dan q,
2
constant. (ii) the range of values of x if x – 7x + 12 . 0.
2
Diberi fungsi kuadratik f(x) = x – 4x + 3 boleh diungkapkan julat nilai x jika x –7x + 12 . 0.
2
2
dalam bentuk f(x) = (x – 2) + m, dengan keadaan m ialah [4]
2
pemalar. (b) By using the values of p and q in 3(a)(i), form a
(a) Find the value of m. quadratic equation which has the roots p – 2 and q + 4.
Cari nilai m. Dengan menggunakan nilai-nilai p dan q di 3(a)(i), bentuk
(b) Sketch the graph of the function f(x). satu persamaan kuadratik yang mempunyai punca-punca
Lakar graf bagi fungsi f(x). p – 2 dan q + 4.
[5] [3]
Ans: (a) m = –1 Ans: (a) (i) p = 4, q = 3; (ii) x , 3 or x . 4
(b) Refer to Answer Section (b) x – 9x + 14 = 0
2


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02 TOP 1 + MATH F4.indd 33 20/12/2019 9:34 AM

Additional Mathematics Form 4 Chapter 2 Quadratic Functions

2
4. A quadratic equation x + 3(x – p) = 0, where p is a (b) Find the minimum or maximum value of the quadratic
constant, has the roots q and –2q, q ≠ 0. function f(x).
Persamaan kuadratik x + 3(x – p) = 0, dengan keadaan p ialah Cari nilai minimum atau maksimum fungsi kuadratik f(x).
2
pemalar, mempunyai punca-punca q dan –2q, q ≠ 0. [1]
2
(a) Find the values of p and q. (c) Sketch the graph of f(x) = x – x – 12 for –4 < x < 5.
2
Cari nilai-nilai p dan q. Lakar graf f(x) = x – x – 12 untuk –4 < x < 5.
[4] [4]
(b) Hence, form a quadratic equation which has the roots (d) State the equation of the curve when the graph is
p – 3 and p + 1. reflected about the x-axis.
Seterusnya, bentuk satu persamaan kuadratik yang Nyatakan persamaan lengkung apabila graf tersebut
mempunyai punca-punca p – 3 dan p + 1. dipantulkan pada paksi-x.
[3]
Ans: (a) p = 6, q = 3 [1]
(b) x – 10x + 21 = 0 1 2 1
2

Ans: (a) x – 2  – 12
4
1
5. It is given the quadratic function f(x) = x – x – 12. (b) Minimum value = –12
2
4
2
Diberi fungsi kuadratik f(x) = x – x – 12. (c) Refer to Answer Section
(a) Express f(x) in the form f(x)= a(x + p) + q. 1 2 1
2

2
Ungkapkan f(x) dalam bentuk f(x) = a(x + p) + q. (d) f(x) = – x – 2  + 12 or f(x) = –x + x + 12
2
4
[2]
HOTS Challenge
1. The diagram shows a right-angled triangle PQR. HOTS Applying
Rajah di sebelah menunjukkan sebuah segi tiga bersudut tegak PQR. P
(a) Find the value of x. Give your answer correct to four significant figures. (2x + 3) cm
Cari nilai x. Berikan jawapan anda betul kepada empat angka bererti. (x – 2) cm
(b) By using the value of x obtained in 1(a), calculate the area, in cm , of
2
the right-angled triangle PQR.
Dengan menggunakan nilai x yang diperoleh di 1(a), hitung luas, dalam cm , Q (2x + 1) cm R
2
segi tiga bersudut tegak PQR.
Answer Guide
Use Pythagoras’ theorem to find the value of x.
Gunakan teorem Pythagoras untuk mencari nilai x.

Ans: (a) x = 12.32
(b) 132.30 cm 2

6q + 1
2. It is given the quadratic function f(x) = px – 3x + q can be expressed in the form f(x) = p(x – ) + .
3 2
2
4 4p
Find HOTS Applying
3 6q + 1
Diberi fungsi kuadratik f(x) = px – 3x + q boleh diungkapkan dalam bentuk f(x) = p(x – 4 ) + 4p . Cari
2
2
(a) the values of p and q,
nilai-nilai p dan q,
(b) the range of values of x if f(x) > 4.
julat nilai x jika f(x) > 4.
Answer Guide
Express the function in vertex form using ‘completing the square’ method.
Ungkapkan fungsi itu dalam bentuk verteks dengan menggunakan kaedah penyempurnaan kuasa dua.

Ans: (a) p = 2, q = 5
1
(b) x < or x > 1
2








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Instruction: Scan the QR codes or visit the website links below to get the answers.
Arahan: Imbas QR code atau layari pautan laman web di bawah untuk mendapatkan jawapan.


Chapter 7 Coordinate
Chapter 1 Functions Geometry
Fungsi Geometri Koordinat
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Chapter 2 Quadratic
Functions Chapter 8 Vectors
Fungsi Kuadratik Vektor
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Chapter 3 Systems of Chapter 9 Solution of
Equations Triangles
Sistem Persamaan Penyelesaian Segi Tiga
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Chapter 4 Indices, Surds
and Logarithms
Indeks, Surd dan Chapter 10 Index Numbers
Logaritma Nombor Indeks
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Chapter 5 Progressions Year-End Assessment
Janjang Kertas Akhir Tahun
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