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π

CHAPTER 2

Quadratic functions

Quadratic Functions
Quadratic functions are in the form of 𝒇(𝒙) = 𝒂𝒙𝟐 + 𝒃𝒙 + 𝒄, where 𝒂 ≠ 𝟎 and
represented by:

Vertex (minimum point)
axis of symmetry

Vertex (minimum point)
𝒂<𝟎

𝒂>𝟎

Roots of a quadratic function:
→ The roots of the function are the x-coordinates of the points of intersection of the
curve with the x-axis
...
53

−𝑏±√𝑏2 −4𝑎𝑐
2𝑎

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CHAPTER 2

Types of roots of a quadratic function:
1) Two different (distinct) real roots:

2) Two equal real roots (one real roots):

3) No real roots:

P
...

Discriminant = 𝛥 = 𝑏 2 − 4𝑎𝑐, we have three cases:
1) 𝛥 = 𝑏 2 − 4𝑎𝑐 > 0

"two distinct "("different")" real roots"

2) 𝛥 = 𝑏 2 − 4𝑎𝑐 = 0

"two equal real roots "("one root")

3) 𝛥 = 𝑏 2 − 4𝑎𝑐 < 0

"no real roots"

Examples:
1) State the nature of the roots for each of the following quadratics:
a) 𝑓(𝑥) = 𝑥 2 − 10𝑥 + 25
∵𝑎=1

𝑏 = −10

𝑐 = 25

∴ 𝛥 = 𝑏 2 − 4𝑎𝑐 = 100 − 100 = 0

It has one real root (two equal roots)

-------------------------------------------------------------------------------------------------------------------------

b) 𝑓(𝑥) = −𝑥 2 + 5𝑥 + 6
∵ 𝑎 = −1

𝑏=5

𝑐=6

∴ 𝛥 = 𝑏 2 − 4𝑎𝑐 = 25 + 24 = 49 > 0

It has two different roots (two distinct roots)

-------------------------------------------------------------------------------------------------------------------------

c) 𝑓(𝑥) = −2𝑥 2 − 5𝑥 − 6
∵ 𝑎 = −2

𝑏 = −5

𝑐 = −6

∴ 𝛥 = 𝑏 2 − 4𝑎𝑐 = 25 − 48 = −23 < 0

It has no real roots

P
...

∵𝑎=3

𝑏=2

𝑐=𝑘

∵ 𝛥 = 𝑏 2 − 4𝑎𝑐 = 0

two equal roots
4

1

∴ 4 − 4 × 3 × 𝑘 = 0 → 4 − 12𝑘 = 0 → 12𝑘 = 4 → 𝑘 = 12 = 3
-------------------------------------------------------------------------------------------------------------------------

3) The equation 𝑘𝑥 2 − 2𝑥 − 7 = 0 has two distinct real roots, find the possible value
of k
...
56

−4
28

→ 𝑘>

−1
7

π

CHAPTER 2

Quadratic functions

 Test yourself:
1) Solve the equation 2𝑥 2 − 2𝑥 − 1 = 0, giving the roots in exact form

------------------------------------------------------------------------------------------------------------------------

2) Solve the equation 3𝑥 2 − 4𝑥 − 9 = 0, giving your answers to 2 d
...


-------------------------------------------------------------------------------------------------------------------------

3) State the nature of the roots for each of the following quadratics:
a) 2𝑥 2 − 3𝑥 − 4 = 0

-------------------------------------------------------------------------------------------------------------------------

b) 2𝑥 2 − 3𝑥 − 5 = 0

P
...


P
...


Example:
Find the coordinates of the vertex of the function𝑓(𝑥) = −2𝑥 2 − 3𝑥 + 4; hence,
state the maximum or minimum value of f(x) and the value of x at which it occurs
...
59

3
4

41
8

π

CHAPTER 2

Quadratic functions

2) Put the function 𝑓(𝑥) = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐 in the form of the completing the square
𝑓(𝑥) = 𝐴(𝑥 + 𝐵)2 + 𝐶
...
The vertex is (-B, C)
...


Solution
−2𝑥 2 − 3𝑥 + 4 ≡ 𝐴(𝑥 + 𝐵)2 + 𝐶 ≡ 𝐴(𝑥 2 + 2𝐵𝑥 + 𝐵 2 ) + 𝐶 ≡ 𝐴𝑥 2 + 2𝐴𝐵𝑥 + 𝐴𝐵 2 + 𝐶

𝐴 = −2
−3

3

2𝐴𝐵 = −3 ⇒ 2(−2)𝐵 = −3 ⇒ −4𝐵 = −3 ⇒ 𝐵 = −4 = 4
3

𝐴𝐵 2 + 𝐶 = 4 ⇒ −2(4)2 + 𝐶 = 4 ⇒
3 2

∴ 𝑓(𝑥) = −2 (𝑥 + 4) +

41

,

8

−18
16

9

+𝐶 =4⇒ 𝐶 =4+8 =
3 41

the vertex is (− 4 ,

8

41
8

)

-------------------------------------------------------------------------------------------------------------------------

2) The quadratic 𝑥 2 − 10𝑥 + 7 is denoted by f(x)
...
Hence, find the least possible value of f(x) and the corresponding value
of x
...
60

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CHAPTER 2

Quadratic functions

3) The equation of a curve is 𝑦 = 8𝑥 − 𝑥 2
...

b) Find the coordinates of the vertex and state whether it is maximum or minimum
...
61

π

CHAPTER 2

Quadratic functions

5) If 𝑓(𝑥) = 2𝑥 2 − 8𝑥 + 10
i) Express 𝑓(𝑥) is the form 𝑎(𝑥 + 𝑏)2 + 𝑐, where a, b and c are constant
ii) hence, state the coordinates of the stationary point of 𝑓(𝑥) and state its type
...


Solution
𝑥 2 + 4𝑥 + 3 ≡ (𝑥 + 2)2 − 4 + 3
𝑥 2 + 4𝑥 + 3 ≡ (𝑥 + 2)2 − 1,

𝑎 = 2 , 𝑏 = −1

-----------------------------------------------------------------------------------------------------------------------

7) Put 2 + 2𝑥 − 𝑥 2 in the form ℎ − (𝑥 + 𝑝)2 stating the values of h and p
...
62

ℎ = 3 𝑎𝑛𝑑 𝑝 = −1

π

CHAPTER 2

Quadratic functions

8) Express 2𝑥 2 − 16𝑥 + 37 in the form𝐴(𝑥 + 𝐵)2 +𝐶, stating the values of A, B
and C
...


Solution
−2𝑥 2 + 4𝑥 = −2(𝑥 2 − 2𝑥) = −2[(𝑥 − 1)2 − 1]
= −2(𝑥 − 1)2 + 2 = 2 − 2(𝑥 − 1)2

𝐴 = 2, 𝐵 = 2 𝑎𝑛𝑑 𝐶 = 1
P
...
64

π

CHAPTER 2

Quadratic functions

 Test yourself:
1) Given that 𝑥 2 − 4𝑥 + 7 ≡ (𝑥 − 𝑎)2 + 𝑏
...


-------------------------------------------------------------------------------------------------------------------------

2) Express each of the following in the form of (𝑥 + 𝑎)2 + 𝑏; stating the value of a
and b
...

a) 𝑥 2 + 2𝑥 + 2

------------------------------------------------------------------------------------------------------------------------

b) 𝑥 2 − 8𝑥 − 3

P
...
66

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CHAPTER 2

Quadratic functions

g) 7 − 8𝑥 − 4𝑥 2

------------------------------------------------------------------------------------------------------------------------

h) −𝑥 2 − 10𝑥 + 7

------------------------------------------------------------------------------------------------------------------------

3) Find the least or the greatest value of each of the following quadratic and the value
of x for which this occurs
...
67

π

CHAPTER 2

Quadratic functions

b) 𝑦 = (𝑥 + 2)2 − 7

------------------------------------------------------------------------------------------------------------------------

c) 𝑦 = 1 + (2𝑥 − 3)2

------------------------------------------------------------------------------------------------------------------------

d) 𝑦 = (5𝑥 + 3)2 + 2

------------------------------------------------------------------------------------------------------------------------

e) 𝑦 = 3 − 2(𝑥 − 4)2

P
...

→ Find the y-intercept (value of y when x = 0)
...

→ Sketch the function
...
69

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CHAPTER 2

Quadratic functions

b) 𝑦 = 7 − 10𝑥 − 𝑥 2

Solution
Roots ⇒ 7 − 10𝑥 − 𝑥 2 = 0 ⇒ 𝑥 2 − 10𝑥 + 7
by using the formula ⇒ 𝑥 = −10
...
657
𝑦-intercept ⇒ 𝑥 = 0 ⇒ 𝑦 = 7 ⇒ (0 , 7)
Vertex ⇒ 𝑥 =

−𝑏
2𝑎

−−10

= 2×−1 = −5

,

𝑦 = 7 − 10(−5) − (−5)2 = 32 ⇒ (−5 , 32)

P
...
71

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CHAPTER 2

Quadratic functions

3) 𝑓(𝑥) = 4𝑥 − 2𝑥 2 + 3

------------------------------------------------------------------------------------------------------------------------

4) 𝑓(𝑥) = 𝑥 2 + 6𝑥

P
...

→ Solve the equation (find the roots)
...

→ State the range of values of x satisfying the inequality
...
73

,

𝑥=3

3

x

π

CHAPTER 2

Quadratic functions

c) 𝑥 2 + 12 < 13𝑥

Solution
1

12

𝑥 2 − 13𝑥 + 12 < 0
𝑥 2 − 13𝑥 + 12 = 0 ⇒ (𝑥 − 1)(𝑥 − 12) = 0 ⇒ 𝑥 = 1

,

x

𝑥 = 12

∴ 1 < 𝑥 < 12
------------------------------------------------------------------------------------------------------------------------

d) 𝑥 2 > 𝑥

Solution
0

𝑥2 − 𝑥 > 0
𝑥 2 − 𝑥 = 0 ⇒ 𝑥(𝑥 − 1) = 0 ⇒ 𝑥 = 0
∴𝑥<0

and

,

x

1

𝑥=1

𝑥>1

------------------------------------------------------------------------------------------------------------------------

2) Find the range of values of k for which the equation 𝑘𝑥 2 + 𝑘𝑥 + 2 = 0 has no real
roots
...
74

𝑘=8

0

8

k

π

CHAPTER 2

Quadratic functions

3) Find the range of values of k for which the equation 𝑘𝑥 2 + 3𝑥 + 𝑘 = 0 has two
distinct real roots
...


Solution
𝑎=1

,

𝑏=𝑘

,

2

𝑐 = 2𝑘 − 3

6

𝛥 = 𝑏 2 − 4𝑎𝑐 ≥ 0 ⇒ 𝑘 2 − 4(1)(2𝑘 − 3) ≥ 0 ⇒ 𝑘 2 − 8𝑘 + 12 ≥ 0
𝑘 2 − 8𝑘 + 12 = 0 ⇒ (𝑘 − 2)(𝑘 − 6) = 0 ⇒ 𝑘 = 2
∴𝑘≤2

,

𝑘≥6

P
...
76

π

CHAPTER 2

Quadratic functions

d) 4 − 9𝑥 2 ≤ 0

------------------------------------------------------------------------------------------------------------------------

2) Find the set of values of k for which the equation 𝑥 2 − 2𝑘𝑥 + 4 = 0 has two real
roots
...
77

π

CHAPTER 2

Quadratic functions

Simultaneous Equations
How to solve a pair of simultaneous equations one of them quadratic and
the other linear:
Examples:
1) Solve simultaneously 𝑥 2 + 2𝑦 2 = 9,

𝑥 + 4𝑦 = 9

Solution
From the line ⇒ 𝑥 = 9 − 4𝑦
Into the curve ⇒ (9 − 4𝑦)2 + 2𝑦 2 = 9 ⇒ 81 − 72𝑦 + 16𝑦 2 + 2𝑦 2 = 9
18𝑦 2 − 72𝑦 + 81 − 9 = 0 ⇒ 18𝑦 2 − 72𝑦 + 72 = 0 → (÷ 18)
𝑦 2 − 4𝑦 + 4 = 0 ⇒ (𝑦 − 2)(𝑦 − 2) = 0 ⇒ 𝑦 = 2
𝑥 = 9 − 4(2) = 1
∴ The line cuts the curve at point (1 , 2)
∴ The line is tangent to the curve and the point of tangency is (1 , 2)
------------------------------------------------------------------------------------------------------------------------

2) Find the point(s) of intersection of the line 𝑥 + 𝑦 = 1 and the curve
𝑥 2 − 𝑥𝑦 + 𝑦 2 = 7

Solution
From the line ⇒ 𝑦 = 1 − 𝑥
Into the curve ⇒ 𝑥 2 − 𝑥(1 − 𝑥) + (1 − 𝑥)2 = 7 ⇒ 𝑥 2 − 𝑥 + 𝑥 2 + 1 − 2𝑥 + 𝑥 2 = 7
3𝑥 2 − 3𝑥 + 1 − 7 = 0 ⇒ 3𝑥 2 − 3𝑥 − 6 = 0 → (÷ 3)
𝑥 2 − 𝑥 − 2 = 0 ⇒ (𝑥 − 2)(𝑥 + 1) = 0 ⇒ 𝑥 = 2
𝑦 = 1 − 2 = −1

,

𝑦 = 1 − −1 = 2

∴ The line cuts the curve at points (2, -1) and (-1, 2)
∴ The points of intersection are (2, -1) and (-1, 2)

P
...

1) 𝛥 > 0

The line cuts the curve at two different (distinct) points
...


3) 𝛥 < 0

The line neither cuts nor touches the curve

Examples:
1) State the relation between the line 𝑥 + 𝑦 = 1and the curve 𝑦 = 𝑥 2 + 2𝑥 − 3

Solution
From the line ⇒ 𝑦 = 1 − 𝑥
Into the curve ⇒ 1 − 𝑥 = 𝑥 2 + 2𝑥 − 3 ⇒ 𝑥 2 + 3𝑥 − 4 = 0
𝑎=1

,

𝑏=3

,

𝑐 = −4

𝛥 = 𝑏 2 − 4𝑎𝑐 = 32 − 4(1)(−4) = 9 + 16 = 25 > 0
∴ The line cuts the curve at two different points
...
79

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CHAPTER 2

Quadratic functions
1

2) Prove that the line 𝑦 = 𝑥 − 1 is a tangent to the curve 𝑦 = 4 𝑥 2
...
111

2

,

𝑚>2

π

CHAPTER 2

Quadratic functions

6) i) 8𝑥 − 𝑥 2 ≡ 𝑎 − (𝑥 + 𝑏)2
≡ 𝑎 − (𝑥 2 + 2𝑏𝑥 + 𝑏 2 )
≡ 𝑎 − 𝑥 2 − 2𝑏𝑥 − 𝑏 2
by equating coefficients:
𝑎 − 𝑏2 = 0

−2𝑏 = 8

𝑎 − (−4)2 = 0

𝑏 = −4

𝑎 − 16 = 0

∴ 𝑎 = 16

∴ 8𝑥 − 𝑥 2 ≡ 16 − (𝑥 − 4)2
-------------------------------------------------------------------------------------------------------------------------

ii) Stationary point of the curve (vertex) is (4, 16)
-------------------------------------------------------------------------------------------------------------------------

iii) 𝑦 ≥ −20
8𝑥 − 𝑥 2 ≥ −20
∴ 𝑥 2 − 8𝑥 − 20 ≤ 0
∴ 𝑥 2 − 8𝑥 − 20 = 0
𝑥 = −2

,

-2

𝑥 = 10

10

∴ −2 ≤ 𝑥 ≤ 10

-------------------------------------------------------------------------------------------------------------------------

7) 𝑦 = 4𝑥 + 𝑘 → (1)

𝑦 = 𝑥 2 → (2)

from (2) into (1):
𝑥 2 = 4𝑥 + 𝑘
𝑥 2 − 4𝑥 − 𝑘 = 0
For the line not to intersect the curve ∆< 0
𝑎=1

𝑏 = −4
𝑐 = −𝑘
𝑏 − 4𝑎𝑐 < 0
16 − 4(1)(−𝑘) < 0
16 + 4𝑘 < 0 → 4𝑘 < −16
2

P
...
113

3 2

9

𝑓(𝑥) = (𝑥 − 2) − 4

π

CHAPTER 2

10) 𝑦 = 𝑥 2 − 4𝑥 + 7 → (1)

Quadratic functions

𝑦 + 3𝑥 = 9 → (2)

from (1) into (2):
𝑥 2 − 4𝑥 + 7 + 3𝑥 = 9
𝑥2 − 𝑥 − 2 = 0
𝑥 = −1

,

𝑥=2

𝑦 = 12

,

𝑦=3

∴Points of intersection of the line and the curve are (-1, 12) and (2, 3)
−1+2

∴ Mid point M = (

2

,

12+3

1

1

2

2

2

) = ( ,7 )

-------------------------------------------------------------------------------------------------------------------------

11) Method (1):
𝑦 = 2𝑥 + 𝑐 → (1)

𝑦 2 = 4𝑥 → (2)

from (1) into (2):
(2𝑥 + 𝑐)2 = 4𝑥
4𝑥 2 + 4𝑐𝑥 + 𝑐 2 = 4𝑥
4𝑥 2 + 4𝑐𝑥 − 4𝑥 + 𝑐 2 = 0
For the line to be a tangent to the curve ∆= 0
𝑎=4

,

𝑏 = 4𝑐 − 4 ,
𝑐 = 𝑐2
𝑏 2 − 4𝑎𝑐 = 0

(4𝑐 − 4)2 − 4(4)(𝑐 2 ) = 0
16𝑐 2 − 32𝑐 + 16 − 16𝑐 2 = 0
∴ −32𝑐 = −16

→

1

𝑐=2

P
...
115

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CHAPTER 2

ii) 𝑦 2 + 2𝑥 = 13 → (1)
𝑓𝑟𝑜𝑚 (2):

Quadratic functions

2𝑦 + 𝑥 = 𝑘 → (2)

𝑥 = 𝑘 − 2𝑦
𝑦 2 + 2(𝑘 − 2𝑦) = 13

𝑖𝑛𝑡𝑜(1):

𝑦 2 + 2𝑘 − 4𝑦 − 13 = 0
∴ 𝑦 2 − 4𝑦 + 2𝑘 − 13 = 0
For the line to be a tangent to the curve ∆= 0
𝑎=1

𝑏 = −4

𝑐 = 2𝑘 − 13

∴ 𝑏 2 − 4𝑎𝑐 = 0
16 − 4(1)(2𝑘 − 13) = 0
16 − 8𝑘 + 52 = 0
68 = 8𝑘 → 𝑘 =

68
8

= 8
...
116

π

CHAPTER 2

Quadratic functions

14) i) Method (1):
𝑥 = −3 & 𝑥 = 5 are roots of the equation, the equation could be written in the form
(𝑥 + 3)(𝑥 − 5) = 0
𝑥 2 − 2𝑥 − 15 = 0 ↔ 𝑥 2 + 𝑝𝑥 + 𝑞 = 0
∴ 𝑝 = −2

,

𝑞 = −15

Method (2): (Longer Method)
𝑥 2 + 𝑝𝑥 + 𝑞 = 0
At 𝒙 = −𝟑 (−3)2 + 𝑝(−3) + 𝑞 = 0

𝑞 − 3𝑝 + 𝑞 = 0

∴ −3𝑝 + 𝑞 = −𝑞

→ (1)

(5)2 + 𝑝(5) + 𝑞 = 0

At 𝒙 = 𝟓

25 + 5𝑝 + 𝑞 = 0

5𝑝 + 𝑞 = −25

→ (2)

Now solve (1) & (2)simultaneously:
−3𝑝 + 𝑞 = −𝑞
−5𝑝 − 𝑞 = 25
−8𝑝 = 16

𝑝 = −2

,

𝑞 = −15

-------------------------------------------------------------------------------------------------------------------------

ii) 𝑥 2 + 𝑝𝑥 + 𝑞 + 𝑟 = 0
𝑥 2 − 2𝑥 − 15 + 𝑟 = 0
𝑎=1

𝑏 = −2

𝑐 = 𝑟 − 15

For the equation to have equal roots ∆= 0
𝑏 2 − 4𝑎𝑐 = 0
(−2)2 − 4(1)(𝑟 − 15) = 0
4 − 4𝑟 + 60 = 0
64 = 4𝑟

→

𝑟 = 16
P
...
75 𝑢𝑛𝑖𝑡𝑠
4
-------------------------------------------------------------------------------------------------------------------------

16) 𝑦 = 𝑘𝑥 2 + 1 → (1) ,

𝑦 = 𝑘𝑥 → (2)

From (1) into (2):
𝑘𝑥 2 + 1 = 𝑘𝑥
𝑘𝑥 2 − 𝑘𝑥 + 1 = 0
For the curve and the line to have no common points
...
118

4

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CHAPTER 2

Quadratic functions

17) i) When 𝒌 = 𝟏𝟏
𝑥𝑦 = 12 → (1) ,

2𝑥 + 𝑦 = 11 → (2)

From (2): 𝑦 = 11 − 2𝑥
Into (1): 𝑥(11 − 2𝑥) = 12
11𝑥 − 2𝑥 2 = 12
∴ 2𝑥 2 − 11𝑥 + 12 = 0
3

𝑥=2

,

𝑦=8

𝑥=4
𝑦=3
3

∴Coordinates o

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