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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)

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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)

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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
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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

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CHAPTER 2

Quadratic functions

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

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


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

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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

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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

)

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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

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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
...
105

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CHAPTER 2

Quadratic functions

33) If 𝑓(𝑥) = 2𝑥 2 − 5𝑥 + 2
i) Express 𝑓(𝑥) in the form 𝐴(𝑥 + 𝐵)2 + 𝐶
ii) State the max/min value of 𝑓(𝑥) and the value of 𝑥 at which it occurs

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34) If 𝑔(𝑥) = 3 − 7𝑥 − 3𝑥 2
i) Express 𝑔(𝑥) in the form 𝐴 − 𝐵(𝑥 + 𝐶)2
ii) State the max/min value of 𝑔(𝑥) and the value of 𝑥 at which it occurs
...
106

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CHAPTER 2

Quadratic functions

35) If 𝑦 = 7 − 10𝑥 − 𝑥 2
i) Express 7 − 10𝑥 − 𝑥 2 in the form 𝑎 − (𝑥 − 𝑏)2
ii) State the max/min value of 𝑦 and the value of 𝑥 at which it occurs
iii) Write down the equation of the line of symmetry of the curve 𝑦 = 7 − 10𝑥 − 𝑥 2

P
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108

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CHAPTER 2

Quadratic functions

Model Answer
1) i) 2𝑥 2 + 8𝑥 − 10 ≡ 𝑎(𝑥 + 𝑏)2 + 𝑐
≡ 𝑎(𝑥 2 + 2𝑏𝑥 + 𝑏 2 ) + 𝑐
≡ 𝑎𝑥 2 + 2𝑎𝑏𝑥 + 𝑎𝑏 2 + 𝑐
by equating coefficients:
∴𝑎=2

2𝑎𝑏 = 8

𝑎𝑏 2 + 𝑐 = −10

4𝑏 = 8

2(2)2 + 𝑐 = −10

𝑏=2

𝑐 = −18

∴ 2𝑥 2 + 8𝑥 − 10 ≡ 2(𝑥 + 2)2 − 18
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ii) for the curve 𝑦 = 2𝑥 2 + 8𝑥 − 10, from the completed square form, vertex (-2,-18)
∴ 𝑙𝑒𝑎𝑠𝑡 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑦 𝑖𝑠 − 18 𝑎𝑛𝑑 𝑜𝑐𝑐𝑢𝑟𝑠 𝑎𝑡 𝑥 = −2
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iii) 𝑦 ≥ 14
2𝑥 2 + 8𝑥 − 10 ≥ 14
2𝑥 2 + 8𝑥 − 24 ≥ 0

2

-6

∴ 2𝑥 2 + 8𝑥 − 24 = 0
∴ 𝑥 ≤ −6

𝑥 = −6 , 𝑥 = 2

P
...
110

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CHAPTER 2

Quadratic functions

2𝑥 2 − 8𝑥 + 14 = 𝑥 + 𝑘
∴ 2𝑥 2 − 9𝑥 + 14 − 𝑘 = 0
for the line not to intersect the curve ∆< 0
𝑎=2

𝑏 = −9

𝑐 = 14 − 𝑘

𝑏 2 − 4𝑎𝑐 < 0
81−4(2)(14 − 𝑘) < 0
81−8(14 − 𝑘) < 0
81 − 112 + 8𝑘 < 0
−31 + 8𝑘 < 0
8𝑘 < 31

∴𝑘<

31
8

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𝑦 = 3𝑥 2 − 4𝑥 + 7 → (2)

5) 𝑦 = 𝑚𝑥 + 4 → (1)
𝑓𝑟𝑜𝑚 (2) 𝑖𝑛𝑡𝑜 (1):
3𝑥 2 − 4𝑥 + 7 = 𝑚𝑥 + 4
3𝑥 2 − 4𝑥 − 𝑚𝑥 + 3 = 0

For the line to intersect the curve at two distinct points ∆> 0
𝑎=3

𝑏 = −4 − 𝑚

𝑐=3

(−4 − 𝑚)2 − 4(3)(3) > 0
16 + 8𝑚 + 𝑚2 − 36 > 0
𝑚2 + 8𝑚 − 20 > 0
-10

∴ 𝑚2 + 8𝑚 − 20 = 0
𝑚 = −10

,

∴ 𝑚 ≤ −10

𝑚=2

P
...
112

𝑘 < −4

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CHAPTER 2

8) 𝑦 = 𝑘𝑥 − 4 → (1)

𝑦 = 𝑥 2 − 2𝑥

Quadratic functions

→ (2)

from (2) into (1):
𝑥 2 − 2𝑥 = 𝑘𝑥 − 4
𝑥 2 − 2𝑥 − 𝑘𝑥 + 4 = 0
For the line to intersect the curve at two distinct points, ∆> 0
𝑎=1

𝑏 = −2 − 𝑘

𝑐=4

𝑏 2 − 4𝑎𝑐 > 0
(−2 − 𝑘 )2 − 4(1)(4) > 0
4 + 4𝑘 + 𝑘 2 − 16 > 0
𝑘 2 + 4𝑘 − 12 > 0

-6

∴ 𝑘 2 + 4𝑘 − 12 = 0

∴ 𝑘 < −6

𝑘 = −6

,

2
𝑘>2
,

𝑘=2

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9) 𝑓(𝑥) = 𝑥 2 − 3𝑥
i) 𝑓(𝑥) > 4
𝑥 2 − 3𝑥 > 4
𝑥 2 − 3𝑥 − 4 > 0

-1

4

∴ 𝑥 2 − 3𝑥 − 4 = 0
𝑥 = −1

,

∴ 𝑥 < −1

𝑥=4

,

𝑥>4

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ii) 𝑥 2 − 3𝑥 ≡ (𝑥 − 𝑎)2 − 𝑏 ≡ 𝑥 2 − 2𝑎𝑥 + 𝑎2 − 𝑏
by equating coefficients:
−2𝑎 = −3
3

𝑎=2

𝑎2 − 𝑏 = 0
9
4

−𝑏 = 0
9

∴𝑏=4
P
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114

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CHAPTER 2

Quadratic functions

Method(2):
𝑦 2 = 4𝑥 → (2)

𝑦 = 2𝑥 + 𝑐 → (1)

𝑓𝑟𝑜𝑚 (2): 𝑦 = √4𝑥 = √4 𝑥 √𝑥 = 2√𝑥
𝑖𝑛𝑡𝑜 (1): 2√𝑥 = 2𝑥 + 𝑐
∴ 2𝑥 − 2√𝑥 + 𝑐 = 0
𝑙𝑒𝑡 𝑚 = √𝑥

∴ 2𝑚2 − 2𝑚 + 𝑐 = 0

For the line to be tangent to the curve ∆= 0
𝑎=2

𝑏 = −2

𝑐=𝑐

4 − 4(2)(𝑐) = 0
4 − 8𝑐 = 0

→

1

𝑐=2

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