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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

π

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
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P
ii) Stationary point (2, 2)

type of stationary point is minimum

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6) put 𝑥 2 + 4𝑥 + 3 in the form (𝑥 + 𝑎)2 + 𝑏 stating the values of a and b
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Solution
−𝑥 2 + 2𝑥 + 2 = −1(𝑥 2 − 2𝑥 − 2)
= −1[(𝑥 − 1)2 − 1 − 2] = −1[(𝑥 − 1)2 − 3]
= −(𝑥 − 1)2 + 3 = 3 − (𝑥 − 1)2
P
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Solution
2𝑥 2 − 16𝑥 + 37 = 2(𝑥 2 − 8𝑥) + 37
= 2[(𝑥 − 4)2 − 16] + 37
= 2(𝑥 − 4)2 − 32 + 37
= 2(𝑥 − 4)2 + 5

A = 2, B = -4 and C = 5

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9) Express 3 + 4𝑥 − 2𝑥 2 in the form ℎ − 𝑘(𝑥 + 𝑝)2 , stating the values of h, k and p

Solution
−2𝑥 2 + 4𝑥 + 3 = −2(𝑥 2 − 2𝑥) + 3
= −2[(𝑥 − 1)2 − 1] + 3
= −2(𝑥 − 1)2 + 2 + 3 = 5 − 2(𝑥 − 1)2
ℎ = 5 , 𝑘 = 2 𝑎𝑛𝑑 𝑝 = −1
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10) Express 𝑥 2 − 3𝑥 in the form (𝑥 − 𝑎)2 +𝑏 stating the values of a and b

Solution
3 2

9

𝑥 2 − 3𝑥 = (𝑥 − 2) − 4

3

9

𝑎 = 2, 𝑏 = −4

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11) Express 4𝑥 − 2𝑥 2 in the form𝐴 − 𝐵(𝑥 − 𝐶)2 , stating the values of A, B and C
...
63

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CHAPTER 2

Quadratic functions

12) Express 8𝑥 − 𝑥 2 in the form 𝑎 − (𝑥 + 𝑏)2 , stating the values of a and b

Solution
−𝑥 2 + 8𝑥 = −1(𝑥 2 − 8𝑥) = −1[(𝑥 − 4)2 − 16]
= 16 − (𝑥 − 4)2

𝑎 = 16, 𝑏 = −4

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P
...
Find the value of a and b
...
Hence, find the coordinates of the vertex and state whether it is minimum or
maximum
...
65

π

CHAPTER 2

Quadratic functions

c) 𝑥 2 + 3𝑥 − 7

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d) 5 − 6𝑥 + 𝑥 2

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e) 2𝑥 2 + 12𝑥 − 5

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f) 3𝑥 2 − 12𝑥 + 3

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a) 𝑦 = (𝑥 − 2)2 + 3

P
...
68

π

CHAPTER 2

Quadratic functions

Graph of quadratic function
How to sketch a quadratic function:
→ Find the roots of the function
...

→ Find the coordinates of the vertex of the function
...


Example:
Sketch each of the following function:
a) 𝑓(𝑥) = 𝑥 2 − 6𝑥 + 8

Solution
Roots ⇒ 𝑥 2 − 6𝑥 + 8 = 0 ⇒ (𝑥 − 4)(𝑥 − 2) = 0 ⇒ 𝑥 = 2 , 𝑥 = 4
𝑦-intercept ⇒ 𝑥 = 0 ⇒ 𝑦 = 8 ⇒ (0 , 8)
Vertex ⇒ 𝑥 =

−𝑏
2𝑎

=

−−6
2×1

=3

,

𝑦 = (3)2 − 6(3) + 8 = −1 ⇒ (3 , − 1)

P
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7 , 𝑥 = 0
...
70

π

CHAPTER 2

Quadratic functions

 Test yourself:
Sketch the graph of the following Function:
1) 𝑓(𝑥) = 𝑥 2 + 2𝑥 − 3

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2) 𝑓(𝑥) = 𝑥 2 − 2𝑥

P
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72

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CHAPTER 2

Quadratic functions

Quadratic Inequalities
How to solve quadratic inequality:
→ Change the inequality to an equation
...

→ Sketch the curve and locate the roots
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Examples:
1) Solve each of the following inequality;
a) 𝑥 2 − 3𝑥 + 2 ≥ 0

Solution
𝑥 2 − 3𝑥 + 2 = 0 ⇒ (𝑥 − 2)(𝑥 − 1) = 0 ⇒ 𝑥 = 2
∴𝑥≤1

and

1

,

2

x

𝑥=1

𝑥≥2

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b) 𝑥 2 − 5𝑥 + 6 ≤ 0
2

Solution
𝑥 2 − 5𝑥 + 6 = 0 ⇒ (𝑥 − 2)(𝑥 − 3) = 0 ⇒ 𝑥 = 2
∴2≤𝑥≤3

P
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Solution
𝑎=𝑘

,

𝑏=𝑘

,

𝑐=2

𝛥 = 𝑏 2 − 4𝑎𝑐 < 0 ⇒ 𝑘 2 − 4(𝑘)(2) < 0 ⇒ 𝑘 2 − 8𝑘 < 0
𝑘 2 − 8𝑘 = 0 ⇒ 𝑘(𝑘 − 8) = 0 ⇒ 𝑘 = 0

,

∴0<𝑘<8

P
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Solution
𝑎=𝑘

,

𝑏=3

,

𝑐=𝑘

−𝟑
𝟐

𝛥 = 𝑏 2 − 4𝑎𝑐 > 0 ⇒ 32 − 4(𝑘)(𝑘) > 0 ⇒ 9 − 4𝑘 2 > 0
9 − 4𝑘 2 = 0 ⇒ (3 − 2𝑘)(3 + 2𝑘) = 0 ⇒ 𝑘 =
∴

−3
2

−3
2

𝟑
𝟐

k

3

,

𝑘=2

3

<𝑘<2

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4) Find the set of values of k for which the equation 𝑥 2 + 𝑘𝑥 + 2𝑘 − 3 = 0 has real
roots
...
75

,

𝑘=6

k

π

CHAPTER 2

Quadratic functions

 Test yourself:
1) Solve each of the following inequality:
a) 𝑥 2 − 7𝑥 + 12 < 0

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b) 2𝑥 2 − 18𝑥 ≥ 0

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c) 2𝑥 2 − 5𝑥 + 3 > 0

P
...


P
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78

,

𝑥 = −1

π

CHAPTER 2

Quadratic functions

Relation between a line and a curve:

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

2) The line is tangent to
the curve
x

3) The line neither cuts
nor touches the curve

How to get the relation between a line and a curve:
We use 𝛥 = 𝑏 2 − 4𝑎𝑐 for the equation formed when we substitute the line in the
curve
...


2) 𝛥 = 0

The line is a tangent to the curve
...


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Solution
From the line ⇒ 𝑦 = 𝑥 − 1
1

Into the curve ⇒ 𝑥For 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 =

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