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QUADRATIC FUNCTIONS The word quadratic comes from the Latin word Quadratus which means square
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Page 1: Quadratic

QUADRATIC FUNCTIONS

The word quadratic comes from the Latin word Quadratus which means square

Page 2: Quadratic

Chapter Objectives

Understand the concept of quadratic functions and their graphs.

Find maximum and minimum values of quadratic Functions.

Sketch graphs of quadratic functions

Understand and use the concept of quadratic Inequalities.

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Recognising quadratic functions

f : x ax2 + bx + c

f (x) =

a , b and c are constants

a 0

The highest power of x is 2

x2 x

c + + a b

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Determine whether each of the following is aquadratic function.

f(x) = 2x 2

h(x) = 4x - 3x + 1 2

g(x) = x + 3x 2

k(x) = 5x + 7

g(x) = 3x + 1 2

x

h(x) = (x + 3) + 4 2

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f(x) = ax + bx + c 2quadratic function

a 0

2 ax + bx + c quadratic expression

2 ax + bx + c = 3 quadratic equation

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Plotting the graphs of quadratic functions

Based on given tabulated values

By constructing the table of values

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xf(x)

xf(x)

-4 -3 0-2 -1 1 2 3 4-7 980 5

The table below shows some values of x and the corresponding values of f(x) of the function f(x) = 9 – x2

8 5 0 -7

Plot the graph of the function

Example

0-2 -1 1 2 5

Select suitable scales on both axes and subsequently plot theGraph.

Given the quadratic function f (x) = x2 – 2x – 4. Plot the graph of the function for -3 ≤ x ≤ 5.

We first construct the table of values of the function.

3 4

-5-44 -1 -4 -1 4 1111

-3

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Shapes of graphs of quadratic functions

a > 0 a < 0

axes of symmetry

Minimum point

Maximum point

f(x) = ax2 + bx + c

If a > 0 , then the graph of the function is a parabola with a min pt.If a < 0 , then the graph of the function is a parabola with a max pt.

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ExampleDescribe the shape of the graph of each of the following quadratic functions.

Solution

(a) f (x) = - 3x2 – 4x + 5 (b) g (x) = 10x2 + 6x + 3

(a) Since a = - 3 < 0 , the graph of the function is a parabola with a maximum pt.

(b) Since a = 10 > 0 , the graph of the function is a parabola with a minimum pt.

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Relating the position of the graph of a quadratic function f(x) = ax2 + bx + c with the for types of roots f(x) = 0

m n

Referring to the graph,

When f(x) = 0 ,x = m and x = n

m and n are the roots of the equation

m and n are also the values of x where the graph intersects the x – axis.

Therefore , the roots of f(x) = ax2 + bx + c are the points where the graph of f(x) intersects the x – axis.

Values of xwhen f(x) = 0

Page 11: Quadratic

In this respect , we have three cases :

(I) If f (x) = ax2 + bx + c has two distinct (different) roots , meaning b2 – 4ac > 0 , then the graph of the function f (x) intersects at two distinct points.

x x

a > 0 a < 0

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Example 1

f (x) = 2x2 –x -10

Points of intersection with the x – axis.f(x) = 0

2x2 –x -10 = 0

(2x – 5)( x + 2) = 0

X = 5/2 , -2

When f(x) = 0

b2 – 4ac = (-1)2 – 4(2)(-10)

= 81

,b2 – 4ac > 0

a > 0

Hence

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Example 2

f (x) = -x2 + 3x +10

Points of intersection with the x – axis.f(x) = 0

-x2 + 3x +10 = 0

(5 - x)( x + 2) = 0

X = 5 , -2

When f(x) = 0

b2 – 4ac = 32 – 4(-1)(10)

= 49

,b2 – 4ac > 0

a < 0

Hence

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In this respect , we have three cases :

(II) If f (x) = ax2 + bx + c has two real and equal roots , meaning b2 – 4ac = 0 , then the graph of the function f (x) intersects at only one point.

x

x

a > 0 a < 0

Page 15: Quadratic

Example 3

f (x) = x2 +6x + 9

Point of intersection with the x – axis.f(x) = 0

x2 + 6x + 9 = 0

(x + 3)( x + 3) = 0

x = -3

When f(x) = 0

b2 – 4ac = 62 – 4(1)(9)

= 0

,b2 – 4ac = 0

a > 0

Hence

Page 16: Quadratic

In this respect , we have three cases :

(III) If f (x) = ax2 + bx + c does not have any real roots , meaning b2 – 4ac < 0 , then the graph of the function f (x) does not intersect the x - axis.

x

x

a > 0 a < 0

Page 17: Quadratic

Example 4

f (x) = 2x2 + 5x + 7

There is NO point of intersection with the x – axis.

When f(x) = 0

b2 – 4ac = (5)2 – 4(2)(7)

= - 31

,b2 – 4ac < 0

a > 0

Hence

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Summary

(I) If b 2 – 4ac > 0 x x

a < 0

(II) If b 2 – 4ac = 0 x

x

a < 0

a > 0

a > 0

(III) If b 2 – 4ac < 0

x

x

a < 0 a > 0

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