Page | 30 3.1 INTRODUCTION
The general form of a quadratic function is
f
(
x
)
=
ax
2+
bx
+
c
; a, b, c are constants and a ≠ 0Similarities and differences between Quadratic Functions and Quadratic Equations
Quadratic Functions Quadratic Equations
General form is
f
(
x
)
=
ax
2+
bx
+
c
General form isax
2+
bx
+
c
=
0
a, b, c are constants a, b, c are constants
x
is independent variable whilef
(x
)
is dependent variablex
is unknownThe highest power of the
x
is 2 The highest power of thex
is 2Involves one variable only Involves one variable only
When
f
(
x
)
=
0
, quadratic functions quadratic equations3.1.1 Recognizing a quadratic function (i)
(
)
=
2
2+
3
−
4
x
x
x
f
is a quadratic function because the highest power of thex
is 2 and Involves onevariable only.
(ii)
f
(
x
)
=
3
x
+
2
is not a quadratic function because the highest power of thex
is 1, not 2.(iii)
f
(
x
)
=
3
x
2+
4
y
−
2
is a not quadratic function because it involves two variables.* Note: The proper way to denote a quadratic function is
c
bx
ax
x
f
:
→
2+
+
.
f
(
x
)
=
ax
2+
bx
+
c
is actually the value (or image) f for a given value of x.
3.2 MINIMUM VALUE AND MAXIMUM VALUE OF A QUADRATIC FUNCTION 1. Do you know that a non-zero number when squared will be always positive. 2. So the minimum value of squared number is 0.
So, what is the minimum value when we find the square of a number? Minimum value of x2 is 0.This is obtained when x = 0.
For example
f
(
x
)
=
(
x
−
1
)
2−
4
.The minimum value for
f
(x
)
is −4when x−1 =0
x=1.
The minimum value of a squared number is zero. When
(...)
2=
0
, we would knowPage | 31 The minimum value of x2+ 3 is 0 + 3 = 3
the minimum value of x2 – 8 is 0 – 8 = – 8 the minimum value of x2+ 100 is 0 + 100 = 100 The minimum value of x2is 0,
It means 2
≥
0
x
So,0
2≤
−
x
Hence the maximum value of – x2 is 0 the maximum value of -x2+ 5 is -0 + 5=5 the maximum value of –x2– 3 is -0 + 3 =– 3. For example
(
)
=
−
(
−
1
)
2−
4
x
x
f
.The maximum value for
f
(x
)
is −4when x−1 =0
x=1.
We can state the minimum value of a quadratic function in the form f(x) = a (x + p)2 + q , a > 0. From that,
We can also state the Maximum Value of a Quadratic Function in the form f(x) = a (x + p)2 + q , a < 0
Example 1:
Express
f
(
x
)
=
x
2−
3
x
−
4
in form f(x) = a (x + p)2+ q. Hence, find the coordinates of the turning point of the graph.
4
3
)
(
=
2−
−
x
x
x
f
) 4 2 3 ( ) 2 3 ( 3 2 2 2 − + − − − − =x x 4 4 9 ) 2 3 ( − 2 − − = x 4 25 ) 2 3 ( − 2 − = xThe minimum value for
f
(x
)
is4 25 − when 0 2 3 = − x ( 2 3 , 4 25 − ) 2 3 = x
When
(...)
2=
0
, we would know theminimum or maximum value for the function. The minimum value of a squared number is zero.
This is in the form a (x + p)2 + q. The value of a is 1.
The coordinates of the turning point means the coordinates of the minimum of maximum point for the function. For this question, it is minimum point because a > 0
f(x) is the value for y. This is because f(x) is the y-axis.
The minimum value of a squared number is zero. When
(...)
2=
0
, we would knowthe minimum value for the function.
If the value of a of a quadratic function is greater than zero, then the function has a minimum point.
If the value of a of a quadratic function is less than zero, then the function has a maximum point.
Page | 32 Example 2:
Express
f
(
x
)
=
−
2
x
2+
8
x
−
14
in form f(x) = a (x + p)2 + q. Hence, find the coordinates of the turningpoint of the graph.
14
8
2
)
(
x
=
−
x
2+
x
−
f
=
−
2
[
x
2−
4
x
+
(
−
2
)
2−
(
−
2
)
2+
7
]
=
−
2
[(
x
−
2
)
2−
4
+
7
]
=
−
2
[(
x
−
2
)
2+
3
]
=
−
2
(
x
−
2
)
2−
6
The maximum value for
f
(
x
)
=
−
6
when x−2 =0 ( 2, -6)
x=2
EXERCISE 3.2
1. State the minimum value of the following function and the corresponding value of x (a)
(
)
=
(
+
1
)
2−
8
x
x
f
(b)(
)
=
2
2+
5
−
1
x
x
x
g
2. State the maximum value of the following function and the corresponding value of x (a)
(
)
=
−
2
2+
4
+
3
x
x
x
f
(b)(
)
=
−
2+
2
−
5
x
x
x
g
3. Given that the minimum value of the quadratic function
f
(
x
)
=
2
x
2+
px
+
q
is 1 when x=−3, findthe value of p and of q.
This is in the form a (x + p)2 + q. The value of a is -2 which is less than 0.
The coordinates of the turning point means the coordinates of the minimum of maximum point for the function. For this question, it is maximum point because a < 0
Page | 33 3.3 GRAPH OF QUADRATIC FUNCTIONS
1. To express quadratic function f(x) = ax2 + bx + c in the form a(x+ p)2 + q, we have to use completing the square that we have learned in chapter 2.
2. From the minimum point of a graph, we can know the equation of the axis of symmetry for the graph. The equation is based on the value of x of the coordinates of the turning point.
3. If the turning point is (2,3), hence the equation of axis of symmetry is x=2.
3.3.1 Basic Equation of quadratic function and its graph Example 1: 2
)
(
x
x
f
=
x
-2
-1
0
1
2
f(x)
4
1
0
1
4
Example 2: 2)
(
x
x
f
=
−
x
-2
-1
0
1
2
f(x)
-4
-1
0
-1
-4
a = 1, b = 0 and c = 0 a =−
1, b = 0 and c = 0By using the information given, sketch the graph
By using the information given, sketch the graph.
Page | 34 Example 3:
Sketch the graph of
(
)
=
2−
4
+
3
x
x
x
f
3
4
)
(
=
2−
+
x
x
x
f
3
)
2
(
)
2
(
4
)
(
=
2−
+
−
2−
−
2+
x
x
x
f
3
4
)
2
(
)
(
x
=
x
−
2−
+
f
1
)
2
(
)
(
=
−
2−
x
x
f
The minimum value of
f
(
x
)
=
3
when x−2 =0 x=2 Minimum point is (2,3). 0 = x ,
3
)
0
(
4
)
0
(
)
0
(
=
2−
+
f
=3 The point is (0,3) Example 4:Express the function
f
(
x
)
=
1
+
4
x
−
2
x
2in the form of a(x+ p)2 + q. Hence(a) state the minimum or maximum point. (b) the equation of axis 0f symmetry
) 2 1 2 ( 2 ) (x =− x2 − x− f ] 2 1 ) 1 ( ) 1 ( 2 [ 2 2 − + − 2 − − 2 − − = x x a = 1, b =
−
4 and c = 3 a =−
2, b = 4 and c = 1By using the information given, sketch the graph.
Page | 35 ] 2 1 1 ) 1 [( 2 − 2 − − − = x ] 2 1 1 ) 1 [( 2 − 2 − − − = x ] 2 1 1 ) 1 [( 2 − 2 − − − = x ] 2 3 ) 1 [( 2 − 2 − − = x
3
)
1
[(
2
−
2+
−
=
x
The maximum value of
f
(
x
)
=
3
when x−1 =0
x=1
So, the maximum point is (1, 3).
The equation of axis of symmetry is x=1.
EXERCISE 3.3
1. Express the quadratic function
y
=
x
2−
12
x
+
13
in the form ofy
=
(
x
+
b
)
2+
c
. State the minimumpoint and the y-intercept. Hence, sketch the graph of the function y.
2. Express the quadratic function
f
(
x
)
=
2
−
x
2−
4
x
in the form off
(
x
)
=
q
−
(
x
+
p
)
2. Determinethe maximum point and the y-intercept. Hence, sketch the graph of the function f(x).
3. Given that the curve
y
=
m
−
(
2
x
+
n
)
2has a maximum point (1, 4), find the values of m and of n.Hence, sketch the curve.
4. For the quadratic function
g x
( )
=
2
x
2−
3
x
+
2
, find the equation of the axis of symmetry.3.4 RELATIONSHIP BETWEEN “b2 – 4ac” AND THE POSITION OF GRAPH f(x)= y 1-If
b
2−
4
ac
>
0
,the graph cuts x-axis at two different pointsf(x) f(x)
x x a < 0, so it is maximum value
Page | 36
f(x) f(x)
x x
1-If
b
2−
4
ac
<
0
,the graph does not touch the x-axisf(x) f(x)
x
x
3.4 QUADRATIC INEQUALITIES
In Form Three, we have learned about linear inequalities. For example when x−1 >0, x>1.
Example:
Given x−1 >0. Find the range of values of x. 0 1 ≥ − x 1 ≥ x
•
1From the line graph, we know the range of values of x.
a > 0 a < 0
a > 0 a < 0
Curve lies below the x- axis because f(x) is always negative Curve lies above the x-axis
Page | 37 To solve the quadratic inequality for example
(
x
+
1
)(
x
−
3
)
>
0
, we use the graph sketching method.If f(x) > 0, the range of values of x: f(x)
x
If f(x) < 0, the range of values of x: f(x)
x
Example 1:
Find the range of x if
4
m
2−
36
>
0
Solution:
0
36
4
2−
>
m
0
9
2−
>
m
Letm
2−
9
=
0
m
2=
9
m=±3 y -3 3 m If 2−
9
>
0
Page | 38 Example 2:
Find the range of the values of k if the equation 2
+
2
−
5
−
6
=
0
k
kx
x
has two real roots.Solution:
two real roots also means two different roots:
b
2−
4
ac
>
0
0
)
6
5
)(
1
(
4
)
2
(
k
2−
−
k
−
>
0
)
6
5
(
4
4
k
2−
−
k
−
>
0
24
20
4
k
2+
k
+
>
0
6
5
2+
+
>
k
k
Letk
2+
5
k
+
6
=
0
(
k
+
2
)(
k
+
3
)
=
0
0 2 = + k or k+3 =0 k =−2 or k =−3 y -3 -2 kIf
k
2+
5
k
+
6
>
0
, the range of values of k is k<−3, k >−2.Example 3:
Find the range of the values of x for which if the equation
x
(
x
+
2
)
≤
15
Solution:
15
)
2
(
x
+
≤
x
0
15
2
2+
−
≤
x
x
Letx
2+
2
x
−
15
=
0
(
x
+
5
)(
x
−
3
)
=
0
0 5 = + x or x−3 =0 k =−5 or x=3 6 5 2 + + =k k yPage | 39 y
-5 -2 x
If
x
2+
2
x
−
15
≤
0
, the range of values of x is −5≤ x≤3.EXERCISE 3.3
1. Find the range of values of x for each of the following inequalities. (a)
x
(
x
+
1
)
<
20
(b)
x
2−
3
x
<
2
x
(c)
x
(
7
−
5
x
)
>
2
2. Given that
y
=
2
x
2+
3
x
, find the range of values of x for which y > 9. 3. Find the range of values of k given that 2
+
(
−
1
)
+
2
+
=
0
k
x
k
x
has two different roots.4. If the straight line
y =
k
, where k is a constant, does not intersect the curvey
=
2
x
2−
6
x
+
5
, showthat
2 1 < k .
CHAPTER REVIEW EXERCISE
1. The graph of the function
f x
( )
=
p
−
2(
q
−
x
)
2 has a maximum point at1
,
5
2
2
−
. Find the valueof p and q.
2. Given the quadratic function
y
=
−
2
x
2+
6
x
−
3
, express f(x) in the formp
(
x
+
q
)
2+
r
, where p, qand r are constants. Hence, sketch the graph of the function y.
3 Given that the straight line
y
=
2
x
−
n
cuts the curvey
=
x
2+
nx
−
2
at two points, find the range ofthe values of n.
4. The quadratic equation
px
2−
4
x
=
p
−
5
, wherep
≠
0
, has real and different roots. Find the rangeof values of p.
5. By expressing the function
(
)
=
3
2−
6
+
5
x
x
x
f
in the forma
(
x
−
p
)
2+
q
, or otherwise, find theminimum value of f(x).
6. Given that the quadratic function
f
(
x
)
=
px
2+
4
x
−
q
, where p and q are constants .Given that the curvey =
f
(x
)
has a maximum point (-1, 5).State the values of p and q.15 2
2 + −
= x x
Page | 40 the curve
y =
f
(x
)
.(a) Write the equation of the axis of symmetry of the curve.
(b) Express f(x) in the form
(
x
+
p
)
2+
q
, where p and q are constants.8. Diagram below shows the graph of the function
y
=
(
p
−
1
)
x
2+
2
x
+
q
.
(a) State the value of q.
(b) Find the range of values of p.
9. Find the range of values of k given that the straight line
