Chapter 3 quadratc functions

Additional Mathematics Module Form 4
Chapter 3- Quadratic Functions SMK Agama Arau, Perlis
Page | 30
CHAPTER 3- QUADRATIC FUNCTIONS
3.1 INTRODUCTION
The general form of a quadratic function is cbxaxxf ++= 2
)( ; a, b, c are constants and a ≠ 0
Similarities and differences between Quadratic Functions and Quadratic Equations
Quadratic Functions Quadratic Equations
General form is cbxaxxf ++= 2
)( General form is 02
=++ cbxax
a, b, c are constants a, b, c are constants
x is independent variable while )(xf is
dependent variable
x is unknown
The highest power of the x is 2 The highest power of the x is 2
Involves one variable only Involves one variable only
When 0)( =xf , quadratic functions quadratic equations
3.1.1 Recognizing a quadratic function
(i) 432)( 2
−+= xxxf is a quadratic function because the highest power of the x is 2 and Involves one
variable only.
(ii) 23)( += xxf is not a quadratic function because the highest power of the x is 1, not 2.
(iii) 243)( 2
−+= yxxf is a not quadratic function because it involves two variables.
* Note: The proper way to denote a quadratic function is
cbxaxxf ++→ 2
: . cbxaxxf ++= 2
)( 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 4)1()( 2
−−= xxf .
The minimum value for )(xf is 4−
when 01 =−x
1=x .
The minimum value of a squared number is
zero. When 0(........)2
= , we would know
the minimum value for the function.

Page | 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 x2
is 0,
It means 02
≥x
So,
02
≤− 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 4)1()( 2
−−−= xxf .
The maximum value for )(xf is 4−
when 01 =−x
1=x .
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 43)( 2
−−= xxxf in form f(x) = a (x + p)2
+ q. Hence, find the coordinates of the turning point
of the graph.
43)( 2
−−= xxxf
4)
2
3
()
2
3
(3 222
−−−−+−= xx
4
4
9
)
2
3
( 2
−−−= x
4
25
)
2
3
( 2
−−= x
The minimum value for )(xf is
4
25
−
when 0
2
3
=−x (
2
3
,
4
25
− )
2
3
=x
When 0(........)2
= , we would know the
minimum 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 0(........)2
= , we would know
the 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 1482)( 2
−+−= xxxf in form f(x) = a (x + p)2
+ q. Hence, find the coordinates of the turning
point of the graph.
1482)( 2
−+−= xxxf
]7)2()2(4[2 222
+−−−+−−= xx
]74)2[(2 2
+−−−= x
]3)2[(2 2
+−−= x
6)2(2 2
−−−= x
The maximum value for 6)( −=xf
when 02 =−x ( 2, -6)
2=x
EXERCISE 3.2
1. State the minimum value of the following function and the corresponding value of x
(a) 8)1()( 2
−+= xxf
(b) 152)( 2
−+= xxxg
2. State the maximum value of the following function and the corresponding value of x
(a) 342)( 2
++−= xxxf
(b) 52)( 2
−+−= xxxg
3. Given that the minimum value of the quadratic function qpxxxf ++= 2
2)( is 1 when 3−=x , find
the 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 2=x .
3.3.1 Basic Equation of quadratic function and its graph
Example 1:
2
)( xxf =
x -2 -1 0 1 2
f(x) 4 1 0 1 4
Example 2:
2
)( xxf −=
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 = 0
By using the information given,
sketch the graph
sketch the graph.

Page | 34
Example 3:
Sketch the graph of 34)( 2
+−= xxxf
34)( 2
+−= xxxf
3)2()2(4)( 222
+−−−+−= xxxf
34)2()( 2
+−−= xxf
1)2()( 2
−−= xxf
The minimum value of 3)( =xf
when 02 =−x
2=x
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 2
241)( xxxf −+= in 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)( 2
−−−= xxxf
]
2
1
)1()1(2[2 222
−−−−+−−= xx
a = 1, b = −4 and c = 3
a = −2, b = 4 and c = 1
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 3)( =xf
when 01 =−x
1=x
So, the maximum point is (1, 3).
The equation of axis of symmetry is 1=x .
EXERCISE 3.3
1. Express the quadratic function 13122
+−= xxy in the form of cbxy ++= 2
)( . State the minimum
point and the y-intercept. Hence, sketch the graph of the function y.
2. Express the quadratic function xxxf 42)( 2
−−= in the form of 2
)()( pxqxf +−= . Determine
the maximum point and the y-intercept. Hence, sketch the graph of the function f(x).
3. Given that the curve 2
)2( nxmy +−= has a maximum point (1, 4), find the values of m and of n.
Hence, sketch the curve.
4. For the quadratic function 2
( ) 2 3 2g x x x= − + , find the equation of the axis of symmetry.
3.4 RELATIONSHIP BETWEEN “b2
– 4ac” AND THE POSITION OF GRAPH f(x)= y
1-If 042
>− acb ,the graph cuts x-axis at two different points
f(x) f(x)
x x
a < 0, so it is maximum value
a > 0 a < 0

Page | 36
1-If 042
=− acb ,the graph touches the x-axis
f(x) f(x)
x
x
1-If 042
<− acb ,the graph does not touch the x-axis
f(x) f(x)
x
x
3.4 QUADRATIC INEQUALITIES
In Form Three, we have learned about linear inequalities. For example when 01 >−x , 1>x .
Example:
Given 01 >−x . Find the range of values of x.
01≥−x
1≥x
•
1
From 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
because f(x) is always positive.

Page | 37
To solve the quadratic inequality for example 0)3)(1( >−+ xx , 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 0364 2
>−m
Solution:
0364 2
>−m
092
>−m
Let
092
=−m
92
=m
3±=m
y
-3 3 m
If 092
>−m , the range of values of m is 3−<m , 3>m .

Page | 38
Example 2:
Find the range of the values of k if the equation 06522
=−−+ kkxx has two real roots.
Solution:
two real roots also means two different roots:
042
>− acb
0)65)(1(4)2( 2
>−−− kk
0)65(44 2
>−−− kk
024204 2
>++ kk
0652
>++ kk
Let
0652
=++ kk
0)3)(2( =++ kk
02 =+k or 03 =+k
2−=k or 3−=k
y
-3 -2 k
If 0652
>++ kk , the range of values of k is 3−<k , 2−>k .
Example 3:
Find the range of the values of x for which if the equation 15)2( ≤+xx
Solution:
15)2( ≤+xx
01522
≤−+ xx
Let
01522
=−+ xx
0)3)(5( =−+ xx
05 =+x or 03 =−x
5−=k or 3=x
652
++= kky

Page | 39
y
-5 -2 x
If 01522
≤−+ xx , the range of values of x is 35 ≤≤− x .
EXERCISE 3.3
1. Find the range of values of x for each of the following inequalities.
(a) 20)1( <+xx
(b) xxx 232
<−
(c) 2)57( >− xx
2. Given that xxy 32 2
+= , find the range of values of x for which y > 9.
3. Find the range of values of k given that 02)1(2
=++−+ kxkx has two different roots.
4. If the straight line ky = , where k is a constant, does not intersect the curve 562 2
+−= xxy , show
that
2
1
<k .
CHAPTER REVIEW EXERCISE
1. The graph of the function 2
( ) 2( )f x p q x= − − has a maximum point at
1 5
,
2 2
 
− 
 
. Find the value
of p and q.
2. Given the quadratic function 362 2
−+−= xxy , express f(x) in the form rqxp ++ 2
)( , where p, q
and r are constants. Hence, sketch the graph of the function y.
3 Given that the straight line nxy −= 2 cuts the curve 22
−+= nxxy at two points, find the range of
the values of n.
4. The quadratic equation 542
−=− pxpx , where 0≠p , has real and different roots. Find the range
of values of p.
5. By expressing the function 563)( 2
+−= xxxf in the form qpxa +− 2
)( , or otherwise, find the
minimum value of f(x).
6. Given that the quadratic function qxpxxf −+= 4)( 2
, where p and q are constants .Given that the
curve )(xfy = has a maximum point (-1, 5).State the values of p and q.
1522
−+= xxy

Page | 40
7. Diagram shows the graph of a quadratic function )(xfy = . The straight line 9−=y is tangent to
the curve )(xfy = .
(a) Write the equation of the axis of symmetry of the curve.
(b) Express f(x) in the form qpx ++ 2
)( , where p and q are constants.
8. Diagram below shows the graph of the function qxxpy ++−= 2)1( 2
.
(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 1=+ yx does not intersect the curve
kyx =+ 22
.
y
x
0
2−
qxxpy ++−= 2)1( 2
0 x
y
1 7
9−=y

Chapter 3 quadratc functions

Empfohlen

Empfohlen

Weitere ähnliche Inhalte

Was ist angesagt?

Was ist angesagt? (20)

Ähnlich wie Chapter 3 quadratc functions

Ähnlich wie Chapter 3 quadratc functions (20)

Chapter 3 quadratc functions