MT T4 (Bab 3: Fungsi Kuadratik)

3.0 QUADRATIC FUNCTION3.0 QUADRATIC FUNCTION
3.1 QUADRATIC FUNCTIONS & THEIR GRAPHS
3.2 THE MAX. & MIN. VALUES OF QUADRATIC FUNCTIONS
3.3 SKETCH GRAPH OF QUADRATIC EQUATIONS
3.4 QUADRATIC INEQUALITIES

SMT PONTIAN, JOHOR
3.1 QUADRATIC FUNCTIONS & THEIR3.1 QUADRATIC FUNCTIONS & THEIR
GRAPHSGRAPHS [2][2]
General form of quadratic function:-
f(x) = ax2
+bx + cf(x) = ax2
+bx + c
a ≠ 0
b
c
constants
Examples: f(x) = 3x2
+ 5x + 2 f(x) = -2x2
- 5x + 4
a = 3
b = 5
c = 2
a = -2
b = -5
c = 4

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [3][3]
 Example 1:
x -2 -1 0 1 2 3 4
f(x) -5 0 3 4 3 0 -5
By using a suitable scale, plot the
tabulated graph below.
f(x)
x
-1 ―
-2 ―
-3 ―
-4 ―
-5 ―
-6 ―
1 ―
2 ―
3 ―
4 ―
5 ―
I
-1
I
-2
I
-3
I
1
I
2
I
3
I
4
I
5
xx
xx
xx
xx
xx
xx
xx
Max. (1 , 4)Max. (1 , 4)
Axis of symmetry, x=1Axis of symmetry, x=1

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [4][4]
 Exercise:
Draw a table and find the value of
f(x) for the following quadratic
function with the given range of x.
By using suitable scale, plot the
graph.
From the graph, state the axis of
symmetry and a maximum or
minimum point.
f(x) = x2
– 3x -10 ; -3 ≤ x ≤ 5
x -3 -2 -1 0 1 2 3 4 5
f(x) 8 0 -6 -10 -12 -12 -10 -6 0
f(x)
x
-2 ―
-4 ―
-6 ―
-8 ―
-10 ―
-12 ―
2 ―
4 ―
6 ―
8 ―
10 ―
I
-2
I
-4
I
2
I
4
I
6
I
8
I
10
0
x
x
x
x
x x
x
x
x
Axis of symmetry, x=1.5Axis of symmetry, x=1.5
Min. (1.5 , -12.25)Min. (1.5 , -12.25)

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [5][5]
Shapes of graphs of quadratic function:-
a > 0a > 0 (+ve)(+ve) a < 0a < 0 (-ve)(-ve)
f(x)
x
0
Minimum graph
(Smile)
Minimum point
along the graph
f(x)
x
0
Maximum point
along the graph
Maximum graph
(Sad)

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [6][6]
Shapes of graphs of quadratic function:-
a > 0a > 0 (+ve)(+ve) a < 0a < 0 (-ve)(-ve)
f(x)
x
0
Minimum graph
(Smile)
Vertex
f(x)
x
0
Maximum graph
(Sad)
Vertex
Axis of
symmetry
Axis of
symmetry

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [7][7]
Identify the shape of each graph of the following.Identify the shape of each graph of the following.Exercise 3.1.3Exercise 3.1.3
a) f(x) = 2x2
+ 3x - 1 a = 2  a > 0
b) f(x) = 5 + 3x - x2
= -x2
+ 3x +5
a = -1  a < 0
Min.
Max.
c) f(x) = x(4 – 2x)
= 4x – 2x2
= -2x2
+4x
a = -2  a < 0 Max.

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [8][8]
f(x)=axf(x)=ax22
+bx+c+bx+c
Value of
discriminant
bb22
-4ac-4ac Type of roots
Shape of
graphs
Points of
intersection with
x-axis
f(x)= x2
-6x+8 = 4 (> 0)
Two distinct
roots
Two x-intercepts
f(x)= x2
-2x+1 = 0 2 equal roots Vertex at x-axis
f(x)= x2
+4x+6 = -8 (< 0) No real roots No x-intercept
f(x)= -x2
+8x-15 = 4 (> 0)
Two distinct
roots
Two x-intercepts
f(x)= -x2
+4x-4 = 0 2 equal roots Vertex at x-axis
f(x)= -x2
+2x-2 = -4 (< 0) No real roots No x-intercept

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [9][9]
Discriminant (bb22
– 4ac– 4ac) a > 0 (+ve)a > 0 (+ve) a < 0 (-ve)a < 0 (-ve)
bb22
– 4ac > 0– 4ac > 0
Two distinct roots
bb22
– 4ac = 0– 4ac = 0
Two equal roots
bb22
– 4ac < 0– 4ac < 0
No real roots
x x x
y
0
x x
y
0
x
y
0
x x x
y
0
x x
y
0
x
y
0
Min.
Min.
Min.
Max.
Max.
Max.

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [10][10]
 Example: Find the values of p for which the x-axis is a
tangent to the graph of f(x) = pxf(x) = px22
+8x + p - 6+8x + p - 6
Solution:Solution:
Informations: •x-axis is a tangent to the graph
•a = p, b = 8, c = (p – 6)a = p, b = 8, c = (p – 6)
 bb22
– 4ac =– 4ac =
00
 bb22
– 4ac =– 4ac =
00
x x
y
0
Min.
x x
y
0
Max.
OR tangent
82
– 4(p)(p – 6) = 0
64 – 4(p2
– 6p) = 0
64 – 4p2
+ 24p = 0
-4p2
+ 24p + 64 = 0
-4(p2
- 6p - 16) = 0
p2
- 6p - 16 = 0
Expand
Rearranged
 General form
Factorised
(p – 8)(p +2) = 0
p – 8 = 0
p = 8p = 8
or; p + 2 = 0
p = -2p = -2

SMT PONTIAN, JOHOR
GRAPHSGRAPHS [11][11]
 Example: Find the values of p for which the x-axis is a
tangent to the graph of f(x) = pxf(x) = px22
+8x + p - 6+8x + p - 6
Solution:Solution:
Informations: •x-axis is a tangent to the graph
•a = p, b = 8, c = (p – 6)a = p, b = 8, c = (p – 6)
 bb22
– 4ac =– 4ac =
00
 bb22
– 4ac =– 4ac =
00
x x
y
0
Min.
x x
y
0
Max.
OR tangent
Check:Check: SubtituteSubtitute p into f(xp into f(x),),
When p = 8;
f(x) = 8x2
+8x + (8-6)
= 8x2
+ 8x + 2
When p = -2;
f(x) = -2x2
+8x + (-2-6)
= -2x2
+8x - 8
SHOW GRAPHSHOW GRAPH SHOW GRAPHSHOW GRAPH

SMT PONTIAN, JOHOR
GRAPHSGRAPHS
3.5
3
2.5
2
1.5
1
0.5
-0.5
-2 -1 1 2 3 4
xA = -0.5
f x( ) = 8⋅x2 +8⋅x+2
A
BACKBACK

SMT PONTIAN, JOHOR
GRAPHSGRAPHS
1
0.5
-0.5
-1
-1.5
-2
-2.5
1 2 3 4
f x( ) = -2⋅x2 +8⋅x( )-8
BACKBACK

SMT PONTIAN, JOHOR
3.2 MAX. & MIN. VALUES OF QUADRATIC3.2 MAX. & MIN. VALUES OF QUADRATIC
FUNCTIONSFUNCTIONS [1][1]
 In this subtopic, we will determine:-
The vertex or the turning pointThe vertex or the turning pointThe vertex or the turning pointThe vertex or the turning point
The axis of symmetryThe axis of symmetryThe axis of symmetryThe axis of symmetry

SMT PONTIAN, JOHOR
Determine the vertex and axis of
symmetry by using completing thecompleting the
square.square.
f(x) = a(x + p)2
+qf(x) = a(x + p)2
+q
Still remember how to solve byStill remember how to solve by
completing the square?????completing the square?????

SMT PONTIAN, JOHOR
f(x) = ax2
+ bx + c






++=
a
c
x
a
b
xa 2






+





−





++=
a
c
a
b
a
b
x
a
b
xa
22
2
22












−+





+=
22
22 a
b
a
c
a
b
xa












−+





+=
22
22 a
b
a
c
a
a
b
xa
a
b
p
2
= 











−=
2
2a
b
a
c
aq f(x) = a(x + p)2
+qf(x) = a(x + p)2
+q
x = -p y = q
Vertex =Vertex = (-p , q)(-p , q)

SMT PONTIAN, JOHOR
Example 1:
State the maximum or minimum value of the following quadratic functions:-
a) f(x) = 3(x + 2)2
- 6
a = 3 (+ve)  Minimum (smile)
x = -2
y = -6
Why x = -2 ????
Why y = -6 ????
f(x) = 3(x + 2)2
- 6
When (x+2)2
= 0, then f(x) = -6
f(x) = 3[0] – 6 = -6
(x+2)2
= 0
x + 2 = 0
x = -2x = -2
Therefore; the vertex is (-2, -6)vertex is (-2, -6)

SMT PONTIAN, JOHOR
 Example 2
Express f(x) = x2
– 10x + 9 in the form of f(x) =(x +a)2
+b , with a and b are
constants. Hence, state the maximum or minimum value of f(x) and the
corresponding value of x.
Solution:-Solution:-
a = 1, b = -10, c = 9
9
2
10
2
10
10)(
22
2
+




 −
−




 −
+−= xxxf
( ) ( ) 95510 222
+−−−+−= xx
( ) 9255 2
+−−= x
( ) 165 2
−−= x
f(x) =(x +a)2
+b Therefore; a = -5, b = -16
a > 0 (+ve)  Minimum (smile)
Min. value of f(x) = -16
Min. (5, -16)Min. (5, -16)
x
y
0 5
-16 -
f(x) = x2
– 10x + 9
( )
5
05
05 2
=
=−
=−
x
x
x

SMT PONTIAN, JOHOR
 Exercises:
1. Find the maximum value of the function f(x) = 5 + x – xf(x) = 5 + x – x22
. State the value
of x, so that f(x) has a maximum value.
2. Express f(x) = 2xf(x) = 2x22
+ 4x +7+ 4x +7 in the form of complete square. Hence, find
the maximum or minimum value of function f(x).
The maximum value of the function f(x) = ??f(x) = ??
x = ??x = ?? 4
21
)( =xf
2
1
=x
f(x) = 2(x + 1)f(x) = 2(x + 1)22
+ 5+ 5
The minimum value of the function f(x) = 5f(x) = 5

SMT PONTIAN, JOHOR
3.3 SKETCH GRAPHS OF QUADRATIC3.3 SKETCH GRAPHS OF QUADRATIC
FUNCTIONFUNCTION [1][1]
 The steps involved in sketching graphs of quadratic
functions f(x) = ax2
+ bx +c are:-
11. Determine the shape of the graph: Value of a.
22. Find the maximum or minimum point by
expressing in f(x) in the form of
f(x) = a(x + p)2
+q.
a > 0 : Min.
(Smile)
a < 0 : Max. (Sad)
a > 0 : Min.
(Smile)
a < 0 : Max. (Sad)
Completing the squareCompleting the square
33. Determine the point of intersection with
x-axis, if its exists, by solving the
equation f(x) = 0f(x) = 0.
xx11 & x& x22xx11 & x& x22
44. Determine the point of intersection with
the y-axis by finding f(x)f(x) when x = 0when x = 0
[value of f(0)f(0)].
f(x) = axf(x) = ax22
+ bx +c+ bx +c
= a(0)= a(0)22
+ b(0) + c+ b(0) + c
= c= c
f(x) = axf(x) = ax22
+ bx +c+ bx +c
= a(0)= a(0)22
+ b(0) + c+ b(0) + c
= c= c
55. Mark the points and draw smooth
parabola through all the points.

SMT PONTIAN, JOHOR
 Example: Sketch the following quadratic function and state
the max. or min. point.
1. a = -3 : Maximum (Sad)
a = -3, b = -18, c = -22
2
.












−
−





−
−−=
3
22
3
18
3)( 2
xxxf
( ) 











−
−−−−=
3
22
63 2
xx












++−=
3
22
63 2
xx












+





−





++−=
3
22
2
6
2
6
63
22
2
xx
( ) ( ) 











+−++−=
3
22
3363 222
xx
( ) 











+−+−=
3
22
933 2
x
( ) 



−+−=
3
5
33 2
x
( ) ( )3
3
5
33 2
−−+−= x
( ) 533 2
++−= x
When (x + 3)2
= 0, x = -3
Max. value of f(x) = y = 5
f(x) = -3x2
-18x - 22f(x) = -3x2
-18x - 22

SMT PONTIAN, JOHOR
 Example: Sketch the following quadratic function and state the
max. or min. point.
a = -3, b = -18, c = -22
3. Find the value of x1 & x2 [f(x) = y = 0]
( ) 5330 2
++−= x
( ) 533 2
=+x
( )
3
5
3 ±=+x
3
5
3 ±−=x
291.4
709.1
2
1
−=
−=
x
x
4. When x = 0; f(x) = -3(0)2
– 18(0) – 22
= -22
y
x
-22 ―
|
-3
― 5
0
f(x) = -3x2
-18x - 22f(x) = -3x2
-18x - 22

SMT PONTIAN, JOHOR
Exercises:
1. Sketch the graph of f(x) = 4 –(xf(x) = 4 –(x
– 3)– 3)22
for the domain 0 ≤ x ≤6.
2. Find the max. or min. value of the
function y = 3(2x – 1)(x + 1) – x(4xy = 3(2x – 1)(x + 1) – x(4x
– 5) + 2.– 5) + 2. Hence, sketch the graph of
function y.
y
x
-5 ―
|
1
(3, 4)
0
|
5
|
6
x
y = 3(2x – 1)(x + 1) – x(4x – 5) + 2
y = 3(2x2
+ x - 1) – 4x2
+ 5x + 2
= 6x2
+ 3x – 3 - 4x2
+ 5x + 2
= 6x2
+ 8x - 1
Minimum value = -9 y
x
-1 ―
(-2, -9)
|
0.1213
x
|
-4.121

SMT PONTIAN, JOHOR
3.4 QUADRATIC INEQUALITIES3.4 QUADRATIC INEQUALITIES [1][1]
The range of quadratic inequalities can
be determined from the shape of the
graph.
a
x
y
0 b
f(x) < 0f(x) < 0 for a < x < b
a
x
y
0 b
f(x) > 0f(x) > 0 for a < x < b
f(x) < 0f(x) < 0 for a < x or x > b
Minimum Maximum
f(x) > 0f(x) > 0 for a < x or x > b

SMT PONTIAN, JOHOR
 Example 1: Find the range of values of x which satisfies the
inequality 0 ≤ x2
– 4x ≤ 5
0 ≤ x2
– 4x ≤ 5
0 ≤ x2
– 4x
x2
– 4x ≤ 5
a > 0 : Minimumy
x
0
x2
– 4x – 5 ≤ 0
y
x
0
a > 0 : Minimum

SMT PONTIAN, JOHOR
inequality 0 ≤ x2
– 4x ≤ 5
0 ≤ x2
– 4x
x2
– 4x = 0
x(x – 4) = 0
x = 0 or; (x – 4) = 0
x = 4
y
x
0 4
0 ≤ f(x) x2
– 4x – 5 ≤ 0 0 ≥ f(x)
x2
– 4x – 5 = 0
(x – 5)(x + 1) = 0
(x – 5) = 0
x = 5
(x + 1) = 0
x = -1
y
x
-1 5

SMT PONTIAN, JOHOR
inequality 0 ≤ x2
– 4x ≤ 5
f(x)
x0
|
4
|
5
|
-1
f(x) = x2
– 4x – 5
f(x) = x2
– 4x
The range of x;
-1 ≤ x ≤ 0 or 4 ≤ x ≤ 5

SMT PONTIAN, JOHOR
 Example 2: Given that f(x) = 2x2
+ px + 30 and that f(x) < 0 only
when 3 < x < k. Find the values of p and k.
3 < x < k a > 0 (Max.)
f(x)
x
0
|
k
|
3
30 -
f(x) < 0
a =2, b = p, c = 30
( )
24
384
38424
024014424
24024144
24012
24012
)30)(2(4)4(3
)2(2
)30)(2(4
3
22
22
22
2
2
2
−
=
−=
=+++−
−=++
−=+
−±=+
−±−=
−±−
=
p
p
ppp
ppp
pp
pp
pp
pp
p = -16
Use formula
method;

SMT PONTIAN, JOHOR
 Example 2: Given that f(x) = 2x2
+ px + 30 and that f(x) < 0 only
when 3 < x < k. Find the values of p and k.
3 < x < k a > 0 (Max.)
f(x)
x
0
|
k
|
3
30 -
f(x) < 0
a =2, b = p, c = 30
Use formula
method;
4
416
4
1616
4
24025616
)2(2
)30)(2(4)16()16( 2
±
=
±
=
−±
=
−−±−−
=k
Substitute p = -16 into formula;
k1 = 5 or; k2 = 3
kk22 = 3= 3 is already exist in the range 3 < x < k. Therefore, k = 5k = 5

SMK SULTAN SULAIMAN SHAH, SELANGOR
 Exercises:
1. Given that f(x) = 5x2
- 4x – 1, find the range of values of x, so that f(x) is
positive.
2. Given that f(x) = x2
+ 4x – 1 and g(x) = 6x + 2, find the range of values
of x if f(x) > g(x).
1,
5
1
>−< xx
3,1 >−< xx

MT T4 (Bab 3: Fungsi Kuadratik)

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