MT T4 (Bab 3: Fungsi Kuadratik)

3.0 QUADRATIC FUNCTION3.0 QUADRATIC FUNCTION

3.1 QUADRATIC FUNCTIONS & THEIR GRAPHS3.2 THE MAX. & MIN. VALUES OF QUADRATIC FUNCTIONS3.3 SKETCH GRAPH OF QUADRATIC EQUATIONS3.4 QUADRATIC INEQUALITIES

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3.1 QUADRATIC FUNCTIONS & THEIR 3.1 QUADRATIC FUNCTIONS & THEIR GRAPHS GRAPHS [2][2]

General form of quadratic function:-

f(x) = ax2 +bx + cf(x) = ax2 +bx + ca ≠ 0bc

constants

Examples: f(x) = 3x2 + 5x + 2 f(x) = -2x2 - 5x + 4a = 3

b = 5

c = 2

a = -2

b = -5

c = 4

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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

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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

100

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)

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Shapes of graphs of quadratic function:-

a > 0a > 0 (+ve) (+ve) a < 0a < 0 (-ve) (-ve)f(x)

x0

Minimum graph(Smile)

Minimum point along the graph

f(x)

x0

Maximum point along the graph

Maximum graph(Sad)

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Shapes of graphs of quadratic function:-

a > 0a > 0 (+ve) (+ve) a < 0a < 0 (-ve) (-ve)f(x)

x0

Minimum graph(Smile)

Vertex

f(x)

x0

Maximum graph(Sad)

VertexAxis of

symmetry

Axis of symmetry

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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 +5a = -1 a < 0

Min.

Max.

c) f(x) = x(4 – 2x) = 4x – 2x2

= -2x2 +4x

a = -2 a < 0 Max.

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f(x)=axf(x)=ax22+bx+c+bx+cValue 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

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Discriminant (bb22 – 4ac – 4ac) a > 0 (+ve)a > 0 (+ve) a < 0 (-ve)a < 0 (-ve)

bb22 – 4ac > 0 – 4ac > 0Two distinct roots

bb22 – 4ac = 0 – 4ac = 0Two equal roots

bb22 – 4ac < 0 – 4ac < 0No 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.

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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

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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: Subtitute Subtitute 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

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3.1 QUADRATIC FUNCTIONS & THEIR 3.1 QUADRATIC FUNCTIONS & THEIR GRAPHSGRAPHS

3.5

3

2.5

2

1.5

1

0.5

-0.5

-3 -2 -1 1 2 3 4 5

x A = -0.5

f x( ) = 8⋅x 2 +8⋅x+2

A

BACKBACK

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3.1 QUADRATIC FUNCTIONS & THEIR 3.1 QUADRATIC FUNCTIONS & THEIR GRAPHSGRAPHS

1

0.5

-0.5

-1

-1.5

-2

-2.5

-2 -1 1 2 3 4 5

f x( ) = -2⋅x 2 +8⋅x( )-8

BACKBACK

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3.2 MAX. & MIN. VALUES OF QUADRATIC 3.2 MAX. & MIN. VALUES OF QUADRATIC FUNCTIONS FUNCTIONS [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

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Determine the vertex and axis of symmetry by using completing the completing the square.square.

f(x) = a(x + p)2 +qf(x) = a(x + p)2 +q

Still remember how to solve by Still remember how to solve by completing the square?????completing the square?????

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f(x) = ax2 + bx + c

++=

ac

xab

xa 2

+

−

++=

ac

ab

ab

xab

xa22

2

22

−+

+=

22

22 ab

ac

ab

xa

−+

+=

22

22 ab

ac

aa

bxa

ab

p2

=

−=

2

2ab

ac

aq f(x) = a(x + p)2 +qf(x) = a(x + p)2 +q

x = -p y = q

Vertex = Vertex = (-p , q)(-p , q)

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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)

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Example 2Express 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

9210

210

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

( )

50505 2

==−=−

x

xx

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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) = 2x2 2 + 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

21=x

f(x) = 2(x + 1)f(x) = 2(x + 1)22 + 5 + 5The minimum value of the function f(x) = 5f(x) = 5

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3.3 SKETCH GRAPHS OF QUADRATIC 3.3 SKETCH GRAPHS OF QUADRATIC FUNCTION FUNCTION [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 & x22 xx11 & x & x22

44 . Determine the point of intersection with the y-axis by finding f(x)f(x) when x = 0 when 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.

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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.

−−

−−−=

322

318

3)( 2 xxxf

( )

−−−−−=

322

63 2 xx

++−=

322

63 2 xx

+

−

++−=

322

26

26

6322

2 xx

( ) ( )

+−++−=

322

3363 222 xx

( )

+−+−=

322

933 2x

( )

−+−=

35

33 2x

( ) ( )335

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

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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

( )35

3 ±=+x

35

3 ±−=x

291.4709.1

2

1

−=−=

xx

4 . When x = 0; f(x) = -3(0)2 – 18(0) – 22

= -22y

x

-22 ―

|

-3

― 5

0

f(x) = -3x2 -18x - 22f(x) = -3x2 -18x - 22

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Exercises:1. Sketch the graph of f(x) = 4 –(x f(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(4x y = 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) + 2y = 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

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3.4 QUADRATIC INEQUALITIES 3.4 QUADRATIC INEQUALITIES [1][1]

The range of quadratic inequalities can be determined from the shape of the graph.

ax

y

0 b

f(x) < 0f(x) < 0 for a < x < b

ax

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

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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

x0

x2 – 4x – 5 ≤ 0

y

x0

a > 0 : Minimum

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0 ≤ x2 – 4xx2 – 4x = 0

x(x – 4) = 0

x = 0 or; (x – 4) = 0

x = 4y

x0 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 = -1y

x-1 5

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f(x)

x0|

4

|

5

|

-1

f(x) = x2 – 4x – 5

f(x) = x2 – 4xThe range of x;

-1 ≤ x ≤ 0 or 4 ≤ x ≤ 5

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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)

x0

|k

|3

30 -

f(x) < 0

a =2, b = p, c = 30

( )

2438438424

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;

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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)

x0

|k

|3

30 -

f(x) < 0

a =2, b = p, c = 30

Use formula method;

4416

416164

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).

1,51 >−< xx

3,1 >−< xx

Education

MT T4 (Bab 3: Fungsi Kuadratik)