Chapter 3 Quadratc Functions

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

CHAPTER 3- QUADRATIC FUNCTIONS

3.1 INTRODUCTION The general form of a quadratic function is f ( x) = ax 2 + bx + c ; a, b, c are constants and a ≠ 0 Similarities and differences between Quadratic Functions and Quadratic Equations Quadratic Functions Quadratic Equations General form is f ( x) = ax 2 + bx + c a, b, c are constants

General form is ax 2 + bx + c = 0 a, b, c are constants

x is independent variable while f ( x ) is

x is unknown

dependent variable 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 f ( x ) = 0 , quadratic functions

quadratic equations

3.1.1 Recognizing a quadratic function (i) f ( x) = 2 x 2 + 3 x − 4 is a quadratic function because the highest power of the x is 2 and Involves one variable only. (ii) f ( x ) = 3x + 2 is not a quadratic function because the highest power of the x 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 f : x → ax 2 + bx + c . 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. The minimum value of a squared number is 2

For example f ( x) = ( x − 1) 2 − 4 . The minimum value for f ( x ) is − 4 when x − 1 = 0 x = 1.

zero. When (........) = 0 , we would know the minimum value for the function.

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

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, 2 It means x 2 ≥ 0 When (........) = 0 , we would know the So, minimum or maximum value for the − x2 ≤ 0 function. The minimum value of a squared Hence the maximum value of – x2 is 0 number is zero. the maximum value of -x2+ 5 is -0 + 5=5 the maximum value of –x2– 3 is -0 + 3 =– 3. For example f ( x ) = −( x − 1) 2 − 4 . The maximum value for f ( x ) is − 4 when x − 1 = 0 x = 1.

The minimum value of a squared number is 2

zero. When (........) = 0 , we would know the minimum value for the function.

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 − 3x − 4 in form f(x) = a (x + p)2 + q. Hence, find the coordinates of the turning point of the graph. 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) = x 2 − 3x − 4 3 3 = x 2 − 3x + ( − ) 2 − ( − ) 2 − 4 2 2 3 9 = (x − )2 − − 4 2 4 3 25 = (x − )2 − 2 4 25 The minimum value for f ( x ) is − 4 3 when x − = 0 2 3 x= 2

This is in the form a (x + p)2 + q. The value of a is 1.

(

3 25 ,− ) 2 4 f(x) is the value for y. This is because f(x) is the y-axis.

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 | 31

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

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 turning point of the graph. The coordinates of the turning point means the coordinates of the minimum of maximum 2 f ( x) = −2 x + 8 x − 14 point for the function. For this question, it is = −2[ x 2 − 4 x + (−2) 2 − (−2) 2 + 7] maximum point because a < 0

= −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 x=2

This is in the form a (x + p)2 + q. The value of a is -2 which is less than 0. ( 2, -6)

EXERCISE 3.2 1. State the minimum value of the following function and the corresponding value of x (a) f ( x) = ( x + 1) 2 − 8 (b) g ( x ) = 2 x 2 + 5 x − 1 2. State the maximum value of the following function and the corresponding value of x (a) f ( x ) = −2 x 2 + 4 x + 3 (b) g ( x) = − x 2 + 2 x − 5 3. Given that the minimum value of the quadratic function f ( x ) = 2 x 2 + px + q is 1 when x = −3 , find the value of p and of q. Page | 32

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

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:

f ( x) = x 2 x f(x)

-2 4

a = 1, b = 0 and c = 0

-1 1

0 0

1 1

2 4 By using the information given, sketch the graph

Example 2:

f ( x) = − x 2 x f(x)

-2 -4

a = − 1, b = 0 and c = 0

-1 -1

0 0

1 -1

2 -4 By using the information given, sketch the graph.

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Chapter 3- Quadratic Functions

Example 3: Sketch the graph of f ( x) = x 2 − 4 x + 3

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

a = 1, b = − 4 and c = 3

f ( x) = x 2 − 4 x + 3 f ( x) = x 2 − 4 x + (−2) 2 − (−2) 2 + 3 f ( x ) = ( x − 2) 2 − 4 + 3 f ( x) = ( x − 2) 2 − 1 The minimum value of f ( x ) = 3 when x − 2 = 0

x=2 Minimum point is (2,3).

x = 0, f (0) = (0) 2 − 4(0) + 3 =3 The point is (0,3) By using the information given, sketch the graph.

Example 4: Express the function f ( x ) = 1 + 4 x − 2 x 2 in the form of a(x+ p)2 + q. Hence (a) state the minimum or maximum point. a = − 2, b = 4 and c = 1 (b) the equation of axis 0f symmetry

1 f ( x) = −2( x 2 − 2 x − ) 2 1 = −2[ x 2 − 2 x + (−1) 2 − (−1) 2 − ] 2 Page | 34

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

1 = −2[( x − 1) 2 − 1 − ] 2 1 = −2[( x − 1) 2 − 1 − ] 2 1 = −2[( x − 1) 2 − 1 − ] 2 3 = −2[( x − 1) 2 − ] 2 2 = −2[( x − 1) + 3 The maximum value of f ( x ) = 3

a < 0, so it is maximum value

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 of y = ( x + b) 2 + c . State the minimum point 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 of f ( x ) = q − ( x + p ) 2 . Determine the 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) 2 has 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 − 4ac > 0 ,the graph cuts x-axis at two different points f(x)

f(x)

x

a>0

x

a0

a 0 Curve lies above the x-axis because f(x) is always positive.

a 0 , x > 1 . Example: Given x − 1 > 0 . Find the range of values of x.

x −1 ≥ 0 x ≥1

• 1 From the line graph, we know the range of values of x.

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

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 4m 2 − 36 > 0 Solution:

4m 2 − 36 > 0 m2 − 9 > 0 Let

m2 − 9 = 0 m2 = 9 m = ±3 y

-3

3

m

If m 2 − 9 > 0 , the range of values of m is m < −3 , m > 3 . Page | 37

Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

Example 2: Find the range of the values of k if the equation x 2 + 2kx − 5k − 6 = 0 has two real roots. Solution: two real roots also means two different roots:

b 2 − 4ac > 0 (2k ) 2 − 4(1)(−5k − 6) > 0

4k 2 − 4(−5k − 6) > 0 4k 2 + 20k + 24 > 0 k 2 + 5k + 6 > 0 Let

k 2 + 5k + 6 = 0 (k + 2)(k + 3) = 0 k + 2 = 0 or k + 3 = 0 k = −2 or k = −3 y

y = k 2 + 5k + 6

-3

-2

k

If k 2 + 5k + 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:

x( x + 2) ≤ 15 x 2 + 2 x − 15 ≤ 0 Let

x 2 + 2 x − 15 = 0 ( x + 5)( x − 3) = 0 x + 5 = 0 or x − 3 = 0 k = −5 or x = 3

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

y

-5

y = x 2 + 2 x − 15

-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 − 3x < 2 x (c) x (7 − 5x ) > 2 2. Given that y = 2 x 2 + 3x , find the range of values of x for which y > 9. 3. Find the range of values of k given that x 2 + (k − 1) x + 2 + k = 0 has two different roots. 4. If the straight line y = k , where k is a constant, does not intersect the curve y = 2 x 2 − 6 x + 5 , show that k <

1 . 2

CHAPTER REVIEW EXERCISE

1 2

1. The graph of the function f ( x ) = p − 2(q − x )2 has a maximum point at  , −

5  . Find the value 2

of p and q. 2. Given the quadratic function y = −2 x 2 + 6 x − 3 , express f(x) in the form p( x + q) 2 + r , where p, q and r are constants. Hence, sketch the graph of the function y. 3 Given that the straight line y = 2 x − n cuts the curve y = x 2 + nx − 2 at two points, find the range of the values of n. 4. The quadratic equation px 2 − 4 x = p − 5 , where p ≠ 0 , has real and different roots. Find the range of values of p. 5. By expressing the function f ( x ) = 3 x 2 − 6 x + 5 in the form a( x − p) 2 + q , or otherwise, find the minimum 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 curve y = f ( x ) has a maximum point (-1, 5).State the values of p and q.

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Additional Mathematics Module Form 4 SMK Agama Arau, Perlis

Chapter 3- Quadratic Functions

7. Diagram shows the graph of a quadratic function y = f ( x ) . The straight line y = − 9 is tangent to the curve y = f ( x ) .

y

0

1

x

7

y = −9 (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 .

y

x 0

−2

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 x + y = 1 does not intersect the curve

x2 + y2 = k .

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