Section 3.3 quadratic functions and their properties

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

Section 3.3

Quadratic Functions and Their Properties

Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.






2( ) 2 4 __ 5 2(__)f x x x

2( ) 4 52 4 2(4)f x x x

2( ) 2 2 3f x x





Without graphing, locate the vertex and axis of symmetry of the parabola defined by . Does it open up or down? 22 3 2f x x x

3 7Vertex is , .4 8

3 32 2 2 4ba

23 3 3 72 3 24 4 4 8

f

3Axis of symmetry is .4

x

Because 2 0, the parabola opens up.a




2(a) Use the information from the previous example and the locations of the intercepts to graph ( ) 2 3 2.f x x x

3 7Vertex is , .4 8

3Axis of symmetry is .4

x

x

y

3 7,4 8

Since a = 2 > 0 the parabola opens up and therefore will have no x-intercepts.

2(0) 2 0 3 0 2 2 so the -intercept = 2.f y

0, 2 3 , 22

By symmetry, the point with the same

3 -value but to the right of the axis of 4

3 3 3symmetry is on the graph. 4 4 2

3so the point , 2 is on the graph.2

y

(b) Determine the domain and the range of .(c) Determine where is increasing and decreasing.

ff

x

y

3 7,4 8

The domain of f is the set of all real numbers.7Based on the graph, the range is the interval , .8

0, 2 3 , 22

The function is 3 from ,4

and3 from , .4

incr

decreasi

g

ng

easin

x

y

2(a) Graph ( ) 2 4 1 by determining whether the graph opens up or down and by finding its vertex, axis of symmetry, and and intercepts if any.

f x x xx y

h

b2a

4

2(2) 1

Since a = 2 > 0 the parabola opens up.

1, 3

-intercepts can be found when ( ) 0.x f x

21 2 1 4( 1) 1 3k f

Vertex = 1, 3

2(0) 2 0 4 0 1 1 so the -intercept = 1.f y

By symmetry, the point ( 2, 1) is also on the graph.

0, 1 2, 1

20 2 4 1 Use the quadratic formula to solve.x x

24 4 4(2)( 1) 4 24 2 6 =2(2) 4 2

x

-intercepts 0.22 and 2.22x Axis of symmetry: 1x

x

y

1, 3

0, 1 2, 1


ff

The domain of f is the set of all real numbers.

Based on the graph, the range is the interval 3, .

The function is from , 1

and from 1, .

inc

decreasing

reasing

x

y

21(a) Graph ( ) 2 2 by determining whether the graph opens up or down 2

and by finding its vertex, axis of symmetry, and - and -intercepts if any.f x x x

x y

h b

2a

2

2 12

2

Since a is negative, the parabola opens down.

2,0

As seen on the graph, the -intercept is 2.x

212 2 2( 2) 2 02

k f

Vertex = 2,0

21(0) 0 2 0 2 2 so the -intercept = 2.2

f y

By symmetry, the point ( 4, 2) is also on the graph. 0, 2 4, 2

Axis of symmetry: 2x

x

y

2,0

0, 2 4, 2


ff

The domain of f is the set of all real numbers.

Based on the graph, the range is the interval ,0 .

The function is from , 2

and from 2, .

dec

increasing

reasing




Determine the quadratic function whose vertex is (–2, 3) and whose y-intercept is 1.

2 22 3f x a x h k a x

Using the fact that the y-intercept is 1: 1a 0 2 23

1 4 3a 12

a

21 2 32

f x x

x

y

21 2 12

f x x x



2Determine whether the quadratic function

4 5has a maximum or minimum value.

Then find the maximum or minimum value.

f x x x

Since a is negative, the graph of f opens down so the function will have a maximum value.

4 2

2 2 1bxa

2So the maximum value is 2 (2) 4(2) 5 9f
