Section 2.3 Quadratic Functions and Their Propertiesfaculty.montgomerycollege.edu/maronne/Ma180/power point/chap-3 … · Copyright © 2013 Pearson Education, Inc. All rights reserved

Section 2.3

Quadratic Functions and Their Properties

Copyright © 2013 Pearson Education, Inc. All rights reserved


S







( )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 = >Copyright © 2013 Pearson Education, Inc. All rights reserved




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

3 7Vertex is ,4 8

3Axis of symmetry is 4

x =

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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 3value 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

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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

easinCopyright © 2013 Pearson Education, Inc. All rights reserved

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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

= + −

4 12 2(2)bhx

= − = − = −

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 = − Copyright © 2013 Pearson Education, Inc. All rights reserved

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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


−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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= − − −

2 212 22

bhx

−= − = − = −

−

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 = −


−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

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= − + = + +

( )2Using the fact that the y-intercept is 1: 1 0 2 3a= + +

1 4 3a= +

12

a = −

( ) ( )21 2 32

f x x= − + +

−4 −3 −2 −1 1 2 3 4 5

−4

−3

−2

−1

1

2

3

4

x

y



( ) 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 = − + + =Copyright © 2013 Pearson Education, Inc. All rights reserved


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Section 2.3 Quadratic Functions and Their Propertiesfaculty.montgomerycollege.edu/maronne/Ma180/power point/chap-3 … · Copyright © 2013 Pearson Education, Inc. All rights reserved