This document contains examples and explanations of properties of quadratic functions, including how to: - Graph quadratic functions by determining if the parabola opens up or down based on the leading coefficient, and finding the vertex, axis of symmetry, and intercepts. - Determine the domain and range of a quadratic function from its graph. - Identify where a quadratic function is increasing or decreasing based on its graph. - Find the quadratic function with given properties like a specific vertex or y-intercept. - Determine if a quadratic function has a maximum or minimum value and calculate its value.

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Section 3.3
Quadratic Functions
and Their Properties

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( )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= + −

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Without graphing, locate the vertex and axis of symmetry of the
parabola defined by . Does it open up or down?( ) 2
2 3 2f x x x= − +
3 7
Vertex is , .
4 8
 
 ÷
 
( )
( )
3 3
2 2 2 4
b
a
− −
− = =
2
3 3 3 7
2 3 2
4 4 4 8
f
     
= − + = ÷  ÷  ÷
     
3
Axis of symmetry is .
4
x =
Because 2 0, the parabola opens up.a = >

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15. 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 7
Vertex is , .
4 8
 
 ÷
 
3
Axis 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
, 2
2
 
 ÷
  By symmetry, the point with the same
3
-value but to the right of the axis of
4
3 3 3
symmetry is on the graph.
4 4 2
3
so the point , 2 is on the graph.
2
y
+ =
 
 ÷
 

16. (b) Determine the domain and the range of .
(c) Determine where is increasing and decreasing.
f
f
−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.
7
Based on the graph, the range is the interval , .
8
 
∞ ÷
 
( )0, 2
3
, 2
2
 
 ÷
 
The function is
3
from ,
4
and
3
from , .
4
 
−∞ ÷
 
 
∞ ÷
 
incr
decreasi
g
ng
easin

17. −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 x
x y
= + −
h = −
b
2a
= −
4
2(2)
= −1
Since a = 2 > 0 the parabola opens up.
( )1, 3− −
-intercepts can be found when ( ) 0.x f x =
( ) ( )
2
1 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− −
2
0 2 4 1 Use the quadratic formula to solve.x x= + −
2
4 4 4(2)( 1) 4 24 2 6
=
2(2) 4 2
x
− ± − − − ± − ±
= =
-intercepts 0.22 and 2.22x ≈ −
Axis of symmetry: 1x = −

18. −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− −
(b) Determine the domain and the range of .
(c) Determine where is increasing and decreasing.
f
f
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

19. −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
= − − −
h = −
b
2a
= −
−2
2 −
1
2




= −2
Since a is negative, the parabola opens down.
( )2,0−
As seen on the graph, the -intercept is 2.x −
( ) ( )
21
2 2 2( 2) 2 0
2
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 = −

20. −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− −
(b) Determine the domain and the range of .
(c) Determine where is increasing and decreasing.
f
f
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

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Determine the quadratic function whose vertex is (–2, 3) and
whose y-intercept is 1.
( ) ( ) ( )
2 2
2 3f x a x h k a x= − + = + +
Using the fact that the y-intercept is 1: 1= a 0 + 2( )
2
+ 3
1 4 3a= + 1
2
a = −
( ) ( )
21
2 3
2
f x x= − + + −4 −3 −2 −1 1 2 3 4 5
−4
−3
−2
−1
1
2
3
4
x
y
( ) 21
2 1
2
f x x x= − − +

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( ) 2
Determine whether the quadratic function
4 5
has 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 1
b
x
a
= − = − =
−
( ) 2
So the maximum value is 2 (2) 4(2) 5 9f = − + + =

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