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Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
Section 3.5
Inequalities Involving
Quadratic Functions
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
2
Solve the inequality 5 4 0 and graph the solution set.x x+ + >
( ) 2
Graph the function = 5 4.f x x x+ +
( ) ( )2
-intercept: 0 = 0 5 0 4 4y f + + =
2
-intercepts: 0 5 4 ( 4)( 1)x x x x x= + + = + +
4 or 1x = − −
5
Vertex:
2 2
b
x
a
= − = −
2
5 5 9
5 4
2 2 4
y
   
= − + − + = − ÷  ÷
   
5 9
,
2 4
 
− − ÷
 
−6 −5 −4 −3 −2 −1 1 2 3
−2
−1
1
2
3
4
5
6
x
y
Where is this function
greater than 0?
( ) ( ), 4 or 1,−∞ − − ∞ −6 −5 −4 −3 −2 −1 0 1 2 3
x
y
) (
2
Solve the inequality 6 and graph the solution set.x x≤ +
( ) 2
Graph the function = 6.f x x x− −
( ) ( )2
-intercept: 0 = 0 0 6 6y f − − = −
2
-intercepts: 0 6 ( 3)( 2)x x x x x= − − = − +
3 or 2x = −
1
Vertex:
2 2
b
x
a
= − =
2
1 1 25
6
2 2 4
y
   
= − − = − ÷  ÷
   
1 25
,
2 4
 
− ÷
 
Where is this function
less than or equal to 0?
[ ]2,3−
2
6 0x x− − ≤
−4 −3 −2 −1 1 2 3 4
−6
−5
−4
−3
−2
−1
1
2
−4 −3 −2 −1 0 1 2 3 4
[ ]
2
Solve the inequality 6 and graph the solution set.x x≤ +
( ) ( )2
Graph and= .6g xf x xx = +
2
Points of intersection: 6x x= +
( ) ( )3 9 2 4f f= − =
3 or 2x = −
Where is f(x) less than
or equal to g(x)?
[ ]2,3−
−4 −3 −2 −1 0 1 2 3 4
[ ]
( ) ( ) ( ) ( )2
If and 6 then we want to solve .f x x g x x f x g x= = + ≤
−5 −4 −3 −2 −1 1 2 3 4 5
−2
−1
1
2
3
4
5
6
7
8
9
10
11
12
2
6 0x x− − = ( 3)( 2) 0x x− + =
( )3,9
( )2,4−
2
Solve the inequality 2 8 9 0 and graph the solution set.x x− + − <
( ) 2
Graph the function = 2 8 9.f x x x− + −
8
Vertex: 2
2 2( 2)
b
x
a
= − = − =
−
( ) ( )
2
2 2 8 2 9 1y = − + − = − ( )2, 1−
( ) ( ) ( )
2
-intercept: 0 = 2 0 8 0 9 9y f − + − = −
By symmetry, the point (4, 9) is also on the graph.−
−2 −1 1 2 3 4 5 6
−11
−10
−9
−8
−7
−6
−5
−4
−3
−2
−1
This function is always less than 0
so the solution set is all real numbers.
−2 −1 0 1 2 3 4 5 6
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
2
Solve the inequality 2 8 9 0 and graph the solution set.x x− + − <
( ) 2
Graph the function = 2 8 9.f x x x− + −
8
Vertex: 2
2 2( 2)
b
x
a
= − = − =
−
( ) ( )
2
2 2 8 2 9 1y = − + − = − ( )2, 1−
( ) ( ) ( )
2
-intercept: 0 = 2 0 8 0 9 9y f − + − = −
By symmetry, the point (4, 9) is also on the graph.−
−2 −1 1 2 3 4 5 6
−11
−10
−9
−8
−7
−6
−5
−4
−3
−2
−1
This function is always less than 0
so the solution set is all real numbers.
−2 −1 0 1 2 3 4 5 6
Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.

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Section 3.5 inequalities involving quadratic functions

  • 1. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall. Section 3.5 Inequalities Involving Quadratic Functions
  • 2. Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 3. 2 Solve the inequality 5 4 0 and graph the solution set.x x+ + > ( ) 2 Graph the function = 5 4.f x x x+ + ( ) ( )2 -intercept: 0 = 0 5 0 4 4y f + + = 2 -intercepts: 0 5 4 ( 4)( 1)x x x x x= + + = + + 4 or 1x = − − 5 Vertex: 2 2 b x a = − = − 2 5 5 9 5 4 2 2 4 y     = − + − + = − ÷  ÷     5 9 , 2 4   − − ÷   −6 −5 −4 −3 −2 −1 1 2 3 −2 −1 1 2 3 4 5 6 x y Where is this function greater than 0? ( ) ( ), 4 or 1,−∞ − − ∞ −6 −5 −4 −3 −2 −1 0 1 2 3 x y ) (
  • 4. 2 Solve the inequality 6 and graph the solution set.x x≤ + ( ) 2 Graph the function = 6.f x x x− − ( ) ( )2 -intercept: 0 = 0 0 6 6y f − − = − 2 -intercepts: 0 6 ( 3)( 2)x x x x x= − − = − + 3 or 2x = − 1 Vertex: 2 2 b x a = − = 2 1 1 25 6 2 2 4 y     = − − = − ÷  ÷     1 25 , 2 4   − ÷   Where is this function less than or equal to 0? [ ]2,3− 2 6 0x x− − ≤ −4 −3 −2 −1 1 2 3 4 −6 −5 −4 −3 −2 −1 1 2 −4 −3 −2 −1 0 1 2 3 4 [ ]
  • 5. 2 Solve the inequality 6 and graph the solution set.x x≤ + ( ) ( )2 Graph and= .6g xf x xx = + 2 Points of intersection: 6x x= + ( ) ( )3 9 2 4f f= − = 3 or 2x = − Where is f(x) less than or equal to g(x)? [ ]2,3− −4 −3 −2 −1 0 1 2 3 4 [ ] ( ) ( ) ( ) ( )2 If and 6 then we want to solve .f x x g x x f x g x= = + ≤ −5 −4 −3 −2 −1 1 2 3 4 5 −2 −1 1 2 3 4 5 6 7 8 9 10 11 12 2 6 0x x− − = ( 3)( 2) 0x x− + = ( )3,9 ( )2,4−
  • 6. 2 Solve the inequality 2 8 9 0 and graph the solution set.x x− + − < ( ) 2 Graph the function = 2 8 9.f x x x− + − 8 Vertex: 2 2 2( 2) b x a = − = − = − ( ) ( ) 2 2 2 8 2 9 1y = − + − = − ( )2, 1− ( ) ( ) ( ) 2 -intercept: 0 = 2 0 8 0 9 9y f − + − = − By symmetry, the point (4, 9) is also on the graph.− −2 −1 1 2 3 4 5 6 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 This function is always less than 0 so the solution set is all real numbers. −2 −1 0 1 2 3 4 5 6 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.
  • 7. 2 Solve the inequality 2 8 9 0 and graph the solution set.x x− + − < ( ) 2 Graph the function = 2 8 9.f x x x− + − 8 Vertex: 2 2 2( 2) b x a = − = − = − ( ) ( ) 2 2 2 8 2 9 1y = − + − = − ( )2, 1− ( ) ( ) ( ) 2 -intercept: 0 = 2 0 8 0 9 9y f − + − = − By symmetry, the point (4, 9) is also on the graph.− −2 −1 1 2 3 4 5 6 −11 −10 −9 −8 −7 −6 −5 −4 −3 −2 −1 This function is always less than 0 so the solution set is all real numbers. −2 −1 0 1 2 3 4 5 6 Copyright © 2012 Pearson Education, Inc. Publishing as Prentice Hall.