Math 20-1 Chapter 7 Absolute Value and Reciprocal Functions

7.2 Absolute Value Function

Teacher Notes

7.2 Graph an Absolute Value Function

Absolute value is defined as the distance from

zero in the number line. Absolute value of -6 is

6, and absolute value of 6 is 6, Both are 6 units

from zero in the number line.

if 0 if 6 then 6

if 0 if –6 then – 6

x xx

x x

Piecewise Definition

7.2.1

Express the distance between the two points, -6 and 6, using

absolute value in two ways.

6 6 6 6

x y = | 2x – 4| y = | 2x – 4|

–2

0

2

4

6

Graph an Absolute Value Function

Method 1: Sketch Using a Table of Values

x y = | 2x – 4| y = | 2x – 4|

–2 |2(-2) - 4| 8

0 |2(0) – 4| 4

2 |2(2) - 4| 0

4 |2(4) – 4| 4

6 |2(6) – 4| 8

:{ | }

:{ | 0, }

Domain x x R

Range y y y R

The x-intercept occurs at the point (2, 0)

The y-intercept occurs at the point (0, 4)

The x-intercept of the linear function is the x-intercept of the

corresponding absolute value function. This point may be called an

invariant point. 7.2.2

y = | 2x – 4|

2 4x2 4x

continuous

Graph an Absolute Value Function

Method 2: Using the Graph of the Linear Function y = 2x - 4

1. Graph y = 2x - 4

2. Reflect in the x-axis the

part of the graph of y = 2x – 4

that is below the x-axis.

3. Final graph y = |2x – 4|

Slope = 2

7.2.3

y = | 2x – 4|

Y-intercept = -4

x-intercept = 2

Express as a piecewise function. y = |2x – 4|

Recall the basic definition of absolute value.

if 0

– if 0

x xx

x x

The expression in the abs is a line

with a slope of +1 and x-intercept of 0.

Note that 0 is the invariant point and can be determined by

making the expression contained within the absolute value

symbols equal to 0.7.2.4

function pieces

domain x < 0

has negative y-values,

It must be reflected in the x-axis,

multiply by -1

0domain x > 0

−(x) x

2 4 if 22 4

(2 4) if 2

x xx

x x

Domain x < 2

Express as a piecewise function. y = |2x – 4|

Invariant point: expression = 0

x- intercept 2x – 4 = 0

x = 2

Domain x > 2

As a piecewise function y = |2x – 4| would be

2

7.2.5

+–2x - 4expression

−(2x - 4) (2x - 4)|2x - 4| function pieces

expression

Slope positive

Be CarefulGraph the absolute value function y = |-2x + 3| and express

it as a piecewise function

x-intercept

2 3 0

3

2

x

x

3

2

-+

– (-2x+3)+ (-2x+3)

Piecewise function:

32 3 if

22 3

3( 2 3) if

2

x x

x

x x

7.2.6

Slope negative

Domain x < 3/2 x > 3/2

:{ | }

:{ | 0, }

Domain x x R

Range y y y R

Suggested Questions:

Page 375:

1a, 2, 3, 5a,b, 6a,c,e, 9a,b, 12, 16,

True or False:

The domain of the function y = x + 2 is always the same as the

domain of the function y = |x + 2|.

The range of the function y = x + 2 is always the same as the range of

the function y = |x + 2|.

True

False

7.2.7

x y = x2 - 4 y = |x2 - 4|

–3 5

-2 0

0 -4

2 0

3 5

5

Create a Table of Values to Compare y = f(x) to y = | f(x) |

0

4

0

5

7.2.8

2x 2x2 2x

Part B: Abs of Quadratic Functions

Compare and contrast the domain and range of the

function and the absolute value of the function.

What is the effect of the vertex of the original function

when the absolute value is taken on the function?

Graph an Absolute Value Function of the Form f(x) = |ax2 + bx + c|

Sketch the graph of the function

Express the function as a piecewise function

2( ) 3 4f x x x

1. Graph y = x2 – 3x - 4

2. Reflect in the x-axis the

part of the graph of

y = x2 – 3x - 4 that is below

the x-axis.

3. Final graph y = |x2 – 3x - 4 |

7.2.9

4 1y x x

( 4)( 1) 0

4 or 1

x x

x x

x-int

vertex

(1.5, -6.25)

:{ | }

:{ | 0, }

Domain x x R

Range y y y R

How does this compare

to the original function?

Express as a piecewise function.

Critical points2 3 4 0

( 4)( 1) 0

4 or 1

x x

x x

x x

-1 4+ – +

2 3 4x x 2 3 4x x2( 3 4)x x

2 3 4x x

Piecewise function:

2

2

2

3 4 if -1 43 4

( 3 4) if -1< 4

x x xx x

x x x

7.2.10

expression = 0

x2 - 3x - 4

expression

2( ) 3 4f x x x

x-intercepts

a > 0, opens up

x < -1 x > 4

-1 < x < 4

Be Careful with DomainGraph the absolute value function y = |-x2 + 2x + 8| and express it as a

piecewise function

Piecewise function:

2

2

2

2 8 if -2 42 8

( 2 8) if -2> 4

x x xx x

x x x

7.2.11

Domain

Critical points

2

2

- 2 8 0

2 8 0

( 4)( 2) 0

4 or 2

x x

x x

x x

x x

-2 4

2 2 8x x

expression = 0

x-intercepts

a < 0, opens down

x < -2 x > 4-2 < x < 4

2 2 8x x2 2 8x x

vertex

(1, 9)

Absolute Value as a Piecewise Function

Match the piecewise definition with the graph of an absolute value function.

7.2.12

Suggested Questions:

Page 375:

4, 7b, 8b,c,d, 10a,c, 11b,d, 13, 15, 20,

7.2.12