Upload
michael-cellini
View
1.403
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Cellini, EDLM, Math, Absolute Value
Citation preview
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