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Chapter 3
Graphs and Functions
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
3.1
Graphing Equations
Learning Objectives:
Plot ordered pairs
Determine whether an ordered pair of numbers is a
solution to an equation in two variables
Graph linear equations
Graph nonlinear equation
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Graphing an Ordered Pair
y-axis
Quadrant: II
Quadrant: I
The origin is the
point (0,0)
x-axis
Quadrant: III
Quadrant: IV
Where would the following points be located:
(2,4)
(-3,-5)
(-4,1)
(2,0)
Quad I
Quad III
Quad II
X-axis
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(0,-5)
Y-axis
Example 1a
Determine whether (3, –2) is a solution of 2x + 5y = −4.
2x + 5y = −4
Replace x with 3 and y with –2
2(3) + 5(–2) = −4
6 + (–10) = −4
−4 = −4
true
So (3, −2) is a solution of 2x + 5y = −4.
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Example 1b
Determine whether (–1, 6) is a solution of 3x – y = 5.
3x – y = 5
Replace x with –1 and y with 6
3(–1) – 6 = 5.
–3 – 6 = 5
–9 = 5
false
So (–1, 6) is not a solution of 3x – y = 5.
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Linear Equations in Two Variables
A linear equation in two variables is an equation that
can be written in the form
Ax + By = C
where A, B, and C are integers. A is positive. The
greatest common factor for A, B, and C is one.
This form is called standard form.
Finding x- and y-Intercepts
To find an x-intercept, find x when y = 0.
To find a y-intercept, find y when x = 0.
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Example 2
Graph the linear equation 2x – y = −4.
X-intercept:
Find x when y = 0
2x – 0 = -4
2x = -4
x = -2
X-int: (-2,0)
Y-intercept:
Find y when x = 0
2(0) – y = -4
– y = -4
y=4
Y-int: (0,4)
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For this example, we will graph
using the x- & y-intercepts
y
x
Example 3
3
4
Graph the linear equation y = x + 3.
For this one, we will use slope
and y-intercept to graph it.
Start with the y-intercept
b = 3,
written as (0,3)
The slope is
m=¾
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y
x
y
Example 4
Graph y  x 2.
x
y
2
4
1
1
0
0
1
1
2
4
5
5
5
x
5
This graph is not a line. Find ordered pairs and plot them on the
graph. Connect the points with a smooth curve.
This curve is given a special name, a parabola. The equation
is known as a quadratic equation.
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Example 5
y
Graph the equation y  x .
5
x
y
2 2
1 1
0
0
1
1
2
2
5
5
5
This graph is not a line. Find ordered pairs and plot them
on the graph. Connect the points.
We see that this graph is V-shaped.
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x
Example 6
y
Graph y = 4
Points on the line will include:
any point where the y-value is 4
(0,4), (1,4), (2,4),…
5
5
5
5
This is a horizontal line
The slope of a horizontal line is zero
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x
Example 7
y
Graph x = 4
Points on the line will include:
any point where the x-value is 4
(4,0), (4,1), (4,2),…
5
5
5
5
This is a vertical line
The slope of a vertical line is undefined
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x
Example 8
Are each of the following equations linear or not:
Y = 2X
Linear
Y = 2X2
Not linear (quadratic)
Y = │2X│
Not linear (absolute value)
Y = 2x – 4
Linear
Y = 2X2 – 4
Not linear (quadratic)
Y = │2X│– 4
Not linear (absolute value)
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3.1 Summary
Learning Objectives:
Plot ordered pairs
Determine whether an ordered pair of numbers is a
solution to an equation in two variables
Graph linear equations
Graph nonlinear equation
Key Vocabulary:
Rectangular coordinate system
Cartesian
Axis
Origin
Quadrant
Ordered pair
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Coordinate
Point
Solution
Standard form
Intercept
3.2
Introduction to Functions
Learning Objectives:
Define relation, domain, and range
Identify functions
Use the vertical line test for functions
Find the domain and range of a function
Use function notation
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Relation, Domain, and Range
A relation is a set of ordered pairs.
The domain of the relation is the set of all first coordinates of
the ordered pairs.
The range of the relation is the set of all second coordinates
of the ordered pairs.
A function is a relation in which each first coordinate in the
ordered pair corresponds to exactly one second coordinate.
I prefer to ask: Are the x-coordinates all different?
Vertical Line Test
If no vertical line can be drawn so that it intersects a graph more
than once, the graph is the graph of a function.
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Example 1
Determine the domain and range of the relation. Is the
relation a function?
{(4,9), (–4,9), (2,3), (10, –5)}
Domain: {4, –4, 2, 10}
Range:
{9, 3, –5}
Ask yourself: “Are the x-values all different?”
Answer: Yes, so yes it is a function
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Example 2
Is the relation y = x2 – 2x a function?
What type of graph is this?
A parabola which opens up
Does it pass the vertical line test?
Answer: yes, so yes it is a function
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Example 3
Is the relation x2 – y2 = 9 a function?
Each element of the domain (the x-values) would correspond with 2
different values of the range (both a positive and negative y-value),
the relation is NOT a function.
For ex: let x = 4 and solve for x
42 – y2 = 9
– y2 = –7
y2 = 7
y=
This would yield two point (4, ) and (4,
) and the xvalues are not all different, the relation is NOT a function
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Example 4
Use the vertical line test to determine whether each graph
is the graph of a function.
y
y
x
This is a function
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x
This is NOT a function
Example 5
Use the vertical line test to determine whether each graph
is the graph of a function.
y
x
This is NOT a function
Also, think about points on the line like: (-3,1), (-3,2), (-3,3)
Are the x-values all different? No
So it is NOT a function
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Example 6
y
Find the domain and
range of the function
graphed to the right.
Domain
x
Domain: 3 ≤ x ≤ 4
Range: 4 ≤ y ≤ 2
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Range
Example 7
y
Find the domain and
range of the function
graphed to the right.
Range
x
Domain: all real numbers
Range: y ≥ – 2
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Domain
Example 8
If y = x2 – 2x, re-write the equation in function
notation and then find f(–3).
f(x) = x2 – 2x
f(–3) = (–3)2 – 2(–3)
= 9 – (–6)
= 15
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Example 9
Given the graph of the following function, find each.
y
f(5) =
7
f(4) =
3
f(x)
f(5) = 1
x
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Example 9
Given the graph of the following function, find each.
y
For how many x-values
does f(x) = 2?
three
x ≈-3.5, x=0, x≈3.2
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f(x)
x
Learning Objectives:
Define relation, domain, and range
Identify functions
Use the vertical line test for functions
Find the domain and range of a function
Use function notation
Key Vocabulary:
Relation
Domain
Range
Function
Vertical line test
Function notation
Dependent/independent variable
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3.2 Summary
3.3
Graphing Linear Functions
Learning Objectives:
Graph linear functions given in slope-intercept form
Graph linear functions by finding intercepts
Graph vertical and horizontal lines
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Finding x- and y-Intercepts
To find an x-intercept
find x when y = 0
To find a y-intercept
find y when x = 0
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Example 1
Graph g(x) = 3x – 1. Compare this graph with the graph
of f(x) = 3x.
y
f(x) = 3x
g(x) = 3x – 1
Notice the graphs are
the parallel.
The graph of g(x) is
f(x) shifted down one
unit.
x
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Example 2
Graph the linear functions f(x) = –4x and g(x) = −4x – 5
on the same set of axis.
y
f(x) = −4x
g(x) = −4x – 5
x
Notice the graphs are
the same except the
graph of g(x) = −4x – 5
is shifted down five
units.
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Example 3
2
1
Find the y-intercept of y  x  .
3
9
2
1
 1
y

x

The y-intercept of
is  0,  .
3
9
 9
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Example 4
The computer will typically want
integer points
Find the intercepts and graph: 4 = x – 3y.
x – 3y = 4
x-intercept: find x, when y = 0
Then 4 = x – 3y becomes
4 = x – 3(0)
4=x
So the x-intercept is (4,0).
y-intercept: find y, when x = 0.
Then 4 = x – 3y becomes
4 = 0 – 3y
4 = – 3y.
4
 y
3
4
So the y-intercept is (0,  3 ).
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Solve for y and use the slope
–3y = –x + 4
y = 1/3 x – 4/3
y
(7, 1)
(4, 0)
4

3
(0, )
•
x
Add on to notes:
Find x-intercept and y-intercept and graph – x + 3y = 5
x-intercept: find x, when y = 0
Then – x + 3y = 5 becomes
– x + 3(0) = 5
x = –5
y
So the x-intercept is (-5,0).
y-intercept: find y, when x = 0.
Then – x + 3y = 5 becomes
(0) + 3y = 5
y = 5 /3
So the y-intercept is
(0,5/
3).
Solve the equation for y to find
the slope of the line
– x + 3y = 5
3y = x + 5
y = 1 /3 x + 5 /3
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•
•
•
x
Example 5
Find the intercepts and graph x = – 3
This line can have any points as long as x = -3
Points like: (-3,1), (-3,2), (-3,3)
y
X-intercept:
(-3,0)
Y-intercept:
none
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(-3, 0)
x
Example 6
Find the intercepts and graph y = 3
This line can have any points as long as y = 3
y
Points like: (1,3), (2,3), (3,3)
(0, 3)
X-intercept:
none
Y-intercept:
(0,3)
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x
Example 7
Homework: system says to graph
with x- and y-intercept
Graph x = 3y
y
Y-intercept:
(0,0)
Solve for y
x = 3y
3 3
1/
3x
=y
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(0, 0)
x
3.3 Summary
Learning Objectives:
Graph linear functions given in slope-intercept form
Graph linear functions by finding intercepts
Graph vertical and horizontal lines
Key Vocabulary:
Linear function
Vertical line
Horizontal line
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3.4
The Slope of a Line
Learning Objectives:
Find the slope given two points
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes
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Slope of a line passing through the points (x1, y1) and (x2, y2) is:
y2  y1
m
x2  x1
Slope-Intercept Form: y = mx + b,
m is the slope and (0,b) is the y-intercept
Slope of a horizontal line is zero
Slope of a vertical line is undefined
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Example 1
Find the slope of the line containing the points
(4, −3) and (2, 2). Graph the line.
y2  y1
2  (3) 3 52  5
m

 m 
x2  x1
2  4 4 22
2
5
Y
4
3
2
1
Notice this is an example of a
negative slope. The graph of the line
moves downward, or decreases, as
we go from left to right.
-5 -4 -3
-2 -1 0
-1
-2
-3
-4
-5
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X
1
2
3
4
5
Example 2
Find the slope and y-intercept of the line –3x + y = −5.
Write the equation in slope-intercept form: y = mx + b
Solve for y.
–3x + y = −5
+ 3x
+ 3x
y = 3x – 5
slope is 3
y-intercept is (0, – 5)
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Example 3
Find the slope and y-intercept of the line 2x – 6y = 12.
Write the equation in slope-intercept form: y = mx + b
Solve for y.
2x – 6y = 12
– 2x
–2x
– 6y = –________
2x + 12
–6
–6
y=
1
3
x–2
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1
3
slope is
y-intercept is (0, –2)
Example 4
Find the slope of the line x = −7.
To find the slope, let’s find two ordered pair solutions.
The solutions must have an x-value of −7.
Y
10
Let’s use (−7, 0) and (−7, 4).
8
y2  y1
m
x2  x1
40
4
m

7   7  0
6
4
2
-10 -8 -6
-4 -2 0
-2
-4
-6
-8
-10
The slope of a vertical line is undefined.
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X
2
4
6
8 10
Example 5
Find the slope of the line y = −4.
To find the slope, let’s find two ordered pair solutions.
The solutions must have an y-value of −4.
Y
Let’s use (0, −4) and (6, –4).
10
8
y2  y1
m
x2  x1
m
4   4 
60
6
4
2
0
 0
6
-10 -8 -6
-4 -2 0
-2
-4
-6
-8
-10
The slope of the horizontal line is zero.
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X
2
4
6
8 10
What type of slope does each of the following
graphs represent:
Negative slope
Down and to the right
Positive slope
Up and to the right
Zero slope
Up zero to the right
Undefined slope
Up and to the right zero
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Parallel Lines:
Have the same slopes but different y-intercepts
Perpendicular Lines:
Have the opposite and reciprocal slopes
The product of slopes of perpendicular lines = 1
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Example 8
Determine whether the following lines are parallel,
perpendicular, or neither.
–5x + y = –6
x + 5y = 5
Solve both equations for y.
Equation 1
Equation 2
x + 5y = 5
–5x + y = –6
y = 5x – 6
5y  x  5
1
y   x 1
5
The first equation has a slope of 5 and the second equation
1

has a slope of 5 , the lines are perpendicular.
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3.4 Summary
Learning Objectives:
Find the slope given two points
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes
Key Vocabulary:
Slope
Rate of Change
Slope-intercept form
Find the slope given an equation of a line
Find the slope of horizontal and vertical lines
Parallel and Perpendicular slopes
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3.5
Equations of Lines
Learning Objectives:
Use the slope-intercept form to write the equation of a line
Graph a line using the slope and y-intercept
Use the point-slope form to write the equation of a line
Write equations of vertical and horizontal lines
Write equations of parallel and perpendicular lines
Parallel lines have the same slope
Perpendicular lines have opposite and reciprocal slopes
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Slope-Intercept Form of an equation of a line is:
y = mx + b
slope of m
y-intercept of (0, b).
Point-Slope Form of an equation of line is:
y – y1 = m(x – x1)
slope of m
the line passes through (x1,y1)
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Example 1
Write the equation of the line with y-intercept (0, 5) and
2
slope of .
3
slope:
2
m
3
y-intercept: b  5
y  mx  b
2
y  x  (5)
3
2
y  x 5
3
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Example 2
1
Graph y  x  3.
4
Y-intercept: (0,-3)
Slope:
¼
Plot: (0, −3)
Rise up 1 unit and run to
the right 4 units.
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y
Example 3
Find an equation of a line with slope –2, through the point
(–11, –12). Write the equation in slope-intercept form.
Substitute the slope and point into the point-slope form of
an equation.
y  y1  m( x  x1 )
y – (–12) = –2(x – (–11))
y + 12 = –2(x + 11)
y + 12 = –2x – 22
– 12
– 12
y = –2x – 34
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Example 4
Find the equation of the line through (–4, 0) and (6, –1).
Write the equation in standard form.
Find the slope.
y2  y1
1  0
1
m


x2  x1 6  (4) 10
Substitute the slope and one of
the points into the point-slope
form.
y  y1  m( x  x1 )
1
y  0   ( x  (4))
10
10 y  1( x  4)
10 y   x  4
x  10 y  4
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Example 5
Find the equation of the line passing through points (–4, 3)
and (2, 5). Write the equation using function notation.
Find the slope.
y2  y1
53
2 1
m

 
x2  x1 2  (4) 6 3
y  y1  m( x  x1 )
1
y  3  ( x  (4))
3
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3(y – 3) = 1(x + 4)
3y – 9 = x + 4
+9
+9
3y = x + 13
3
3
1 13
f  x  x 
3
3
Example 6
Find an equation of the
horizontal line containing
the point (4, 5).
10
9
8
7
6
5
4
3
2
1
Y
0
-10-9 -8 -7 -6 -5 -4 -3 -2 -1
-1 1 2 3 4 5 6 7 8 9 10
-2
-3
-4
-5
-6
-7
-8
-9
-10
The equation is y = 5.
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X
Find an equation of the
line containing the point
(4, 5) with undefined
slope.
10
9
8
7
6
5
4
3
2
1
Y
X
0
-10-9 -8 -7 -6 -5 -4 -3 -2 -1
-1 1 2 3 4 5 6 7 8 9 10
-2
-3
-4
-5
-6
-7
-8
-9
-10
The equation is x = 4.
Example 7
Find an equation of a line that contains the point (– 2, 4)
and is parallel to the line x + 3y = 6.
Write the equation in standard form.
First, we need to find the slope of the given line.
x+ 3y = 6
3y =  x + 6
y = - 1/ 3x + 2
The slope of the given line is -1/3.
Parallel lines have the same slope. So we will
use the slope of -1/3 for our new equation,
together with the given point of (-2,4).
y – y1 = m(x – x1)
y–4=
- 1/
3(x
– (-2))
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3(y – 4) = -1(x + 2)
3y – 12 = – x – 2
+x
+x
x + 3y – 12 = – 2
+12 +12
x + 3y = 10
Example 8
Write a function that describes the line containing the
point (3, 5) and is perpendicular to the line 3x + 2y = 7.
First, we need to find the slope of the given line.
3x + 2y = 7
2y =  3x + 7
3
7

y = 2x + 2
Since perpendicular lines have slopes that are
negative reciprocals of each other, we use the
slope of 2/3 for our new equation, together with
the given point of (3,-5).
2
y  (5)  ( x  3)
3
3(y + 5) = 2(x – 3)
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3y + 15 = 2x – 6
– 15
– 15
3y = 2x – 21
3y = 2x – 21
3
3
2
f  x  x  7
3
Example 9
Find the equation of the perpendicular bisector to the
segment with endpoints at (-3,4) and (5,-6). Write the
equation in standard form.
Find the slope of the line segment
Find the midpoint of the line segment
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Use point-slope formula, but
remember to take the opposite and
reciprocal slope and use the midpoint.
5(y + 1) = 4(x – 1)
5y + 5 = 4x – 4
-4x
-4x
-4x + 5y + 5 = -4
-5 -5
-4x + 5y = -9
4x – 5y = 9
3.5 Summary
Learning Objectives:
Use the slope-intercept form to write the equation of a line
Graph a line using the slope and y-intercept
Use the point-slope form to write the equation of a line
Write equations of vertical and horizontal lines
Write equations of parallel and perpendicular lines
Parallel lines have the same slope
Perpendicular lines have opposite and reciprocal slopes
Formulas:
Slope-intercept: y = mx + b
Point-slope:
y – y1 = m(x – x1)
Standard form: Ax + By = C,
where A,B,&C are integers
A is positive
where A,B,&C’s GCD is 1
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UIL Math:
Show shortcuts
for the last two
examples
3.6
Graphing Piecewise-Defined
Functions and Shifting and
Reflecting Graphs of Functions
Learning Objectives:
Graph piecewise-defined functions
Vertical and horizontal shifts
Reflect graphs
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Example 1
Evaluate f(2), f(−4), and f(0) for the function
3x  1 if x  0
f (x)  
.
 x  4 if x  0
Then write your results in ordered-pair form.
f(2)
f(−4)
f(0)
f(2) = x – 4
f(−4) = 3x +1
f(0) = 3x + 1
f(2) = 2 – 4
f(−4) = 3(−4) +1
f(0) = 3(0) + 1
f(2) = –2
f(−4) = −11
f(0) = 1
(2, –2)
(−4, –11)
(0, 1)
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 2
3x  1 if x  0
Graph f (x)  
 x  3 if x  0
Graph each “piece” separately.
Values  0
Values > 0
x
f (x) = 3x – 1
0
– 1 (closed circle)
What is the domain and
range for the graph?
Domain: all real numbers
Range: y < -1 or y > 3
y
–1 – 4
–2 – 7
x
f (x) = x + 3
0
3 (open circle)
1
4
2
5
3
6
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x
Example 3
Graph
X
2X
X X–4
-2
-4
-2
-6
-3
-6
0
-4
-4
-8
1
-3
X
3X
1
3
2
6
3
9
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
y
x
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Vertical Shifts (Upward or Downward)
Let k be a Positive Number
Graph of
Same As
Moved
g(x) = f(x) + k
f(x)
k units upward
g(x) = f(x)  k
f(x)
k units downward
Horizontal Shifts (To the Left or Right)
Let h be a Positive Number
Graph of
Same As
Moved
g(x) = f(x  h)
f(x)
h units to the right
g(x) = f(x + h)
f(x)
h units to the left
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 4
Graph f  x   x 2 and g(x)  x 2  3.
y
Begin with the
graph of f(x) = x2.
Shift the original
graph downward
3 units.
What is the domain and range for
g(x) (the red graph)?
Domain: all real numbers
Range: y > –3
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
5
5
5
5
x
Example 5
Graph f (x)  x and g  x   x  2 .
Begin with the
graph of f(x) = |x|.
Shift the original
graph to the left
2 units.
What is the domain and range for
g(x) (the red graph)?
Domain: all real numbers
Range: y > 0
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
y
5
5
5
5
x
Reflections about the x-axis
The graph of g(x) = – f(x) is the graph of f(x)
reflected about the x-axis.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 6
Graph f (x)   x  2 + 14
y
5
The graph is reflected
about the x-axis, then
moved two units left and
four units up.
5
5
5
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
x
Example 7
Given the function
What is the domain?
What is the range?
Think about the parent function graph, and how it has been shifted
Right 20 and up 12 units
(20,12)
Domain: x > 20
Range: y > 12
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Learning Objectives:
Graph piecewise-defined functions
Vertical and horizontal shifts
Reflect graphs
Vocabulary:
Vertical shift
Horizontal shift
reflection
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
3.6 Summary
3.7
Graphing Linear Inequalities
Learning Objectives:
Graph linear inequalities
Graph the intersection or union of two linear inequalities
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 1
Graph 7x + y > –14.
• Graph y = –7x – 14 as a
dashed line.
•I used x-int of (0,-2) and
the slope of -7
• Pick a point not on the
graph: (0,0)
y
(0, 0)
x
• Test it in the original
inequality.
7(0) + 0 > –14,
0 > –14
• True, so shade the side
containing (0,0).
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 2
Graph 3x + 5y  –2.
• Graph
solid line.
y
as a
Points on the line (0,-2/5) (1, -1)
I used (1,-1) with a slope of -3/5
• Pick a point not on the
graph: (0,0), but just barely
• Test it in the original
inequality.
3(0) + 5(0) > –2
0 > –2
• False, so shade the side that
does not contain (0,0).
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
(0, 0)
x
Example 3
Graph 3x < 15.
y
• Graph 3x = 15 as a
dashed line. (x = 5)
• Pick a point not on
the graph: (0,0)
• Test it in the original
inequality.
3(0) < 15
0 < 15
• True, so shade the
side containing (0,0).
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
(0, 0)
x
Example 3
Graph the intersection of x  1 and y  2x – 1.
Graph each inequality. The intersection of the two graphs
is all points common in both regions.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 4
1
Graph the union of x  2 y  4 or y  2.
Graph each inequality. The union of both inequalities is
all the shaded regions.
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Warning!
3.7 Summary
Note that although all of our examples allowed us to select
(0, 0) as our test point, that will not always be true.
If the boundary line contains (0,0), you must select another
point that is not contained on the line as your test point.
Learning Objectives:
Graph linear inequalities
Graph the intersection or union of two linear inequalities
Vocabulary:
Boundary
Half-plane
Solution region
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
3.8
Stretching and Compression
Graphs of Absolute Value
Learning Objectives:
Graph absolute value functions
Write an equation of an absolute value function graph
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 1
1
Graph h(x) = 4|x| and g  x   x .
4
Find and plot orderedpairs solutions for the
functions.
x h(x) g(x)
−2
8
1/2
−1
1
1/4
0
0
0
1
4
1/4
2
8
1/2
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y
h(x) = 4|x|
f  x  x
g  x 
1
x
4
x
The Graph of the Absolute Value Function
f(x) = a│x – h│ + k
•Vertex @ (h,k) and is V-shaped
•Opens up if a is positive and down if a is negative
•Shifts h units right/left and k units up/down
•The larger the │a│ the narrower the graph
•The smaller the │a│ the wider the graph
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall
Example 2
1
Graph f  x    x  3  1.
4
y
What type of graph is this?
V-shape
Opens which direction?
downward
Vertex?
(−3, 1)
Steepness?
–¼
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Example 3
Write an equation of the absolute value function graphed.
5
Y
The vertex is (2,−1)
4
3
h= 2
2
1
X
-5
-4
-3
-2
-1
0
1
2
3
-1
4
5
k = −1
a= 2
-2
-3
y  a xh k
-4
-5
The equation of the graph is
y  2 x  2  1.
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3.8 Summary
Learning Objectives:
Graph absolute value functions
Write an equation of an absolute value function graph
Copyright © 2011 Pearson Education, Inc. Publishing as Prentice Hall