§ 2.4
Linear Functions and Slope
x- and y-Intercepts
Ax + By = C is called the standard form
of the equation of the line.
Using Intercepts to Graph Ax + By = C
1) Find the x-intercept. Let y = 0 and solve for x.
2) Find the y-intercept. Let x = 0 and solve for y.
3) Find a checkpoint, a third ordered-pair solution.
4) Graph the equation by drawing a line through the three points.
Blitzer, Intermediate Algebra, 5e – Slide #2 Section 2.4
127
Graphing Using Intercepts
128
EXAMPLE
Graph the equation using intercepts.
 2 x  4 y  12
SOLUTION
1) Find the x-intercept. Let y = 0 and then solve for x.
 2 x  4 y  12
 2 x  4  0  12
 2 x  12
x  6
Replace y with 0
Multiply and simplify
Divide by -2
The x-intercept is -6, so the line passes through (-6,0).
Blitzer, Intermediate Algebra, 5e – Slide #3 Section 2.4
Graphing Using Intercepts
CONTINUED
2) Find the y-intercept. Let x = 0 and then solve for y.
 2 x  4 y  12
 2  0  4 y  12
4 y  12
y3
Replace x with 0
Multiply and simplify
Divide by 4
The y-intercept is 3, so the line passes through (0,3).
Blitzer, Intermediate Algebra, 5e – Slide #4 Section 2.4
Graphing Using Intercepts
CONTINUED
3) Find a checkpoint, a third ordered-pair solution. For our
checkpoint, we will let x = 1 (because x = 1 is not the xintercept) and find the corresponding value for y.
 2 x  4 y  12
Replace x with 1
 2  1  4 y  12
 2  4 y  12
Multiply
Add 2 to both sides
4 y  14
y
14
4

7
2
 3 .5
Divide by 4 and simplify
The checkpoint is the ordered pair (1,3.5).
Blitzer, Intermediate Algebra, 5e – Slide #5 Section 2.4
Graphing Using Intercepts
128
CONTINUED
4) Graph the equation by drawing a line through the three
points. The three points in the figure below lie along the
same line. Drawing a line through the three points results in
the graph of  2 x  4 y  12 .
(0,3)
(1,3.5)
(-6,0)
Blitzer, Intermediate Algebra, 5e – Slide #6 Section 2.4
Slope of a Line
129
The slope m of the line through the two distinct
points ( x , y ) and ( x , y )
1
m 
1
2
is m, where
2
Change
in y
Change
in x

y 2  y1
x 2  x1
Note that you may call either point ( x1 , y 1 )
However, don’t subtract in one order in the numerator and use another order for
subtraction in the denominator - or the sign of your slope will be wrong.
Blitzer, Intermediate Algebra, 5e – Slide #7 Section 2.4
Slopes of Lines
130
Check Point 2 a
Find the slope of the line passing through the pair of points.
Then explain what the slope means.
(-3,4) and (-4,-2)
SOLUTION
Let
 x1 , y1     3 , 4  and  x 2 , y 2     4 ,  2 .
The slope is obtained as follows:
m 
Change in y
Change in x

y 2  y1
x 2  x1

 2  4 
 4   3 

 2  (4)
43

6
1
6
This means that for every six units that the line (that passes through
both points) travels up, it also travels 1 units to the right.
Blitzer, Intermediate Algebra, 5e – Slide #8 Section 2.4
Slopes of Lines
130
Check Point 2 b
Find the slope of the line passing through the pair of points.
Then explain what the slope means.
(4,-2) and (-1,5)
SOLUTION
Let  x1 , y 1    4 ,  2  and
 x 2 , y 2     1,5 .
The slope is obtained as follows:
m 
Change in y
Change in x

y 2  y1
x 2  x1

5   2 
 1  4 

5  (2)
 1  (4)
Blitzer, Intermediate Algebra, 5e – Slide #9 Section 2.4

7
5

7
5
Slopes of Lines
130
Possibilities for a Line’s Slope
Positive Slope
Negative Slope
Zero Slope
Undefined Slope
Blitzer, Intermediate Algebra, 5e – Slide #10 Section 2.4
Slopes of Lines
131
Slope-Intercept Form of the Equation of a Line
y = mx + b with slope m, and y-intercept b
Examples
y = .968x – 17.3
f (x) = 3-5x
m = .968; b = -17.3
m = -5; b = 3
Blitzer, Intermediate Algebra, 5e – Slide #11 Section 2.4
Slopes of Lines
131
EXAMPLE
Give the slope and y-intercept for the line whose equation is:
3x - 5y = 7
SOLUTION
Hint: You should solve for y so as to have the equation
in slope intercept form.
3x - 5y = 7
3x - 3x - 5y = -3x+7
-5y = -3x+7
-5y

 3x  7
5
-5
y
3
5
x
7
5
Subtract 3x from both sides
Simplify
Divide both sides by -5
Simplify
Therefore, the slope is 3/5 and the y-intercept is -7/5.
Blitzer, Intermediate Algebra, 5e – Slide #12 Section 2.4
Slopes of Lines
132
Check Point 3
Give the slope and y-intercept for the line whose equation is:
8x - 4y = 20
SOLUTION
Solve for y so as to have the equation in slope intercept form.
8x - 4y = 20
8x - 8x - 4y = -8x+20
-4y = -8x+20
-4y
-4

 8 x  20
4
y  2x  5
Subtract 8x from both sides
Simplify
Divide both sides by -4
Simplify
Therefore, the slope is 2 and the y-intercept is -5.
Blitzer, Intermediate Algebra, 5e – Slide #13 Section 2.4
Slopes of Lines
132
Graphing y = mx + b Using the Slope and y-Intercept
1) Plot the point containing the y-intercept on the y-axis. This is the point
(0,b).
2) Obtain a second point using the slope, m. Write m as a fraction, and use
rise-over-run, starting at the point containing the y-intercept, to plot this point.
3) Use a straightedge to draw a line through the two points. Draw arrowheads
at the ends of the line to show that the line continues indefinitely in both
directions
Blitzer, Intermediate Algebra, 5e – Slide #14 Section 2.4
Graphing Lines
EXAMPLE
Graph the line whose equation is f (x) = 2x + 3.
SOLUTION
The equation f (x) = 2x + 3 is in the form y = mx + b.
The slope is 2 and the y-intercept is 3.
Now that we have identified the slope and the y-intercept,
we use the three steps in the box to graph the equation.
Blitzer, Intermediate Algebra, 5e – Slide #15 Section 2.4
132
Graphing Lines
CONTINUED
1) Plot the point containing the y-intercept on the y-axis.
The y-intercept is 3. We plot the point (0,3), shown below.
(0,3)
Blitzer, Intermediate Algebra, 5e – Slide #16 Section 2.4
Graphing Lines
CONTINUED
2) Obtain a second point using the slope, m. Write m as a
fraction, and use rise over run, starting at the point
containing the y-intercept, to plot this point. We express
the slope, 2, as a fraction.
m 
2
1

Rise
Run
We plot the second point on the line by starting
at (0,3), the first point. Based on the slope, we
move 2 units up (the rise) and 1 unit to the
right (the run). This puts us at a second point
on the line, (1,-1), shown on the graph.
Blitzer, Intermediate Algebra, 5e – Slide #17 Section 2.4
(1,5)
(0,3)
Graphing Lines
CONTINUED
3) Use a straightedge to draw a line through the two
points. The graph of y = 2x + 3 is show below.
(1,5)
(0,3)
Blitzer, Intermediate Algebra, 5e – Slide #18 Section 2.4
Graphing Lines
Check Point 4
Graph the line whose equation is f (x) = 4x - 3.
SOLUTION
The equation f (x) = 4x - 3 is in the form y = mx + b.
The slope is 4 and the y-intercept is -3.
With the slope and the y-intercept, graph the equation.
Blitzer, Intermediate Algebra, 5e – Slide #19 Section 2.4
133
Horizontal and Vertical Lines
134
EXAMPLE 6 on page 134
Graph the linear equation: y = -4.
SOLUTION
All ordered pairs that are solutions of y = -4 have a value of y
that is always -4. Any value can be used for x. Let’s select
three of the possible values for x: -3, 1, 6:
x
y = -4
(x,y)
-3
-4
(-3,-4)
1
-4
(1,-4)
6
-4
(6,-4)
(1,-4)
(-3,-4)
Upon plotting the three resultant points and connecting the
points with a line, the graph to the right is the solution.
Blitzer, Intermediate Algebra, 5e – Slide #20 Section 2.4
(6,-4)
Horizontal and Vertical Lines
135
EXAMPLE – similar to Ex 7 on page 135
Graph the linear equation: x = 5.
SOLUTION
All ordered pairs that are solutions of x = 5 have a value of x
that is always 5. Any value can be used for y. Let’s select three
of the possible values for y: -3, 1, 5:
(5,5)
x=5
y
(x,y)
5
-3
(5,-3)
5
1
(5,1)
5
5
(5,5)
(5,1)
(5,-3)
Upon plotting the three resultant points and connecting the
points with a line, the graph to the right is the solution.
Blitzer, Intermediate Algebra, 5e – Slide #22 Section 2.4
Slope as Rate of Change
p 136
A linear function that models data is described. Slope may be
interpreted as the rate of change of the dependent variable per unit
change in the independent variable.
• Find the slope of each model.
• Then describe what this means in terms of the rate of change of
the dependent variable y per unit change in the independent
variable x.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change
p 136
EXAMPLE 8
Using the Figure 2.27 graph on page 136 which show the percentage of
Americans in two age groups who reported using illegal drugs in the previous
month.
SOLUTION
• Find the slope of the line for the 12-17 age group.
m 
y 2  y1
x 2  x1

9 . 9  11 . 6
2005  2002

 1 .7
  0 . 57
3
• The slope indicates that for the 12-17 age group, the percentage
reporting using illegal drugs in the previous month decreased by
approximately 0.57% per year. The rate of change is a decrease
of approximately 0.57%.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change
p 136
Check Point 8
Using the Figure 2.27 graph on page 136 to find the slope of the line segment for
the 50-59 age group. Express the slope correct to two decimal places and
describes what it represents.
SOLUTION
• Find the slope of the line for the 50-59 age group.
m 
4 .4  2 .7
2005  2002

1 .7
 0 . 57
3
• For the 50-67 age group, the percentage reporting using illegal
drugs in the previous month increased by approximately 0.57%
per year.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Modeling with Slope-Intercept
p 139
EXAMPLE 10
Using the Figure 2.31(b) scatter plot on page 138 find a function in the form
T(x)=mx+b.that models the average ticket price for a movie T(x), x years after
1990 and use the model to predict the average ticket price in 2010.
SOLUTION
• Find the slope.
m 
y 2  y1
x 2  x1

6 . 41  4 . 23
15  0
 0 . 15
• The average ticket price for a movie, T(x), x years after 1990 can be
modeled by the linear function. The slope indicates that ticket prices
increase $0.15 per year.
T ( x )  0 . 15 x  4 . 23
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Modeling with Slope-Intercept
p 139
EXAMPLE 10, continued
• Because 2010 is 20 years after 1990, substitute 20 for x and
evaluate.
T ( 20 )  0 . 15 ( 20 )  4 . 23  7 . 23
• The model predicts an average ticket price of $7.23 in 2010.
• Try doing Check Point 10 on your own.
Note that vertical intercept and y-intercept are the same.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
DONE
Graphing Using Intercepts
129
Check Point 1
Graph the equation using intercepts.
3x  2 y  6
1) Find the x-intercept. Let y = 0 and then solve for x.
3x  2  0  6
3x  6 x  2
The x-intercept is 2, so the line passes through (2,0).
2) Find the y-intercept. Let x = 0 and then solve for y.
30  2y  6
- 2y  6 y   3
The y-intercept is -3, so the line passes through (0,-3).
3) Find a checkpoint, a third ordered-pair solution. Let x=4.
34  2y  6
- 2y   6 y  3
The checkpoint is the ordered pair (4,3).
Blitzer, Intermediate Algebra, 5e – Slide #28 Section 2.4
Slopes of Lines
EXAMPLE
Find the slope of the line passing through the pair of points.
Then explain what the slope means.
(-7,-8) and (4,-2)
SOLUTION
Let  x1 , y 1     7 ,  8  and
 x 2 , y 2    4 ,  2 .
Remember, the slope is obtained as follows:
m 
Change in y
Change in x

y 2  y1
x 2  x1

 2   8 
4   7 

28
47

6
11
This means that for every six units that the line (that passes through
both points) travels up, it also travels 11 units to the right.
Blitzer, Intermediate Algebra, 5e – Slide #29 Section 2.4
Slope as Rate of Change
p 136
EXAMPLE
The linear function y = 2x + 24 models the average cost in dollars of
a retail drug prescription in the United States, y, x years after 1991.
SOLUTION
• The slope of the linear model is 2.
• This means that every year (since 1991) the average cost in
dollars of a retail drug prescription in the U.S. has increased
approximately $2.
Blitzer, Intermediate Algebra, 5e – Slide #23 Section 2.4
Slope as Rate of Change
130
The change in the y values is the vertical change or the “Rise”.
The change in the x values is the horizontal change or the “Run.”
m is commonly used to denote the slope of a line. The absolute
value of a line is related to its steepness.
When the slope is positive, the line rises from left to right.
When the slope is negative, the line falls from left to right.
Blitzer, Intermediate Algebra, 5e – Slide #31 Section 2.4
Slope Intercept Form
131
In the form of the equation of the line where the
equation is written as y = mx + b,
it is true that if x = 0, then y = b.
Therefore, b is the y intercept for the equation of the
line. Furthermore, the x coefficient m is the slope of the
line.
Blitzer, Intermediate Algebra, 5e – Slide #32 Section 2.4