10TH
EDITION
COLLEGE
ALGEBRA
LIAL
HORNSBY
SCHNEIDER
2.4 - 1
2.4 Linear Functions
Graphing Linear Functions
Standard Form Ax + By = C
Slope
Average Rate of Change
Linear Models
2.4 - 2
Linear Function
A function  is a linear function if, for
real numbers a and b,
( x )  ax  b.
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Example 1
GRAPHING A LINEAR FUNCTION
USING INTERCEPTS
Graph (x) = –2x + 6. Give the domain and
range.
Solution The x-intercept is found by letting
(x) = 0 and solving for x.
( x )  2 x  6
0  2 x  6
x 3
Add 2x; divide by 2.
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Example 1
GRAPHING A LINEAR FUNCTION
USING INTERCEPTS
Graph (x) = –2x + 6. Give the domain and
range.
Solution
The x-intercept is 3, so we plot (3, 0). The
y-intercept is
(0)  2(0)  6  6
Plot this point and connect the two points
with a straight line. Find a check point.
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Example 1
GRAPHING A LINEAR FUNCTION
USING INTERCEPTS
Graph (x) = –2x + 6. Give the domain and
range.
y
Solution
(0, 6)
y-intercept
( x )  2 x  6
check point
(2, 2)
x
The domain and range
are both(–, ).
(3, 0)
x-intercept
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Example 2
GRAPHING A HORIZONTAL LINE
Graph (x) = –3. Give the domain and
range.
y
Solution Since (x),
or y, always equals –3,
the value of y can
never be 0.
A line with no xintercept is parallel to
the x-axis.
The domain is (–, ).
The range is {–3}.
( x )  3
x
Horizontal line
(0, –3)
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Example 3
GRAPHING A VERTICAL LINE
Graph x = –3. Give the domain and range.
y
Solution Since x
always equals –3,
Vertical
the value of x can
never be 0, and the
( –3, 0)
graph has no yintercept and is
parallel to the yx  3
axis
line
x
2.4 - 8
Example 3
GRAPHING A VERTICAL LINE
Graph x = –3. Give the domain and range.
y
The domain of
this relation,
which is not a
function, is {–3}.
The range is
(–, ).
Vertical line
( –3, 0)
x
x  3
2.4 - 9
Motion Problems
Note In this text we will agree that if the
coefficients and constant in a linear equation are
rational numbers, then we will consider the
standard form to be Ax  By  C
where A ≥ 0, A, B, and C are integers, and the
greatest common factor of A, B, and C is 1. (If
two or more integers have a greatest common
factor of 1, they are said to be relatively prime.)
2.4 - 10
Example 4
GRAPHING Ax + By = C WITH C = 0
Graph 4x –5y = 0. Give the domain and
range.
Solution Find the intercepts.
4(0)  5y  0
y 0
Let x = 0.
y-intercept
4 x  5(0)  0
x 0
Let y = 0.
x-intercept
This graph has one intercept-at the origin.
2.4 - 11
Example 4
GRAPHING Ax + By = C WITH C = 0
Graph 4x –5y = 0. Give the domain and
range.
y
Solution Graph the
intercept (0, 0) and find
4 x  5y  0
another point.
( 5, 4)
4 x  5y  0
( 0, 0)
4(5)  5y  0
x
20  5 y  0
4y
2.4 - 12
Slope
An important characteristic of a straight line
is its slope, a numerical measure of the
steepness of a line. Geometrically it may be
interpreted as the ratio of rise to run.
Use two distinct points. The change in the
horizontal distance, x2 – x1, is denoted as ∆x
(delta x) and the change in the vertical
distance, y2 – y1, is denoted as ∆y.
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Slope
The slope m of a line through points (x1, y1) and
(x2, y2) is
rise y y 2  y1
m


,
run x x2  x1
where ∆x ≠ 0.
2.4 - 14
Caution When using the slope formula, it
makes no difference which point is used (x1, y1)
or (x2, y2); however, be consistent . Start with the
x- and y-values of one point (either one) and
subtract the corresponding values of the other
point.y  y
y1  y 2
y 2  y1
y1  y 2
2
1
Use
or
, not
or
.
x2  x1
x1  x2
x1  x2
x2  x1
Be sure to write the difference of the y-values
in the numerator and the difference of the xvalues in the denominator.
2.4 - 15
Undefined Slope
The slope of a vertical line is
undefined.
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Example 5
FINDING SLOPES WITH THE
SLOPE FORMULA
Find the slope of the line through the given
points.
a. (  4,8), (2, 3)
Solution Let x1 = –4, y1 = 8, and x2 = 2,
y2 = –3. Then, rise  y  3  8  11
and run  x  2  ( 4)  6.
rise y 11
11
m



run x
6
6
2.4 - 17
FINDING SLOPES WITH THE
SLOPE FORMULA
Example 5
Find the slope of the line through the given
points.
b. (2,7), (2,  4)
Solution
 4  7 11
m

22
0
Undefined
The slope of a vertical line is undefined.
2.4 - 18
Example 5
FINDING SLOPES WITH THE
SLOPE FORMULA
Find the slope of the line through the given
points.
c. (5, 3), ( 2, 3)
Solution
3  ( 3) 0
m

0
2  5
7
Drawing a graph through these two points
would produce a horizontal line.
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Zero Slope
The slope of a horizontal line is 0.
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Eample 6
FINDING THE SLOPE FROM AN
EQUATION
Find the slope of the line.
y   4x  3
Solution Find any two ordered pairs that are
solutions of the equation.
If x  2, then y   4( 2)  3  5,
and if x  0, then y   4(0)  3  3.
so two ordered pairs are ( 2,5) and (0, 3).
3  5 8
m

  4.
0  ( 2) 2
2.4 - 21
Example 7
GRAPHING A LINE USING A
POINT AND THE SLOPE
Graph the line passing through
5
( 1,5) and having slope  .
3
Solution Locate the point and move 5 units
down and three units horizontally to the right.
This gives a second point (2, 0) which can be
used to complete the graph.
2.4 - 22
Example 7
GRAPHING A LINE USING A
POINT AND THE SLOPE
y
(– 1, 5)
Down
5
(2, 0)
x
Right
3
2.4 - 23
Slopes
1. A line with a positive slope rises from left
to right.
2. A line with a negative slope falls from left
to right.
3. When the slope is positive, the function is
increasing.
4. When the slope is negative, the function
is decreasing.
2.4 - 24
Average Rate of Change
We know that the slope of a line is the ratio
of the vertical change in y to the horizontal
change in x. So, the slope gives the rate of
change in y per unit of change in x, where
the value of y depends on the value of x. If
 is a linear function defined on [a, b], then
( b )  ( b )
average rate of change on a, b  
.
ba
2.4 - 25
Example 8
INTERPRETING SLOPE AS
AVERAGE RATE OF CHANGE
In 2001, sales of DVD players numbered 12.7
million. In 2006, estimated sales of DVD
players were 19.8 million. Find the average
rate of change in DVD players, in millions, per
year.
Solution If x = 2001 with y = 12.7 and x =
2006 with y = 19.8, then the ordered pairs are
(2001, 12.7) and (2006, 19.8)
2.4 - 26
Example 8
INTERPRETING SLOPE AS
AVERAGE RATE OF CHANGE
In 2001, sales of DVD players numbered 12.7
million. In 2006, estimated sales of DVD players
were 19.8 million. Find the average rate of change
in DVD players, in millions, per year.
Solution (2001, 12.7) and (2006, 19.8)
19.8  12.7 7.1
average rate of change 

 1.42
2006  2001 5
The line through the ordered pair rises from left to
right and therefore has positive slope. The sales of
DVD players increased by an average of 1.42
million each year from 2001 to 2006.
2.4 - 27
Linear Models
A linear cost function has the form
C( x )  mx  b
where x represents the number of items produced,
m represents the variable cost per item, and b
represents the fixed costs.
The fixed costs do not change as more items are
made. The variable cost per item increases as
more product is made. The revenue function for
selling depends on the price per item p and the
number of items sold x.
2.4 - 28
Linear Models
A linear cost function has the form
C( x )  mx  b
The revenue function has the form
R( x )  px
Profit is described by the profit function
defined as
P ( x )  R( x )  C( x ).
2.4 - 29
Example 9
WRITING LINEAR COST, REVENUE,
AND PROFIT FUNCTIONS
Assume that the cost to produce an item is a
linear function and all items produced are
sold. The fixed cost is $1500, the variable
cost per item is $100, and the item sells for
$125. Write linear functions to model
a. cost
Solution C( x )  mx  b
C( x )  100 x  1500
2.4 - 30
Example 9
WRITING LINEAR COST, REVENUE,
AND PROFIT FUNCTIONS
Assume that the cost to produce an item is a
linear function and all items produced are
sold. The fixed costs is $1500, the variable
cost per item is $100, and the item sells for
$125. Write linear functions to model
b. revenue
Solution
R( x )  px
R( x )  px  125 x
2.4 - 31
Example 9
WRITING LINEAR COST, REVENUE,
AND PROFIT FUNCTIONS
Assume that the cost to produce an item is a
linear function and all items produced are
sold. The fixed costs is $1500, the variable
cost per item is $100, and the item sells for
$125. Write linear functions to model
c. profit
P ( x )  R( x )  C( x ).
Solution  125 x  (100 x  1500)
Use
parentheses
here.
 125 x  100 x  1500
 25 x  1500
2.4 - 32
Example 9
WRITING LINEAR COST, REVENUE,
AND PROFIT FUNCTIONS
Assume that the cost to produce an item is a
linear function and all items produced are
sold. The fixed costs is $1500, the variable
cost per item is $100, and the item sells for
$125.
d. How many items must be sold for the
company to make a profit?
Solution To make a profit, P(x) must be
positive.
2.4 - 33
Example 9
WRITING LINEAR COST, REVENUE,
AND PROFIT FUNCTIONS
Solution To make a profit, P(x) must be
positive.
P ( x )  25 x  1500
P( x )  0
25 x  1500  0
25 x  1500
x  60
P ( x )  25 x  1500
Add 1500 to each side.
Divide by 25.
Since the number of items must be a whole number,
at least 61 items must be sold for the company to
make a profit.
2.4 - 34