Chapter 8
More on Functions
and Graphs
§ 8.1
Graphing and Writing
Linear Functions
Linear Functions
Identifying Linear Functions
By
the vertical line test, we know that all
linear equations except those whose graphs
are vertical lines are functions.
Thus, all linear equations except those of the
form x = c (vertical lines) are linear functions.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
3
Graphing Linear Functions
Example:
Graph the linear function f (x) =
3
4
x + 3.
Let x = 4.
f (4) =
3
4
(4) + 3
f (4) = 3 + 3 = 6
Replace x with 4.
Simplify the right side.
One solution is (4, 6).
Continued.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
4
Graphing Linear Functions
Example continued:
Graph the linear function f (x) =
3
4
x + 3.
For the second solution, let x = 0.
f (0) =
3
4
(0) + 3
f (0) = 0 + 3 = 3
Replace x with 0.
Simplify the right side.
So a second solution is (0, 3).
Continued.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
5
Graphing Linear Functions
Example continued:
Graph the linear function f (x) =
3
4
x + 3.
For the third solution, let x = – 4.
f (– 4) =
3
4
(– 4) + 3
f (– 4) = – 3 + 3 = 0
Replace x with – 4.
Simplify the right side.
The third solution is (– 4, 0).
Continued.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
6
Graphing Linear Functions
y
Example continued:
(4, 6)
Plot all three of the
solutions (4, 6), (0, 3) and
(– 4, 0).
Draw the line that
contains the three
points.
(0, 3)
(– 4, 0)
Martin-Gay, Beginning and Intermediate Algebra, 4ed
x
7
Writing Linear Functions
Example:
Find an equation of the line whose slope is 5 and contains the point
(4, 3). Write the equation using function notation.
m = 5, x1 = 4, y1 = 3
y – y1 = m(x – x1)
y – (– 3) = 5(x – 4)
y + 3 = 5x – 20
y = 5x – 23
f (x) = 5x – 23
Substitute the values for m, x1, and y1.
Simplify and distribute.
Subtract 3 from both sides.
Replace y with f (x).
Martin-Gay, Beginning and Intermediate Algebra, 4ed
8
Writing Linear Functions
Example:
Write a function that describes the line containing the point
(4, 1) and is perpendicular to the line 5x – y = 20
 y =  5x + 20
y = 5x  20
Solve the equation for y to find the slope
from the slope-intercept form.
5 is the slope of the line perpendicular to the one needed.
As perpendicular lines have slopes that are negative
reciprocals of each other, the slope of the line we want is
-
1
.
5
Continued.
Martin-Gay, Beginning and Intermediate Algebra, 4ed
9
Writing Linear Functions
Example:
Write a function that describes the line containing the point
(4, 1) and is perpendicular to the line 5x – y = 20
m=
-
1
5
,. x1 = 4, y1 = 1
y – y1 = m(x – x1)
y – (1) =
-
1
5
y+1=
-
y =
-
f (x) =
-
(x – 4)
1
5
1
5
1
5
x
+
x
-
x
Substitute the values for m, x1, and y1.
4
5
1
-
5
1
5
Simplify and distribute.
Subtract 1 from both sides.
Replace y with f (x).
Martin-Gay, Beginning and Intermediate Algebra, 4ed
10