College Algebra, 2013a

2.2 Absolute Value Functions 179
Example 2.2.3. Graph the following functions starting with the graph of f(x) =|x| and using
transformations.
1. g(x) =|x − 3| 2. h(x) =|x|−3 3. i(x) =4− 2|3x +1|
Solution. We begin by graphing f(x) =|x| and labeling three points, (−1, 1), (0, 0) and (1, 1).
y
4
3
2
(−1, 1) 1 (1, 1)
−3 −2 −1 (0, 0) 1 2 3 x
f(x) =|x|
1. Since g(x) =|x − 3| = f(x − 3), Theorem 1.7 tells us to add 3toeachofthex-values of the
points on the graph of y = f(x) to obtain the graph of y = g(x). This shifts the graph of
y = f(x) totheright 3 units and moves the point (−1, 1) to (2, 1), (0, 0) to (3, 0) and (1, 1)
to (4, 1). Connecting these points in the classic ‘∨’ fashion produces the graph of y = g(x).
y
y
4
4
3
3
2
2
(−1, 1)
1
(1, 1)
1
(2, 1)
(4, 1)
−3 −2 −1 (0, 0) 1 2 3
f(x) =|x|
x
shift right 3 units
−−−−−−−−−−−−→
add3toeachx-coordinate
1 2 (3, 0) 4 5 6 x
g(x) =f(x − 3) = |x − 3|
2. For h(x) =|x|−3=f(x) − 3, Theorem 1.7 tells us to subtract 3fromeachofthey-values of
the points on the graph of y = f(x) to obtain the graph of y = h(x). This shifts the graph of
y = f(x) down 3 units and moves (−1, 1) to (−1, −2), (0, 0) to (0, −3) and (1, 1) to (1, −2).
Connecting these points with the ‘∨’ shape produces our graph of y = h(x).
y
y
4
1
(−1, 1)
3
2
1
(1, 1)
−3 −2 −1 1 2 3
−1
(−1, −2) −2
−3
(1, −2)
(0, −3)
x
−3 −2 −1 (0, 0) 1 2 3
f(x) =|x|
x
shift down 3 units
−−−−−−−−−−−−→
subtract 3 from each y-coordinate
−4
h(x) =f(x) − 3=|x|−3