College Algebra, 2013a

180 Linear and Quadratic Functions
3. We re-write i(x) =4− 2|3x +1| =4− 2f(3x +1)=−2f(3x + 1) + 4 and apply Theorem
1.7. First, we take care of the changes on the ‘inside’ of the absolute value. Instead of |x|,
we have |3x +1|, so, in accordance with Theorem 1.7, wefirstsubtract 1fromeachofthe
x-values of points on the graph of y = f(x), then divide each of those new values by 3. This
effects a horizontal shift left 1 unit followed by a horizontal shrink by a factor of 3. These
transformations move (−1, 1) to ( − 2 3 , 1) ,(0, 0) to ( − 1 3 , 0) and (1, 1) to (0, 1). Next, we take
care of what’s happening ‘outside of’ the absolute value. Theorem 1.7 instructs us to first
multiply each y-value of these new points by −2 thenadd 4. Geometrically, this corresponds
to a vertical stretch by a factor of 2, a reflection across the x-axis and finally, a vertical shift
up 4 units. These transformations move ( − 2 3 , 1) to ( − 2 3 , 2) , ( − 1 3 , 0) to ( − 1 3 , 4) ,and(0, 1) to
(0, 2). Connecting these points with the usual ‘∨’ shape produces our graph of y = i(x).
y
(−1, 1)
4
3
2
1
−3 −2 −1 (0, 0) 1 2 3
y
(1, 1)
x
−−−−−−−−−−−−→
, 4) (
−
1
3
3
(
−
2
3 , 2) 2 (0, 2)
1
−1 1 x
−1
f(x) =|x| i(x) =−2f(3x +1)+4
= −2|3x +1| +4
While the methods in Section 1.7 can be used to graph an entire family of absolute value functions,
not all functions involving absolute values posses the characteristic ‘∨’ shape. As the next example
illustrates, often there is no substitute for appealing directly to the definition.
Example 2.2.4. Graph each of the following functions. Find the zeros of each function and the
x- andy-intercepts of each graph, if any exist. From the graph, determine the domain and range
of each function, list the intervals on which the function is increasing, decreasing or constant, and
find the relative and absolute extrema, if they exist.
1. f(x) = |x|
x
2. g(x) =|x +2|−|x − 3| +1
Solution.
1. We first note that, due to the fraction in the formula of f(x), x ≠ 0. Thus the domain is
(−∞, 0) ∪ (0, ∞). To find the zeros of f, we set f(x) = |x|
x
= 0. This last equation implies
|x| =0,which,fromTheorem2.1, implies x = 0. However, x = 0 is not in the domain of f,