College Algebra, 2013a

178 Linear and Quadratic Functions
4
y
1
y
3
2
1
1 2 3 4 5
g(x) =|x − 3|
x
−3 −2 −1 1 2 3
−1
−2
−3
−4
h(x) =|x|−3
x
4. As before, we set i(x) = 0 to find the zeros of i and get 4−2|3x+1| = 0. Isolating the absolute
value term gives |3x+1| =2,soeither3x+1 = 2 or 3x+1 = −2. We get x = 1 3
or x = −1, so
our x-intercepts are ( 1
3 , 0) and (−1, 0). Substituting x = 0 gives y = i(0) = 4−2|3(0)+1| =2,
for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives
i(x) =
{
{
4 − 2(−(3x +1)), if (3x +1)< 0 6x +6, if x