## Calculus: Early Transcendentals 8th Edition

The standard absolute value graph is shifted 2 units to the left due to $h=-2.$
The general form of the absolute value function is $f(x)=a|x-h|+k.$ In this problem, $h=-2,$ $a=1,$ and $k=0.$ Since $h=-2,$ we shift the absolute value graph 2 units to the left.