## College Algebra (6th Edition)

$j(x)=(x-1)^{2}-1$
$f(x)=(x+1)^{2}-1$ $g(x)=(x+1)^{2}+1$ $h(x)=(x-1)^{2}+1$ $j(x)=(x-1)^{2}-1$ First, you will notice the graph has the vertex $(1,-1)$. The $-1$ inside the parentheses of $j(x)$ moves to the right one unit and the outside number moves down one unit. Therefore, $j(x)=(x-1)^{2}-1$ is the correct answer.