Answer
False.
Work Step by Step
This function is simply a transformation of the function $y=x^2$, which has a vertex at $(0,0)$. The $(x-h)^2$ term pushes the function $h$ units right, and the $+k$ term pushes it up $k$ units, making the new vertex located at $(h,k)$, not $(-h,k)$. The statement must be false.
