Answer
$(2, 0)$
Work Step by Step
The vertex-form of a quadratic function is $y=a(x-h)^{2}+k$ where the vertex is the opposite of $h$ and $k$, or $(h, k)$. Given the function from the problem, $f(x)=(x-2)^{2}$, written in vertex form this function looks like $f(x)=1(x-2)^{2}+0$. Therefore the vertex is at $(2, 0)$.