Answer
The vertex form of the function is $y=(x-2)^{2}+5.$ The vertex is $(2,5)$.
Work Step by Step
$ y=x^{2}-4x+9\qquad$ ...write in form of $x^{2}+bx=c$ (add $-9$ to each side).
$ y-9=x^{2}-4x\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{-4}{2})^{2}=(-2)^{2}=4\qquad$ ...complete the square by adding $4$ to each side of the expression
$ y-9+4=x^{2}-4x+4\qquad$ ... write $x^{2}-4x+4$ as a binomial squared.
$ y-5=(x-2)^{2}\qquad$ ...add $5$ to each side of the expression
$ y=(x-2)^{2}+5\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
The vertex form of a quadratic function is $y=a(x-h)^{2}+k$ where $(h,k)$ is the vertex of the function's graph.
Here, $h=2,\ k=5$, so the vertex is $(2,5)$
