Answer
The vertex form of the function is $y=(x-(-6))^{2}+1.$ The vertex is $(-6,1)$.
Work Step by Step
$ y=x^{2}+12x+37\qquad$ ...prepare to complete the square.
$ y+?=x^{2}+12x+?+37\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{12}{2})^{2}=(6)^{2}=36\qquad$ ...complete the square by adding $36$ to each side of the expression
$ y+36=x^{2}+12x+36+37\qquad$ ... write $x^{2}+12x+36$ as a binomial squared.
$ y+36=(x+6)^{2}+37\qquad$ ...add $-36$ to each side of the expression
$ y+36-36=(x+6)^{2}+37-36\qquad$ ...simplify.
$ y=(x+6)^{2}+1\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$y=(x-(-6))^{2}+1$
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, $h=-6,\ k=1$, so the vertex is $(-6,1)$
