Answer
The vertex form of the function is $y=(x-(-3))^{2}+(-6).$ The vertex is $(-3,-6)$.
Work Step by Step
$ y=x^{2}+6x+3\qquad$ ...prepare to complete the square.
$ y+(?)=x^{2}+6x+(?)+3\qquad$ ...square half the coefficient of $x$.
$(\displaystyle \frac{6}{2})^{2}=3^{2}=9\qquad$ ...add $9$ to each side of the expression
$ y+9=x^{2}+6x+9+3\qquad$ ... write $x^{2}+6x+9$ as a binomial squared.
$ y+9=(x+3)^{2}+3\qquad$ ...add $-9$ to each side of the expression
$ y+9-9=(x+3)^{2}+3-9\qquad$ ...simplify.
$ y=(x+3)^{2}-6\qquad$ ...write in vertex form $y=a(x-h)^{2}+k$.
$y=(x-(-3))^{2}+(-6)$
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, $h=-3,\ k=-6$, so the vertex is $(-3,-6)$
