Answer
Vertex form: $y=(x+3)^2-2$
Work Step by Step
The vertex form of a quadratic function is $f(x) = a (x – h)^2 + k$, where $a, h, k$ are constants of the parabola at $(h, k)$.
Now completing the square for $y=x^2+6x+7$ we get, $y=(x+3)^2-2$
Thus $(-3,-2)$ is the vertex.
