The parabola will show the vertex in the format: y-k = (x-h)^2, where the vertex point
lies at (h, k).

let's first put it in "y =" standard format:

Since we cannot get a perfect square out of this, we complete the square: a=1, b=2, c=3
(b/2)^2 = (2/2)^2 = 1, so

So there's +2 leftover, since 3-1=2; so:

Now we'll subtract the 2 from both sides to show our vertex:

where our vertex (h, k) is at (-1, 2)