The parabola will show the vertex in the format: y-k = (x-h)^2, where the vertex point
lies at (h, k).
[tex]so \: for \: {x}^{2} + 2x - y + 3 = 0[/tex]
let's first put it in "y =" standard format:
[tex]y = {x}^{2} + 2x + 3[/tex]
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
[tex]y = {(x +1)}^{2} ...\: is \: y = {x}^{2} + 2x + 1[/tex]
So there's +2 leftover, since 3-1=2; so:
[tex]y = {(x + 1)}^{2} + 2[/tex]
Now we'll subtract the 2 from both sides to show our vertex:
[tex]y - 2 = {(x + 1)}^{2} [/tex]
where our vertex (h, k) is at (-1, 2)
