What is the vertex of this parabola?
y-5 = x^2?

Work Shown:

y-5 = x^2

y = x^2 + 5 ... add 5 to both sides

y = 1x^2 + 5

y = 1(x-0)^2 + 5 ... x replaced with (x-0)

The equation is now in the form y = a(x-h)^2 + k

We have a = 1, h = 0, k = 5.

The vertex is (h,k) = (0,5)

The value of 'a' determines how vertically stretched or compressed the parabola is. Also, if a > 0, then the parabola opens upward; or if a < 0, then the parabola opens downward.