Question

What is the y coordinate of the vertex of a parabola `y=x^2-4x+12`?

Answer 1

Vertex is `(2, 8)`

Method
Arrange the equation in the form `y - k = 4a(x - h)^2`

where `(h, k)` is your vertex.

So we have,

`=>y = x^2 - 4x + 12`

By completing the square,

`=> y = x^2 - 4x + (-4/2)^2 - (-4/2)^2 + 12`

`=> y = (x - 2)^2 - 4 + 12`

`=> y - 8 = (x - 2)^2`

So we get our vertex to be `(2, 8)`