Question

What is the the vertex of `x = (y + 6)^2 - 3`?

Answer 1

The vertex is `(-3,-6)`.

Explanation:

Expand the parabola:

`(y+6)^2-3 = y^2+12y+36-3=y^2+12y+33`

The vertex is the minimum of a parabola, so we can derive it and set the derivative to zero:

`2y+12=0 \iff y=-6`.

So, the vertex has `y`-coordinate `-6`. To find the `x`-coordinate, simply compute

`f(-6)=(-6+6)-3=-3`