Question

How do you find the vertex of `y = x^2 - 4x - 12`?

Answer 1

It is in `(2, -16)`.

Explanation:

The vertex is the minimum or the maximum of a parabola. Then it is enough to search the zero of the derivative. We want to solve

`d/dx (x^2-4x-12)=0`
that is
`2x-4=0`
and
`x=2`.
To find the `y` it is enough to substitute in the original equation

`y=2^2-4*2-12`
`y=-16`.

The vertex is then in the point `(2, -16)` and we can verify it looking at the plot.

graph{x^2-4x-12 [-9.255, 10.735, -17.9, -7.89]}

Answer 2

Vertex at `(2,-16)`

Explanation:

The general vertex form for a parabola is
`color(white)("XXX")y=color(green)(m)(x-color(red)(a))^2+color(blue)(b)`
with vertex at `(color(red)(a),color(blue)(b))`

Converting the given equation `y=x^2-4x-12` into vertex form:

(extracting the `color(green)(m)` and completing the square:
`color(white)("XXX")y=color(green)(1)(x^2-4x+4)-12-4`

(into vertex form)
`color(white)("XXX")y=color(green)(m)(x-color(red)(2))^2+color(blue)(""(-16))`
with vertex at `(color(red)(2),color(blue)(""(-16)))`

For verification purposes, here is the graph of the given equation:
graph{x^2-4x-12 [-13.07, 18.96, -16.62, -0.59]}