Question

What is the vertex of the parabola `y=3(x-4)^2-22`?

Answer 1

`(4, -22)`

Explanation:

The equation:

`y = 3(x-4)^2-22`

is in vertex form:

`y = a(x-h)+k`

with multiplier `a = 3` and vertex `(h, k) = (4, -22)`

The nice thing about vertex form is that you can immediately read the vertex coordinates from it.

Notice that `(x-4)^2 >= 0`, taking its minimum value `0` when `x=4`. When `x=4` we have `y = 3(4-4)^2-22 = 0-22 = -22`.

So the vertex is at `(4, -22)`.