Question

How do you find the vertex of a parabola `y=2(x+3)^2+6`?

Answer 1

The vertex of `y=2(x+3)^2+6` is at `(-3,6)`

Explanation:

Any parabolic equation of the form:
`color(white)("XXXX")` `y=m(x-a)^2+b`
is said to be in vertex form with it's vertex at `(a,b)`

`y=2(x+3)^2+6` is almost in vertex form.

It can easily be converted to vertex form as:
`color(white)("XXXX")` `y=2(x-(-3))^2+6`