Question

How do you find the vertex for `2x^-16x+4`?

Answer 1

I think you mean `2x^2-16x+4`

`2x^2-16x+4 = 2(x-4)^2-28`

This is in vertex form allowing us to read the vertex as `(4, -28)`

Explanation:

Let `f(x) = 2x^2-16x+4`

Then `f(x) = 2(x^2-8x+16)-32+4 = 2(x-4)^2-28`

`f(x) = 2(x-4)^2-28` is in vertex form, being of the form:

`f(x) = a(x-x_1)+y_1`,

where `(x_1, y_1) = (4, -28)` is the vertex and `a` is some constant.

More explicitly:

The minimum value of `f(x)` occurs when `(x-4) = 0`, that is when `x = 4`.

`f(4) = 2(4-4)^2-28 = 0-28 = -28`

So the vertex is at `(4, -28)`