Question

How do you find the coordinates for the vertex of `f(x)=2x^2-8x+1`?

Answer 1

The vertex is at `(2,-7)`.

Explanation:

`f(x)=2x^2-8x+1`

We have to put the equation into vertex form by completing the square.
`f(x)=2[x^2-4x+1/2]`

`f(x)=2[(x-2)^2-7/2]`

`f(x)=2(x-2)^2-7`

The vertex is at `(2,-7)`.

Check using the graph:
graph{2x^2-8x+1 [-10, 10, -7.76, 2.24]}