Question

How do you write `f(x) = x² - 8x +11` in vertex form?

Answer 1

`f(x)=(x-4)^2-5`

Explanation:

To go from Standard Form to Vertex Form, we need to complete the square. That means we need to find the constant that makes `x^2-8x` a perfect square.

`y=x^2-8x+11`

To find that constant, we use this equation: `c=(1/2*b)^2` or `(1/2*-8)^2` or `16`.

We now have the constant that completes the square, but we can't just add a value to an equation. We can either add `16` to both sides or add it and the immediately subtract it. Either way works

`y=x^2-8xcolor(blue)(+16-16)+11`
`y=(x^2-8x+16)-16+11`

`x^2-8x+16` is a perfect square, so let's simplify:

`y=(x-4)^2-16+11`
`y=(x-4)^2-5`
`f(x)=(x-4)^2-5`

That's our vertex form, so now we're done. Good job!