Question

What is the vertex form of `y=x^2-16x+63`?

Answer 1

`y=(x-8)^2 - 1`

Explanation:

`y=x^2-16x+63`

We need to convert our equation to the form `y=a(x-h)^2+k`

Let us use completing the square.

`y=(x^2-16x) + 63`

We need to write `x^2-16x` as a perfect square.

For this divide coefficient of `x` by `2` and square the result and add and subtract with the expression.

`x^2-16x +64 - 64`

This would become `(x-8)^2 - 64`

Now we can write our equation as

`y=(x-8)^2-64 + 63`

`y=(x-8)^2 - 1`

This is the vertex form.