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

1 Answer
Jan 24, 2016

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.