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.