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

Jan 24, 2016

$y = {\left(x - 8\right)}^{2} - 1$

Explanation:

$y = {x}^{2} - 16 x + 63$

We need to convert our equation to the form $y = a {\left(x - h\right)}^{2} + k$

Let us use completing the square.

$y = \left({x}^{2} - 16 x\right) + 63$

We need to write ${x}^{2} - 16 x$ as a perfect square.

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

${x}^{2} - 16 x + 64 - 64$

This would become ${\left(x - 8\right)}^{2} - 64$

Now we can write our equation as

$y = {\left(x - 8\right)}^{2} - 64 + 63$

$y = {\left(x - 8\right)}^{2} - 1$

This is the vertex form.