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

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Answer 1 · 2016-01-24T05:33:24

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

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

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

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

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

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

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

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

Now we can write our equation as

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

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

This is the vertex form.

What is the vertex form of y=x^2-16x+63y=x2−16x+63y=x^2-16x+63?