What is the vertex form of y=x^2-16x+72 ?

Nov 15, 2015

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

Explanation:

The vertex form of a parabola is in the form $y = a {\left(x - h\right)}^{2} + k$, where the vertex is at the point $\left(h , k\right)$.

In order to find the vertex, we must complete the square. When we have $y = {x}^{2} - 16 x + 72$, we should think about it as y=color(red)(x^2-16x+?)+72, so that color(red)(x^2-16x+?) is a perfect square.

Perfect squares appear in the form ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$. We already have an ${x}^{2}$ in both, and we know that $- 16 x = 2 a x$, that is, $2$ times $x$ times some other number. If we divide $- 16 x$ by $2 x$, we see that $a = - 8$. Therefore, the completed square is ${x}^{2} - 16 x + 64$, which is equivalent to ${\left(x - 8\right)}^{2}$.

However, we're not done. If we plug $64$ into our equation, we must counteract that somewhere else to keep both sides equal. So, we can say that $y = \textcolor{red}{{x}^{2} - 16 x + 64} + 72 - 64$. This way, we have added and subtracted $64$ to the same side, so the equation has not actually been changed because $64 - 64 = 0$.

We can rewrite $y = \textcolor{red}{{x}^{2} - 16 x + 64} + 72 - 64$ to resemble the form $y = a {\left(x - h\right)}^{2} + k$.

$y = \textcolor{red}{{x}^{2} - 16 x + 64} + 72 - 64$
$y = \textcolor{red}{{\left(x - 8\right)}^{2}} + 72 - 64$
color(blue)(y=(x-8)^2+8

With this equation, we can determine that the vertex $\left(h , k\right)$ is at the point $\left(8 , 8\right)$.