# How do you write y=-x^2+12x-11 in vertex form?

Oct 10, 2016

$x = \left(- 1\right) {\left(x - 6\right)}^{2} + 25$

#### Explanation:

Note that the general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = - {x}^{2} + 12 x - 11$

Extract the $\textcolor{g r e e n}{m}$ factor from the first two terms:
$\textcolor{w h i t e}{\text{XXX")y=color(green)(} \left(- 1\right)} \left({x}^{2} - 12 x\right) - 11$

If $\left({x}^{2} - 12 x\right)$ are the first two terms of a squared binomial the third term must be ${6}^{2}$
[since (x-a)^2=(x^2-2ax+a^2)]
and anything we add to complete the square must also be subtracted.

Completing the square
color(white)("XXX")y=color(green)(""(-1))(x^2-12xcolor(magenta)(+6^2))-11 color(magenta)(-color(green)(""(-1)) * (6^2))

Simplifying:
$\textcolor{w h i t e}{\text{XXX")y=color(green)(} \left(- 1\right)} {\left(x - \textcolor{red}{6}\right)}^{2} + \textcolor{b l u e}{25}$

Here is a graph of the original equation to help verify this result:
graph{-x^2+12x-11 [-3.106, 12.697, 18.77, 26.67]}