# How do you write the Vertex form equation of the parabola y = –x^2 +12x – 4 ?

May 8, 2017

$y = - {\left(x - 6\right)}^{2} + 32$

#### Explanation:

The general vertex form of a parabola is:
$\textcolor{w h i t e}{\text{XXX")y=color(green)m(x-color(red)a)^2+color(blue)bcolor(white)("XXX}}$with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given:
$\textcolor{w h i t e}{\text{XXX}} y = - {x}^{2} + 12 x - 4$

Extract the $\textcolor{g r e e n}{m}$ factor from the first 2 terms:
$\textcolor{w h i t e}{\text{XXX}} y = \left(\textcolor{g r e e n}{- 1}\right) \left({x}^{2} - 12 x\right) - 4$

Complete the square
color(white)("XXX")y=(color(green)(-1))(x^2-12xcolor(magenta)(+(12/2)^2))-4color(magenta)(-(color(green)(-1))*((12/2)^2)

Simplify
$\textcolor{w h i t e}{\text{XXX}} y = \left(\textcolor{g r e e n}{- 1}\right) \left({x}^{2} - 12 x \textcolor{m a \ge n t a}{+ {6}^{2}}\right) - 4 \textcolor{m a \ge n t a}{+ 36}$

Write as a squared binomial and a simplified constant
$\textcolor{w h i t e}{\text{XXX}} y = \left(- 1\right) {\left(x - \textcolor{red}{6}\right)}^{2} + \textcolor{b l u e}{32}$
which is a specific vertex form with vertex at $\left(\textcolor{red}{6} , \textcolor{b l u e}{32}\right)$

Here is a graph of the original equation for verification purposes: 