# What is the vertex of  y= (x-3)^2-2x^2-x-2?

Aug 31, 2017

vertex at: $\left(- 3 \frac{1}{2} , + 19 \frac{1}{4}\right)$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{m a g \ne t a}{{\left(x - 3\right)}^{2}} - 2 {x}^{2} - x - 2$

Expanding
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{m a \ge n t a}{{x}^{2} - 6 x + 9} - 2 {x}^{2} - x - 2$
and simplifying
$\textcolor{w h i t e}{\text{XXX}} y = - {x}^{2} - 7 x + 7$

We would like to convert this into vertex form: $y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

First extract the $\textcolor{g r e e n}{m}$ factor from the first 2 terms
$\textcolor{w h i t e}{\text{XXX")y=color(green)(} \left(- 1\right)} \left({x}^{2} + 7 x\right) + 7$
Complete the square
color(white)("XXX")y=color(green)(""(-1))(x^2+7xcolor(brown)(+(7/2)^2))+7color(brown)(-color(green)(""(-1))(7/2)^2)

$\textcolor{w h i t e}{\text{XXX")y=color(green)(} \left(- 1\right)} {\left(x + \frac{7}{2}\right)}^{2} + 7 + \frac{49}{4}$

color(white)("XXX")y=color(green)(""(-1))(x-color(red)(""(-7/2)))^2+color(blue)(77/4)
which is the vertex form with vertex at $\left(\textcolor{red}{- \frac{7}{2}} , \textcolor{b l u e}{\frac{77}{4}}\right) = \left(\textcolor{red}{- 3 \frac{1}{2}} , \textcolor{b l u e}{19 \frac{1}{4}}\right)$