# What is the vertex form of y= 2x^2 - 9x – 18 ?

Jul 5, 2017

$y = 2 {\left(x - \frac{9}{4} x\right)}^{2} - 28 \frac{1}{8}$

$a {\left(x + b\right)}^{2} + c$

This is vertex form, giving the vertex as $\left(- b , c\right)$ which is:

$\left(2 \frac{1}{4} , - 28 \frac{1}{8}\right)$

#### Explanation:

Write it in the form $a {\left(x + b\right)}^{2} + c$

$y = 2 \left[{x}^{2} \textcolor{b l u e}{- \frac{9}{2}} x - 9\right] \text{ } \leftarrow$ factor out $2$ to to get $1 {x}^{2}$

Complete the square by adding and subtracting $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2}}$

$\textcolor{b l u e}{{\left(\left(- \frac{9}{2}\right) \div 2\right)}^{2} = {\left(- \frac{9}{4}\right)}^{2} = \frac{81}{16}}$

$y = 2 \left[{x}^{2} \textcolor{b l u e}{- \frac{9}{2}} x \textcolor{b l u e}{+ \frac{81}{16} - \frac{81}{16}} - 9\right]$

Group to create a perfect square.

$y = 2 \left[\textcolor{red}{\left({x}^{2} - \frac{9}{2} x + \frac{81}{16}\right)} + \left(- \frac{81}{16} - 9\right)\right]$

$y = 2 \left[\textcolor{red}{{\left(x - \frac{9}{4} x\right)}^{2}} + \left(- 5 \frac{1}{16} - 9\right)\right] \text{ } \leftarrow$ distribute the $2$

$y = 2 {\left(x - \frac{9}{4} x\right)}^{2} + 2 \left(- 14 \frac{1}{16}\right)$

$y = 2 {\left(x - \frac{9}{4} x\right)}^{2} - 28 \frac{1}{8}$

This is now vertex form, giving the vertex at $\left(2 \frac{1}{4} , - 28 \frac{1}{8}\right)$