# What is the vertex form of y= 2x^2 + 9x - 5 ?

Mar 29, 2016

$y = 2 {\left(x + \frac{9}{4}\right)}^{2} - \frac{121}{8}$

#### Explanation:

Given:$\text{ } y = 2 {x}^{2} + 9 x - 5$......................(1)

Write as:$\text{ } y = 2 \left({x}^{\textcolor{m a \ge n t a}{2}} + \frac{9}{2} x\right) - 5 + k$

Where $k$ is a correction factor for an unfortunate consequence of what we are about to do.

Take the power of 2 from ${x}^{2}$ and move it to outside the brackets

$\text{ } y = 2 {\left(x + \frac{9}{2} \textcolor{b l u e}{x}\right)}^{\textcolor{m a \ge n t a}{2}} - 5 + k$

'Get rid' of the $\textcolor{b l u e}{x}$ from $\frac{9}{2} \textcolor{b l u e}{x}$

$\text{ } y = 2 {\left(x + \frac{9}{2}\right)}^{2} - 5 + k$

Apply $\left(- \frac{1}{2}\right) \times \frac{9}{2} = - \frac{9}{4}$

$\text{ } y = 2 {\left(x + \frac{9}{4}\right)}^{2} - 5 + k$ .....................................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The error comes from the $\frac{9}{4}$ being squared. It introduces an extra value that we not there before. By the way, do not forget to multiply it by the constant of 2 outside the bracket.

So the error is $2 {\left(\frac{9}{4}\right)}^{2}$

Consequently it has to be the case that: $2 {\left(\frac{9}{4}\right)}^{2} + k = 0$

so we have $2 \left(\frac{81}{16}\right) + k = 0$

$\implies k = - \frac{81}{8}$

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
So equation (2) becomes

$\text{ } y = 2 {\left(x + \frac{9}{4}\right)}^{2} - 5 - \frac{81}{8}$ .....................................(2_a)

Giving:

$\textcolor{b l u e}{\text{ } y = 2 {\left(x + \frac{9}{4}\right)}^{2} - \frac{121}{8}}$