# How do you write y=2x^2-5x into vertex form?

##### 1 Answer
Jun 19, 2018

$y = 2 {\left(x - \frac{5}{4}\right)}^{2} + \left(- \frac{25}{8}\right)$

#### Explanation:

Here is one way to approach converting the given form $y = 2 {x}^{2} - 5 x$ into vertex form: $m {\left(x - a\right)}^{2} + b$

$y = 2 {x}^{2} - 5 x$

$\Rightarrow \frac{y}{2} = {x}^{2} - \frac{5}{2} x$

Remember that the expansion perfect square binomial will have the form ${x}^{2} + 2 a x + {a}^{2}$

If ${x}^{2} - \frac{5}{2} x$ are the first two terms of the expansion of a perfect square binomial, then
$\textcolor{w h i t e}{\text{XXX}} a = - \frac{5}{4}$
and
$\textcolor{w h i t e}{\text{XXX}} {a}^{2} = {\left(- \frac{5}{4}\right)}^{2}$

So we will need to add ${\left(- \frac{5}{4}\right)}^{2}$ to "complete the square" (of course if we add this to one side we will also need to add it to the other to keep the equation valid).

$\frac{y}{2} + {\left(- \frac{5}{4}\right)}^{2} = {x}^{2} - \frac{5}{2} x + {\left(- \frac{5}{4}\right)}^{2}$

$\textcolor{w h i t e}{\text{XXXXXXXX}} = {\left(x - \frac{5}{4}\right)}^{2}$

To get this into its proper form we need to isolate the $y$ on the left side:
Subtracting ${\left(- \frac{5}{4}\right)}^{2} = \frac{25}{16}$ from both sides:
$\textcolor{w h i t e}{\text{XXX}} \frac{y}{2} = {\left(x - \frac{5}{4}\right)}^{2} - \left(\frac{25}{16}\right)$
then multiplying both sides by $2$
$\textcolor{w h i t e}{\text{XXX}} y = 2 {\left(x - \frac{5}{4}\right)}^{2} + \left(- \frac{25}{8}\right)$

Here is the graph of $y = 2 {x}^{2} - 5 x$ to indicate that this result is reasonable:
graph{2x^2-5x [-3.03, 5.742, -4.165, 0.217]}