# What is the vertex form of 2y=5x^2-3x+11 ?

Sep 12, 2017

see explanation

#### Explanation:

...I can never remember it, so I always have to look it up.

$f \left(x\right) = a {\left(x - h\right)}^{2} + k$

So, for your original equation $2 y = 5 {x}^{2} - 3 x + 11$, you have to do some algebraic manipulation.

Firstly, you need the ${x}^{2}$ term to have a multiple of 1, not 5.
So divide both sides by 5:

$\frac{2}{5} y = {x}^{2} - \frac{3}{5} x + \frac{11}{5}$

...now you have to perform the infamous "complete the square" maneuver. Here's how I go about it:

Say that your $- \frac{3}{5}$ coefficient is $2 a$. Then $a = - \frac{3}{5} \cdot \frac{1}{2} = - \frac{3}{10}$

And ${a}^{2}$ would be $\frac{9}{100}$.

So, if we add and subract this from the quadratic equation, we'd have:

$\frac{2}{5} y = {x}^{2} - \frac{3}{5} x + \frac{9}{100} - \frac{9}{100} + \frac{11}{5}$

...and now the 1st 3 terms of the right side are a perfect square in form ${\left(x - a\right)}^{2} = {x}^{2} - 2 a x + {a}^{2}$

...so you can write:

$\frac{2}{5} y = {\left(x - \frac{3}{10}\right)}^{2} + \left(\frac{11}{5} - \frac{9}{100}\right)$

$\frac{2}{5} y = {\left(x - \frac{3}{10}\right)}^{2} + \frac{220 - 9}{100}$

$\frac{2}{5} y = {\left(x - \frac{3}{10}\right)}^{2} + \frac{211}{100}$

So now, all you gotta do is multiply through by $\frac{5}{2}$, giving:

$y = \frac{5}{2} {\left(x - \frac{3}{10}\right)}^{2} + \frac{5}{2} \cdot \frac{211}{100}$

$y = \frac{5}{2} {\left(x - \frac{3}{10}\right)}^{2} + \frac{211}{40}$

which is vertex form, $y = a {\left(x - h\right)}^{2} + k$

where $a = \frac{5}{2}$, $h = \frac{3}{10}$, and $k = \frac{211}{40}$