# What is the vertex form of #3y=8x^2 + 17x - 13?

##### 1 Answer
Nov 25, 2015

The vertex form is $y = \frac{8}{3} {\left(x + \frac{17}{16}\right)}^{2} - \frac{235}{32}$.

#### Explanation:

First, let's rewrite the equation so the numbers are all on one side:

$3 y = 8 {x}^{2} + 17 x - 13$
$y = \frac{8 {x}^{2}}{3} + \frac{17 x}{3} - \frac{13}{3}$

To find the vertex form of the equation, we must complete the square:

$y = \frac{8 {x}^{2}}{3} + \frac{17 x}{3} - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x\right) - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x + {\left(\frac{17}{8} \div 2\right)}^{2} - {\left(\frac{17}{8} \div 2\right)}^{2}\right) - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x + {\left(\frac{17}{8} \cdot \frac{1}{2}\right)}^{2} - {\left(\frac{17}{8} \cdot \frac{1}{2}\right)}^{2}\right) - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x + {\left(\frac{17}{16}\right)}^{2} - {\left(\frac{17}{16}\right)}^{2}\right) - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x + \left(\frac{289}{256}\right) - \left(\frac{289}{256}\right)\right) - \frac{13}{3}$

$y = \frac{8}{3} \left({x}^{2} + \frac{17}{8} x + \left(\frac{289}{256}\right)\right) - \frac{13}{3} - \left(\frac{289}{256} \cdot \frac{8}{3}\right)$

$y = \frac{8}{3} {\left(x + \frac{17}{16}\right)}^{2} - \frac{13}{3} - \frac{289}{96}$

$y = \frac{8}{3} {\left(x + \frac{17}{16}\right)}^{2} - \frac{235}{32}$