# What is the vertex form of y= 2x^2+7x-15?

Mar 23, 2016

$y = 2 {\left(x + \frac{7}{4}\right)}^{2} + \frac{169}{8}$

#### Explanation:

Given -

$y = 2 {x}^{2} + 7 x - 15$

Find the vertex

$x = \frac{- b}{2 a} = \frac{- 7}{2 \times 2} = - \frac{7}{4}$

$y = 2 {\left(- \frac{7}{4}\right)}^{2} + 7 \left(- \frac{7}{4}\right) - 15$
$y = 2 \left(\frac{49}{16}\right) - \frac{49}{4} - 15$
$y = \frac{49}{8} - \frac{49}{4} - 15 = \frac{169}{8}$

$y = a {\left(x - h\right)}^{2} + k$
$a$ is the co-efficient of ${x}^{2}$
$h$ is $x$coordinate of the vertex
$k$ is the $y$ coordinate of the vertex
$y = 2 {\left(x - \left(- \frac{7}{4}\right)\right)}^{2} + \frac{169}{8}$
$y = 2 {\left(x + \frac{7}{4}\right)}^{2} + \frac{169}{8}$