# What is the vertex form of y=2x^2 + 13x + 6?

$\frac{1}{2} \left(y - \left(- \frac{121}{8}\right)\right) = {\left(x - - \frac{13}{4}\right)}^{2}$

#### Explanation:

From the given
$y = 2 {x}^{2} + 13 x + 6$

perform completing the square method

factor out 2 first

$y = 2 \left({x}^{2} + \frac{13 x}{2}\right) + 6$

$y = 2 \left({x}^{2} + \frac{13 x}{2} + \frac{169}{16} - \frac{169}{16}\right) + 6$

$y = 2 \left({\left(x + \frac{13}{4}\right)}^{2} - \frac{169}{16}\right) + 6$

$y = 2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{169}{8} + 6$

$y = 2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{169}{8} + \frac{48}{8}$

$y = 2 {\left(x + \frac{13}{4}\right)}^{2} - \frac{121}{8}$

$\frac{1}{2} \left(y - \left(- \frac{121}{8}\right)\right) = {\left(x - - \frac{13}{4}\right)}^{2}$

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