# How do I use the vertex formula to determine the vertex of the graph of the function and write it in standard form for 2x^2+6x-1?

Jun 2, 2015

$y = 2 {x}^{2} + 6 x - 1$ is already in standard form

$\textcolor{w h i t e}{\text{XXXXX}}$General standard form for a parabola is $y = a {x}^{2} + b x + c$

Vertex form for a parabola is $y = m \left(x - a\right) + b$
$\textcolor{w h i t e}{\text{XXXXX}}$Note: the $a \mathmr{and} b$ are not the same as in the standard form but they are constants.

$y = 2 {x}^{2} + 6 x + 1$
$\textcolor{w h i t e}{\text{XXXXX}}$extract $m$
$y = 2 \left({x}^{2} + 3 x\right) + 1$
$\textcolor{w h i t e}{\text{XXXXX}}$complete the square
$y = 2 \left({x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2}\right) + 1 - \frac{9}{2}$

$y = 2 {\left(x + \frac{3}{2}\right)}^{2} - \frac{7}{2}$

$y = 2 {\left(x - \left(- 3\right)\right)}^{2} + \left(- \frac{7}{2}\right)$
$\textcolor{w h i t e}{\text{XXXXX}}$which is in vertex form with a vertex at $\left(- 3 , - \frac{7}{2}\right)$