# What is the vertex form of 2y=10x^2+7x-3?

May 5, 2018

$\textcolor{b l u e}{y = 5 {\left(x + \frac{7}{20}\right)}^{2} - \frac{169}{80}}$

#### Explanation:

$2 y = 10 {x}^{2} + 7 x - 3$

Divide by 2:

$y = 5 {x}^{2} + \frac{7}{2} x - \frac{3}{2}$

We now have the form:

$\textcolor{red}{y = a {x}^{2} + b x + c}$

We need the form:

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

Where:

$\boldsymbol{a} \textcolor{w h i t e}{8888}$ is the coefficient of ${x}^{2}$

$\boldsymbol{h} \textcolor{w h i t e}{8888}$ is the axis of symmetry.

$\boldsymbol{k} \textcolor{w h i t e}{8888}$ is the maximum or minimum value of the function.

It can be shown that:

$h = - \frac{b}{2 a} \textcolor{w h i t e}{8888}$ and $\textcolor{w h i t e}{8888} k = f \left(h\right)$

$\therefore$

$h = - \frac{\frac{7}{2}}{2 \left(5\right)} = - \frac{7}{20}$

$k = f \left(h\right) = 5 {\left(- \frac{7}{20}\right)}^{2} + \frac{7}{2} \left(- \frac{7}{20}\right) - \frac{3}{2}$

$\textcolor{w h i t e}{8888} = \frac{245}{400} - \frac{49}{40} - \frac{3}{2}$

$\textcolor{w h i t e}{8888} = \frac{49}{80} - \frac{49}{40} - \frac{3}{2}$

$\textcolor{w h i t e}{8888} = \frac{49 - 98 - 120}{80} = - \frac{169}{80}$

Vertex form:

$y = 5 {\left(x + \frac{7}{20}\right)}^{2} - \frac{169}{80}$