# What is the vertex form of 6y=(x + 13)(x - 3) ?

Jan 8, 2018

$y = \frac{1}{6} {\left(x + 5\right)}^{2} - \frac{32}{3}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form }$is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a }$
$\text{is a multiplier}$

$6 y = \left(x + 13\right) \left(x - 3\right) = {x}^{2} + 10 x - 39$

$\Rightarrow y = \frac{1}{6} \left({x}^{2} + 10 x - 39\right)$

$\text{using the method of "color(blue)"completing the square}$

$\text{on } {x}^{2} + 10 x - 39$

• " the coefficient of the "x^2" term must be 1"

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} + 10 x$

${x}^{2} + 2 \left(5\right) x \textcolor{red}{+ 25} \textcolor{red}{- 25} - 39 = {\left(x + 5\right)}^{2} - 64$

$\Rightarrow y = \frac{1}{6} {\left(x + 5\right)}^{2} - \frac{32}{3}$