# What is the vertex form of y= x^2/10 + x/4 + 1/6 ?

Jan 23, 2018

$y = \frac{1}{10} {\left(x + \frac{5}{4}\right)}^{2} + \frac{1}{96}$

#### Explanation:

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

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

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

$\text{to obtain this form use "color(blue)"completing the square}$

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

$\Rightarrow y = \frac{1}{10} \left({x}^{2} + \frac{5}{2} x + \frac{5}{3}\right)$

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} + \frac{5}{2} x$

$y = \frac{1}{10} \left({x}^{2} + 2 \left(\frac{5}{4}\right) x \textcolor{red}{+ \frac{25}{16}} \textcolor{red}{- \frac{25}{16}} + \frac{5}{3}\right)$

$\textcolor{w h i t e}{y} = \frac{1}{10} {\left(x + \frac{5}{4}\right)}^{2} + \frac{1}{10} \left(- \frac{25}{16} + \frac{5}{3}\right)$

$\textcolor{w h i t e}{y} = \frac{1}{10} {\left(x + \frac{5}{4}\right)}^{2} + \frac{1}{96} \leftarrow \textcolor{red}{\text{in vertex form}}$