What is the vertex form of y=1/2x^2-1/6x+6/13?

Jun 11, 2016

$y = \frac{1}{2} {\left(x - \frac{1}{6}\right)}^{2} + \frac{409}{936}$ (assuming I managed the arithmetic correctly)

Explanation:

The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
for a parabola with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given:
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} {x}^{2} - \frac{1}{6} x + \frac{6}{13}$

$\Rightarrow$
$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} \left({x}^{2} - \frac{1}{3} x\right) + \frac{6}{13}$

$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} \left({x}^{2} - \frac{1}{3} x + {\left(\frac{1}{6}\right)}^{2}\right) + \frac{6}{13} - \frac{1}{2} \cdot {\left(\frac{1}{6}\right)}^{2}$

$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} {\left(x - \frac{1}{6}\right)}^{2} + \frac{6}{13} - \frac{1}{72}$

$\textcolor{w h i t e}{\text{XXX}} y = \frac{1}{2} {\left(x - \frac{1}{6}\right)}^{2} + \frac{6 \cdot 72 - 1 \cdot 13}{13 \cdot 72}$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{\frac{1}{2}} {\left(x - \textcolor{red}{\frac{1}{6}}\right)}^{2} + \textcolor{b l u e}{\frac{409}{936}}$
which is the vertex form with vertex at $\left(\textcolor{red}{\frac{1}{6}} , \textcolor{b l u e}{\frac{409}{936}}\right)$

The graph below of the original equation indicates that our answer is at least approximately correct.

graph{1/2x^2-1/6x+6/13 [-0.6244, 1.0606, -0.097, 0.7454]}