# What is the vertex form of y=1/3x^2+5/6x+7/8?

Nov 30, 2015

$y = \frac{1}{3} {\left(x + \frac{5}{4}\right)}^{2} - \frac{11}{16}$
Have a look at the explanation to see how it is done!

#### Explanation:

Given:$\textcolor{w h i t e}{\ldots .} y = \frac{1}{3} {x}^{2} + \frac{5}{6} x + \frac{7}{8}$

Consider the part inside the brackets:$\textcolor{w h i t e}{\ldots .} y = \left(\frac{1}{3} {x}^{2} + \frac{5}{6} x\right) + \frac{7}{8}$

Write as: $\frac{1}{3} \left({x}^{2} + \left\{\frac{5}{6} \div \frac{1}{3}\right\} x\right)$

$\frac{1}{3} \left(\textcolor{red}{{x}^{2}} + \textcolor{b l u e}{\frac{5}{2} \textcolor{g r e e n}{x}}\right)$

If we halve $\frac{5}{2}$ we get $\frac{5}{4}$

Change the bracketed bit so that have

$\frac{1}{3} {\left(\textcolor{red}{x} + \textcolor{b l u e}{\frac{5}{4}}\right)}^{2}$

We have changed $\textcolor{red}{{x}^{2}}$ to just $\textcolor{red}{x}$; halved the coefficient of $\textcolor{g r e e n}{x} \to \textcolor{b l u e}{\frac{1}{2} \times \frac{5}{2} = \frac{5}{4}}$ and totally removed the single $\textcolor{g r e e n}{x}$

So we know write the equation as:

$y \to \frac{1}{3} {\left(x + \frac{5}{4}\right)}^{2} + \frac{7}{8}$

The thing is; we have introduced an error that results from squaring the bracket. The error is when we square the $\left(+ \frac{5}{4}\right)$ bit. This error means that the right no longer ='s the left. That is why I have used $y \to$

$\textcolor{b l u e}{\text{To correct for this we write:}}$

$y \to \frac{1}{3} {\left(x + \frac{5}{4}\right)}^{2} \textcolor{b l u e}{- {\left(\frac{5}{4}\right)}^{2}} + \frac{7}{8}$

The correction now means that the $\textcolor{red}{\text{left does = right.}}$

$y \textcolor{red}{=} \frac{1}{3} {\left(x + \frac{5}{4}\right)}^{2} \textcolor{b l u e}{- {\left(\frac{5}{4}\right)}^{2}} + \frac{7}{8}$

So the arithmetic now gives:

$y = \frac{1}{3} {\left(x + \frac{5}{4}\right)}^{2} - \frac{11}{16}$