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

Mar 14, 2016

$\textcolor{red}{y = \frac{1}{3} {\left(x - 1\right)}^{2} - \frac{1}{6}}$

#### Explanation:

Given:$\text{ } y = \frac{1}{3} {x}^{2} - \frac{2}{3} x + \frac{1}{6}$..........................(1)

Write as:$\text{ } y = \frac{1}{3} \left({x}^{2} - 2 x\right) + \frac{1}{6}$

What we are about to do will introduce an error. Compensate for this error by adding a constant

Let $k$ be a constant

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

$\frac{1}{2}$ the coefficient of $x$

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

'Get rid' of the single $x$ leaving its coefficient of 1

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

Move the index (power) of 2 to outside the brackets

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

$\textcolor{b r o w n}{\text{This is your basic form. Now we need to find } k}$

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Consider the form 1/3(?-1)^2. It produces the error of

$\frac{1}{3} \times {\left(- 1\right)}^{2} = + \frac{1}{3}$

To 'get rid' of this error we make $k = - \frac{1}{3}$

So equation (2) becomes

$y = \frac{1}{3} {\left(x - 1\right)}^{2} - \frac{1}{3} + \frac{1}{6} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left({2}_{a}\right)$
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$\textcolor{red}{y = \frac{1}{3} {\left(x - 1\right)}^{2} - \frac{1}{6}}$