# What is the vertex form of 3y=-(x-2)(x-1) ?

May 11, 2017

$y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} + \frac{1}{12}$

#### Explanation:

Given: $3 y = - \left(x - 2\right) \left(x - 1\right)$

Vertex form is: y = a(x - h)^2 + k;  where the vertex is $\left(h , k\right)$ and $a$ is a constant.

Distribute the two linear terms:$\text{ } 3 y = - \left({x}^{2} - 3 x + 2\right)$

Divide by $3$ to get $y$ by itself: $y = - \frac{1}{3} \left({x}^{2} - 3 x + 2\right)$

One method is to use completing of the square to put in vertex form:

Only work with the $x$ terms: $\text{ } y = - \frac{1}{3} \left({x}^{2} - 3 x\right) - \frac{2}{3}$

Half the coefficient of the $x$ term: $- \frac{3}{2}$

Complete the square: $y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} - \frac{2}{3} + \frac{1}{3} {\left(\frac{3}{2}\right)}^{2}$

Simplify: $y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} - \frac{2}{3} + \frac{1}{3} \cdot \frac{9}{4}$

$y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} - \frac{8}{12} + \frac{9}{12}$

$y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} + \frac{1}{12}$

A second method is to put the equation in $y = A {x}^{2} + B x + C$:

Distribute the given equation: $3 y = - {x}_{2} + 3 x - 2$

Divide by $3$: $\text{ } y = - \frac{1}{3} {x}^{2} + x - \frac{2}{3}$

Find the vertex $x = - \frac{B}{2 A} = - \frac{1}{- \frac{2}{3}} = - \frac{1}{1} \cdot - \frac{3}{2} = \frac{3}{2}$

Find the $y$ of the vertex: $y = - \frac{1}{3} \cdot {\left(\frac{3}{2}\right)}^{2} + \frac{3}{2} - \frac{2}{3}$

$y = - \frac{1}{3} \cdot \frac{9}{4} + \frac{9}{6} - \frac{4}{6} = - \frac{9}{12} + \frac{5}{6} = - \frac{9}{12} + \frac{10}{12} = \frac{1}{12}$

Vertex form is: y = a(x - h)^2 + k;  where the vertex is $\left(h , k\right)$ and $a$ is a constant.

$y = a {\left(x - \frac{3}{2}\right)}^{2} + \frac{1}{12}$

Find $a$ by inputting a point into the equation. Use the original equation to find this point:

Let $x = 2 , \text{ " 3y = -(2-2)(2-1); " " 3y =0; " } y = 0$

Use $\left(2 , 0\right)$ and substitute it into $y = a {\left(x - \frac{3}{2}\right)}^{2} + \frac{1}{12}$:

$0 = a {\left(2 - \frac{3}{2}\right)}^{2} + \frac{1}{12}$

$- \frac{1}{12} = a {\left(\frac{1}{2}\right)}^{2}$

$- \frac{1}{12} = a \frac{1}{4}$

$a = \frac{- \frac{1}{12}}{\frac{1}{4}} = - \frac{1}{12} \cdot \frac{4}{1} = - \frac{1}{3}$

vertex form: $y = - \frac{1}{3} {\left(x - \frac{3}{2}\right)}^{2} + \frac{1}{12}$