# What is the vertex form of 7y=3x^2 − 2x + 12 ?

Oct 7, 2016

We will have to complete the square for this quadratic which will put the equation in vertex form.

First lets solve for the y variable by dividing both sides by 7

$\frac{\cancel{7} y}{\cancel{7}} = \frac{3}{7} {x}^{2} - \frac{2}{7} x + \frac{12}{7}$

Set the equation equal to zero.

$0 = \frac{3}{7} {x}^{2} - \frac{2}{7} x + \frac{12}{7}$

Subtract $\frac{12}{7}$ from both sides

$0 \textcolor{red}{- \frac{12}{7}} = \frac{3}{7} {x}^{2} - \frac{2}{7} x + \frac{12}{7} \textcolor{red}{- \frac{12}{7}}$

Simplify

$\textcolor{red}{- \frac{12}{7}} = \frac{3}{7} {x}^{2} - \frac{2}{7} x$

Factor out $\frac{3}{7}$

$- \frac{12}{7} = \frac{3}{7} \left({x}^{2} - \frac{2}{\cancel{7}} \left(\frac{\cancel{7}}{3}\right) x\right)$

Simplify

$- \frac{12}{7} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x\right)$

Take the coefficient of x and divide it by 2 and then square it

${\left(\frac{- \frac{2}{3}}{2}\right)}^{2} = {\left(- \frac{2}{3} \cdot \frac{1}{2}\right)}^{2} = {\left(- \frac{2}{6}\right)}^{2} = {\left(- \frac{1}{3}\right)}^{2} = \frac{1}{9}$

Add $\frac{1}{9}$ to the right side and add $\frac{3}{7} \left(\frac{1}{9}\right)$ to the left side because we factored out $\frac{3}{7}$ in the beginning. This process will keep the equation balanced.

$\textcolor{red}{\frac{3}{7} \left(\frac{1}{9}\right)} - \frac{12}{7} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x + \textcolor{red}{\frac{1}{9}}\right)$

Simply

$\textcolor{red}{\frac{\cancel{3}}{7} \left(\frac{1}{\cancel{9} 3}\right)} - \frac{12}{7} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x + \textcolor{red}{\frac{1}{9}}\right)$

$\frac{1}{21} - \frac{12}{7} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x + \textcolor{red}{\frac{1}{9}}\right)$

Find Common Denominator

$\frac{1}{21} - \frac{12}{7} \cdot \textcolor{red}{\frac{3}{3}} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x + \textcolor{red}{\frac{1}{9}}\right)$

$\frac{1}{21} - \frac{36}{21} = \frac{3}{7} \left({x}^{2} - \frac{2}{3} x + \textcolor{red}{\frac{1}{9}}\right)$

The right side is a perfect square trinomial

$\frac{1}{21} - \frac{36}{21} = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2}$

$- \frac{35}{21} = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2}$

$- \frac{\cancel{35} 5}{\cancel{21} 3} = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2}$

$- \frac{5}{3} = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2}$

Add $\frac{5}{3}$ from both sides

$\textcolor{red}{\frac{5}{3}} - \frac{5}{3} = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2} \textcolor{red}{+ \frac{5}{3}}$

$0 = \frac{3}{7} {\left(x - \frac{1}{3}\right)}^{2} \textcolor{red}{+ \frac{5}{3}}$

Vertex form $\implies y = {\left(x - h\right)}^{2} + k$

Vertex $\implies \left(h , k\right) \implies \left(\frac{1}{3} , \frac{5}{3}\right)$