# What is the vertex form of 3y=4x^2 + 9x - 1?

Oct 18, 2017

y=color(green)(4/3)(x-color(red)((-9/8)))^2+color(blue)(""(-81/48))#
with vertex at $\left(\textcolor{red}{- \frac{9}{8}} , \textcolor{b l u e}{- \frac{81}{48}}\right)$

#### Explanation:

Remember our target form is $y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

$3 y = 4 {x}^{2} + 9 x - 1$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} {x}^{2} + 3 x - \frac{1}{3}$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} \left({x}^{2} + \frac{9}{4} x\right) - \frac{1}{3}$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} \left({x}^{2} + \frac{9}{4} x \textcolor{m a \ge n t a}{+ {\left(\frac{9}{8}\right)}^{2}}\right) - \frac{1}{3} \textcolor{w h i t e}{\text{xx}} \textcolor{m a \ge n t a}{- \textcolor{g r e e n}{\frac{4}{3}} \cdot {\left(\frac{9}{8}\right)}^{2}}$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} {\left(x + \frac{9}{8}\right)}^{2} - \frac{1}{3} - \frac{27}{16}$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} {\left(x - \textcolor{red}{\left(- \frac{9}{8}\right)}\right)}^{2} - \frac{16}{48} - \frac{81}{48}$

$\rightarrow y = \textcolor{g r e e n}{\frac{4}{3}} {\left(x - \textcolor{red}{\left(- \frac{9}{8}\right)}\right)}^{2} + \textcolor{b l u e}{\left(- \frac{97}{48}\right)}$

which is the vertex form with vertex at $\left(\textcolor{red}{- \frac{9}{8}} , \textcolor{b l u e}{- \frac{97}{48}}\right)$

I admit this is not very pretty, so here is a graph of the given equation to demonstrate that this answer is at least reasonable: 