# What is the vertex form of 4y=5x^2 -7x +3?

Jan 12, 2018

$y = \textcolor{g r e e n}{\frac{5}{4}} {\left(x - \textcolor{red}{\frac{7}{10}}\right)}^{2} + \textcolor{b l u e}{\frac{11}{80}}$

#### Explanation:

Remember that the vertex form (our target) is in general
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} 4 y = 5 {x}^{2} - 7 x + 3$

We will need to divide everything by $4$ to isolate $y$ on the right side
$\textcolor{w h i t e}{\text{XXX}} y = \frac{5}{4} {x}^{2} - \frac{7}{4} x + \frac{3}{4}$

We can now extract the $\textcolor{g r e e n}{m}$ factor from the first two terms:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{\frac{5}{4}} \left({x}^{2} - \frac{7}{5} x\right) + \frac{3}{4}$

We want to write $\left({x}^{2} - \frac{7}{5} x\right)$ as a squared binomial by inserting some constant (which will need to be subtracted somewhere else).

Remember that the squared binomial
$\textcolor{w h i t e}{\text{XXX}} {\left(x + p\right)}^{2} = \left({x}^{2} + \left(2 p\right) x + {p}^{2}\right)$
since the coefficient of the $x$ term of $\left({x}^{2} - \frac{7}{5} x\right)$ is $\left(- \frac{7}{5}\right)$
our value for $2 p = - \frac{7}{5} \rightarrow p = - \frac{7}{10} \rightarrow {p}^{2} = \frac{49}{100}$
So we need to insert a term of $\textcolor{m a \ge n t a}{{\left(- \frac{7}{10}\right)}^{2}} = \textcolor{m a \ge n t a}{\frac{49}{100}}$ into the factor $\left({x}^{2} - \frac{7}{5} x\right)$ making it $\left({x}^{2} - \frac{7}{5} + \textcolor{m a \ge n t a}{{\left(- \frac{7}{10}\right)}^{2}}\right)$

...but remember that this factor is multiplied by $\textcolor{g r e e n}{\frac{5}{4}}$
so to balance thing out we will need to subtract $\textcolor{g r e e n}{\frac{5}{4}} \times \textcolor{m a \ge n t a}{\frac{49}{100}} = \textcolor{b r o w n}{\frac{49}{80}}$

Our equation now looks like
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{\frac{5}{4}} \left({x}^{2} - \frac{7}{5} + \textcolor{m a \ge n t a}{{\left(- \frac{7}{10}\right)}^{2}}\right) + \frac{3}{4} - \textcolor{b r o w n}{\frac{49}{80}}$

Writing this with a squared binomial and simplifying the constant terms:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{\frac{5}{4}} {\left(x - \textcolor{red}{\frac{7}{10}}\right)}^{2} + \textcolor{b l u e}{\frac{11}{80}}$
which is our required vertex form with vertex at $\left(\textcolor{red}{\frac{7}{10}} , \textcolor{b l u e}{\frac{11}{80}}\right)$

For verification purposes here is a graph of the original equation:

Jan 12, 2018

$y = \frac{5}{4} {\left(x - \frac{7}{10}\right)}^{2} + \frac{11}{80}$

#### Explanation:

$\text{the equation of a parabola in "color(blue)"vertex form}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$

$\text{to express "5x^2-7x+3" in this form}$

$\text{use the method of "color(blue)"completing the square}$

• "the coefficient of the "x^2" term must be 1"

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

• " add/subtract "(1/2"coefficient of x-term")^2" to"
${x}^{2} - \frac{7}{5} x$

$5 \left({x}^{2} + 2 \left(- \frac{7}{10}\right) x \textcolor{red}{+ \frac{49}{100}} \textcolor{red}{- \frac{49}{100}} + \frac{3}{5}\right)$

$= 5 {\left(x - \frac{7}{10}\right)}^{2} + 5 \left(- \frac{49}{100} + \frac{3}{5}\right)$

$= 5 {\left(x - \frac{7}{10}\right)}^{2} + \frac{11}{20}$

$\Rightarrow 4 y = 5 {\left(x - \frac{7}{10}\right)}^{2} + \frac{11}{20}$

$\Rightarrow y = \frac{1}{4} \left[5 {\left(x - \frac{7}{10}\right)}^{2} + \frac{11}{20}\right]$

$\textcolor{w h i t e}{\Rightarrow y} = \frac{5}{4} {\left(x - \frac{7}{10}\right)}^{2} + \frac{11}{80}$