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

Dec 19, 2017

$y = \frac{4}{7} {\left(x + \frac{1}{4}\right)}^{2} - \frac{13}{28}$

Explanation:

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

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

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

$\text{given the parabola in "color(blue)"standard form}$

•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0

$\text{then the x-coordinate of the vertex is}$

•color(white)(x)x_(color(red)"vertex")=-b/(2a)#

$7 y = 4 {x}^{2} + 2 x - 3 \leftarrow \textcolor{b l u e}{\text{divide all terms by 7}}$

$\Rightarrow y = \frac{4}{7} {x}^{2} + \frac{2}{7} x - \frac{3}{7} \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{with } a = \frac{4}{7} , b = \frac{2}{7}$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{\frac{2}{7}}{\frac{8}{7}} = - \frac{1}{4}$

$\text{substitute this value into equation for y-coordinate}$

${y}_{\textcolor{red}{\text{vertex}}} = \frac{4}{7} {\left(- \frac{1}{4}\right)}^{2} + \frac{2}{7} \left(- \frac{1}{4}\right) - \frac{3}{7}$

$\textcolor{w h i t e}{\times \times} = \frac{1}{28} - \frac{2}{28} - \frac{12}{28} = - \frac{13}{28}$

$\text{here "a=4/7" and } \left(h , k\right) = \left(\frac{1}{4} , - \frac{13}{28}\right)$

$\Rightarrow y = \frac{4}{7} {\left(x + \frac{1}{4}\right)}^{2} - \frac{13}{28} \leftarrow \textcolor{red}{\text{in vertex form}}$