What is the vertex form of y= -5x^2+x-2 ?

Sep 11, 2017

$y = - 5 {\left(x - \frac{1}{10}\right)}^{2} - \frac{39}{20}$

Explanation:

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

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where (h , k ) are the coordinates of the vertex and a is a multiplier.

$\text{for a parabola in standard form } y = a {x}^{2} + b x + c$

"the x-coordinate of the vertex is " x_(color(red)"vertex")=-b/(2a)

$y = - 5 {x}^{2} + x - 2 \text{ is in standard form}$

$\text{with } a = - 5 , b = 1 , c = - 2$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{1}{- 10} = \frac{1}{10}$

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

${y}_{\textcolor{red}{\text{vertex}}} = - 5 {\left(\frac{1}{10}\right)}^{2} + \frac{1}{10} - 2 = - \frac{39}{20}$

$\text{here "(h,k)=(1/10,-39/20)" and } a = - 5$

$\Rightarrow y = - 5 {\left(x - \frac{1}{10}\right)}^{2} - \frac{39}{20} \leftarrow \textcolor{red}{\text{ in vertex form}}$