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

Dec 28, 2017

$y = - 3 {\left(x + \frac{1}{6}\right)}^{2} + \frac{109}{12}$

Explanation:

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

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

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

$\text{given the equation in standard form } y = a {x}^{2} + b x + c$

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

${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$

$y = - 3 {x}^{2} - x + 9 \text{ is in standard form}$

$\text{with } a = - 3 , b = - 1 , c = 9$

$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{- 1}{- 6} = - \frac{1}{6}$

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

${y}_{\textcolor{red}{\text{vertex}}} = - 3 {\left(- \frac{1}{6}\right)}^{2} + \frac{1}{6} + 9 = \frac{109}{12}$

$\Rightarrow \left(h , k\right) = \left(- \frac{1}{6} , \frac{109}{12}\right) \text{ and } a = - 3$

$\Rightarrow y = - 3 {\left(x + \frac{1}{6}\right)}^{2} + \frac{109}{12} \leftarrow \textcolor{red}{\text{in vertex form}}$