# How do you find the coordinates of the vertex y= -3x^2 + 8x?

Feb 22, 2017

$\left(\frac{4}{3} , \frac{16}{3}\right)$

#### Explanation:

The $\textcolor{b l u e}{\text{standard quadratic function}}$ is.

color(red)(bar(ul(|color(white)(2/2)color(black)(y=ax^2+bx+c; a≠0)color(white)(2/2)|)))

$\text{for } y = - 3 {x}^{2} + 8 x$

$a = - 3 , b = 8 \text{ and } c = 0$

The x-coordinate of the vertex can be found using.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{x}_{\text{vertex}} = - \frac{b}{2 a}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\Rightarrow {x}_{\text{vertex}} = - \frac{8}{- 6} = \frac{4}{3}$

Substitute this value into the equation to obtain the corresponding y-coordinate.

${y}_{\text{vertex}} = - 3 {\left(\frac{4}{3}\right)}^{2} + 8 \left(\frac{4}{3}\right)$

$\textcolor{w h i t e}{{y}_{\text{vertex}}} = \left(- 3 \times \frac{16}{9}\right) + \frac{32}{3}$

$\textcolor{w h i t e}{{y}_{\text{vertex}}} = - \frac{16}{3} + \frac{32}{3}$

$\textcolor{w h i t e}{{y}_{\text{vertex}}} = \frac{16}{3}$

$\Rightarrow \text{coordinates of vertex } = \left(\frac{4}{3} , \frac{16}{3}\right)$
graph{-3x^2+8x [-20, 20, -10, 10]}