# How do you find the vertex and the intercepts for f(x)= -6x^2+ 5x + 18?

Jul 31, 2018

$\text{vertex } = \left(\frac{5}{12} , \frac{457}{24}\right) , x = \frac{5}{12} \pm \frac{\sqrt{457}}{12}$

#### Explanation:

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

•color(white)(x)y=a(x-h)^2+k

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

$\text{to obtain this form "color(blue)"complete the square}$

$y = - 6 \left({x}^{2} - \frac{5}{6} x - 3\right)$

$\textcolor{w h i t e}{y} = - 6 \left({x}^{2} + 2 \left(- \frac{5}{12}\right) x + \frac{25}{144} - \frac{25}{144} - 3\right)$

$\textcolor{w h i t e}{y} = - 6 \left(x - \frac{5}{12}\right) + \frac{457}{24} \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{5}{12} , \frac{457}{24}\right)$

$\text{to obtain the x-intercepts set y = 0}$

$- 6 {\left(x - \frac{5}{12}\right)}^{2} + \frac{457}{24} = 0$

${\left(x - \frac{5}{12}\right)}^{2} = \frac{457}{144}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$x - \frac{5}{12} = \pm \sqrt{\frac{457}{144}} = \pm \frac{\sqrt{457}}{12}$

$\text{add "5/12" to both sides}$

$x = \frac{5}{12} \pm \frac{\sqrt{457}}{12} \leftarrow \textcolor{red}{\text{exact values}}$