# How do you find the vertex and intercepts for f(x)=-3x^2 +7x +4?

Aug 9, 2018

$\text{vertex } = \left(\frac{7}{6} , \frac{97}{12}\right)$

#### Explanation:

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

•color(white)(x)y=a(x-h)^2+k

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

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

$y = - 3 \left({x}^{2} - \frac{7}{3} x - \frac{4}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 \left({x}^{2} + 2 \left(- \frac{7}{6}\right) x + \frac{49}{36} - \frac{49}{36} - \frac{4}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{7}{6}\right)}^{2} - 3 \left(- \frac{49}{36} - \frac{4}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 {\left(x - \frac{7}{6}\right)}^{2} + \frac{97}{12} \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{7}{6} , \frac{97}{12}\right)$

$\text{for y-intercept let x = 0}$

$y = 4 \leftarrow \textcolor{red}{\text{y-intercept}}$

$\text{for x-intercepts let y = 0}$

$- 3 {\left(x - \frac{7}{6}\right)}^{2} + \frac{97}{12} = 0$

$- 3 {\left(x - \frac{7}{6}\right)}^{2} = - \frac{97}{12}$

${\left(x - \frac{7}{6}\right)}^{2} = \frac{97}{36}$

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

$x - \frac{7}{6} = \pm \frac{97}{36} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\text{add "7/6" to both sides}$

$x = \frac{7}{6} \pm \frac{\sqrt{97}}{6} \leftarrow \textcolor{red}{\text{exact values}}$

$x \approx - 0.47 , x \approx 2.81 \text{ to 2 dec. places}$