# How do you find the vertex and intercepts for f(x)=-3x^2+5x+9?

Nov 29, 2017

$\text{see explanation}$

#### Explanation:

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

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

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

$\text{to obtain this form use the method of "color(blue)"completing the square}$

• " ensure the coefficient of the "x^2" term is 1"

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

•" add/subtract "(1/2"coefficient of x-term")^2"to"
${x}^{2} - \frac{5}{3} x$

$f \left(x\right) = - 3 \left({x}^{2} + 2 \left(- \frac{5}{6}\right) x \textcolor{red}{+ \frac{25}{36}} \textcolor{red}{- \frac{25}{36}} - 3\right)$

$\textcolor{w h i t e}{f \left(x\right)} = - 3 {\left(x - \frac{5}{6}\right)}^{2} + \frac{133}{12} \leftarrow \textcolor{b l u e}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{5}{6} , \frac{133}{12}\right)$

$\textcolor{b l u e}{\text{Intercepts}}$

• " let x = 0, in the equation for y-intercept"

• " let y = 0, in the equation for x-intercepts"

$x = 0 \to y = 9 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to - 3 {\left(x - \frac{5}{6}\right)}^{2} + \frac{133}{12} = 0$

$\Rightarrow - 3 {\left(x - \frac{5}{6}\right)}^{2} = - \frac{133}{12}$

$\Rightarrow {\left(x - \frac{5}{6}\right)}^{2} = \frac{133}{36}$

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

$\Rightarrow x - \frac{5}{6} = \pm \sqrt{\frac{133}{36}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = \frac{5}{6} \pm \frac{\sqrt{133}}{6} \leftarrow \textcolor{red}{\text{exact solutions}}$

$\Rightarrow x \approx - 1.09 \text{ or "x~~2.76larrcolor(red)"x-intercepts}$