# How do you find the vertex and the intercepts for  y= -3x^2 + 12x - 8?

Feb 8, 2018

$\text{see explanation}$

#### Explanation:

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

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

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

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

• " the coefficient of the "x^2" term must be 1"

$\Rightarrow y = - 3 \left({x}^{2} - 4 x + \frac{8}{3}\right)$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} - 4 x$

$y = - 3 \left({x}^{2} + 2 \left(- 2\right) x \textcolor{red}{+ 4} \textcolor{red}{- 4} + \frac{8}{3}\right)$

$\textcolor{w h i t e}{y} = - 3 {\left(x - 2\right)}^{2} - 3 \left(- 4 + \frac{8}{3}\right)$

$\Rightarrow y = - 3 {\left(x - 2\right)}^{2} + 4 \leftarrow \textcolor{red}{\text{in vertex form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(2 , 4\right)$

$\text{to find the intercepts}$

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

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

$x = 0 \to y = - 3 {\left(- 2\right)}^{2} + 4 = - 8 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to - 3 {\left(x - 2\right)}^{2} + 4 = 0$

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

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

$x - 2 = \pm \sqrt{\frac{4}{3}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 2 \pm \frac{2}{\sqrt{3}}$

$\Rightarrow x = 2 \pm \frac{2 \sqrt{3}}{3} \leftarrow \textcolor{red}{\text{x-intercepts}}$