# How do you find the vertex and intercepts for 2(x-3)^2=6y+72?

Feb 8, 2018

$\text{see explanation}$

#### Explanation:

$\text{expressing in translated form}$

•color(white)(x)(x-h)^2=4p(y-k)

$\text{where "(h,k)" are the coordinates of the vertex and p}$
$\text{is the distance betwee the vertex and the focus/directrix}$

$\Rightarrow 2 {\left(x - 3\right)}^{2} = 6 \left(y + 12\right)$

$\Rightarrow {\left(x - 3\right)}^{2} = 3 \left(y + 12\right) \leftarrow \textcolor{b l u e}{\text{in translated form}}$

$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(3 , - 12\right)$

$\text{to obtain the intercepts}$

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

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

$x = 0 \to 9 = 3 y + 36 \Rightarrow y = - 9 \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to {\left(x - 3\right)}^{2} = 36$

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

$\Rightarrow x - 3 = \pm \sqrt{36} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 3 \pm 6$

$\Rightarrow x = - 3 , x = 9 \leftarrow \textcolor{red}{\text{x-intercepts}}$