# How do you find the vertex and intercepts for (x − 2)^2 = -12(y − 2)?

Feb 8, 2018

So the curve has:

Vertex: maximum at $\left(2 , 2\right)$

Y-intercept: $y = \frac{13}{6}$

X-intercepts: $x = 2 + \sqrt{6}$ and $x = 2 - \sqrt{6}$

#### Explanation:

${\left(x - 2\right)}^{2} = - 12 \left(y - 2\right)$

Rearrange for y.

$- \frac{1}{12} {\left(x - 2\right)}^{2} = y - 2$

$y = - \frac{1}{12} {\left(x - 2\right)}^{2} + 2$

From the completed square form, the maximum turning point is at $\left(2 , 2\right)$

For the y-intercept, let $x = 0$

$y = - \frac{1}{12} {\left(- 2\right)}^{2} + 2$
$y = \frac{13}{6}$

For the x-intercepts, let $y = 0$

$0 = - \frac{1}{12} {\left(x - 2\right)}^{2} + 2$
$0 = {\left(x - 2\right)}^{2} - 6$
$6 = {\left(x - 2\right)}^{2}$

$x - 2 = \pm \sqrt{6}$

$x = 2 \pm \sqrt{6}$

So the curve has:

Vertex: maximum at $\left(2 , 2\right)$

Y-intercept: $y = \frac{13}{6}$

X-intercepts: $x = 2 + \sqrt{6}$ and $x = 2 - \sqrt{6}$

graph{(x-2)^2=-12(y-2) [-10, 10, -5, 5]}

Feb 8, 2018

$\text{see explanation}$

#### Explanation:

$\text{the quadratic is in standard translated form}$

•color(white)(x)(x-h)^2=4p(y-k)

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

$\text{is the distance from the vertex to the focus/directrix}$

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

$\text{to find intercepts}$

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

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

$x = 0 \to 4 = - 12 y + 24 \Rightarrow y = \frac{5}{3} \leftarrow \textcolor{red}{\text{y-intercept}}$

$y = 0 \to {\left(x - 2\right)}^{2} = 24$

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

$x - 2 = \pm \sqrt{24} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x = 2 \pm 2 \sqrt{6} \leftarrow \textcolor{red}{\text{x-intercepts}}$