# How do you find the vertex and intercepts for  y^2+6y+12x+33=0?

Feb 13, 2017

$\text{Vertex" -> (x,y)=(-2,-3) larr" graph of shape } \supset$

${x}_{\text{intercept}} \to \left(x , y\right) = \left(- \frac{11}{4} , 0\right)$

y_("intercept")-> "none"

#### Explanation:

This is a quadratic in $y$ instead of $x$

So $y$ is the independent variable. The consequence is that the graph of type $\cup$ is rotated clockwise by ${90}^{o}$

Being given this question suggests that your mathematical manipulation skills are more than basic. So not a lot of explanation is given.

$12 x = - {y}^{2} - 6 y - 33$

$\textcolor{red}{\text{Correction:}}$ as ${y}^{2}$ is negative so the graph type of $\cup$ would be rotated ${90}^{o}$ anticlockwise giving shape $\supset$

$x = - \frac{1}{12} {y}^{2} - \frac{1}{2} y - \frac{11}{4}$

$\textcolor{b l u e}{\text{Determine the x-intercept}}$
Instead of $- \frac{11}{4}$ being the intercept of the y-axis it is the intercept of the x-axis

$\textcolor{b l u e}{\text{Determine the y-intercept}}$

The determinant of form
${b}^{2} - 4 a c \text{->} {\left(- \frac{1}{2}\right)}^{2} - 4 \left(- \frac{1}{12}\right) \left(- \frac{11}{4}\right) = + \frac{1}{4} - \frac{11}{12}$

$= - \frac{2}{3}$

As the determinant is negative there is no y-intercept
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Determine vertex}}$

Write as:$\text{ } x = - \frac{1}{12} \left({y}^{2} + \frac{12}{2} y\right) - \frac{11}{4}$

Consider the $\frac{12}{2}$ from $\frac{12}{2} y$

$\implies {y}_{\text{vertex}} = \left(- \frac{1}{2}\right) \times \frac{12}{2} = - 3$

Substituting $y = - 3$

${x}_{\text{vertex}} = - \frac{1}{12} {\left(- 3\right)}^{2} - \frac{1}{2} \left(- 3\right) - \frac{11}{4}$

${x}_{\text{vertex}} = - \frac{3}{4} + \frac{3}{2} - \frac{11}{4} = - 2$

$\text{Vertex} \to \left(x , y\right) = \left(- 2 , - 3\right)$ 