Chapter 11.4, Problem 26E

To determine

To calculate: the vertex, focus and directrix of the parabola and sketch the graph

Answer to Problem 26E

The vertex is $\left(-3,0\right)$ , foci is $\left(-3,-3\right)$ , directrix is $y=3$

Explanation of Solution

Given information: The equation is $x^2+6x+12y+9=0$

Calculation: for the vertex foci and directrix of $x^2+6x+12y+9=0$

  $\begin{array}{l}a{x}^2+bx+c=0\\ y=a{\left(x-h\right)}^2+k\\ y=\frac{-x^2+6x-9}{12}\\ \Rightarrow -\frac{1}{12}{x}^2-\frac{x}{2}-\frac{9}{12}\\ vertex=\left(-3,0\right)\\ foci=\left(-3,-3\right)\\ directrix:y=3\end{array}$

The graph of $x^2+6x+12y+9=0$ shown below