College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 6

Determine the vertex, focus and directrix of the parabola $\displaystyle 2x^2 + 6x + 5y + 10 = 0$ and sketch the graph.


$
\begin{equation}
\begin{aligned}

2x^2 + 6x + 5y =& -10
&& \text{Subtract } 10
\\
\\
2(x^2 + 3x + \quad ) + 5y =& -10
&& \text{Factor and group terms}
\\
\\
2 \left(x^2 + 3x + \frac{9}{4} \right) + 5y =& -10 + \frac{9}{2}
&& \text{Complete the square: add } \left( \frac{3}{2} \right)^2 = \frac{9}{4} \text{ on the left side and } \frac{9}{2} \text{ on the right side}
\\
\\
2 \left(x + \frac{3}{2} \right)^2 + 5y =& \frac{-11}{2}
&& \text{Perfect Square}
\\
\\
2 \left( x + \frac{3}{2} \right)^2 =& -5y - \frac{11}{2}
&& \text{Subtract } 5y
\\
\\
\left(x + \frac{3}{2} \right)^2 =& \frac{-5}{2} y - \frac{11}{4}
&& \text{Factor out } \frac{5}{2}
\\
\\
\left( x + \frac{3}{2} \right)^2 =& \frac{-5}{2} \left( y + \frac{22}{20} \right)
&&

\end{aligned}
\end{equation}
$


Now, the parabola has the form $(x - h)^2 = -4p(y - k)$ with vertex at $\displaystyle \left( \frac{-3}{2}, \frac{-22}{20} \right)$ that opens downward. Since $\displaystyle 4p = \frac{5}{2}$, we have $\displaystyle p = \frac{5}{8}$. It means that the focus is $\displaystyle \frac{5}{8}$ below to the vertex and the directrix is $\displaystyle \frac{5}{8}$ above the vertex. Thus, by using transformations, the focus is at

$\displaystyle \left( \frac{-3}{2}, \frac{-22}{20} \right) \to \left( \frac{-3}{2}, \frac{-22}{20} - \frac{5}{8} \right) = \left( \frac{-3}{2}, \frac{-69}{40} \right)$

and the directrix is the line $\displaystyle y = \frac{-22}{20} + \frac{5}{8} = \frac{-19}{40}$

Therefore, the graph is