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

Determine the vertex, focus and directrix of the parabola $\displaystyle x^2 = 3(x + y)$ and sketch the graph.


$
\begin{equation}
\begin{aligned}

x^2 =& 3x + 3y
&& \text{Distribute } 3
\\
\\
x^2 - 3x =& 3y
&& \text{Subtract } 3x
\\
\\
x^2 - 3x + \frac{9}{4} =& 3y + \frac{9}{4}
&& \text{Complete the square: add } \left( \frac{-3}{2} \right)^2 = \frac{9}{4}
\\
\\
\left(x - \frac{3}{2} \right)^2 =& 3 \left(y + \frac{9}{12} \right)
&& \text{Perfect Square, factor } 3

\end{aligned}
\end{equation}
$


Now, the parabola has the form $(x - h)^2 = 4p(y - k)$ with vertex at $\displaystyle \left( \frac{3}{2}, \frac{-9}{12} \right)$ that opens upward. Since $\displaystyle 4p = 3$, we have $\displaystyle p = \frac{3}{4}$. It means that the focus is $\displaystyle \frac{3}{4}$ above the vertex and the directrix is $\displaystyle \frac{3}{4}$ below the vertex. Thus, by using transformations, the focus is at

$\displaystyle \left( \frac{3}{2}, \frac{-9}{12} \right) \to \left( \frac{3}{2}, \frac{-9}{12} + \frac{3}{4} \right) = \left( \frac{3}{2}, 0 \right)$

and the directrix is the line $\displaystyle y = \frac{-9}{12} - \frac{3}{4} = \frac{-3}{2}$

Therefore, the graph is