College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 24

Determine the center, vertices, foci and asymptotres of the hyperbola $\displaystyle y^2 = x^2 + 6y$. Then, sketch its graph

$\begin{equation} \begin{aligned} x^2 - y^2 + 6y &= 0 && \text{Subtract } y^2\\ \\ x^2 - (y^2 - 6y) &= 0 && \text{Group terms}\\ \\ x^2 - (y^2 - 6y + 9) &= -9 && \text{Complete the square: Add $\left( \frac{6}{2} \right)^2 = 9$ on the left side and subtract 9 on the right side. }\\ \\ x^2 - (y - 3)^2 &= -9 && \text{Perfect square}\\ \\ \frac{(y - 3)^2}{9} - \frac{x^2}{9} &= 1 && \text{Divide both sides by } -9 \end{aligned} \end{equation}$


Now, the hyperbola has the form $\displaystyle \frac{(y - k)^2}{a^2} = \frac{(x-h)^2}{b^2} = 1$ with center at $(h,k)$ and vertical transverse axis.
Since the denominator $y^2$ is positive. It is derived from the hyperbola $\displaystyle \frac{y^2}{9} - \frac{x^2}{9} = 1$ by shifting it $3$ units upward.
This gives $a^2 = 9$ and $b^2 = 9$, so $a = 3, b = 3$ and $c = \sqrt{a^2 + b^2} = \sqrt{9+9} = 3\sqrt{2}$
Then, by applying transformations

$\begin{equation} \begin{aligned} \text{center } & (h,k) && \rightarrow && (0,3)\\ \\ \text{vertices } & (0,a)&& \rightarrow && (0,3) && \rightarrow && (0,3+3) && = && (0,6)\\ \\ & (0,-a)&& \rightarrow && (0,-3) && \rightarrow && (0,-3+3) && = && (0,0)\\ \\ \text{foci } & (0,c)&& \rightarrow && (0,3\sqrt{2}) && \rightarrow && (0,3\sqrt{2}+3) && = && (0,3\sqrt{2}+3)\\ \\ & (-c,0)&& \rightarrow && (0,-3\sqrt{2}) && \rightarrow && (0,-3\sqrt{2}+3) && = && (0,-3\sqrt{2}+3)\\ \\ \text{asymptote } &y = \pm \frac{a}{b}x && \rightarrow && y = \pm x && \rightarrow && y - 3 = \pm x\\ \\ &&&&&&&&& y = \pm x + 3\\ \\ &&&&&&&&& y = x + 3 \text{ and } y = - x + 3 \end{aligned} \end{equation}$

Therefore, the graph is