College Algebra, Chapter 8, Review Exercises, Section Review Exercises, Problem 22

Determine the center, vertices, foci and asymptotres of the hyperbola $\displaystyle \frac{(x-2)^2}{8} - \frac{(y+2)^2}{8} = 1$. Then, sketch its graph

The shifted hyperbola was the form $\displaystyle \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ with center at $(h,k)$ and horizontal transverse axis.
It is derived from the hyperbola $ \displaystyle \frac{x^2}{8} - \frac{y^2}{8} = 1$ by shifting it 2 units to the right and 2 units downward. Thus gives
$a^2 = 8$ and $b^2 = 8$. So, $ a = 2\sqrt{2}, b = 2\sqrt{2}$ and $c = \sqrt{a^2 + b^2} = \sqrt{8+8} = 4$
Then, by applying transformations

$
\begin{equation}
\begin{aligned}
\text{center } & (h,k) && \rightarrow && (2,-2)\\
\\
\text{vertices } & (a,0)&& \rightarrow && (2 \sqrt{2},0) && \rightarrow && (2\sqrt{2}+2,0-2) && = && (2\sqrt{2}+2,-2)\\
\\
& (-a,0)&& \rightarrow && (-2\sqrt{2},0) && \rightarrow && (-2\sqrt{2}+2,0-2) && = && (-2\sqrt{2}+2,-2)\\
\\
\text{foci } & ( c, 0)&& \rightarrow && (4,0) && \rightarrow && (4 +2, 0-2) && = && (6,-2)\\
\\
& (-c,0)&& \rightarrow && (-4,0) && \rightarrow && (-4+2,0-2) && = && (-2,-2)\\
\\
\text{asymptote } &y = \pm \frac{b}{a}x && \rightarrow && y = \pm x && \rightarrow && y + 2 = \pm (x - 2)\\
\\
&&&&&&&&& y + 2 = \pm x \mp 2\\
\\
&&&&&&&&& y = x -4 \text{ and } y = -x
\end{aligned}
\end{equation}
$

Therefore, the graph is