Answer
(a) center $(2, -2)$, vertices $V(2\pm2\sqrt 2,-2)$, foci $F_1(-2,-2)$ and $F_2(6,-2)$, asymptotes $y=\pm (x-2)-2$
(b) See graph.
Work Step by Step
(a) Step 1. Write the given equation as $\frac{(x-2)^2}{8}-\frac{(y+2)^2}{8}=1$
Step 2. Identify the center as $(2, -2)$
Step 3. Identify $a=b=2\sqrt 2, c=\sqrt {8+8}=4$
Step 4. The original vertices are $(\pm2\sqrt 2, 0)$, vertices after the shift $V(2\pm2\sqrt 2,-2)$
Step 5. Original foci $(\pm 4, 0)$, foci after shift $F(2\pm4, -2)$ or $F_1(-2,-2)$ and $F_2(6,-2)$
Step 6. Original asymptotes $y=\pm x$, new asymptotes $y=\pm (x-2)-2$
(b) See graph.
