Answer
Hyperbola, center $C(-3,0)$, foci $F(-3\pm2\sqrt 2, 0)$, vertices $V_1(-6,0)$ and $V_2(0,0)$, asymptotes $y=\pm \frac{1}{3}(x+3)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $(x+3)^2-9y^2=9$ which gives $\frac{(x+3)^2}{9}-y^2=1$, thus the equation is of the form of a hyperbola.
Step 2. The center can be found as $C(-3,0)$
Step 3. With $a=3, b=1, c=\sqrt {9-1}=2\sqrt 2$, the foci can be found as $F(-3\pm2\sqrt 2, 0)$
Step 4. The vertices can be found as $V(-3\pm3,, 0)$ or $V_1(-6,0)$ and $V_2(0,0)$
Step 5. The original asymptotes can be found as $y=\pm \frac{1}{3}(x)$ and the shifted asymptotes are $y=\pm \frac{1}{3}(x+3)$
Step 6. See graph.
