Answer
Hyperbola, center $C(\frac{1}{2},-1)$, foci $F(\frac{1}{2}\pm\sqrt {10},-1)$, vertices $V_1(-\frac{1}{2},-1)$ and $V_2(\frac{3}{2},-1)$, asymptotes $y=\pm 3(x-\frac{1}{2})-1$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $36(x^2-x+\frac{1}{4})-4(y^2+2y+1)=31+9-4=36$ which gives $(x-\frac{1}{2})^2-\frac{(y+1)^2}{9}=1$, thus the equation is of the form of a hyperbola.
Step 2. The center can be found as $C(\frac{1}{2},-1)$
Step 3. With $a=1, b=3, c=\sqrt {1+9}=\sqrt {10}$, the foci can be found as $F(\frac{1}{2}\pm\sqrt {10},-1)$
Step 4. The vertices can be found as $V(\frac{1}{2}\pm1,-1)$ or $V_1(-\frac{1}{2},-1)$ and $V_2(\frac{3}{2},-1)$
Step 5. The original asymptotes can be found as $y=\pm 3(x)$ and the shifted asymptotes are $y=\pm 3(x-\frac{1}{2})-1$
Step 6. See graph.
