Answer
asymptotes: $y=\pm(x-1)+2$
foci: $(1\pm\sqrt {6},2)$
Work Step by Step
Step 1. Rewriting the equation as $\frac{(x-1)^2}{3}-\frac{(y-2)^2}{3}=1$, we have $a=b=\sqrt 3, c=\sqrt {a^2+b^2}=\sqrt {6}$ centered at $(1,2)$ with a horizontal transverse axis.
Step 2. We can find the vertices as $(1\pm\sqrt 3,2)$ and asymptotes as $y-2=\pm\frac{b}{a}(x-1)$ or $y=\pm(x-1)+2$
Step 3. We can graph the equation as shown in the figure with foci at $(1\pm\sqrt {6},2)$
