Answer
See below
Work Step by Step
Given: $\frac{(x-6)^2}{25}-(y+1)^2=1$
Compare the given equation to the standard form of an equation of a hyperbola. You can see that the graph is a circle with its center at $(h,k)=(6,-1)$ and $a=5,b=1$.
Hence, the center of the hyperbola is at $(6,-1)$. The vertices are at $(1,-1)$ and $(11,-1)$.
Find asymptotes: $y-k=\pm \frac{b}{a}(x-h)\\y+1=\pm\frac{1}{5}(x-6)$
Hence, $y_1=\frac{1}{5}x-\frac{11}{5}$
and $y_2=-\frac{1}{5}x+\frac{1}{5}$
