Answer
(a) $\frac{2x-5}{x-3}$
(b) $ \frac{5}{2}$
(c) vertical asymptote $x=3$, horizontal asymptote $y=2$
Work Step by Step
(a) translating the graph of $y=\frac{1}{x}$ to the right 3 units and up 2 units will result an equation of $f(x)=\frac{1}{x-3}+2=\frac{2x-5}{x-3}$
(b) The zero(s) of $f(x)$ can be found by letting $f(x)=0$ which gives $x=\frac{5}{2}$
(c) We can identify a vertical asymptote $x=3$ and a horizontal asymptote $y=2$
