Answer
domain $\{x|x\ne3 \}$.
horizontal asymptote $none$,
vertical asymptote $x=3$,
oblique asymptote $y=x+9$.
Work Step by Step
Step 1. $R(x)=\frac{x^2+6x+5}{x-3}=\frac{x^2-3x+9x-27+32}{x-3}=x+9+\frac{32}{x-3}$, domain $\{x|x\ne3 \}$.
Step 2. We can find horizontal asymptote $none$, vertical asymptote $x=3$, oblique asymptote $y=x+9$.
