Answer
$(-\infty,3)U(3,\infty)$
Vertical asymptote $x=3$
Oblique asymptote $y=x+9$
Work Step by Step
Step 1. The domain requirement is $x-3\ne0$, thus $x\ne3$ or $(-\infty,3)U(3,\infty)$
Step 2. We have $R(x)=\frac{(x^2-3x)+(9x-27)+32}{x-3}=x+9+\frac{32}{x-3}$
Step 3. We can identify a vertical asymptote as $x=3$ and an oblique asymptote as $y=x+9$
