Answer
vertical asymptote $x=3$,
horizontal asymptote $none$,
oblique asymptote $y=x+5$.
Work Step by Step
Step 1. Simplify the function $R(x)=\frac{x^3-8}{x^2-5x+6}=\frac{(x-2)(x^2+2x+4)}{(x-2)(x-3)}=\frac{x^2+2x+4}{x-3}=x+5+\frac{19}{x-3}, (x\ne2)$,
Step 2. we can find the vertical asymptote $x=3$,
Step 3. we can find the horizontal asymptote $none$,
Step 4. we can find the oblique asymptote $y=x+5$.
