Answer
$\{x|x\ne2 \}$.
H.A. $none$, V.A. $x=2$, O.A. $y=x+2$.
Work Step by Step
Step 1. Given $R(x)=\frac{x^2+4}{x-2}=\frac{x^2-4+8}{x-2}=x+2+\frac{8}{x-2}$, we can find the domain $x\ne2$ or $\{x|x\ne2 \}$.
Step 2. We can find horizontal asymptote(s) H.A. $none$, vertical asymptote(s) V.A. $x=2$, oblique asymptote(s) O.A. $y=x+2$.
