Answer
domain $\{x|x\ne-10,4 \}$.
H.A. $y=2$, V.A. $x=-10$, O.A. $none$.
Work Step by Step
Step 1. For $g(x)=\frac{2x^2-14x+24}{x^2+6x-40}=\frac{2(x-3)(x-4)}{(x+10)(x-4)}=\frac{2(x-3)}{x+10}, x\ne4$, we can find the domain $\{x|x\ne-10,4 \}$.
Step 2. We can find the horizontal asymptote(s) H.A. $y=2$, vertical asymptote(s) V.A. $x=-10$, oblique asymptote(s) O.A. $none$.
