Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Section 13.3 One-sided Limits; Continuous Functions - 13.3 Assess Your Understanding - Page 910: 79

hole at $x=2$, asymptote at $x=-3$.

1. Factor to get $R(x)=\frac{x^3-2x^2+4x-8}{x^2+x-6}=\frac{x^2(x-2)+4(x-2)}{(x+3)(x-2)}=\frac{(x-2)(x^2+4)}{(x+3)(x-2)}=\frac{x^2+4}{x+3}, x\ne2$, thus the rational function is undefined at $x=2$ and $x=-3$. 2. We can determine a hole at $x=2$ and an asymptote at $x=-3$.