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: 81

hole at $x=-1$, asymptote at $x=-\sqrt[3] 2$.

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