Answer
See graph, hole at $x=3$, asymptote at $x=-1$.
Work Step by Step
1. See graph for $R(x)=\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$
2. We can confirm a hole at $x=3$ and an asymptote at $x=-1$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
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: 88
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