Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.3 - Polynomial Functions and Their Graphs - Exercise Set - Page 349: 41

Answer

a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to\infty$; see explanations. b. $x=-2,-1,1$, crosses the x-axis at each intercept. c. $y=-2$. d. neither e. See graph and explanations.

Work Step by Step

Given the function $f(x)=x^3+2x^2-x-2$, we have: a. The leading term is $x^3$ with an odd power; when $x\to-\infty, y\to-\infty$ and when $x\to\infty, y\to\infty$, thus the curve will rise as $x$ increases (right end) and it will fall as $x$ decreases (left end). b. Factor the equation as $f(x)=x^2(x+2)-(x+2)=(x+2)(x^2-1)=(x+2)(x+1)(x-1)$; thus the x-intercepts are $x=-2,-1,1$ and the graph crosses the x-axis at each intercept. c. We can find the y-intercept by letting $x=0$, which gives $y=-2$. d. Test $f(-x)=(-x)^3+2(-x)^2-(-x)-2=-x^3+2x^2+x-2$. As $f(-x)\ne f(x)$ and $f(-x)\ne -f(x)$, the graph is neither symmetric with respect to the y-axis nor to the origin. e. See graph; as $n=3$, the maximum number of turning points will be $2$ which agrees with the graph.
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