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: 51

Answer

a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$, b. $x=0,\pm\sqrt 3$, crosses the x-axis at $x=0$; will touch and turn around at $x=\pm\sqrt 3$ c. $y=0$. d. symmetric with respect to the origin. e. See graph

Work Step by Step

Given the function $f(x)=-x^5+6x^3-9x$, we have: a. The leading term is $-x^5$ with a coefficient of $-1$ and an odd power; thus $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$, and the end behaviors are that the curve will fall as $x$ increases (right end) and it will rise as $x$ decreases (left end). b. Factor the equation as $f(x)=-x(x^4-6x^2+9)=-x(x^2-3)^2=-x(x+\sqrt 3)^2(x-\sqrt 3)^2$; thus the x-intercepts are $x=0,\pm\sqrt 3$ and the graph crosses the x-axis at $x=0$ (odd multiplicity), but will touch and turn around at $x=\pm\sqrt 3$ (even multiplicity) . c. We can find the y-intercept by letting $x=0$ which gives $y=0$. d. Test $f(-x)=-(-x)^5+6(-x)^3-9(-x)=x^5-6x^3+9x$. As $f(-x)= -f(x)$, the graph is symmetric with respect to the origin. e. See graph; as $n=5$, the maximum number of turning points will be $4$, which agrees with the graph.
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