Precalculus (6th Edition) Blitzer

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

Chapter 2 - Review Exercises - Page 430: 39


See graph and explanations.

Work Step by Step

Step 1. Given the function $f(x)=2x^2(x-1)^3(x+2)$, we can identify the zeros and their multiplicities as $x=0$ (multiplicity 2), $x=1$ (multiplicity 3), $x=-2$ (multiplicity 1). Step 2. The curve will cross the x-axis at $x=1$ and $x=-2$, but will touch and turn around at $x=0$ Step 3. The leading term is $2x^6$ with a coefficient of $+2$ and an even power. Thus, we can identify its end behaviors as $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$. That is, the curve rises to the left and also rises to the right. Step 4. We test $f(-x)=2(-x)^2(-x-1)^3(-x+2)=2x^2(x+1)^3(x-2)$ as $f(-x)\ne f(x)$ and $f(-x)\ne -f(x)$, the function is neither symmetric with respect to the y-axis nor with the origin. Step 5. Let $x=0$; we can find the y-intercept as $y=0$ Step 6. Based on the above results, we can graph the function as shown in the figure.
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