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

Answer

See graph and explanations.
1582465633

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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.