## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 2 - Review Exercises - Page 430: 40

#### Answer

See graph and explanations. #### Work Step by Step

Step 1. Given the function $f(x)=-x^3(x+4)^2(x-1)$, we can identify the zeros and their multiplicities as $x=0$ (multiplicity 3), $x=-4$ (multiplicity 2), and $x=1$ (multiplicity 1). Step 2. The curve will cross the x-axis at $x=0$ and $x=1$, but will touch and turn around at $x=-4$. Step 3. The leading term is $-x^6$, with a coefficient of $-1$ 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 falls to the left and also falls to the right. Step 4. We test: $f(-x)=-(-x)^3(-x+4)^2(-x-1)=-x^3(x-4)^2(x+1)$ 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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