#### 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.