Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$
b. $x=-3,0,1$, crosses the x-axis at $x=-3,1$, touches and turns around at $x=0$
c. $y=0$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=-x^2(x-1)(x+3)$, we have:
a. The leading term is $-x^4$ with a coefficient of $-1$ and an even 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 also fall as $x$ decreases (left end).
b. The equation is factored; thus the x-intercepts are $x=-3,0,1$ and the graph crosses the x-axis at $x=-3,1$ (odd multiplicity), but touches and turns around at $x=0$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$, which gives $y=0$.
d. Test $f(-x)=-(-x)^2(-x-1)(-x+3)=-x^2(x+1)(x-3)$. As $f(-x)\ne f(x)$ and $f(-x)\ne -f(x)$, the graph is neither symmetric with respect to the y-axis nor with the origin.
e. See graph; as $n=4$, the maximum number of turning points will be $3$, which agrees with the graph.
