Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$,
b. $x=0,3$, crosses the x-axis at $x=3$, will touch and turn around at $x=0$
c. $y=0$.
d. neither
e. See graph,
Work Step by Step
Given the function $f(x)=-x^3+3x^2$, we have:
a. The leading term is $-x^3$ with a coefficient of $-1$ and an odd 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 rise as $x$ decreases (left end).
b. Factor the equation as $f(x)=-x^2(x-3)$; thus the x-intercepts are $x=0,3$ and the graph crosses the x-axis at $x=3$ (odd multiplicity), but will touch and turn around at $x=0$ (even multiplicity),
c. We can find the y-intercept by letting $x=0$, which gives $y=0$.
d. Test $f(-x)=-(-x)^3+3(-x)^2=x^3+3x^2$. 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 $n=3$, the maximum number of turning points will be $2$, which agrees with the graph.
