Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$,
b. $x=0,2$, crosses the x-axis.
c. $y=0$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=-2x^4+4x^3$, we have:
a. The leading term is $-2x^4$ with a coefficient of $-2$ and an even power; when $x\to-\infty, y\to-\infty$ and when $x\to\infty, y\to-\infty$, thus 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. Factor the equation as $f(x)=-2x^3(x-2)$; thus the x-intercepts are $x=0,2$ and the graph crosses the x-axis at $x=0,2$ (odd multiplicity).
c. We can find the y-intercept by letting $x=0$, which gives $y=0$.
d. Test $f(-x)=-2(-x)^4+4(-x)^3=-2x^4-4x^3$. 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.
