Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$,
b. $x=0,1$, crosses the x-axis
c. $y=0$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=-2x^4+2x^3$, we have:
a. The leading term is $-2x^4$ with a coefficient of $-2$ 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. Factor the equation as $f(x)=-2x^3(x-1)$; thus the x-intercepts are $x=0,1$ and the graph crosses the x-axis at $x=0,1$ (odd multiplicity).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=-2(-x)^4+2(-x)^3=-2x^4-2x^3$. 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.
