Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$,
b. $x=0,\pm\sqrt 2$, crosses the x-axis
c. $y=0$.
d. symmetric with respect to the origin.
e. See graph,
Work Step by Step
Given the function $f(x)=-x^5-x^3+6x$, we have:
a. The leading term is $-x^5$ 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 $f(x)=-x(x^4+x^2-6)=-x(x^2+3)(x^2-2)=-x(x^2+3)(x+\sqrt 2)(x-\sqrt 2)$; thus the x-intercepts are $x=0,\pm\sqrt 2$ and the graph crosses the x-axis at $x=0,\pm\sqrt 2$ (odd multiplicity),
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=-(-x)^5-(-x)^3+6(-x)=x^5+x^3-6x$. As $f(-x)= -f(x)$, the graph is symmetric with respect to the origin.
e. See graph; as $n=5$, the maximum number of turning points will be $4$, which agrees with the graph.
