Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to-\infty$,
b. $x=0,\pm\sqrt 3$, crosses the x-axis at $x=0$; will touch and turn around at $x=\pm\sqrt 3$
c. $y=0$.
d. symmetric with respect to the origin.
e. See graph
Work Step by Step
Given the function $f(x)=-x^5+6x^3-9x$, 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 $f(x)=-x(x^4-6x^2+9)=-x(x^2-3)^2=-x(x+\sqrt 3)^2(x-\sqrt 3)^2$; thus the x-intercepts are $x=0,\pm\sqrt 3$ and the graph crosses the x-axis at $x=0$ (odd multiplicity), but will touch and turn around at $x=\pm\sqrt 3$ (even multiplicity) .
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=-(-x)^5+6(-x)^3-9(-x)=x^5-6x^3+9x$. 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.
