Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$
b. $x=-2,-1,0$, crosses the x-axis at $x=-1,0$, touches and turns around at $x=-2$
c. $y=0$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=x^3(x+2)^2(x+1)$, we have:
a. The leading term is $x^6$ with a coefficient of $+1$ 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 rise as $x$ increases (right end) and it will also rise as $x$ decreases (left end).
b. The equation is factored; thus the x-intercepts are $x=-2,-1,0$ and the graph crosses the x-axis at $x=-1,0$ (odd multiplicity), but touches and turns around at $x=-2$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$ which gives $y=0$.
d. Test $f(-x)=(-x)^3(-x+2)^2(-x+1)=x^3(x-2)^2(x-1)$. 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 $n=6$, the maximum number of turning points will be $5$, which agrees with the graph.
