Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$
b. $x=-2,0,2$, crosses the x-axis at $x=-2,2$, touches and turns around at $x=0$
c. $y=0$.
d. symmetric with respect to the y-axis
e. See graph
Work Step by Step
Given the function $f(x)=-x^2(x+2)(x-2)$, we have:
a. The leading term is $-x^4$ 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 fall as $x$ increases (right end) and it will also fall as $x$ decreases (left end).
b. The equation is factored; thus the x-intercepts are $x=-2,0,2$ and the graph crosses the x-axis at $x=-2,2$ (odd multiplicity), but touches and turns around at $x=0$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$, which gives $y=0$.
d. Test $f(-x)=-(-x)^2(-x+2)(-x-2)=-x^2(x-2)(x+2)=f(x)$
Thus the graph is symmetric with respect to the y-axis.
e. See graph; as $n=4$, the maximum number of turning points will be $3$, which agrees with the graph.
