Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$
b. $x=1,\pm2$, crosses the x-axis at $x=\pm2$, touches and turns around at $x=1$
c. $y=12$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=-3(x-1)^2(x^2-4)$, we have:
a. The leading term is $-3x^4$ with a coefficient of $-3$ 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 $f(x)=-3(x-1)^2(x^2-4)=-3(x-1)^2(x+2)(x-2)$; thus the x-intercepts are $x=1,\pm2$ and the graph crosses the x-axis at $x=\pm2$ (odd multiplicity), but touches and turns around at $x=1$ (even multiplicity).
c. We can find the y-intercept by letting $x=0$, which gives $y=12$.
d. Test $f(-x)= -3(-x-1)^2((-x)^2-4)=-3(x+1)^2(x^2-4)$. 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.
