Answer
a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$
b. $x=-4,1,2$, crosses the x-axis at $x=-4,1$, touches and turns around at $x=2$
c. $y=-16$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=(x-2)^2(x+4)(x-1)$, 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 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 $x=-4,1,2$ and the graph crosses the x-axis at $x=-4,1$ (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 $y=-16$.
d. Test $f(-x)=(-x-2)^2(-x+4)(-x-1)=(x+2)^2(x-4)(x+1)$. 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 $n=4$, the maximum number of turning points will be $3$, which agrees with the graph.
