Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to\infty$
b. $x=-4,-3,-1$, crosses the x-axis at $x=-4,-3,-1$
c. $y=12$.
d. neither
e. See graph
Work Step by Step
Given the function $f(x)=(x+3)(x+1)^3(x+4)$, 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 rise as $x$ increases (right end) and it will fall as $x$ decreases (left end).
b. The equation is factored; thus the x-intercepts are $x=-4,-3,-1$ and the graph crosses the x-axis at $x=-4,-3,-1$ (odd multiplicity),
c. We can find the y-intercept by letting $x=0$, which gives $y=12$.
d. Test $f(-x)=(-x+3)(-x+1)^3(-x+4)=-(x-3)(x-1)^3(x-4)$. 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 $n=5$, the maximum number of turning points will be $4$, which agrees with the graph.
