Answer
a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to\infty$, see explanations.
b. $x=-2,-1,2$, crosses the x-axis at each intercept.
c. $y=-4$.
d. neither
e. See graph and explanations.
Work Step by Step
Given the function $f(x)=x^3+x^2-4x-4$, we have:
a. The leading term is $x^3$ with an odd power, when $x\to-\infty, y\to-\infty$ and when $x\to\infty, y\to\infty$, thus the curve will rise as $x$ increases (right end) and it will fall as $x$ decreases (left end).
b. Factor the equation as $f(x)=x^2(x+1)-4(x+1)=(x+1)(x^2-4)=(x+2)(x+1)(x-2)$
Thus the x-intercepts are $x=-2,-1,2$ and the graph crosses the x-axis at each intercept.
c. We can find the y-intercept by letting $x=0$ which gives $y=-4$.
d. Test $f(-x)=(-x)^3+(-x)^2-4(-x)-4=-x^3+x^2+4x-4$. As $f(-x)\ne f(x)$ and $f(-x)\ne -f(x)$, the graph is neither symmetric with respect to the y-axis nor to the origin.
e. See graph; as $n=3$, the maximum number of turning points will be $2$, which agrees with the graph.
