## Precalculus (6th Edition) Blitzer

a. $x\to-\infty, y\to-\infty$ and $x\to\infty, y\to-\infty$ b. $x=-5,0,1$, crosses the x-axis at $x=-5,0$, touches and turns around at $x=1$ c. $y=0$. d. neither e. See graph
Given the function $f(x)=-2x^3(x-1)^2(x+5)$, we have: a. The leading term is $-2x^6$ with a coefficient of $-2$ 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 $x=-5,0,1$ and the graph crosses the x-axis at $x=-5,0$ (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 $y=0$. d. Test $f(-x)=-2(-x)^3(-x-1)^2(-x+5)=-2x^3(x+1)^2(x-5)$. 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 $n=6$, the maximum number of turning points will be $5$, which agrees with the graph.