## Precalculus (6th Edition) Blitzer

a. $x\to-\infty, y\to\infty$ and $x\to\infty, y\to\infty$ b. $x=-2,0,1$, crosses the x-axis at $x=-2,1$, touches and turns around at $x=0$ c. $y=0$. d. neither e. See graph Given the function $f(x)=x^2(x-1)^3(x+2)$, we have: a. The leading term is $x^6$ 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 $x=-2,0,1$ and the graph crosses the x-axis at $x=-2,1$ (odd multiplicity), but touches and turns around at $x=0$ (even multiplicity). c. We can find the y-intercept by letting $x=0$ which gives $y=0$. d. Test $f(-x)=(-x)^2(-x-1)^3(-x+2)=x^2(x+1)^3(x-2)$. 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.