Answer
(a) See graph.
(b) $(-\infty,-3]U(-2,-\frac{3}{2}]U(2,\infty)$
Work Step by Step
(a) Factor the function to get $f(x)=\frac{(2x+3)(x+3)}{(x-2)(x+2)}$. We have V.A. $x=\pm2$, H.A. $y=2$, x-intercept $x=-3,-\frac{3}{2}$, y-intercept $f(0)=-\frac{9}{4}$. Use additional test values to plot. See graph.
(b) Use the graph, for $f(x)\ge0$, the solution intervals are $(-\infty,-3]U(-2,-\frac{3}{2}]U(2,\infty)$
