Answer
(a) See graph.
(b) domain $(-\infty,\infty)$ and range $[\frac{1}{2},\infty)$.
(c) increasing on $(-\frac{1}{3},\infty)$ and decreasing on $(-\infty,-\frac{1}{3})$.
Work Step by Step
(a) Given $f(x)=\frac{9}{2}x^2+3x+1=\frac{9}{2}(x+\frac{1}{3})^2+\frac{1}{2}$ with $a=\frac{9}{2}$, thus its graph opens up, its vertex is $(-\frac{1}{3},\frac{1}{2})$, axis of symmetry is $x=-\frac{1}{3}$, y-intercept is $(0,1)$ (let x=0), and x-intercepts are $none$ (let f=0). See graph.
(b) Based on the graph, we can find the domain $(-\infty,\infty)$ and range $[\frac{1}{2},\infty)$.
(c) Based on the graph, we can find the function is increasing on $(-\frac{1}{3},\infty)$ and decreasing on $(-\infty,-\frac{1}{3})$.
