Answer
(a) See graph
(b) domain $(-\infty,\infty)$, range $[\frac{1}{2},\infty)$.
(c) decreasing on $(-\infty,-\frac{1}{3})$, increasing on $(-\frac{1}{3},\infty)$.
Work Step by Step
(a) For $y=\frac{9}{2}x^2+3x+1=\frac{1}{2}(9x^2+6x+1)+\frac{1}{2}=\frac{1}{2}(3x+1)^2+\frac{1}{2}$, we can determining that the graph opens up, vertex $(-\frac{1}{3},\frac{1}{2})$, axis of symmetry $x=-\frac{1}{3}$, y-intercept $(0,1)$, x-intercept(s) $none$. See graph
(b) We can determine the domain $(-\infty,\infty)$, range $[\frac{1}{2},\infty)$.
(c) We can determine the function is decreasing on $(-\infty,-\frac{1}{3})$, increasing on $(-\frac{1}{3},\infty)$.
