Answer
(a) opens up.
(b) vertex $(2,-8)$.
(c) $x=2$.
(d) x-intercepts $x=\frac{6\pm2\sqrt {6}}{3}$, y-intercept $f(0)=4$.
(e) See graph.
(f) domain $(-\infty,\infty)$ range $[-8,\infty)$
(g) decreasing $(-\infty,2)$, increasing $(2,\infty)$.
Work Step by Step
(a) From the given function $f(x)=3x^2-12x+4$ with $a=3\gt0$, we can determine the graph opens up.
(b) $f(x)=3x^2-12x+4=3(x^2-4x+4)-8=3(x-2)^2-8$ and we can determine the vertex $(2,-8)$.
(c) We can determine the axis of symmetry $x=2$.
(d) $3x^2-12x+4=0\Longrightarrow x=\frac{12\pm\sqrt {12^2-4(3)(4)}}{2(3)}=\frac{6\pm2\sqrt {6}}{3}$ and we can determine the x-intercepts $x=\frac{6\pm2\sqrt {6}}{3}$, y-intercept $f(0)=4$.
(e) See graph.
(f) Based on the graph, we can determine the domain $(-\infty,\infty)$ range $[-8,\infty)$
(g) Based on the graph, we can determine the function is decreasing $(-\infty,2)$, increasing $(2,\infty)$.
