Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $2$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (2+3x^2-x^3) = \infty$
$\lim\limits_{x \to \infty} (2+3x^2-x^3) = -\infty$
There are no asymptotes.
E. The function is increasing on the interval $(0, 2)$
The function is decreasing on the intervals $(-\infty,0)\cup (2, \infty)$
F. $(0,2)$ is a local minimum.
$(2,6)$ is a local maximum.
G. The graph is concave down on the interval $(1, \infty)$
The graph is concave up on the interval $(-\infty, 1)$
The point of inflection is $(1, 4)$
H. We can see a sketch of the curve below.

Work Step by Step
$y = 2+3x^2-x^3$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = 2+3(0)^2-(0)^3 = 2$
The y-intercept is $2$
C. The function is not odd or even.
D. $\lim\limits_{x \to -\infty} (2+3x^2-x^3) = \infty$
$\lim\limits_{x \to \infty} (2+3x^2-x^3) = -\infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = 6x-3x^2 = 0$
$3x(2-x) = 0$
$x = 0, 2$
The function is increasing on the interval $(0, 2)$
The function is decreasing on the intervals $(-\infty,0)\cup (2, \infty)$
F. When $x = 0$, then $y = 2+3(0)^2-(0)^3 = 2$
$(0,2)$ is a local minimum.
When $x = 2$, then $y = 2+3(2)^2-(2)^3 = 6$
$(2,6)$ is a local maximum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = 6-6x = 0$
$1-x = 0$
$x = 1$
When $x \gt 1$, then $y'' \lt 0$
The graph is concave down on the interval $(1, \infty)$
When $x \lt 1$, then $y'' \gt 0$
The graph is concave up on the interval $(-\infty, 1)$
When $x= 1$, then $y = 2+3(1)^2-(1)^3 = 4$
The point of inflection is $(1, 4)$
H. We can see a sketch of the curve below.
