Answer
See graph
Work Step by Step
$y$ = $f(x)$ = $x^{3}+3x^{2}$ = $x^{2}(x+3)$
$A.$ $f$ is a polynomial so $D$ = $R$
$B.$ $y$-intercept = $f(0)$ = $0$, $x$-intercept are $0$ and $-3$
$C.$ No symmetry
$D.$ No asymptote
$E.$ $f'(x)$ = $3x^{2}+6x$ = $3x(x+2)$ $\gt$ $0$ => $x$ $\lt$ $-2$ or $x$ $\gt$ $0$ so $f$ is increasing on $(-\infty,-2)$ and $(0,\infty)$ and decreasing on $(-2,0)$
$F.$ Local maximum value $f(-2)$ = $4$, local minimum value $f(0)$ = $0$
$G.$ $f''(x)$ = $6x+6$ = $6(x+1)$ $\gt$ $0$ => $x$ $\gt$ $-1$ so $f$ is concave upward on $(-1,\infty)$ and concave downward on $(-\infty,-1)$
There is inflection point at $(-1,2)$
