#### Answer

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

#### Work Step by Step

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