Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is an even function.
D. $\lim\limits_{x \to -\infty} (\frac{x^2}{x^2+3}) = 1$
$\lim\limits_{x \to \infty} (\frac{x^2}{x^2+3}) = 1$
$y = 1$ is a horizontal asymptote.
E. The function is decreasing on the interval $(-\infty, 0)$
The function is increasing on the interval $(0, \infty)$
F. The local minimum is $(0, 0)$
G. The graph is concave down on the intervals $(-\infty,-1)\cup (1,\infty)$
The graph is concave up on the intervals $(-1,1)$
The points of inflection are $(-1,\frac{1}{4})$ ans $(1, \frac{1}{4})$
H. We can see a sketch of the curve below.

Work Step by Step
$y = \frac{x^2}{x^2+3}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x = 0$, then $y = \frac{0^2}{0^2+3} = 0$
The y-intercept is $0$
When $y = 0$:
$\frac{x^2}{x^2+3} = 0$
$x^2 = 0$
$x = 0$
The x-intercept is $0$
C. $\frac{(-x)^2}{(-x)^2+3} = \frac{x^2}{x^2+3}$
The function is an even function.
D. $\lim\limits_{x \to -\infty} (\frac{x^2}{x^2+3}) = 1$
$\lim\limits_{x \to \infty} (\frac{x^2}{x^2+3}) = 1$
$y = 1$ is a horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{(2x)(x^2+3)-(x^2)(2x)}{(x^2+3)^2} = \frac{6x}{(x^2+3)^2} = 0$
$6x = 0$
$x = 0$
When $x \lt 0,$ then $y' \lt 0$
The function is decreasing on the interval $(-\infty, 0)$
When $x \gt 0$, then $y' \gt 0$
The function is increasing on the interval $(0, \infty)$
F. When $x = 0,$ then $y = \frac{(0)^2}{(0)^2+3} = 0$
The local minimum is $(0, 0)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{6(x^2+3)^2-(6x)(2)(x^2+3)(2x)}{(x^2+3)^4}$
$y'' = \frac{6(x^2+3)-(6x)(2)(2x)}{(x^2+3)^3}$
$y'' = \frac{6x^2+18-24x^2}{(x^2+3)^3}$
$y'' = \frac{18-18x^2}{(x^2+3)^3} = 0$
$18-18x^2 = 0$
$x^2-1 = 0$
$x = -1, 1$
When $x \lt -1~~$ or $x \gt 1$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty,-1)\cup (1,\infty)$
When $-1 \lt x \lt 1$, then $y'' \gt 0$
The graph is concave up on the intervals $(-1,1)$
When $x = -1,$ then $y = \frac{(-1)^2}{(-1)^2+3} = \frac{1}{4}$
When $x = 1,$ then $y = \frac{(1)^2}{(1)^2+3} = \frac{1}{4}$
The points of inflection are $(-1,\frac{1}{4})$ ans $(1, \frac{1}{4})$
H. We can see a sketch of the curve below.
