Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $1$
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{(x-1)^2}{x^2+1} = 1$
$\lim\limits_{x \to \infty} \frac{(x-1)^2}{x^2+1} = 1$
$y = 1$ is a horizontal asymptote.
E. The function is decreasing on the interval $(-1, 1)$
The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$
F. The local maximum is $(-1, 2)$
The local minimum is $(1, 0)$
G. The graph is concave down on the intervals $(-1.73,0)\cup (1.73,\infty)$
The graph is concave up on the intervals $(-\infty,-1.73)\cup (0, 1.73)$
The points of inflection are $(-1.73,1.87),(0,1),$ and $(1.73,0.13)$
H. We can see a sketch of the curve below.

Work Step by Step
$y = \frac{(x-1)^2}{x^2+1}$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x = 0$, then $y = \frac{(0-1)^2}{0^2+1} = 1$
The y-intercept is $1$
When $y = 0$:
$\frac{(x-1)^2}{x^2+1} = 0$
$(x-1)^2 = 0$
$x = 1$
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} \frac{(x-1)^2}{x^2+1} = 1$
$\lim\limits_{x \to \infty} \frac{(x-1)^2}{x^2+1} = 1$
$y = 1$ is a horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{2(x-1)(x^2+1)-(x-1)^2(2x)}{(x^2+1)^2}$
$y' = \frac{2x^3+2x-2x^2-2-2x^3+4x^2-2x}{(x^2+1)^2}$
$y' = \frac{2x^2-2}{(x^2+1)^2} = 0$
$2x^2-2 = 0$
$x^2 = 1$
$x = -1, 1$
When $-1 \lt x \lt 1$ then $y' \lt 0$
The function is decreasing on the interval $(-1, 1)$
When $x \lt -1$ or $x \gt 1$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -1)\cup (1, \infty)$
F. When $x = -1,$ then $y = \frac{((-1)-1)^2}{(-1)^2+1} = 2$
The local maximum is $(-1, 2)$
When $x = 1,$ then $y = \frac{((1)-1)^2}{(1)^2+1} = 0$
The local minimum is $(1, 0)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' =\frac{(4x)(x^2+1)^2-(2x^2-2)(2)(x^2+1)(2x)}{(x^2+1)^4}$
$y'' =\frac{(4x)(x^2+1)-(2x^2-2)(2)(2x)}{(x^2+1)^3}$
$y'' =\frac{4x^3+4x-8x^3+8x}{(x^2+1)^3}$
$y'' =\frac{-4x^3+12x}{(x^2+1)^3} = 0$
$-4x^3 +12x= 0$
$-4x(x^2-3) = 0$
$x = -\sqrt{3}, 0, \sqrt{3}$
$x = -1.73, 0, 1.73$
When $-1.73 \lt x \lt 0~~$ or $x \gt 1.73$, then $y'' \lt 0$
The graph is concave down on the intervals $(-1.73,0)\cup (1.73,\infty)$
When $x \lt -1.73$ or $0 \lt x \lt 1.73$, then $y'' \gt 0$
The graph is concave up on the intervals $(-\infty,-1.73)\cup (0, 1.73)$
When $x = -1.73$ then $y = \frac{(-1.73-1)^2}{(-1.73)^2+1} = 1.87$
When $x = 0$ then $y = \frac{(0-1)^2}{(0)^2+1} = 1$
When $x = 1.73$ then $y = \frac{(1.73-1)^2}{(1.73)^2+1} = 0.13$
The points of inflection are $(-1.73,1.87),(0,1),$ and $(1.73,0.13)$
H. We can see a sketch of the curve below.
