#### Answer

A. The domain is $(-\infty,1) \cup(1, \infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} (\frac{x}{x-1}) = 1$
$\lim\limits_{x \to \infty} (\frac{x}{x-1}) = 1$
$y = 1$ is a horizontal asymptote.
$x = 1$ is a vertical asymptote.
E. The function is decreasing on the intervals $(-\infty, 1)\cup (1,\infty)$
F. There is no local maximum or local minimum.
G. The graph is concave down on the interval $(-\infty, 1)$
The graph is concave up on the interval $(1, \infty)$
There no points of inflection.
H. We can see a sketch of the curve below.

#### Work Step by Step

$y = \frac{x}{x-1}$
A. The function is defined for all real numbers except $x = 1$.
The domain is $(-\infty,1) \cup(1, \infty)$
B. When $x = 0,$ then $y = \frac{0}{0-1}= 0$
The y-intercept is $0$
When $y = 0$:
$\frac{x}{x-1} = 0$
$x = 0$
The x-intercept is $0$
C. The function is not an odd function or an even function.
D. $\lim\limits_{x \to -\infty} (\frac{x}{x-1}) = 1$
$\lim\limits_{x \to \infty} (\frac{x}{x-1}) = 1$
$y = 1$ is a horizontal asymptote.
$x = 1$ is a vertical asymptote.
E. We can try to find the values of $x$ such that $y' = 0$:
$y' = \frac{(x-1)-(x)}{(x-1)^2} = -\frac{1}{(x-1)^2} = 0$
There are no values of $x$ such that $y'=0$
When $x \lt 1$ or $x \gt 1$, then $y' \lt 0$
The function is decreasing on the intervals $(-\infty, 1)\cup (1,\infty)$
F. There is no local maximum or local minimum.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = -\frac{-2(x-1)}{(x-1)^4} = \frac{2}{(x-1)^3}$
There are no values of $x$ such that $y''=0$
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 $(1, \infty)$
There no points of inflection.
H. We can see a sketch of the curve below.