Answer
A. The domain is $(-\infty,-1) \cup (1, \infty)$
B. There is no y-intercept.
There are no x-intercepts.
C. The function is an odd function.
D. $y = -1$ is a horizontal asymptote.
$y = 1$ is a horizontal asymptote.
$x = -1$ is a vertical 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 minimum or local maximum.
G. The graph is concave down on the interval $(-\infty,-1)$
The graph is concave up on the interval $(1, \infty)$
There is no point of inflection.
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{x}{\sqrt{x^2-1}}$
A. The function is defined for all real numbers such that $x^2-1 \gt 0$
$x^2 \gt 1$
$x \lt -1$ or $x \gt 1$
The domain is $(-\infty,-1) \cup (1, \infty)$
B. Since $x \neq 0$, there is no y-intercept.
When $y = 0$:
$\frac{x}{\sqrt{x^2+1}} = 0$
There are no values of $x$ in the domain such that $y = 0$
There are no x-intercepts.
C. $\frac{-x}{\sqrt{(-x)^2-1}} = -\frac{x}{\sqrt{x^2-1}}$
The function is an odd function.
D. $\lim\limits_{x \to -\infty} \frac{x}{\sqrt{x^2-1}} = -1$
$y = -1$ is a horizontal asymptote.
$\lim\limits_{x \to \infty} \frac{x}{\sqrt{x^2-1}} = 1$
$y = 1$ is a horizontal asymptote.
$\lim\limits_{x \to -1^-} \frac{x}{\sqrt{x^2-1}} = -\infty$
$x = -1$ is a vertical asymptote.
$\lim\limits_{x \to 1^+} \frac{x}{\sqrt{x^2-1}} = \infty$
$x = 1$ is a vertical asymptote.
E. We can try to find values of $x$ such that $y' = 0$:
$y' =\frac{\sqrt{x^2-1}-(x)(\frac{x}{\sqrt{x^2-1}})}{x^2-1}$
$y' =\frac{x^2-1-x^2}{(x^2-1)^{3/2}}$
$y' =\frac{-1}{(x^2-1)^{3/2}}$
There are no values of $x$ such that $y' = 0$
$y' \lt 0$ for all values of $x$ in the domain.
The function is decreasing on the intervals $(-\infty,-1)\cup (1, \infty)$
F. There is no local minimum or local maximum.
G. We can try to find the values of $x$ such that $y'' = 0$:
$y'' = \frac{-(-1)(\frac{3x}{\sqrt{x^2-1}})}{(x^2-1)^3}$
$y'' = \frac{3x}{(x^2-1)^{5/2}}$
There are no values of $x$ in the domain 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 is no point of inflection.
H. We can see a sketch of the curve below.
