Question 1200106
<pre>
{{{"f(x)"}}}{{{""=""}}}{{{x^""/sqrt(x^2-1)}}}

To find vertical asymptotes, set the denominator equal to 0:

{{{sqrt(x^2-1)}}}{{{""=""}}}{{{0}}}

{{{x^2-1}}}{{{""=""}}}{{{0}}}

{{{(x-1)(x+1)}}}{{{""=""}}}{{{0}}}

x-1 = 0;  x+1 = 0
  x = 1;    x = -1

So the vertical asymptotes are the vertical lines whose equations are
x = 1  and x = -1

To find the horizontal asymptotes we substitute large positive and negative
numbers for x, and see if they approach any finite number.

Substituting x = 1000,

{{{"f(1000)"}}}{{{""=""}}}{{{(1000^"")/sqrt(1000^2-1)}}}{{{""=""}}}{{{1.0000005}}}

That is very close to 1, so we assume y = 1 is a horizontal asymptote.

Substituting x = -1000,

{{{"f(-1000)"}}}{{{""=""}}}{{{(-1000)^""/sqrt((-1000)^2-1)}}}{{{""=""}}}{{{-1.0000005}}}

That is very close to -1, so we assume y = -1 is also a horizontal asymptote.

So we draw the vertical asymptotes x = 1 and x= -1 and horizontal asymptotes 
are y = 1 and y = -1

{{{drawing(400,400,-5,5,-5,5,

graph(400,400,-5,5,-5,5), green(line(-11,-1,11,-1), line(-11,1,11,1),

line(-1,-11,-1,11), line(1,-11,1,11)))}}} 

Since the function contains a square root, we must ensure that what's under the
square root is greater than 0.

{{{x^2-1>=0}}}
{{{(x-1)(x+1)>=0}}}

They have zeros 1 and -1.  We make a number line

----------o-----o---------
-4 -3 -2 -1  0  1  2  3  4 

Choose test point x=-2 in interval {{{(matrix(1,3,-infinity, ",", -1))}}}
{{{(x-1)(x+1)>=0}}}
{{{(-2-1)(-2+1)>=0}}}
{{{(-3)(-1)>=0}}}
{{{3>=0}}}
That is a true inequality, so {{{(matrix(1,3,-infinity, ",", -1))}}} is part of
the domain.

Choose test point x=0 in interval {{{(matrix(1,3,-1, ",", 1))}}}
{{{(x-1)(x+1)>=0}}}
{{{(0-1)(0+1)>=0}}}
{{{(-1)(1)>=0}}}
{{{-1>=0}}}
That is a false inequality, so {{{(matrix(1,3,-1, ",", 1))}}} is NOT part of
the domain.  Therefore there is no graph between where x=-1 and where x=1.

Choose test point x=2 in interval {{{(matrix(1,3,1, ",", infinity))}}}
{{{(x-1)(x+1)>=0}}}
{{{(2-1)(2+1)>=0}}}
{{{(1)(3)>=0}}}
{{{3>=0}}}
That is a true inequality, so {{{(matrix(1,3,1, ",", infinity))}}} is part of
the domain.  So the shaded number line is

<==========o-----o=========>
 -4 -3 -2 -1  0  1  2  3  4 

So the domain is {{{matrix(1,3,(matrix(1,3,-infinity, ",", -1)),U,(matrix(1,3,1, ",", infinity)))}}} 

We find a couple points in both parts of the domain, say the points 
(-2,-1.2) and (2,1.2).  Then sketch the graph:

{{{drawing(400,400,-5,5,-5,5,

graph(400,400,-5,5,-5,5,x/sqrt(x^2-1)), green(line(-11,-1,11,-1), line(-11,1,11,1),

line(-1,-11,-1,11), line(1,-11,1,11)))}}} 


Edwin</pre>