SOLUTION: Find all horizontal and vertical asymptotes of the function: f(x) = x/sqrt(x^2-1)

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Question 1200106: Find all horizontal and vertical asymptotes of the function: f(x) = x/sqrt(x^2-1)
Found 2 solutions by Edwin McCravy, greenestamps:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!


To find vertical asymptotes, set the denominator equal to 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,



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

Substituting x = -1000,



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

 

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




That is a true inequality, so  is part of
the domain.

Choose test point x=0 in interval 




That is a false inequality, so  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 




That is a true inequality, so  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  

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:

 


Edwin
Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!


The denominator is , so the radicand must be non-negative.

Furthermore, since that square root is in the denominator,the radicand can't be zero.

, so the function is undefined on [-1,1].

For x values a tiny bit greater than 1, the denominator is close to 0, and the numerator is positive, so the function value is large positive. That makes x=1 a vertical asymptote.

Similarly, for x values a tiny bit less than -1, the denominator is close to 0, and here the numerator is negative, so the function value is large negative. So x=-1 is also a vertical asymptote.

For either large positive or large negative values of x, is very close to , so the function is very nearly equal to .

For large positive values of x, , so y=1 is a horizontal asymptote for x>1.

For large negative values of x, , so y=-1 is a horizontal asymptote for x<-1.

ANSWERS:
vertical asymptotes at x=-1 and x=1;
horizontal asymptotes at y=1 (for positive x values) and y=-1 (for negative x values).

A graph of the function and the two horizontal asymptotes....
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