Line Crossing Horizontal and Vertical Asymptotes

Date: 03/16/99 at 02:58:11
From: Kristin
Subject: Horizontal Asymptotes--why would a line cross the horizontal 
asymptote?

I encountered a function that had two vertical asymptotes and one 
horizontal asymptote. The function is 

5x + 1
-------
x^2 - 1

The line between the two vertical asymptotes crosses the horizontal 
asymptote. Why does it do that?  

I have thought about the concept of range vs. domain. Does this 
function really not have a horizontal asymptote? Or is there a hole?  
Does y ever equal 0? Or, maybe the function has a horizontal asymptote 
that does not count between -1 and 1 (the vertical asymptotes). It 
does not seem that the denominator and the numerator have a common 
factor, so that cannot be the problem. These are some of my 
speculations. Neither my teacher nor can I figure this out.  
 
Thanks!

Date: 03/16/99 at 08:26:43
From: Doctor Jerry
Subject: Re: Horizontal Asymptotes--why would a line cross the 
horizontal asymptote?

I think you must be assuming something that just is not true. Namely, 
you believe that horizontal asymptotes cannot share points with the 
curve. As this graph (which you appear to have graphed correctly) 
shows, the curve can cross a horizontal asymptote. A horizontal 
asymptote is a line or a curve that the curve approaches for large 
values of |x|. For small |x|, the horizontal asymptote has nothing 
much to do with the curve.

Now, to answer the other questions.

It happens that the range of this function is the set of all real 
numbers. Let y be any real number. Look at the equation 

   y = (5x + 1)/(x^2 - 1)

If y = 0, we can choose x = -1/5. So, 0 is in the range. Now assume 
that y is not zero.

Expand the above equation:

   y*x^2 - y = 5x + 1

   y*x^2 - 5x + (-y - 1) = 0

Solve the quadratic formula:

   y = [5+/- sqrt(25 + 4(y^2 + y))]/(2y)

Since y^2 + y = y(y + 1), which is positive outside of [-1, 0], there 
are two solutions for all such y. For y in [-1,0) (y is not zero in 
this part of the argument), 25 + 4(y^2 + 1) is positive. So, there is 
at least one root here too.

There is no hole in the graph. Of course, x cannot be -1 or 1; that 
is, the domain is all real except for these two values.

Here is something that may help explain. In most textbook graphing 
problems, the graph does not cross any horizontal asymptote. 

- Doctor Jerry, The Math Forum
  http://mathforum.org/dr.math/