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### Line Crossing Horizontal and Vertical Asymptotes

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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

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/
```
Associated Topics:
High School Equations, Graphs, Translations
High School Functions

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