# Re: asymtope

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Posted by Joel on October 30, 2002 at 18:28:01:

In Reply to: Re: asymtope posted by T.Gracken on October 30, 2002 at 10:10:30:

: : : I'm haveing trouble finding the horz. asymtpoe. problem 2x/x-1. I can find the vert. asymtope easy enough but I don't understand how to find the Horz. one. Help please I have an exam tomorrow.

: : You have the equation y = 2x/(x-1). Solve it for x to see if you end up with a horizontal asymptote.

: : y = 2x/(x-1)
: : xy - y = 2x
: : xy - y - 2x = 0
: : xy -2x = y
: : x(y-2) = y
: : x = y/(y-2)

: I like that approach.

: In the case where it seems too difficult to solve for x, here's another (memory) approach.

: if y = polynomial/polynomial, then

: 1. if the degree (highest power of variables) of the numerator is greater than degree of denominator, then no horizontal asymptote

: 2. if degree of numerator is equal to degree of denominator, then y=(# multiplied to highest power variable in top)/(# multiplied to highest power in bottom) is the horiz. asy.

: 3. if degree of numerator is less than degree of denominator, then x-axis (y=0) is horiz. asy.

: your equation: y=2x/(x-1), degrees are same so y=2/1 (coefficients of highest powers of x) is the horiz. asy.

:
: ...and, if you know a little calculus, you can also determine horiz. asy. by using what are called "limits"

: : So you have a horiz. asymptote at y=2 (where the denominator = 0).

I like that approach. Must try to remember it.

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