# Re: asymtope

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Posted by T.Gracken on October 30, 2002 at 10:10:30:

In Reply to: Re: asymtope posted by Joel on October 30, 2002 at 09:37:01:

: : 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).

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