Posted by Joel on October 30, 2002 at 09:37:01:
In Reply to: asymtope posted by Bill on October 30, 2002 at 08:32:51:
: 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)
So you have a horiz. asymptote at y=2 (where the denominator = 0).
Post a Followup