How do you find the asymptotes for #f(x)=(1-5x) /( 1+2x)#?

1 Answer
Feb 19, 2016

Answer:

The vertical may be found by setting the lower part #=0#

Explanation:

So you may expect an asymptote at #1+2x=0->x=-1/2#

For the horizontal asymptote we make #x# very large (both pos. and neg.). The function will begin to look more and more like:

#(-5cancelx)/(2cancelx)# as the #1#'s won't make any difference.

So the horizontal asymptote will be #y=-2 1/2#
graph{(1-5x)/(1+2x) [-10, 10, -5, 5]}