How do you find the asymptotes for #y= (x + 1 )/ (2x - 4)#?

1 Answer
Feb 3, 2016

Answer:

There is a vertical asymptote at #x=2#.
There is a horizontal asymptote at #y=1/2#.

Explanation:

First step is always to simplify what you have.

#frac{x+1}{2x-4} -= 1/2(1+frac{3}{x-2})#

As you know, dividing by a very large number results in almost zero. Therefore,

#lim_{x->-oo} frac{x+1}{2x-4} = lim_{x->-oo} 1/2(1+frac{3}{x-2}) #
#= 1/2(1+0) = 1/2 #

Similarly,

#lim_{x->oo} frac{x+1}{2x-4} = 1/2#.

Hence there is a horizontal asymptote at #y=1/2#.

Also note that #x!=2#, as that will result in division by zero.

There is a vertical asymptote at #x=2#.

Please refer to the graph below.

graph{(x+1)/(2x-4) [-10, 10, -5, 5]}