How do you find the asymptotes for #(2x-4)/(x^2-4)#?

1 Answer
May 11, 2016

Answer:

vertical asymptote x = -2
horizontal asymptote y = 0

Explanation:

The first step is to factorise and simplify the function

#rArr(2 cancel((x-2)))/(cancel((x-2)) (x+2))=2/(x+2 #

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve: x + 2 = 0 → x = -2 is the asymptote

Horizontal asymptotes occur as #lim_(xto+-oo) , f(x) to 0#

divide terms on numerator/denominator by x

#rArr (2/x)/(x/x+2/x)=(2/x)/(1+2/x)#

as #xto+-oo, y to 0/(1+0)#

#rArry=0" is the asymptote"#
graph{2/(x+2) [-10, 10, -5, 5]}