How do you find the vertical, horizontal or slant asymptotes for #(x-2)/(x^2-4)#?

1 Answer
Apr 24, 2016

Answer:

vertical asymptote x = -2
horizontal asymptote y = 0

Explanation:

The first step we have to take here is to factorise and simplify the function.

#rArr (x-2)/(x^2-4) = (x-2)/((x-2)(x+2) ) =cancel((x-2))/(cancel((x-2)) (x+2) #

# = 1/(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_(x to+-oo) , f(x) to 0 #

divide terms on numerator/denominator by x

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

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

#rArr y = 0 " is the asymptote " #

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here hence there are no slant asymptotes.
graph{(x-2)/(x^2-4) [-10, 10, -5, 5]}