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

1 Answer
Jul 3, 2016

Answer:

vertical asymptote x = -2
horizontal asymptote y = 0

Explanation:

The first step is to factorise and simplify the function.

#rArrf(x)=(x-2)/(x^2-4)=cancel((x-2))/(cancel((x-2))(x+2))=1/(x+2)#

The denominator of this function cannot be zero. This is undefined.Setting the denominator equal to zero and solving for x gives us the value that x cannot be and if the numerator is non-zero for this x then it is a vertical asymptote.

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

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by x

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

as #xto+-oo,f(x)to0/(1+0)#

#rArry=0" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator.This is not the case here (numerator-degree 0 , denominator-degree 1) Hence there are no slant asymptotes.
graph{1/(x+2) [-10, 10, -5, 5]}