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

Question

Benjamin Achtenberg · Accepted Answer

vertical asymptote x = -2
horizontal asymptote y = 0

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]}

Xavier Alonso · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $ x $. In this case, $ x^2 - 4 = 0 $ yields $ x = 2 $ and $ x = -2 $ as vertical asymptotes.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $. If the degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. In this case, since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is $ y = 0 $.

To find the slant asymptote, perform polynomial long division. Divide the numerator by the denominator. If the degree of the numerator is exactly one greater than the degree of the denominator after division, the quotient represents the equation of the slant asymptote. Otherwise, there is no slant asymptote. In this case, the degree of the numerator is one less than the degree of the denominator, so there is no slant asymptote.