How do you find the vertical, horizontal or slant asymptotes for #(x^2-4) /( x^2-2x-3)#?

Question

Benjamin Achtenberg · Accepted Answer

vertical asymptotes x = -1 , x = 3
horizontal asymptote y = 1

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes. solve:# x^2-2x-3=0rArr(x-3)(x+1)=0# #rArrx=-1" and " x=3" are the asymptotes"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)toc" (a constant)"# divide terms on numerator/denominator by the highest power of x.that is #x^2# #(x^2/x^2-4/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(1-4/x^2)/(1-2/x-3/x^2)# as #xto+-oo,f(x)to(1-0)/(1-0-0)# #rArry=1" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. Tis is not the case here ( both of degree 2) Hence there are no slant asymptotes. graph{(x^2-4)/(x^2-2x-3) [-10, 10, -5, 5]}

Cora Alexander · Answer

undefined To find the vertical asymptotes, factor the denominator and find the values of x that make the denominator zero. For horizontal asymptotes, 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. If the degree of the numerator is greater, there is no horizontal asymptote. For slant asymptotes, if the degree of the numerator is exactly one more than the degree of the denominator, perform polynomial long division to find the equation of the slant asymptote.