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

1 Answer
Aug 4, 2016

Answer:

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

Explanation:

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]}