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

1 Answer
Jul 16, 2016

Answer:

vertical asymptotes x = -1 , x = 3
horizontal asymptote y = 0

Explanation:

The denominator of the rational function cannot equal 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 of x then they are the asymptotes.

solve: #x^2-2x-3=0rArr(x-3)(x+1)=0#

#rArrx=-1,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#

#(2/x^2)/(x^2/x^2-(2x)/x^2-3/x^2)=(2/x^2)/(1-2/x-3/x^2)#

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

#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 2) Hence there are no slant asymptotes.
graph{(2)/(x^2-2x-3) [-10, 10, -5, 5]}