What are the asymptotes for #(2)/(x^2-2x-3)#?

1 Answer
Jul 4, 2016

Answer:

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

Explanation:

The denominator of this rational function cannot be zero as this would give division by zero which is undefined. Setting the denominator equal to zero and solving for x will give us the values that x cannot be and if the numerator is non-zero for these values of x then they are vertical 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 term on numerator/denominator by highest exponent of x, that is #x^2#

#f(x)=(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"#
graph{(2)/(x^2-2x-3) [-10, 10, -5, 5]}