How do you identify all asymptotes for #f(x)=(4x)/(x^2-1)#?

1 Answer
Dec 22, 2016

Answer:

#"vertical asymptotes at " x=+-1#
#"horizontal asymptote at " y=0#

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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^2-1=0rArrx^2=1rArrx=+-1#

#rArrx=-1" and " x=1" 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#

#f(x)=((4x)/x^2)/(x^2/x^2-1/x^2)=(4/x)/(1-1/x^2)#

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

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2) Hence there are no oblique asymptotes.
graph{(4x)/(x^2-1) [-10, 10, -5, 5]}