How do you find the vertical, horizontal or slant asymptotes for #f(x)=( x-4)/ (x^2-1)#?

1 Answer
Jun 16, 2016

vertical asymptotes x = ± 1
horizontal asymptote y = 0

Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve : #x^2-1=0rArrx^2=1rArrx=±1#

#rArrx=-1,x=1" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" (a constant)"#

divide terms on numerator/denominator by #x^2#

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

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

#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 1, denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x-4)/(x^2-1) [-10, 10, -5, 5]}