How do you find the vertical, horizontal or slant asymptotes for #(6x^2+2x-1) /( x^2-1)#?

1 Answer
Sep 15, 2016

Answer:

vertical asymptotes at #x=+-1#
horizontal asymptote at y = 6

Explanation:

The denominator of the function cannot be zero as this would make the function 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=+-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)=((6x^2)/x^2+(2x)/x^2-1/x^2)/(x^2/x^2-1/x^2)=(6+2/x-1/x^2)/(1-1/x^2)#

as #xto+-oo,f(x)to(6+0-0)/(1-0)#

#rArry=6" is the asymptote"#

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 2 ) Hence there are no slant asymptotes.
graph{(6x^2+2x-1)/(x^2-1) [-20, 20, -10, 10]}