How do you find the vertical, horizontal and slant asymptotes of: #f(x)=(3x^2+2) / (x^2 -1)#?

1 Answer
Sep 1, 2016

Answer:

vertical asymptotes at x = ± 1
horizontal asymptote at y = 3

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

as #xto+-oo,f(x)to(3+0)/(1-0)#

#rArry=3" 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{(3x^2+2)/(x^2-1) [-20, 20, -10, 10]}