How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x) = (2x - 3)/( x^2 - 1)#?

1 Answer
Aug 2, 2016

Answer:

vertical asymptotes x = ± 1
horizontal asymptote y = 0

Explanation:

The denominator of f(x) cannot be zero as this is 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=0rArr(x-1)(x+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#

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

as #xto+-oo,f(x)to(0-0)/(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{(2x-3)/(x^2-1) [-10, 10, -5, 5]}