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

1 Answer
Sep 4, 2016

Answer:

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 there values then they are vertical asymptotes.

solve: #x^2+2=0rArrx^2=-2#

x has no real solutions hence there are no vertical 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)=((2x)/x^2-3/x^2)/(x^2/x^2+2/x^2)=(2/x-3/x^2)/(1+2/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+2) [-10, 10, -5, 5]}