Question

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

Answer 1

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]}