Question

How do you find the Vertical, Horizontal, and Oblique Asymptote given `(x+3)/ (x^2+4x+3)`?

Answer 1

vertical asymptote x = -1
horizontal asymptote y = 0

Explanation:

The first step is to factorise and simplify f(x).

`f(x)=cancel((x+3))^1/(cancel((x+3))^1(x+1))=1/(x+1)`

The denominator of f(x) cannot be zero as this would be undefined. Equating the denominator and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: x + 1 = 0 → x = -1 is the asymptote

Horizontal asymptotes occur as

`lim_(xto+-oo),f(x)toc" (a constant)"`

divide terms on numerator/denominator by x

`(1/x)/(x/x+1/x)=(1/x)/(1+1/x)`

as `xto+-oo,f(x)to0/(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 0 , denominator-degree 1 ) Hence there are no oblique asymptotes.
graph{(1)/(x+1) [-10, 10, -5, 5]}