Question

How do you find vertical, horizontal and oblique asymptotes for `f(x) = (x+1) / (x+2)`?

Answer 1

vertical asymptote x = -2
horizontal asymptote y = 1

Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero 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 + 2 = 0 → x = -2 is the asymptote

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

`(x/x+1/x)/(x/x+2/x)=(1+1/x)/(1+2/x)`

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

`rArry=1" is the asymptote"`

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) Hence there are no oblique asymptotes.
graph{(x+1)/(x+2) [-10, 10, -5, 5]}