Question

How do you find the vertical, horizontal or slant asymptotes for `(x + 2)/( x + 3)`?

Answer 1

`(x+2)/(x+3)` has a vertical asymptote `x=-3` and a horizontal asymptote `y=1`

Explanation:

`f(x) = (x+2)/(x+3)`

When `x=-3`, the denominator is zero and the numerator non-zero, so `f(-3)` is undefined and `f(x)` has a vertical asymptote there.

Note that:

`(x+2)/(x+3) = ((x+3)-1)/(x+3) = 1-1/(x+3)`

As `x->+-oo`, the expression `-1/(x+3)->0`

So `f(x)->1+0 = 1` as `x->+-oo` and `f(x)` has a horizontal asymptote `y = 1`.

graph{(x+2)/(x+3) [-13.21, 6.79, -4.24, 5.76]}