Question

How do you find all the asymptotes for `(x^2+5x+6)/(x+4)`?

Answer 1

There is only one vertical asymptote, as `x+4!=0`

Explanation:

Therefore `x=-4` is the vertical asymptote
There is no horizontal asymptote, as the function can have any value.
There is a local minimum though at `(-2 1/2,-1/4)`
graph{(x^2+5x+6)/(x+4) [-10, 10, -5, 5]}

Answer 2

There is also a slant (oblique, skew) asymptote: y = x+1#

Explanation:

In addition to vertical and horizontal asymptotes, is you are interested in slant asymptotes (also known as 'oblique' or 'skew' asymptotes), there is one for this function.

The degree of the numerator is one more than that of the denominator. That implies that when we do the division, we will get a linear quotient and a remainder with degree on the numerator less that that of the denominator.

Do the division (or group the terms of the numerator), to get:

`(x^2+5x+6)/(x+4) = ((x^2+4x)+(x+4)+2)/(x+4)`

`= x+1+2/(x+4)`

The asymptote is the line `y=x+1`.

(The difference between the full expression and the line `y=x+1` is `2/(x+4)` which gets arbitrarily close to `0` as `x` increases or decreases without bound.)