Question

How do you find all the asymptotes for function `f(x) = (2x^2+3x+8)/(x+3)`?

Answer 1

Vertical asymptote `x=3` and slanting or oblique asymptote `y=2x+3`

Explanation:

There are three types of asymptotes.

Vertical asymptotes are indicated by denominator. Here we have `(x+3)` in denominator, so vertical asymptote is given by `x+3=0` or `x=-3`. Observe that in case `x+3` is a factor of numerator, it will cancel out and we will not have an asymptote, rather we will havea hole at `x=-3`. But here `x+3` is not a factor of numerator, we do have a vertical asymptote as

`lim_(x->-3)(2x^2+3x+8)/(x+3)=oo`

Horizontal asymptotes in such cases (algebraic expressions) are there when degree of numerator is equal to that of denominator. Here it is not so we do not a horizontal asymptote. Stll assume numerator as only `3x+8` and then

`lim_(x->oo)(3x+8)/(x+3)=lim_(x->oo)(3+8/x)/(1+3/x)=3/1=3` and we would have `y=3` as horizontal asymptote.

Here we have `(2x^2+3x+8)/(x+3)` and we have

`lim_(x->oo)(2x^2+3x+8)/(x+3)=lim_(x->oo)(2x+3+8/x)/(1+3/x)`

= `2x+3`

Hence we have a slanting or oblique asymptote `y=2x+3`

It is apparent that we have this only when degree of numerator is one more than that of denominator.