Question

How do you find vertical, horizontal and oblique asymptotes for `(x+2)/sqrt(6x^2+5x+4)`?

Answer 1

The horizontal asymptote is `y=1/sqrt6`
There are no oblique or vertical asymptote

Explanation:

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

`AA x in RR`, `(6x^2+5x+4 )>0`, `=>`, no vertical asymptote

Let rewrite the function,

`f(x)=(cancel(x)(1+2/x))/(cancel(x)sqrt(6+5/x+4/x^2))`

Therefore,

`lim_(x->+-oo)f(x)=lim_(x->+-oo)((1+2/x))/(sqrt(6+5/x+4/x^2))=1/sqrt6`

The horizontal asymptote is `y=1/sqrt6`

graph{(y-(x+2)/sqrt(6x^2+5x+4))(y-1/sqrt6)=0 [-5.546, 5.555, -2.773, 2.774]}