Question

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

Answer 1

Deleted - see the other posting.

Explanation:

incorrect original.

Answer 2

We have two horizontal asymptotes given by `y=+-1/sqrt6`

Explanation:

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

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

Hence when `x->oo`, `f(x)->1/sqrt6=sqrt6/6~=0.41`

However, when `x->-oo`, while numerator is negative, denominator is positive and hence `f(x)->-1/sqrt6`

Hence we have two horizontal asymptotes given by `y=+-1/sqrt6`

graph{(x+2)/sqrt(6x^2+5x+4) [-5.647, 4.353, -0.84, 4.16]}