How do you find the asymptotes for f(x)=(x+2)/sqrt(6x^2+5x+4) f(x)=x+26x2+5x+4?

2 Answers
Sep 25, 2017

Deleted - see the other posting.

Explanation:

incorrect original.

Sep 25, 2017

We have two horizontal asymptotes given by y=+-1/sqrt6y=±16

Explanation:

f(x)=(x+2)/sqrt(6x^2+5x+4)f(x)=x+26x2+5x+4

= (1+2/x)/sqrt(6+5/x+4/x^2)1+2x6+5x+4x2

Hence when x->oox, f(x)->1/sqrt6=sqrt6/6~=0.41f(x)16=660.41

However, when x->-oox, while numerator is negative, denominator is positive and hence f(x)->-1/sqrt6f(x)16

Hence we have two horizontal asymptotes given by y=+-1/sqrt6y=±16

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