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

2 Answers
Sep 25, 2017

Answer:

Deleted - see the other posting.

Explanation:

incorrect original.

Sep 25, 2017

Answer:

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]}