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

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Answer 1 · 2017-09-25T07:36:33

Deleted - see the other posting.

incorrect original.

Answer 2 · 2017-09-25T08:25:26

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

#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]}