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

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Answer 1 · 2017-07-28T19:59:47

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

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]}