How do you find the domain, vertical, and horizontal asymptotes of #f( x ) = \frac { 2x } { 3x ^ { 2} + 1}#?

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Answer 1 · 2017-11-08T19:59:27

See below.

The function is valid for all #x#, so domain is:

#{x in RR }#

Because the function is valid for all #x# there are no vertical asymptotes:

#lim_(x->+oo)(2x)/(3x^2+1)=0#

#lim_(x->-oo)(2x)/(3x^2+1)=0#

This shows the x axis is a horizontal asymptote.

The curve crosses the x axis where y = 0.

#(2x)/(3x^2+1)=0=>x=0#

Graph:

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