How do you find the asymptotes for #h(x) = (2x+3)/(3x+1 )#?

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Answer 1 · 2016-04-13T20:11:35

#color(blue)("Vertical asymptote at "x= -1/3)#

#color(blue)("Horizontal asymptote at "y= 2/3 ->0.66bar6)#

The equation becomes undefined at #x=-1/3#
The denominator is not 'allowed' to become 0

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As absolute #x # becomes increasingly larger the +3 and +1 become insignificant.

So #lim_(xto+-oo) (2x+3)/(3x+1) -> 2/3xx x/x = 2/3#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Vertical asymptote at "x= -1/3)#

#color(blue)("Horizontal asymptote at "y= 2/3 ->0.66bar6)#

Tony B