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

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Answer 1 · 2016-11-23T06:29:38

The vertical asymptote is #x=1/2#
The horizontal asymptote is #y=0#
No slant asymptote

The domain of #y# is #D_y=RR-{1/2}#

As you cannot divide by #0#, #x!=1/2#

So, #x=1/2# is a vertical asymptote.

The degree of the numerator is #<# the degree of the denominator, so there is no slant asymptote.

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

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

So #y=0# is a horizontal asymptote.

graph{(y-3/(2x-1))(y)=0 [-10, 10, -5, 5]}