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

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Answer 1 · 2017-02-04T17:46:49

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

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

The vertical asymptote is #x=1#

As the degree of the numerator #=# the degree of the denominator,

there is no slant asymptote.

#lim_(x->+-oo)f(x)=lim_(x->+-oo)(2x)/x=2#

The horizontal asymptote is #y=2#

graph{(y-(2x+1)/(x-1))(y-2)(y-100x+100)=0 [-18.02, 18.03, -9.01, 9.01]}