How do you find the Vertical, Horizontal, and Oblique Asymptote given #y=(3x+1)/(2-x)#?

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Answer 1 · 2017-02-03T13:59:55

The vertical asymptote is #x=2#
The horizontal asymptote is #y=-3#
No oblique asymptote

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

The vertical asymptote is #x=2#

As the degree of the numerator is #=# to the degree of the denominator, there is nooblique asymptote.

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

Therefore,

the horizontal asymptote is #y=-3#

graph{(y-(3x+1)/(2-x))(y+3)(y+50x-100)=0 [-20.62, 19.94, -10.68, 9.6]}