How do you find oblique asymptote of #(x^2 + x + 3) /(x-1)#?

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Answer 1 · 2015-06-20T20:15:26

Partially divide the numerator by the denominator to get:

#(x^2+x+3) / (x-1) = x+2+5/(x-1)#

Hence the oblique asymptote is #y = x+2#

#(x^2+x+3) / (x-1) = ((x-1)(x+2) + 5) / (x-1) = x+2+5/(x-1)#

with exclusion #x != 1#

#5/(x-1) -> 0# as #x->+-oo#

So the oblique asymptote is #y = x + 2#

graph{(y - (x^2+x+3) / (x-1))(y - x - 2) = 0 [-40, 40, -20, 20]}