How do you find all the asymptotes for #(x^2 + x + 3) /( x-1)#?

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Answer 1 · 2015-10-06T19:46:09

Simplify:

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

to find the oblique asymptote #y = x + 2# and vertical asymptote #x = 1#

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

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

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

So #y = x+2# is an oblique asymptote.

#f(x)# also has a vertical asymptote #x=1#, where the denominator of #5/(x-1)# becomes #0# while the numerator is non-zero.

graph{(y - (x^2+x+3)/(x-1))(y - x - 2)(0.999x - 1) = 0 [-21.19, 24.42, -9.2, 13.63]}