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How do you identify the oblique asymptote of #f(x) = (x^3-6x^2+12x-2)/(x^2-2x+2)#?

Answer 1

Divide numerator by denominator to get a polynomial quotient and a remainder. The quotient is the oblique asymptote #x - 4#

#(x^3-6x^2+12x-2)/(x^2-2x+2)#

#=((x^3-2x^2+2)-(4x^2-8x+8)+(4x+4))/(x^2-2x+2)#

#=((x-4)(x^2-2x+2)+4(x+1))/(x^2-2x+2)#

#=(x-4)+(4(x+1))/(x^2-2x+2)#

It's a little nicer to use synthetic division to do this, but it's not easy to typeset clearly.

Then:

#(4(x+1))/(x^2-2x+2) -> 0# as #x->+-oo#

So the oblique asymptote is #x-4#

graph{(y - (x^3-6x^2+12x-2)/(x^2-2x+2))(y - x + 4) = 0 [-17.75, 22.25, -10.96, 9.04]}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.