Oblique Asymptotes: What happens to the Remainder?

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To find the oblique asymptote of the function f(x) = (-3x^2 + 2) / (x - 1), polynomial long division yields a quotient of -3x - 3 and a remainder of -1. The remainder is disregarded when determining the oblique asymptote because, as x approaches infinity or negative infinity, the term involving the remainder approaches zero. This means the function effectively behaves like the quotient, which is the line -3x - 3. Understanding this concept clarifies why the remainder does not affect the asymptotic behavior of the function. Thus, the oblique asymptote is represented solely by the quotient.
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Let's say I'm trying to find the oblique asymptote of the function:

f(x)=
-3x2 + 2
x-1

Forgive my poor formatting.

So because the denominator isn't linear, we do polynomial long division of the function and ultimately get -3x - 3 as our quotient, with a remainder of -1. For the sake of the oblique asymptote, we disregard the remainder. But why? I haven't found a website that explains it; they all simply say to ignore the remainder. It bothers me that we just forget about it, regardless of what it is.
Can you explain to me why we disregard the remainder, and where it "goes?"
 
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An "asymptote" is a line that a curve approaches as x goes to, in this case, negative infinity and infinity. Yes, long division gives a quotient of -3x- 3 with a remaider of -1. That means that
\frac{-3x^2+ 2}{x- 1}= -3x- 3- \frac{1}{x- 1}
As x goes to either infinity or negative infinity, that last fraction goes to 0.
 
That makes perfect sense! Thank you so much!
 
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