• anonymous
How do you find the slant asymptote? I already used long division and synthetic division but there is always a remainder. Or, am I just dividing incorrectly? (2x^2 - 5x + 5)/(x-2)
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
do u get 2x-1
  • AccessDenied
In general, a slant asymptote occurs when we have a rational function that is reduced (all like-factors have been removed) whose numerator is one degree higher than the denominator (and they are both of degree 1 or higher). As an example, lets say we have a rational function like this: \(\displaystyle y = \frac{Ax^2 + Bx + C}{Dx+E}\) It just happens to divide out to be: \(\displaystyle y = Px + M + \frac{R}{Dx+E}\). The remainder is simply added on to the end after we figure out the quotient by taking it as remainder/divisor, sort of like how 12/5 divides with remainder 2, so it is 2 + [2]/5 This is a slant asymptote because logically, as x gets extraordinarily big in either positive or negative direction, the rational term at the end tends to approach 0 (1/ extremely large #)., so it tends to behave at its ends like the line, hence the slant.
  • anonymous
@aditya96 Yeah, that's what I got! @AccessDenied I got 3 as the remainder. So I just add it to my quotient, and that's my slant asymptote?

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