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How to find the slope of the slant asymptote for x^3/(x^2+3)?
Having a bit of difficulty with the polynomial division, wondering if anyone can explain it or show a simpler way.
Cheers
3 Answers
- JaredLv 79 years agoFavorite Answer
Do the long division:
===== x
--------------
x² + 3 | x³
____ - (x³ + 3x)
---------------------
______ 0 - 3x <-- here's your remainder:
x³ / (x² + 3) = x - 3x/(x² + 3)
Therefore x is the oblique asymptote, so the slope of the oblique asymptote is 1
- condonLv 44 years ago
For the vertical asympotes, think of of a place the place the function won't be able to exist for that x-fee. for the reason that there is branch, this may be the place the denominator of the fraction is 0. What values make the denominator 0? For the slant asymptote, what line is the function drawing near as x gets quite super (or small)? think of roughly simplifying the equation that can assist you you.
- 9 years ago
x^3/(x^2+3)= (x^3+3x)/(x^2+3)-3x/(x^2+3)=x-3x/(x^2+3)
when x is near infinity -3x/(x^2+3) is near 0
