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Horizontal asymptote

Mathematics
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\[\[\huge y=\frac{ x^3-5x^2-4x }{ x^3-8x}\]
1. Look at the x values for the numerator and denominator separately. 2. If the power associated with the x (i.e. x^2, x^5) in the numerator is greater than the power of the x in the denominator then the horizontal asymptote does not exist. So if your example changed to (x^2+2)/(2x-5) the asymptote would not exist. 3. If the power associated with the x in the numerator is equal to the power of the x in the denominator then you divide the factors. So in your example since the power of x is the same on top and bottom, then the 1 in (x+2) and 2 in (2x-5) are the fraction 1/2, so y=1/2. 4. If the power associated with the x in the numerator is less than the power of the x in the denominator then the asymptote is always y=0. So your example would need to change to (x+2)/(2x^2-5).
lol i used another example btw

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