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How do you find the horizontal asymptote.. It says in my textbook that " to find horizontal asymptote divide numerator and denominator by x & investigate x --> +/- infinite .. I dont understand the investigation part :S

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1) Put equation or function in standard form. 2) Remove everything except the biggest exponents of x found in the numerator and denominator.
One way to find the horizontal asymptote: Take the highest degree of the numerator and the denominator and divide. For example, -3x^2/x^2 = -3 Horizontal asymptote is y=-3. Now if you do that and it turns out to be x/x^2, (simplifying to 1/x) your horizontal asymptote is y=0. If it ends up being x^2/x, (simplifying to x/1) then you have no horizontal asymptote.
your book is stupid, if you don't mind me saying so

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Other answers:

LOOOLL agreed
if the degree of the numerator is larger than the degree of the denominator, there is no horizontal asymptote if the degree of the numerator is smaller than the degree of the denominator, the horizontal asymptote is \(y=0\) if the degrees are the same, the horizontal asymptote is the ratio of the leading coefficients
yaah I know that way, but on my test I have to show how I got the horizontal asymptotes by doing what it says inthe textbook
then i guess you have to do it
set your limit to +/- infinity and solve, if you come out with a number that is your horizontal asymptote

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