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Thread: Slant asymptotes (proving it?)

  1. #1
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    Slant asymptotes (proving it?)

    Hey, so I know how to find whether there is a slant asymptotes by doing long division of the numberators degree > denominator, but why does the limit of it as x goes to infinity and negative infity of f (x) -mx-b =0.

    That's the only thing I don't get. What and how do it tell us about slant asymptotes?
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  2. #2
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    Re: Slant asymptotes (proving it?)

    Quote Originally Posted by lc99 View Post
    That's the only thing I don't get. What and how do it tell us about slant asymptotes?
    Have a look at this reference
    Thanks from topsquark and lc99
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    Re: Slant asymptotes (proving it?)

    Oh. So take the difference in f (x) and the slant line gives us the limit ad f (x) approaches the slant which should be 0..
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    Re: Slant asymptotes (proving it?)

    A "slant asymptote" is line such that the graph gets closer and closer to the graph as x goes to positive or negative infinity. If the graph is given by y= f(x) and the line is y= mx+ b. For any x, the vertical distance between the graph and the line is f(x)- (mx+ b)= f(x)- mx- b. That must go to 0 as x goes to positive or negative infinity.
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  5. #5
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    Re: Slant asymptotes (proving it?)

    Thank you! I was wondering what taking the different of f (x) and line of slant asymptotes did, and it makes sense now! Thanks a bunch.
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