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Math Help - Limit

  1. #1
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    Limit

    [IMG]www.0zz0.com][/url][/IMG]
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    What ideas have you had so far?
    I am trying to find the equation of the line that pass through (a/2 ,a^2/2) and (0,b)
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    Quote Originally Posted by BAHADEEN View Post
    I am trying to find the equation of the line that pass through (a/2 ,a^2/2) and (0,b)
    Not a bad way to start, except that the line will go through \left(\frac{a}{2},\frac{a^{2}}{4}\right) not \left(\frac{a}{2},\frac{a^{2}}{2}\right). What do you get?
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  5. #5
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    Quote Originally Posted by Ackbeet View Post
    Not a bad way to start, except that the line will go through \left(\frac{a}{2},\frac{a^{2}}{4}\right) not \left(\frac{a}{2},\frac{a^{2}}{2}\right). What do you get?
    i get y=(2a^2-8b/4a)x+b
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  6. #6
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    Quote Originally Posted by BAHADEEN View Post
    i get y=(2a^2-8b/4a)x+b
    I don't think you've quite got it yet. If you plug in x=a/2, you don't get y=a^{2}/4, like you should. Can you show your work?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Not a bad way to start, except that the line will go through \left(\frac{a}{2},\frac{a^{2}}{4}\right) not \left(\frac{a}{2},\frac{a^{2}}{2}\right). What do you get?
    Actually it is \left(\frac{a}{2},\frac{a^{2}}{2}\right)..
    Because we want the midpoint of the line segment between (0,0),~\&~(a,a^2).
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  8. #8
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    Quote Originally Posted by Plato View Post
    Actually it is \left(\frac{a}{2},\frac{a^{2}}{2}\right)..
    Because we want the midpoint of the line segment between (0,0),~\&~(a,a^2).
    Ach! You're right. The midpoint should be on the line, not the function. BAHADEEN's line is still incorrect, though, because I get a^{3} for y when I plug in x=a/2.

    Quote Originally Posted by topsquark
    Ack! Beet beat me again!
    How about "Ack! Beet-red beat me again!"? Like Deveno, you could even put that in your signature if you wanted.
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