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Math Help - tangent curve question

  1. #1
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    tangent curve question

    y= 4x^3-4x^2+x . Find the tangent at the zeros?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by jamman790 View Post
    y= 4x^3-4x^2+x . Find the tangent at the zeros?
    What I think its suggesting is first find the zeros of y=4x^3-4x^2+x and then evaluate y' at those values.


    To find the zeros, 4x^3-4x^2+x=0\implies x\left(4x^2-4x+1\right)=0\implies x\left(2x-1\right)^2=0\implies x=0 or x=\tfrac{1}{2}.

    Now, find y'\left(0\right) and y'\left(\tfrac{1}{2}\right)

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    What I think its suggesting is first find the zeros of y=4x^3-4x^2+x and then evaluate y' at those values.


    To find the zeros, 4x^3-4x^2+x=0\implies x\left(4x^2-4x+1\right)=0\implies x\left(2x-1\right)^2=0\implies x=0 or x=\tfrac{1}{2}.

    Now, find y'\left(0\right) and y'\left(\tfrac{1}{2}\right)

    Can you take it from here?
    im so sorry man i really cant..its just this is a question on my assignment due tomo and i cant figure out what to do, he has not given us an example like this or even a question like this....but let me ask when ur finding the tangent, you use the formula f (x) - f (a) / x-a so would that be?....:s im so confused..blahhh sorry
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  4. #4
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    Quote Originally Posted by jamman790 View Post
    im so sorry man i really cant..its just this is a question on my assignment due tomo and i cant figure out what to do, he has not given us an example like this or even a question like this....but let me ask when ur finding the tangent, you use the formula f (x) - f (a) / x-a so would that be?....:s im so confused..blahhh sorry
    If r is a zero of the function then the tangent is a line passing through the point (r,0) , with slope f'(r) .

    So the tangent at the root r is:

    y=f'(r)x - f'(r)r

    CB
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  5. #5
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    Quote Originally Posted by CaptainBlack View Post
    If r is a zero of the function then the tangent is a line passing through the point (r,0) , with slope f'(r) .

    So the tangent at the root r is:

    y=f'(r)x - f'(r)r

    CB
    Yessss that helps a lot thatnks i got it
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