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Math Help - Tangent Line in Point..

  1. #1
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    Tangent Line in Point..

    Hello MHF, i hv a new problemn, the question is:

    knowing that 2y+4x-6=0 is the equation which is one of the lines tangent to the curve y= 2x^3-x^2+cx+d.
    Find the derivative of this function in one of the points of the curve.

    How can i find the common point?
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  2. #2
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    Re: Tangent Line in Point..

    Rewrite the tangent line \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*} as \displaystyle \begin{align*} y = -2x + 3 \end{align*}, so you can see that the tangent line has gradient = -2.

    This means the derivative of your curve must be -2. So take the derivative, set it equal to -2, solve for x, then substitute this value into the equation for the curve to find y.
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    Re: Tangent Line in Point..

    But now how can i find the value of "C"?
    Quote Originally Posted by Prove It View Post
    Rewrite the tangent line \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*} as \displaystyle \begin{align*} y = -2x + 3 \end{align*}, so you can see that the tangent line has gradient = -2.

    This means the derivative of your curve must be -2. So take the derivative, set it equal to -2, solve for x, then substitute this value into the equation for the curve to find y.
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  4. #4
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    Re: Tangent Line in Point..

    You don't need to find the value of c. It is a parameter. Your point of tangency will be in terms of c and d.
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  5. #5
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    Re: Tangent Line in Point..

    Quote Originally Posted by Prove It View Post
    You don't need to find the value of c. It is a parameter. Your point of tangency will be in terms of c and d.
    This would be my X??
    y'= 6x^2-x+c
    6x^2-2x+c+2
    x'= (2+ \sqrt{-44-24c})/12 and x''= (2- \sqrt{-44-24c})/12
    Last edited by Chipset3600; November 6th 2012 at 03:34 PM.
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