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Math Help - Find the equation of tangent and normal to the curve

  1. #1
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    Find the equation of tangent and normal to the curve

    Question : Find the eq of tangent and normal to the curve
    y(x-2)(x-3)-x+7 = 0
    at the point where the curve meets the x-axis
    -----------------------------------------------------------------
    Solution:

    y = \frac{x-7}{x^2 - 5x +6 }.............Is that correct?

    \therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}...................Is that correct?????


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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : Find the eq of tangent and normal to the curve
    y(x-2)(x-3)-x+7 = 0
    at the point where the curve meets the x-axis
    -----------------------------------------------------------------
    Solution:

    y = \frac{x-7}{x^2 - 5x +6 }.............Is that correct?

    \therefore \frac{dy}{dx} = \frac{-x^2 + 14x -29}{(x^2 - 5x +6)^2}...................Is that correct?????

    differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

    You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.
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  3. #3
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    what should i do nxt

    Quote Originally Posted by mr fantastic View Post
    differentiate (x - 7)/(x^2 - 5x + 6) - Wolfram|Alpha

    You can do the simplifying of the Wolfram expression and see if it's equivalent to what you got.

    \frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}
    what should i do??????
    Now eg of tangent is at (x_0,y_0) is given by

     (y-y_0) = \frac{dy}{dx} (x-x_0)

    but i dont know (x_0,y_0).................what should i do??????
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  4. #4
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    Quote Originally Posted by zorro View Post
    \frac{dy}{dx} = \frac{- (x^2 - 14x + 31)}{(x^2 - 5x + 6)^2}
    what should i do??????
    Now eg of tangent is at (x_0,y_0) is given by

     (y-y_0) = \frac{dy}{dx} (x-x_0)

    but i dont know (x_0,y_0).................what should i do??????
    When the curve meets the x-axis, y = 0. Therefore y_0 = 0. To get x_0, solve -x + 7 = 0 (I hope you see why).
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