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Math Help - Equation of Tangent(2) [stucked @ b]

  1. #1
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    Equation of Tangent(2) [stucked @ b]

    The equation of a curve is y=2x^2-kx+3, where k is a constant.

    a) Find in terms of k, the gradient of the tangent at point A where x=1.

    b) Given that the tangent at point A passes through the point B(5,1), find the value of k.

    Attempt

    a) \frac{dy}{dx}= 4x-k
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  2. #2
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    Quote Originally Posted by Punch View Post
    The equation of a curve is y=2x^2-kx+3, where k is a constant.

    a) Find in terms of k, the gradient of the tangent at point A where x=1.

    b) Given that the tangent at point A passes through the point B(5,1), find the value of k.

    Attempt

    a) \frac{dy}{dx}= 4x-k
    hi

    for (a) , you are correct, then substitute x=1 .

    (b) it's similar to your previous question . Find the point (A) where the tangent touches the curve in terms of k , then find the gradient of AB which is also the gradient of tangent .
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  3. #3
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    I can't find the value of k after finding the gradient in terms of k... please note that the question is asking for the value of k...
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  4. #4
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    Quote Originally Posted by Punch View Post
    I can't find the value of k after finding the gradient in terms of k... please note that the question is asking for the value of k...
    ok , you are not getting what i mean .

    point A = (1,5-k) , B (5,1)

    so the gradient of AB(tangent) = \frac{k-4}{4} which is also what you found in (1) , 4-k so

    4-k=\frac{k-4}{4}

    16-4k=k-4

    5k=20

    k=4
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