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Thread: Finding tangent to function parallell with line

  1. #1
    Newbie Bato91's Avatar
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    Finding tangent to function parallell with line

    Hello all! I just found this forum and registered. I would like some help with the following problem:

    What is the function of the tangent to the parable $\displaystyle y=x^2$ , parallell with the line $\displaystyle y=6x$ ? I can see where the tangent is graphically and go from there, but how can you solve it solely by equating?


    (feel free not to copy my terminology; I'm not good at English and may use some odd names)
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  2. #2
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    Derivative

    Your function
    $\displaystyle
    y = 6 x
    $

    has slope 6.

    Your function
    $\displaystyle
    y = x ^ 2
    $

    has derivative:
    $\displaystyle
    \frac { d y } { dx } = 2 x
    $

    which has slope 6 at x=3. Thus y = 9. The tangent line must be y=6x-9.
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    Newbie Bato91's Avatar
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    Thank you!
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  4. #4
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    Quote Originally Posted by Bato91 View Post
    Hello all! I just found this forum and registered. I would like some help with the following problem:

    What is the function of the tangent to the parable $\displaystyle y=x^2$ , parallell with the line $\displaystyle y=6x$ ? I can see where the tangent is graphically and go from there, but how can you solve it solely by equating?


    (feel free not to copy my terminology; I'm not good at English and may use some odd names)
    Here is a different approach:

    The tangent line has the equation: $\displaystyle t: y = 6x + b$

    Now calculate the coordinates of the intersection between the parabola and the straight line:

    $\displaystyle x^2=6x+b~\implies~x^2-6x-b=0$

    Use the formula to solve this quadratic equation:

    $\displaystyle x = 3\pm\sqrt{9+b}$

    You get only one intersection point (=tangent point) if the radicand equals zero:

    $\displaystyle 9+b = 0~\implies~\boxed{b=-9}$ . Consequently the tangentpoint is T(3, 9)

    Therefore the equation of the tangent line becomes: $\displaystyle t: y = 6x-9$
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