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Math Help - Tricky Differentiation Problem

  1. #1
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    Tricky Differentiation Problem

    Find the equation of the line having a negative slope, that passes through the point (2,-3), and is tangent to the curve y = x^2 + x

    dy/dx = 2x^2 +1

    Can someone point me in the right direction?
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  2. #2
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    The equation of a line of slope m passing through (x_0,y_0) is

    y-y_0=m(x-x_0).

    Therefore, we seek a tangent line of y=x^2+x of the form

    y-(x_0^2+x_0)=(2x_0+1)(x-x_0)

    that passes through (2,-3). Here, we have used the fact that

    \frac{d}{dx}(x^2+x)=\frac{d}{dx}(x^2)+\frac{d}{dx}  (x)=2x+1.

    Edit: Changed index to 0 for clarity.
    Last edited by Scott H; December 8th 2009 at 04:08 PM.
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  3. #3
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    Quote Originally Posted by millerst View Post
    Find the equation of the line having a negative slope, that passes through the point (2,-3), and is tangent to the curve y = x^2 + x

    dy/dx = 2x^2 +1

    Can someone point me in the right direction?
    correction ... \frac{dy}{dx} = 2x+1


    y - (-3) = m(x - 2)

    y + 3 = (2x+1)(x - 2)

    (x^2+x)+3 = (2x+1)(x-2)

    x^2+x+3 = 2x^2-3x-2

    0 = x^2-4x-5

    solve for and choose the value of x that meets the given requirements.
    I leave finding the tangent line equation to you.
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  4. #4
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    Is the answer:

    y = -x - 1?
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