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Math Help - Find another point on the curve with the same tangent.

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    Find another point on the curve with the same tangent.

    Question:

    Find the equation of the tangent to y = 8x^4 + 40x^3 + 26x^2 - 63x + 24 at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.


    Answer:

    f'(x) = 32x^3 + 120x^2 + 52x - 63
    f'(-3) = m
    m = -3

    y - y_1 = m(x - x_1)
    y = -3x + 6

    Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for x, or is there a simpler / more correct way of doing such?
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  2. #2
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    Quote Originally Posted by RogueDemon View Post
    Question:

    Find the equation of the tangent to y = 8x^4 + 40x^3 + 26x^2 - 63x + 24 at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.


    Answer:

    f'(x) = 32x^3 + 120x^2 + 52x - 63
    f'(-3) = m
    m = -3

    y - y_1 = m(x - x_1)
    y = -3x + 6

    Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for x, or is there a simpler / more correct way of doing such?
    To have the same tangent line, the graph must have the same slope at that points. This gives f'(x)=-3

    f'(x)=32x^3+120x^2+52x-63=-3 \iff 32x^3+120x^2+52x-60=0

    Since you already know that x=-3 is a zero you can divide the cubic by (x+3)to get

    (x+3)(32x^2+24x-20)

    The quadratic factor will give the possibilities for the x coordinate of the possible points. Can you finish from here?
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    Yes, thanks very much!
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  4. #4
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    Quote Originally Posted by RogueDemon View Post
    Question:

    Find the equation of the tangent to y = 8x^4 + 40x^3 + 26x^2 - 63x + 24 at the point (3,-15). Find another point on the curve which has exactly the same equation for the tangent at the point.
    (3, -15) is not on the curve. I assume you mean (-3, 15).


    Answer:

    f'(x) = 32x^3 + 120x^2 + 52x - 63
    f'(-3) = m
    Okay, here you are using x= -3 so the (3, -15) was a typo.

    m = -3

    y - y_1 = m(x - x_1)
    Better to write this out completely: y- (-15)= -3(x- 3) gives y+ 15= -3x+ 9 or y= -3x+ 6.

    y = -3x + 6

    Now that I have the equation of the tangent, how do I go about finding the point? Do I have to set the equation of the tangent equal to the original equation and solve for x, or is there a simpler / more correct way of doing such?
    You have to show that there is another x where that same tangent line is tangent to the curve. That means both that the values are the same and that they have the same slope there. Specifically, you need to find an x such that
    8x^4+ 40x^3+ 26x^2- 63x+ 24= -3x+ 6
    and 32x^3+ 120x^2- 52x- 63= -3 which is the same as
    32x^3+ 120x^2- 52x- 60= 0.

    Since you already know that x= -3 is a solution to the second equation, you can divide by x+ 3 to get a quadratic equation for two more values of x where the slope is -3. Check those in the first equation to see if one of them gives a point where the line and curve intersect.
    Last edited by HallsofIvy; February 21st 2011 at 05:28 AM.
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  5. #5
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    Hello, RogueDemon!

    There is something wrong with the problem . . . please check.


    \text{Find the equation of the tangent to:}
    . . y \:=\: 8x^4 + 40x^3 + 26x^2 - 63x + 24\,\text{ at the point }(3,\text{-}15).

    \text{Find another point on the curve which has exactly the same equation}
    . . \text{for the tangent at the point.}

    The point (3,\text{-}15) is not on the curve.

    . . The point (3,1797) is on the curve.

    . . The point (\text{-}3,15) is on the curve.


    What's going on?

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