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Math

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Show that the tangent to the curve 25x^5+5x^4-45x^3-5x^2+2x+6y-24=0 at the point (-1,1) is also a normal at two points of the curve.

  • Math -

    whewww!

    125x^4 + 20x^3 - 135x^2 - 10x + 2 + 6dy/dx = 0
    at (-1,1)
    125 - 20 - 135 + 10 - 2 + 6dy/dx = 0

    6dy/dx = 22, so the slope of the tangent at (-1,1) is 22/6 = 11/3
    equation of tangent:
    1= (11/3)(-1) + b
    b = 1 + 11/3 = 14/3

    y = (11/3)x + 14/3 = (11x+14)/3

    sub that into the original equation
    25x^5+5x^4-45x^3-5x^2+2x+6y-24=0
    25x^5+5x^4-45x^3-5x^2+2x+6(11x+14)/3-24=0
    25x^5+5x^4-45x^3-5x^2 + 24x + 4=0

    after dividing out the factor (x+1) we are left with
    25x^4 - 20x^3 - 25x^2 +20x+4=0

    tried to solve this, even had Wolfram page try it
    http://www.wolframalpha.com/input/?i=25x%5E4+-+20x%5E3+-+25x%5E2+%2B20x%2B4%3D0

    it gave me x = appr -.9507 and appr -.169872

    If we sub those values into the derivative equation we should get -3/11 or -.2727..
    I did not get that.

    Can't find my error
    OR,
    those two points don't exist
    (you might print out my solution and go through it to see if find any errors)

  • Math -

    -6y = 25x^5+5x^4-45x^3-5x^2+2x+24
    y' = -1/6 (125x^4 + 20x^3 - 135x^2 - 10x + 2)
    y'(-1) = 3
    normal has slope -1/3

    Go to http://rechneronline.de/function-graphs/

    and plot

    -1/6 * (25x^5+5x^4-45x^3-5x^2+2x-24)
    and its derivative
    as well as
    3x+4

    The line is not normal to the curve anywhere. Typo somewhere here?

  • Math - PS -

    When plotting the graphs, set
    x: -2 to 2
    y: -10 to 10

    that will show things quite clearly.

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