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Math Help - witch of Maria Agnesi

  1. #1
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    witch of Maria Agnesi

    The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).



    y =

    How do I do this? Please help me out
    It looks so weird
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  2. #2
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    Quote Originally Posted by miss.strw View Post
    The curve below is called a witch of Maria Agnesi. Find an equation of the tangent line to this curve at the point (-2, 1/5).


    slope, m = y'(-2)

    now use the point-slope form of a linear equation ...

    y - y_1 = m(x - x_1) where (x_1,y_1) is the given point.
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    Quote Originally Posted by skeeter View Post
    slope, m = y'(-2)

    now use the point-slope form of a linear equation ...

    y - y_1 = m(x - x_1) where (x_1,y_1) is the given point.
    Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon!

    i do not know how to get the y` like i don't know how to do tat.
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    Quote Originally Posted by miss.strw View Post
    Do i need to plug in -2 into the equation to get m? or u're telling me that m = -2? thanks for helpin! please reply soon!

    i do not know how to get the y` like i don't know how to do tat.
    the slope is the derivative of the function evaluated at x = -2

    what does this mean? ...
    like i don't know how to do tat.
    you don't know how to find a derivative?
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  5. #5
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    Quote Originally Posted by skeeter View Post
    the slope is the derivative of the function evaluated at x = -2

    what does this mean? ...

    you don't know how to find a derivative?
    Not for those one. i don't know is it like 1/1+x^2 = 2x?

    so it's like

    y-(1/5)=(2x)(x-(-2)) ?

    thank you so much for helping!
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    Quote Originally Posted by miss.strw View Post
    Not for those one. i don't know is it like 1/1+x^2 = 2x? no

    so it's like

    y-(1/5)=(2x)(x-(-2)) ? no
    y = \frac{1}{1+x^2}

    use the quotient rule to find y' , or ...

    y = (1+x^2)^{-1}

    ... take the derivative using the chain rule.
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  7. #7
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    Quote Originally Posted by skeeter View Post
    y = \frac{1}{1+x^2}

    use the quotient rule to find y' , or ...

    y = (1+x^2)^{-1}

    ... take the derivative using the chain rule.




    g`(-2)h(-2)-g(-2)h`(-2)
    ---------------------------
    [h(-2)]^2

    this doesn't make sense to me

    if i do the derivative

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

    is derivative is x^2 = 2x
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  8. #8
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    you need to find the derivative first, then evaluate it at x = -2
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