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Math Help - Horizontal Tangents

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    Horizontal Tangents

    I am having trouble figuring out the horizontal tangents for this problem y^2=x^3+3x^2. I know that one of the horizontal tangents is (-2,2), but I cannot seem to find the other one.
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    Quote Originally Posted by xxju1anxx View Post
    I am having trouble figuring out the horizontal tangents for this problem y^2=x^3+3x^2. I know that one of the horizontal tangents is (-2,2), but I cannot seem to find the other one.
    Use implicit differentiation:

    2y \frac{dy}{dx} = 3x^2 + 6x \Rightarrow \frac{dy}{dx} = \frac{3x^2 + 6x}{2y} = \frac{3x^2 + 6x}{\sqrt{x^3+3x^2}}.

    Solve (with care) \frac{dy}{dx} = 0 for x. Use this value of x to solve for y.
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    Quote Originally Posted by xxju1anxx View Post
    I am having trouble figuring out the horizontal tangents for this problem y^2=x^3+3x^2. I know that one of the horizontal tangents is (-2,2), but I cannot seem to find the other one.
    \frac{d}{dx}(y^2) = \frac{d}{dx}(x^3 + 3x^2)

    2y\,\frac{dy}{dx} = 3x^2 + 6x

    \frac{dy}{dx} = \frac{1}{2y}(3x^2 + 6x)

    \frac{dy}{dx} = \frac{3x^2 + 6x}{2\sqrt{x^3 + 3x^2}}


    To find the horizontal tangents, let \frac{dy}{dx} = 0.


    0 = \frac{3x^2 + 6x}{2\sqrt{x^3 + 3x^2}}

    Note that the denominator can not be 0, i.e. x \neq 0 and x \neq -3


    So 3x^2 + 6x = 0

    3x(x + 2) = 0

    x = 0 or x = -2.


    Since x \neq 0, let x = -2 and solve for y.
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