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Thread: 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:

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

    Solve (with care) $\displaystyle \frac{dy}{dx} = 0$ 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.
    $\displaystyle \frac{d}{dx}(y^2) = \frac{d}{dx}(x^3 + 3x^2)$

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

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

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


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


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

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


    So $\displaystyle 3x^2 + 6x = 0$

    $\displaystyle 3x(x + 2) = 0$

    $\displaystyle x = 0$ or $\displaystyle x = -2$.


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