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 $\displaystyle y$.