# Math Help - Horizontal Tangents

1. ## 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.

2. Originally Posted by xxju1anxx
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.

3. Originally Posted by xxju1anxx
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 $y$.