For what value(s) of x is the line tangent to the curve y = xV6 - x horizontal? Vertical?

Given curve is:

$y=x\sqrt{6-x}$

Differentiating with respect to x we get,

$$\begin{gathered} \frac{dy}{dx}=\sqrt{6-x}-\frac{x}{2\sqrt{6-x}} \\ \frac{dy}{dx}=\frac{2\left(6-x\right)-x}{2\sqrt{6-x}} \\ \frac{dy}{dx}=\frac{12-2x-x}{2\sqrt{6-x}} \\ \frac{dy}{dx}=\frac{12-3x}{2\sqrt{6-x}} \end{gathered}$$

This is the slope of the tangent line to the given curve.

The tangent line becomes horizontal when:

$\frac{dy}{dx}=0$

That is when:

$$\begin{gathered} \frac{12-3x}{2\sqrt{6-x}}=0 \\ 12-3x=0 \\ 3x=12 \\ x=\frac{12}{3} \\ x=4 \end{gathered}$$

Hence the answer.