Answer
\[\begin{align}
& \text{Horizontal tangent line for }x=4 \\
& \text{Vertical tangent line for }x=6 \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \text{Let }y=x\sqrt{6-x} \\
& \text{Differentiate} \\
& \frac{dy}{dx}=\frac{d}{dx}\left[ x\sqrt{6-x} \right] \\
& \frac{dy}{dx}=x\frac{d}{dx}\left[ \sqrt{6-x} \right]+\sqrt{6-x}\frac{d}{dx}\left[ x \right] \\
& \frac{dy}{dx}=x\left( -\frac{1}{2\sqrt{6-x}} \right)+\sqrt{6-x}\left( 1 \right) \\
& \frac{dy}{dx}=-\frac{x}{2\sqrt{6-x}}+\sqrt{6-x} \\
& \text{Let }\frac{dy}{dx}=0\text{ }\left( \text{Horizontal tangent line} \right) \\
& -\frac{x}{2\sqrt{6-x}}+\sqrt{6-x}=0 \\
& \frac{x}{2\sqrt{6-x}}=\sqrt{6-x} \\
& x=2\left( 6-x \right) \\
& x=12-2x \\
& 3x=12 \\
& x=4 \\
\text{Vertical Tangent Lines:}
\\ & \frac{dy}{dx}=-\frac{x}{2\sqrt{6-x}}+\sqrt{6-x} \text{ is undefined when} \\
& 2\sqrt{6-x}=0 \\
& x=6 \\
\end{align}\]
