The problem asks to sketch the graph of the function $y = x\sqrt{4 - x^2}$.

1. Problem Description

The problem asks to sketch the graph of the function $y = x\sqrt{4 - x^2}$.

2. Solution Steps

First, we determine the domain of the function. Since we have a square root, we must have $4 - x^2 \ge 0$, which means $x^2 \le 4$, or $-2 \le x \le 2$. Therefore, the domain of the function is $[-2, 2]$.

Next, we find the derivative of the function with respect to $x$:

$y = x\sqrt{4 - x^2}$

Using the product rule, we have:

$\frac{dy}{dx} = \sqrt{4 - x^2} + x \cdot \frac{1}{2\sqrt{4 - x^2}} \cdot (-2x) = \sqrt{4 - x^2} - \frac{x^2}{\sqrt{4 - x^2}}$

$\frac{dy}{dx} = \frac{4 - x^2 - x^2}{\sqrt{4 - x^2}} = \frac{4 - 2x^2}{\sqrt{4 - x^2}}$

To find the critical points, we set $\frac{dy}{dx} = 0$:

$\frac{4 - 2x^2}{\sqrt{4 - x^2}} = 0$

$4 - 2x^2 = 0$

$2x^2 = 4$

$x^2 = 2$

$x = \pm\sqrt{2}$

Since both values are within the domain $[-2, 2]$, they are both critical points.

Now, we determine the sign of $\frac{dy}{dx}$ in the intervals $[-2, -\sqrt{2})$, $(-\sqrt{2}, \sqrt{2})$, and $(\sqrt{2}, 2]$.

- For $x \in [-2, -\sqrt{2})$, say $x = -1.5$, we have $\frac{dy}{dx} = \frac{4 - 2(2.25)}{\sqrt{4 - 2.25}} = \frac{4 - 4.5}{\sqrt{1.75}} = \frac{-0.5}{\sqrt{1.75}} < 0$.

- For $x \in (-\sqrt{2}, \sqrt{2})$, say $x = 0$, we have $\frac{dy}{dx} = \frac{4 - 0}{\sqrt{4}} = \frac{4}{2} = 2 > 0$.

- For $x \in (\sqrt{2}, 2]$, say $x = 1.5$, we have $\frac{dy}{dx} = \frac{4 - 2(2.25)}{\sqrt{4 - 2.25}} = \frac{4 - 4.5}{\sqrt{1.75}} = \frac{-0.5}{\sqrt{1.75}} < 0$.

So, the function is decreasing on $[-2, -\sqrt{2})$, increasing on $(-\sqrt{2}, \sqrt{2})$, and decreasing on $(\sqrt{2}, 2]$.

This means that there is a local minimum at $x = -\sqrt{2}$ and a local maximum at $x = \sqrt{2}$.

Let us find the y-values at these points.

- When $x = -\sqrt{2}$, $y = -\sqrt{2} \sqrt{4 - (-\sqrt{2})^2} = -\sqrt{2} \sqrt{4 - 2} = -\sqrt{2} \sqrt{2} = -2$.

- When $x = \sqrt{2}$, $y = \sqrt{2} \sqrt{4 - (\sqrt{2})^2} = \sqrt{2} \sqrt{4 - 2} = \sqrt{2} \sqrt{2} = 2$.

At the endpoints of the domain, $x = -2$ and $x = 2$, we have $y = 0$.

- When $x = -2$, $y = -2 \sqrt{4 - (-2)^2} = -2 \sqrt{0} = 0$.

- When $x = 2$, $y = 2 \sqrt{4 - (2)^2} = 2 \sqrt{0} = 0$.

The function is odd since $y(-x) = (-x) \sqrt{4 - (-x)^2} = -x \sqrt{4 - x^2} = -y(x)$. Therefore the graph is symmetric with respect to the origin.

3. Final Answer

The graph is a curve that passes through $(-2, 0)$, $(-\sqrt{2}, -2)$, $(0, 0)$, $(\sqrt{2}, 2)$, and $(2, 0)$. There is a local minimum at $(-\sqrt{2}, -2)$ and a local maximum at $(\sqrt{2}, 2)$. The function is odd, and the domain is $[-2, 2]$. The curve increases from $x=-2$ to $x=-\sqrt{2}$, decreases from $x=-\sqrt{2}$ to $x=0$ and $x=0$ to $x=\sqrt{2}$, increases from $x=0$ to $x=\sqrt{2}$ and decreases from $x=\sqrt{2}$ to $x=2$.

The graph looks like a sideways figure-eight.

Final Answer: (Sketch of the graph is required for the actual answer)