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The amplitude of a sinusoidal function is the absolute value of the coefficient of sine or cosine in the equation. In our case, the amplitude A is equal to the coefficient of the sine function, which is 2; thus, \(A = 2\).

The period of a sinusoidal function can be found by dividing \(2 \pi\) by the absolute value of the coefficient of x inside the sine function's argument. In our case, the coefficient of x is 2, so the period T can be calculated as: \[T = \frac{2 \pi}{\left|2\right|} = \pi\]

To sketch the graph, we need to know some key points within one period of the sine function. The following steps will help us find these points: 1. Start with the standard sine function \(y = \sin(x)\), which has key points at \(x = 0, \frac{\pi}{2}, \pi, \frac{3 \pi}{2}, 2 \pi\), corresponding to y-values of 0, 1, 0, -1, and 0, respectively. 2. Modify the amplitude by multiplying the y-values by 2: 0, 2, 0, -2, and 0. 3. Modify the frequency by dividing the x-values by 2: 0, \(\frac{\pi}{4}\), \(\frac{\pi}{2}\), \(\frac{3 \pi}{4}\), and \(\pi\). 4. Shift the graph 90 degrees (\(\frac{\pi}{2}\) radians) to the left by subtracting \(\frac{\pi}{2}\) from the x-values. This results in the final points: \(-\frac{\pi}{2}, 0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}\) with corresponding y-values of 0, 2, 0, -2, 0. Now we can sketch the graph of the given function over one period, using the key points and the amplitude and period we found. The graph starts at \(\left(-\frac{\pi}{2}, 0\right)\), rises to a maximum value of 2 at \((0, 2)\), returns to 0 at \(\left(\frac{\pi}{4}, 0\right)\), reaches a minimum value of -2 at \((\frac{\pi}{2}, -2)\), and returns to 0 at \(\left(\frac{3 \pi}{4}, 0\right)\), completing one period.