To analyze the given function:

$y = \sin \left( \frac{\pi}{3}x + \frac{5\pi}{3} \right)$

Step 1: Identify key parameters of the sine function

The general form of a sine function is:

$y = A \sin(Bx + C) + D$

Where:

• $A$ is the amplitude
• $B$ affects the period
• $C$ affects the phase shift
• $D$ is the vertical shift (if any)

Step 2: Determine the period

The period $T$ of a sine function is given by:

$T = \frac{2\pi}{|B|}$

For the function $y = \sin \left( \frac{\pi}{3}x + \frac{5\pi}{3} \right)$, the value of $B$ is $\frac{\pi}{3}$, so the period is:

$T = \frac{2\pi}{\left| \frac{\pi}{3} \right|} = \frac{2\pi}{\frac{\pi}{3}} = 6$

Thus, the period of the function is 6.

Step 3: Determine the phase shift

The phase shift is given by:

$\text{Phase shift} = -\frac{C}{B}$

In the given function, $C = \frac{5\pi}{3}$ and $B = \frac{\pi}{3}$. Therefore, the phase shift is:

$\text{Phase shift} = -\frac{\frac{5\pi}{3}}{\frac{\pi}{3}} = -5$

Thus, the phase shift is 5 units to the left.

Step 4: Determine the range

The range of the sine function is determined by its amplitude. Since no vertical shift is indicated (the value of $D$ is 0), the range remains the same as a standard sine function, i.e.,:

[ \text{Range}