The diagram shows the curve $y = p \sin \frac{x}{q} + r$ for $0 \le x \le 8\pi$ radians. The curve has a minimum point at $(2\pi, -3)$ and a maximum point at $(6\pi, 1)$.

1. Calculate the vertical shift $r$ using the maximum and minimum $y$-values

The vertical shift $r$ is the average of the maximum and minimum values of the curve.
$r = \frac{\text{Maximum value} + \text{Minimum value}}{2}$
$r = \frac{1 + (-3)}{2}$
$r = \frac{-2}{2}$
$r = -1 \text{ (shown)}$

1. Determine the period of the function to find $q$.

The distance between the minimum point at $x = 2\pi$ and the maximum point at $x = 6\pi$ represents half a period.
$\text{Half period} = 6\pi - 2\pi = 4\pi$
$\text{Period} = 2 \times 4\pi = 8\pi$
The period of $y = p \sin \frac{x}{q} + r$ is given by $2\pi q$.
$2\pi q = 8\pi \implies q = 4$

2. Use the coordinates of the minimum point to find $p$.

Substitute $x = 2\pi$, $y = -3$, $q = 4$, and $r = -1$ into the equation:
$-3 = p \sin\left(\frac{2\pi}{4}\right) - 1$
$-2 = p \sin\left(\frac{\pi}{2}\right)$
Since $\sin\left(\frac{\pi}{2}\right) = 1$,
$p = -2$

1. Substitute the values of $p$, $q$, and $r$ found in previous parts into the general equation

From (a), $r = -1$.
From (b), $p = -2$ and $q = 4$.
Substituting these into $y = p \sin \frac{x}{q} + r$ gives:
$y = -2 \sin \frac{x}{4} - 1$