The diagram shows the curve $y = a\cos bx + c$ for $0 \le x \le \tfrac{3\pi}{2}$ radians. A minimum point $(\tfrac{\pi}{2}, -2)$ and a maximum point $(\tfrac{5\pi}{4}, 4)$ are indicated on the diagram.

1. Find the horizontal distance between the given minimum and maximum points.

Horizontal distance $= \frac{5\pi}{4} - \frac{\pi}{2} = \frac{3\pi}{4}$

2. Relate the distance to the period of the function to find $b$.

From the graph, this distance spans $1.5$ cycles (from a minimum to the second maximum after it).
$1.5 \times \text{Period} = \frac{3\pi}{4}$
$\text{Period} = \frac{\pi}{2}$
Since $\text{Period} = \frac{2\pi}{b}$, we have $\frac{2\pi}{b} = \frac{\pi}{2} \implies b = 4$.

1. Relate the constant $c$ to the maximum and minimum values of the curve

The constant $c$ represents the vertical shift (principal axis) of the curve, which is the average of the maximum and minimum $y$-values.
$c = \frac{\text{Maximum value} + \text{Minimum value}}{2}$

2. Substitute the given maximum and minimum values to show $c = 1$

$c = \frac{4 + (-2)}{2}$
$c = \frac{2}{2} = 1$

1. Substitute the known values of $b$, $c$, and a point on the curve to find $a$.

Using $b = 4$ and $c = 1$ from parts (a) and (b), the equation is $y = a\cos 4x + 1$.
Substitute the minimum point $(\frac{\pi}{2}, -2)$ into the equation:
$-2 = a\cos\left(4\left(\frac{\pi}{2}\right)\right) + 1$
$-2 = a\cos(2\pi) + 1$
$-2 = a(1) + 1$
$a = -3$

2. State the final equation of the curve.

The equation of the curve is $y = -3\cos 4x + 1$.