The function $f$ is defined, for $x \ge 0^\circ$, by $f(x) = 3\cos\left(\frac{2x}{3}\right) - 1$.

1. Find the maximum value of the function

The maximum value of $\cos\left(\frac{2x}{3}\right)$ is $1$.
Maximum value of $f(x) = 3(1) - 1 = 2$

2. Find the minimum value of the function

The minimum value of $\cos\left(\frac{2x}{3}\right)$ is $-1$.
Minimum value of $f(x) = 3(-1) - 1 = -4$

1. Identify the amplitude from the function's equation

For a function of the form $f(x) = a\cos(bx) + c$, the amplitude is given by $|a|$.
Here, $a = 3$, so the amplitude is $3$.

1. Calculate the period of the function

The period of $\cos(bx)$ in degrees is given by $\frac{360^\circ}{b}$.
Here, $b = \frac{2}{3}$.
$\text{Period} = \frac{360^\circ}{\frac{2}{3}} = 360^\circ \times \frac{3}{2} = 540^\circ$

1. Identify the key points of the graph within the given domain.

At $x = 0^\circ$, $y = 3\cos(0^\circ) - 1 = 3(1) - 1 = 2$.
At $x = 270^\circ$, $y = 3\cos\left(\frac{2(270^\circ)}{3}\right) - 1 = 3\cos(180^\circ) - 1 = 3(-1) - 1 = -4$.
The period is $540^\circ$, so $270^\circ$ corresponds to half a period, which is the first minimum point.

2. Sketch the curve.

Draw a smooth, decreasing curve starting from the maximum point at $(0^\circ, 2)$, crossing the $x$-axis, and ending at the minimum point at $(270^\circ, -4)$.