Question

graph the function over a two - period interval.

$y=-2+\frac{2}{3}\cot(4x - 8\pi)$

Explanation:

Step1: Find the period

For the cotangent function \(y = A\cot(Bx - C)+D\), the period is \(\frac{\pi}{|B|}\).
Here \(B = 4\), so the period \(T=\frac{\pi}{4}\).

Step2: Simplify the function

We know that \(\cot(4x - 8\pi)=\cot(4(x - 2\pi))\). Since the cotangent function has a period of \(\pi\), \(\cot(4x-8\pi)=\cot(4x)\)

Step3: Analyze the key points

The standard cotangent function \(y = \cot(x)\) has vertical asymptotes at \(x = n\pi\), \(n\in\mathbb{Z}\), and passes through \((\frac{\pi}{2},0)\).
For \(y=\frac{2}{3}\cot(4x)- 2\), the vertical asymptotes are \(4x=n\pi\Rightarrow x=\frac{n\pi}{4}\), \(n\in\mathbb{Z}\).
When \(4x=\frac{\pi}{2}\Rightarrow x = \frac{\pi}{8}\), \(y=\frac{2}{3}\cot(\frac{\pi}{2})-2=- 2\)

Step4: Graph over two - period interval

For two - period interval, we can consider the interval \((0,\frac{\pi}{2})\) (since the period is \(\frac{\pi}{4}\), two periods give a length of \(\frac{\pi}{2}\)).
Plot the vertical asymptotes \(x = 0,x=\frac{\pi}{4},x=\frac{\pi}{2}\)
Take some points in between:
Let \(x=\frac{\pi}{16}\), \(y=\frac{2}{3}\cot(\frac{\pi}{4})-2=\frac{2}{3}-2=-\frac{4}{3}\)
Let \(x=\frac{3\pi}{16}\), \(y=\frac{2}{3}\cot(\frac{3\pi}{4})-2=-\frac{2}{3}-2=-\frac{8}{3}\)

Answer:

Graph the function \(y = - 2+\frac{2}{3}\cot(4x)\) (equivalent to the given function) with vertical asymptotes at \(x=\frac{n\pi}{4}\), \(n\in\mathbb{Z}\), passing through \((\frac{\pi}{8},-2)\) and plot points in the interval \((0,\frac{\pi}{2})\) (two - period interval) using the key - point analysis above.