91Ó°ÊÓ

The function given is \(y = \frac{1}{2} \cot 2x\). The cotangent function, \(\cot x\), has a period of \(\pi\) and an undefined value at x = 0. Here, the coefficient of 2 next to x will cause the frequency or the number of cycles in the function in a given range to double. The \(\frac{1}{2}\) in front of the \(\cot(2x)\) will affect the amplitude or the 'height' of the function.

For a normal \(\cot\) function, key points typically include where the function starts and ends its period, which are usually multiples of \(\pi\), and where the function is undefined, typically at points like 0. Thus, the key points to consider for this function will be \(x = 0\), \(x = \frac{\pi}{4}\), \(x = \frac{\pi}{2}\), \(x = \frac{3\pi}{4}\), and \(x = \pi\).

Begin the sketch by marking the key points identified in the previous step on the x-axis. Draw a vertical dotted line at \(x = \frac{\pi}{4}\) and \(x = \frac{3\pi}{4}\) while these will be asymptotes here. Start sketching the function from \(-\infty\) at \(x = 0\) and go up to \(\infty\) at \(x = \frac{\pi}{4}\). After \(x = \frac{\pi}{4}\), go from \(-\infty\) and rise up to \(\infty\) at \(x = \frac{\pi}{2}\), continue this pattern. Don't forget to keep the amplitude to \(\frac{1}{2}\) so the peaks should be at \(\frac{1}{2}\) and valleys should be at \(-\frac{1}{2}\) instead of 1 and -1 respectively. The graph between any two vertical asymptotes should look like the 'u' shape of a standard cotangent function after considering the amplitude.