1) Sketch one cycle for the following trigonometric function and complete the properties below. \[ y=-2 \sin \left[\frac{1}{4}\left(x+\frac{2 \pi}{3}\right)\right]-5 \]

To sketch one cycle of the function \[ y = -2 \sin \left[\frac{1}{4}\left(x+\frac{2 \pi}{3}\right)\right]-5, \] we need to analyze its properties, including amplitude, period, phase shift, and vertical shift. ### Step 1: Identify the properties 1. **Amplitude**: The amplitude of a sine function is given by the coefficient in front of the sine. Here, the amplitude is \[ |A| = |-2| = 2. \] 2. **Period**: The period of a sine function is calculated using the formula \[ \text{Period} = \frac{2\pi}{B}, \] where \(B\) is the coefficient of \(x\) inside the sine function. In this case, we have \[ B = \frac{1}{4}, \] so the period is \[ \text{Period} = \frac{2\pi}{\frac{1}{4}} = 8\pi. \] 3. **Phase Shift**: The phase shift can be found from the expression inside the sine function. The phase shift is given by \[ -\frac{C}{B}, \] where \(C\) is the constant added to \(x\). Here, \[ C = \frac{2\pi}{3}, \] so the phase shift is \[ -\frac{\frac{2\pi}{3}}{\frac{1}{4}} = -\frac{2\pi}{3} \cdot 4 = -\frac{8\pi}{3}. \] This means the graph shifts to the left by \(\frac{8\pi}{3}\). 4. **Vertical Shift**: The vertical shift is given by the constant added at the end of the function. Here, the vertical shift is \[ -5. \] ### Step 2: Sketch the function 1. **Starting Point**: The sine function starts at 0. With the phase shift of \(-\frac{8\pi}{3}\), the starting point of the sine wave will be at \[ x = -\frac{8\pi}{3}. \] 2. **Key Points**: The sine function has key points at \(0\), \(\frac{\pi}{2}\), \(\pi\), \(\frac{3\pi}{2}\), and \(2\pi\). We will adjust these points based on the period and phase shift. - The first key point (maximum) occurs at \(x = -\frac{8\pi}{3}\). - The second key point (zero) occurs at \[ -\frac{8\pi}{3} + \frac{8\pi}{4} = -\frac{8\pi}{3} + 2\pi = -\frac{8\pi}{3} + \frac{6\pi}{3} = -\frac{2\pi}{3}. \] - The third key point (minimum) occurs at \[ -\frac{8\pi}{3} + \frac{3 \cdot 8\pi}{4} = -\frac{8\pi}{3} + 6\pi = -\frac{8\pi}{3} + \frac{18\pi}{3} = \frac{10\pi}{3}. \] - The fourth key point (zero) occurs at \[ -\frac{8\pi}{3} + 8\pi = -\frac{8\pi}{3} + \frac{24\pi}{3} = \frac{16\pi}{3}. \] 3. **Graphing**: The sine function will start at \(-5\) (the vertical shift), reach a maximum of \(-5 + 2 = -3\) at \(-\frac{8\pi}{3}\), return to \(-5\) at \(-\frac{2\pi}{3}\), reach a minimum of \(-5 - 2 = -7\) at \(\frac{10\pi}{3}\), and return to \(-5\) at \(\frac{16\pi}{3}\). ### Summary of Properties - **Amplitude**: 2 - **Period**: \(8\pi\) - **Phase Shift**: \(-\frac{8\pi}{3}\) (to the left) - **Vertical Shift**: \(-5\) ### Sketch To sketch the graph, plot the key points and connect them smoothly, keeping in mind the amplitude and vertical shift. The graph will oscillate between \(-7\) and \(-3\) with a period of \(8\pi\). The sketch would look like a sine wave starting at \(-5\), peaking at \(-3\), dipping to \(-7\), and returning to \(-5\) over the interval of one period.