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

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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.
The function $$y=-2 \sin \left[\frac{1}{4}\left(x+\frac{2 \pi}{3}\right)\right]-5$$ has an amplitude of 2, a period of $$8\pi$$, a phase shift of $$-\frac{8\pi}{3}$$ to the left, and a vertical shift of $$-5$$. Sketch the graph with these properties, starting at $$-5$$, peaking at $$-3$$, dipping to $$-7$$, and returning to $$-5$$ over one period.

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

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