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 $$|-2| = 2$$.

2. **Period**: The period of the 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:
   $$
   B = \frac{1}{4} \implies \text{Period} = \frac{2\pi}{\frac{1}{4}} = 8\pi.
   $$

3. **Phase Shift**: The phase shift can be found from the expression inside the sine function. The phase shift is given by:
   $$
   \text{Phase Shift} = -\frac{C}{B},
   $$
   where $$C$$ is the constant added to $$x$$. Here, $$C = \frac{2\pi}{3}$$, so:
   $$
   \text{Phase Shift} = -\frac{\frac{2\pi}{3}}{\frac{1}{4}} = -\frac{2\pi}{3} \cdot 4 = -\frac{8\pi}{3}.
   $$
   This means the graph shifts to the left by $$\frac{8\pi}{3}$$.

4. **Vertical Shift**: The vertical shift is given by the constant term outside the sine 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:
   - $$0$$ (maximum)
   - $$\frac{\pi}{2}$$ (zero)
   - $$\pi$$ (minimum)
   - $$\frac{3\pi}{2}$$ (zero)
   - $$2\pi$$ (maximum)

However, due to the phase shift and the period, we will calculate these points based on the new starting point and the period of $$8\pi$$.

3. **Calculate Key Points**:
   - Starting point: $$x = -\frac{8\pi}{3}$$ (value of $$y$$ is $$-5$$)
   - Next point (maximum): 
     $$
     x = -\frac{8\pi}{3} + 2\pi = -\frac{8\pi}{3} + \frac{6\pi}{3} = -\frac{2\pi}{3} \quad (y = -5 + 2 = -3)
     $$
   - Next point (zero):
     $$
     x = -\frac{8\pi}{3} + 4\pi = -\frac{8\pi}{3} + \frac{12\pi}{3} = \frac{4\pi}{3} \quad (y = -5)
     $$
   - Next point (minimum):
     $$
     x = -\frac{8\pi}{3} + 6\pi = -\frac{8\pi}{3} + \frac{18\pi}{3} = \frac{10\pi}{3} \quad (y = -5 - 2 = -7)
     $$
   - Next point (zero):
     $$
     x = -\frac{8\pi}{3} + 8\pi = -\frac{8\pi}{3} + \frac{24\pi}{3} = \frac{16\pi}{3} \quad (y = -5)
     $$

### Step 3: Summary of Key Points
- $$(- \frac{8\pi}{3}, -5)$$
- $$(- \frac{2\pi}{3}, -3)$$
- $$(\frac{4\pi}{3}, -5)$$
- $$(\frac{10\pi}{3}, -7)$$
- $$(\frac{16\pi}{3}, -5)$$

### Step 4: Sketch the Graph
1. Start at $$(- \frac{8\pi}{3}, -5)$$.
2. Move to $$(- \frac{2\pi}{3}, -3)$$ (maximum).
3. Move to $$(\frac{4\pi}{3}, -5)$$ (zero).
4. Move to $$(\frac{10\pi}{3}, -7)$$ (minimum).
5. Move to $$(\frac{16\pi}{3}, -5)$$ (zero).

### Final Properties
- **Amplitude**: 2
- **Period**: $$8\pi$$
- **Phase Shift**: $$-\frac{8\pi}{3}$$ (to the left)
- **Vertical Shift**: $$-5$$

This completes the analysis and sketch of one cycle of the function.
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.

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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