Worksheet for graphing cosine functions with six different equations and corresponding coordinate grids.

Let's solve each of these cosine graphing problems step by step. We'll analyze the transformations applied to the basic cosine function $ y = \cos(x) $, and then sketch the graph accordingly.

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General Form of a Cosine Function:

$$
y = A \cdot \cos(Bx + C) + D
$$

Where:
- $ A $: amplitude (vertical stretch/compression)
- $ B $: affects period ($ \text{Period} = \frac{2\pi}{|B|} $)
- $ C $: phase shift (horizontal shift): $ -\frac{C}{B} $
- $ D $: vertical shift (up/down)

We’ll use this form to analyze each function.

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## 1. $ y = \cos\left(x + \frac{\pi}{4}\right) - 2 $

Step-by-step Analysis:

- $ A = 1 $ → amplitude = 1
- $ B = 1 $ → period = $ 2\pi $
- Phase shift: $ x + \frac{\pi}{4} = 0 $ → $ x = -\frac{\pi}{4} $ → shift left by $ \frac{\pi}{4} $
- Vertical shift: $ D = -2 $ → shift down by 2 units

Key Points (based on standard cosine):

Standard cosine has max at $ x=0 $, min at $ x=\pi $, zero crossings at $ \frac{\pi}{2}, \frac{3\pi}{2} $, etc.

Apply shifts:
- Max: $ x = -\frac{\pi}{4} $, $ y = 1 - 2 = -1 $
- Zero: $ x = -\frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4} $, $ y = -2 $
- Min: $ x = -\frac{\pi}{4} + \pi = \frac{3\pi}{4} $, $ y = -1 - 2 = -3 $
- Zero: $ x = -\frac{\pi}{4} + \frac{3\pi}{2} = \frac{5\pi}{4} $, $ y = -2 $
- Max: $ x = -\frac{\pi}{4} + 2\pi = \frac{7\pi}{4} $, $ y = -1 $

Graph: Start at $ (-\frac{\pi}{4}, -1) $, go down to $ (\frac{3\pi}{4}, -3) $, up to $ (\frac{7\pi}{4}, -1) $, etc., all shifted down by 2.

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## 2. $ y = \cos(x + \pi) + 5 $

Analysis:

- $ A = 1 $
- $ B = 1 $ → period = $ 2\pi $
- Phase shift: $ x + \pi = 0 $ → $ x = -\pi $ → shift left by $ \pi $
- Vertical shift: $ D = 5 $ → shift up by 5 units

Key Points:

- Max: $ x = -\pi $, $ y = 1 + 5 = 6 $
- Zero: $ x = -\pi + \frac{\pi}{2} = -\frac{\pi}{2} $, $ y = 5 $
- Min: $ x = -\pi + \pi = 0 $, $ y = -1 + 5 = 4 $
- Zero: $ x = -\pi + \frac{3\pi}{2} = \frac{\pi}{2} $, $ y = 5 $
- Max: $ x = -\pi + 2\pi = \pi $, $ y = 6 $

So it starts at $ (-\pi, 6) $, goes down to $ (0, 4) $, up to $ (\pi, 6) $, etc.

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## 3. $ y = \cos\left(x - \frac{\pi}{2}\right) + 3 $

Analysis:

- $ A = 1 $
- $ B = 1 $ → period = $ 2\pi $
- Phase shift: $ x - \frac{\pi}{2} = 0 $ → $ x = \frac{\pi}{2} $ → shift right by $ \frac{\pi}{2} $
- Vertical shift: $ D = 3 $ → shift up by 3 units

Key Points:

- Max: $ x = \frac{\pi}{2} $, $ y = 1 + 3 = 4 $
- Zero: $ x = \frac{\pi}{2} + \frac{\pi}{2} = \pi $, $ y = 3 $
- Min: $ x = \frac{\pi}{2} + \pi = \frac{3\pi}{2} $, $ y = -1 + 3 = 2 $
- Zero: $ x = \frac{\pi}{2} + \frac{3\pi}{2} = 2\pi $, $ y = 3 $
- Max: $ x = \frac{\pi}{2} + 2\pi = \frac{5\pi}{2} $, $ y = 4 $

Starts at $ (\frac{\pi}{2}, 4) $, down to $ (\frac{3\pi}{2}, 2) $, up to $ (\frac{5\pi}{2}, 4) $, etc.

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## 4. $ y = \cos(x - \pi) + 4 $

Analysis:

- $ A = 1 $
- $ B = 1 $ → period = $ 2\pi $
- Phase shift: $ x - \pi = 0 $ → $ x = \pi $ → shift right by $ \pi $
- Vertical shift: $ D = 4 $ → shift up by 4 units

Key Points:

- Max: $ x = \pi $, $ y = 1 + 4 = 5 $
- Zero: $ x = \pi + \frac{\pi}{2} = \frac{3\pi}{2} $, $ y = 4 $
- Min: $ x = \pi + \pi = 2\pi $, $ y = -1 + 4 = 3 $
- Zero: $ x = \pi + \frac{3\pi}{2} = \frac{5\pi}{2} $, $ y = 4 $
- Max: $ x = \pi + 2\pi = 3\pi $, $ y = 5 $

So the wave starts at $ (\pi, 5) $, goes down to $ (2\pi, 3) $, up to $ (3\pi, 5) $, etc.

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## 5. $ y = -3 \cdot \cos\left(-\frac{2}{3}x\right) $

Note: $ \cos(-\theta) = \cos(\theta) $, so $ \cos(-\frac{2}{3}x) = \cos(\frac{2}{3}x) $

So this becomes:
$$
y = -3 \cdot \cos\left(\frac{2}{3}x\right)
$$

Analysis:

- $ A = -3 $ → amplitude = 3, but negative → reflection over x-axis
- $ B = \frac{2}{3} $ → period = $ \frac{2\pi}{2/3} = 3\pi $
- No phase shift (no $ C $)
- No vertical shift ($ D = 0 $)

Key Points:

Standard cosine: max at 0, min at $ \pi $, etc.

But since $ B = \frac{2}{3} $, period is $ 3\pi $, so:

- Max: $ x = 0 $, $ y = -3 \cdot \cos(0) = -3 \cdot 1 = -3 $
- But wait: because of the negative sign, it starts at a minimum, not maximum.

So:
- At $ x = 0 $: $ y = -3 \cdot \cos(0) = -3 $ → minimum
- First zero: when $ \frac{2}{3}x = \frac{\pi}{2} $ → $ x = \frac{3\pi}{4} $, $ y = 0 $
- Maximum: $ \frac{2}{3}x = \pi $ → $ x = \frac{3\pi}{2} $, $ y = -3 \cdot (-1) = 3 $
- Second zero: $ \frac{2}{3}x = \frac{3\pi}{2} $ → $ x = \frac{9\pi}{4} $, $ y = 0 $
- Minimum again: $ \frac{2}{3}x = 2\pi $ → $ x = 3\pi $, $ y = -3 $

So the graph starts at $ (0, -3) $, rises to $ (\frac{3\pi}{2}, 3) $, back to $ (3\pi, -3) $, with zeros in between.

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## 6. $ y = -2 \cdot \cos\left(\frac{2}{3}x\right) $

This one is similar to #5, but no negative inside cosine.

Analysis:

- $ A = -2 $ → amplitude = 2, reflected over x-axis
- $ B = \frac{2}{3} $ → period = $ \frac{2\pi}{2/3} = 3\pi $
- No phase shift
- No vertical shift

Key Points:

- $ x = 0 $: $ y = -2 \cdot \cos(0) = -2 \cdot 1 = -2 $ → minimum
- First zero: $ \frac{2}{3}x = \frac{\pi}{2} $ → $ x = \frac{3\pi}{4} $, $ y = 0 $
- Maximum: $ \frac{2}{3}x = \pi $ → $ x = \frac{3\pi}{2} $, $ y = -2 \cdot (-1) = 2 $
- Second zero: $ \frac{2}{3}x = \frac{3\pi}{2} $ → $ x = \frac{9\pi}{4} $, $ y = 0 $
- Minimum: $ x = 3\pi $, $ y = -2 $

So same shape as #5, but smaller amplitude (2 instead of 3).

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Summary Table:

| Problem | Amplitude | Period | Phase Shift | Vertical Shift | Reflection? |
|--------|-----------|--------|-------------|----------------|-------------|
| 1 | 1 | $2\pi$ | $-\frac{\pi}{4}$ | -2 | No |
| 2 | 1 | $2\pi$ | $-\pi$ | +5 | No |
| 3 | 1 | $2\pi$ | $+\frac{\pi}{2}$ | +3 | No |
| 4 | 1 | $2\pi$ | $+\pi$ | +4 | No |
| 5 | 3 | $3\pi$ | 0 | 0 | Yes (negative A) |
| 6 | 2 | $3\pi$ | 0 | 0 | Yes (negative A) |

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How to Graph Each:

For each problem:
1. Draw the midline (if any vertical shift).
2. Mark the phase shift (starting point).
3. Plot key points: max, min, zeros.
4. Use the period to repeat the pattern.

You can now draw the graphs based on these descriptions.

Let me know if you'd like me to generate the actual plots or provide coordinate tables!