Let f(x) = ln(3-x) + 2. Describe how to use function transformation to get from ln(x) to f(x). Then sketch the graph of f(x) by starting with a graph ln(x) and transforming it.

To describe how to use function transformations to get from \( \ln(x) \) to \( f(x) = \ln(3-x) + 2 \), we can break down the process into several steps, each corresponding to a standard transformation of the function \( \ln(x) \).

### Step-by-Step Function Transformations

1. **Start with \( \ln(x) \):** This is the basic natural logarithm function with a domain of \( (0, \infty) \).

2. **Horizontal Reflection and Translation ( \( \ln(-x) \) ):**
   - Reflect \( \ln(x) \) across the \( y \)-axis to get \( \ln(-x) \).
   - The domain changes to \( (-\infty, 0) \), which includes all negative values of \( x \).

3. **Horizontal Shift ( \( \ln(3-x) \) ):**
   - Shift \( \ln(-x) \) to the right by 3 units. This gives the function \( \ln(3-x) \).
   - The domain now changes to \( (-\infty, 3) \), so the function is defined for all \( x \) less than 3.

4. **Vertical Translation ( \( \ln(3-x) + 2 \) ):**
   - Shift \( \ln(3-x) \) upward by 2 units. This results in the function \( f(x) = \ln(3-x) + 2 \).

### Transformations Summary
- **Reflection:** \( \ln(x) \) → \( \ln(-x) \) (across the \( y \)-axis).
- **Horizontal Shift:** \( \ln(-x) \) → \( \ln(3-x) \) (right by 3 units).
- **Vertical Shift:** \( \ln(3-x) \) → \( \ln(3-x) + 2 \) (up by 2 units).

### Sketching the Graph of \( f(x) = \ln(3-x) + 2 \)

1. **Graph of \( \ln(x) \):**
   - The function \( \ln(x) \) has a vertical asymptote at \( x = 0 \) and increases without bound as \( x \) becomes larger.
   - It passes through \( (1, 0) \) and approaches \( -\infty \) as \( x \) approaches 0 from the right.

2. **Transform to \( \ln(-x) \):**
   - Reflect the graph of \( \ln(x) \) across the \( y \)-axis.
   - Now there is a vertical asymptote at \( x = 0 \), but the function is defined for \( x < 0 \).

3. **Transform to \( \ln(3-x) \):**
   - Shift the graph of \( \ln(-x) \) 3 units to the right.
   - The vertical asymptote moves to \( x = 3 \).
   - The function passes through \( (2, 0) \) and is defined for \( x < 3 \).

4. **Transform to \( \ln(3-x) + 2 \):**
   - Shift the graph of \( \ln(3-x) \) upwards by 2 units.
   - Now the vertical asymptote remains at \( x = 3 \), and the graph passes through \( (2, 2) \).
   - The graph approaches 2 as \( x \) becomes very small (approaching \( -\infty \)).

### Final Graph of \( f(x) = \ln(3-x) + 2 \)
- **Domain:** \( (-\infty, 3) \)
- **Vertical Asymptote:** \( x = 3 \)
- **Horizontal Shift:** Moves right by 3 units.
- **Vertical Shift:** Moves up by 2 units.

**Would you like a detailed sketch of the graph or have any further questions on this topic?**

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Here are 5 related questions for further exploration:

1. What is the range of the function \( f(x) = \ln(3-x) + 2 \)?
2. How would you find the x-intercept of \( f(x) \)?
3. What is the behavior of \( f(x) \) as \( x \) approaches 3 from the left?
4. Can you find the inverse of the function \( f(x) \)?
5. How would you apply transformations if the function were \( f(x) = -\ln(3-x) + 2 \)?

**Tip:** Understanding transformations of the basic logarithmic function is crucial for sketching more complex log functions effectively.

Explore how to transform the basic logarithmic function ln(x) to f(x) = ln(3-x) + 2 using horizontal and vertical shifts. This detailed guide covers logarithmic properties, transformations, and sketching the graph of the function. Suitable for students in Grades 10-12.

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