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1. **Description of the problem:**
   The depreciation \( D(t) \) in million dollars of a company's assets from 2009 through 2013 can be approximated by the function:
   \[
   D(t) = -1.2 \ln(t) + 3
   \]
   where \( t \) represents the year since 2009 (i.e., \( t = 0 \) corresponds to 2009, \( t = 1 \) to 2010, etc.).

2. **Questions in the problem:**
   a) Describe the transformation of the parent function \( f(x) = \ln(x) \). Options include:
   - Translate left or right.
   - Vertical stretch or shrink.
   - Reflection.
   - Vertical shift.

   b) Use a graphing utility to graph the model over the interval \( 0 < t < 6 \). There are four graphs shown.

   c) Determine the year when the depreciation will be approximately 1 million dollars.

   d) Rewrite the function to set \( t = 0 \) to represent 2011.

### Step-by-Step Analysis:

**a) Describe the transformation of the parent function \( f(x) = \ln(x) \).**

The given function is:
\[
D(t) = -1.2 \ln(t) + 3
\]

- The term \(-1.2 \ln(t)\) indicates a vertical reflection across the x-axis and a vertical stretch by a factor of 1.2.
- The constant \(+3\) represents a vertical shift upward by 3 units.

So, the correct transformation includes:
- A vertical reflection.
- A vertical stretch by a factor of 1.2.
- A vertical shift up 3 units.

**b) Use a graphing utility to graph the model over the interval \(0 < t < 6\).**

The correct graph would be one that:
- Starts at a high value when \( t \) is close to 0.
- Decreases as \( t \) increases, since the function has a negative coefficient in front of the logarithm, indicating a downward trend.

By observing the graphs in the image, the correct graph will have these characteristics:
- A curve that starts at around 3 (because of the +3 vertical shift) and decreases as \( t \) increases.

**c) Determine the year when the depreciation will be approximately 1 million dollars.**

We need to solve for \( t \) when \( D(t) = 1 \):

\[
1 = -1.2 \ln(t) + 3
\]

Rearrange to solve for \( \ln(t) \):

\[
-1.2 \ln(t) = 1 - 3 \\
-1.2 \ln(t) = -2 \\
\ln(t) = \frac{-2}{-1.2} \\
\ln(t) = \frac{5}{3}
\]

Now, solve for \( t \):

\[
t = e^{5/3}
\]

Calculate \( t \):

\[
t \approx 4.48
\]

Since \( t = 0 \) corresponds to 2009, \( t = 4.48 \) corresponds to the year \( 2009 + 4.48 \approx 2013 \).

Thus, in approximately 2013, the depreciation will be around 1 million dollars.

**d) Rewrite the function to set \( t = 0 \) to represent 2011.**

If \( t = 0 \) represents 2011, then:
- 2009 corresponds to \( t = -2 \).
  
So, the new function \( D(t') \) where \( t' = t - 2 \) will be:

\[
D(t') = -1.2 \ln(t' + 2) + 3
\]

Would you like any specific details or clarifications on any part of this solution?

### Related Questions:

1. How do you determine vertical and horizontal transformations for logarithmic functions?
2. What is the effect of a vertical reflection on the graph of a logarithmic function?
3. How can you rewrite logarithmic functions to change their starting point or domain?
4. What is the importance of understanding function transformations in real-world contexts?
5. How do you use logarithmic equations to model exponential decay or growth?

**Tip:** Always remember that logarithmic functions are undefined for non-positive values of their argument. Make sure to adjust the domain accordingly when applying transformations.

The depreciation in millions of dollars of a company's assets from 2009 through 2013 can be approximated by the function D(t) = -1.2ln(t) + 3. Answer various questions related to function transformations, graphing, and calculating specific depreciation amounts.

Analyze a company's asset depreciation using the logarithmic function D(t) = -1.2ln(t) + 3 over the years 2009 to 2013. Solve for transformations, graph the function, and determine when depreciation reaches 1 million dollars. Suitable for Grades 10-12 students.

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