Math Problem Statement

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.

Solution

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

  2. Questions in the problem: a) Describe the transformation of the parent function f(x)=ln⁡(x)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<60 < 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=0t = 0 to represent 2011.

Step-by-Step Analysis:

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

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

  • The term −1.2ln⁡(t)-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+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<60 < t < 6.

The correct graph would be one that:

  • Starts at a high value when tt is close to 0.
  • Decreases as tt 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 tt increases.

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

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

1=−1.2ln⁡(t)+31 = -1.2 \ln(t) + 3

Rearrange to solve for ln⁡(t)\ln(t):

-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.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Functions
Function Transformations
Depreciation Modeling
Natural Logarithms

Formulas

D(t) = -1.2ln(t) + 3
ln(t) = -b/a
t = e^(result)

Theorems

Properties of Logarithms
Graph Transformations (Vertical Stretch, Reflection, Vertical Shift)

Suitable Grade Level

Grades 10-12

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