| Module 3 - Answers |
| Lesson 1 |
| Answer 1 |
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3.1.1
The graph of
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| Answer 2 |
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3.1.2
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| Lesson 2 |
| Answer 1 |
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3.2.1
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| Answer 2 |
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3.2.2
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| Answer 3 |
3.2.3
The graph of
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| Lesson 3 |
| Answer 1 |
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3.3.1
If
If
If a < 0, the transformed graph will be a reflection across the x-axis. |
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| Answer 2 |
| 3.3.2 According to the table the investment will have grown to approximately $1418.52 in six years. |
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| Answer 3 |
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3.3.3
The investment will double in approximately 12 years, which agrees with the solution found from the table of values. |
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| Answer 4 |
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3.3.4
About 3.15 grams remain after 50 days. |
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| Answer 5 |
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3.3.5
There will be 1 gram remaining after about 100 days. |
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| Answer 6 |
| 3.3.6 Y2 = log(2x) appears to be a vertical shift of Y1 = log(x). |
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| Answer 7 |
| 3.3.7 The third graph suggests that the difference in the first two functions is a constant. In other words: Y2 Y1 = C so Y2 = Y1 + C. This last equation supports the conjecture that Y2 is a vertical shift of Y1. |
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| Answer 8 |
| 3.3.8 For positive values of a and x, log(ax) = log(x) + log(a). |
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| Lesson 4 |
| Answer 1 |
3.4.1
The x-value that yields an output close to 30 is x = 40. So the transformed regression equation predicts that the water will cool to approximately 30°C in a bit less than 40 minutes, which corresponds well with the data. |
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| Self Test |
| Answer 1 |
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Any ordering of these 4 transformations will result in the correct graph as long as the vertical stretch and reflection are done before the vertical shift. ReflectVertical stretch by factor of 2 Shift left 1 Shift down 3
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| Answer 2 |
| The investment will double in about 10.2 years |
| Answer 3 |
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x
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| Answer 4 |
| Y2 is a vertical shift of Y1 downward and the shift is equal to log(2). |
| Answer 5 |
| y = 64(0.93)x + 23 |
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