Question
The amount of money in a savings account over time is given in the table below. Answer the questions below to explain what kind of function would better model the data, linear or exponential.
Number of Years Since Account Opened, x 11 22 33 44
Money in account ($), f(x) 127, point, 7, 9127.79 149, point, 9, 6149.96 175, point, 1, 0175.10 204, point, 4, 6204.46
(a linear or an exponential) function would better model the data because as x increases, the y values change (additively or multiplicatively) The (common difference/slope or common ratio/multiplier/base) of this function is approximately (?)
Number of Years Since Account Opened, x 11 22 33 44
Money in account ($), f(x) 127, point, 7, 9127.79 149, point, 9, 6149.96 175, point, 1, 0175.10 204, point, 4, 6204.46
(a linear or an exponential) function would better model the data because as x increases, the y values change (additively or multiplicatively) The (common difference/slope or common ratio/multiplier/base) of this function is approximately (?)
Answers
GPT-4o mini
To determine whether a linear or exponential function better models the data, we need to analyze how the money in the account changes with the number of years since the account was opened.
Linear functions have a constant rate of change (constant difference), while exponential functions have a variable rate of change (constant ratio or multiplier) that increases or decreases multiplicatively.
To examine the data, let's look at the values provided for \( f(x) \) (the money in the account) at each \( x \):
1. \( f(11) = 127.79 \)
2. \( f(22) = 149.96 \)
3. \( f(33) = 175.10 \)
4. \( f(44) = 204.46 \)
Now, let's calculate the differences between successive values to see if we have a constant change:
- \( f(22) - f(11) = 149.96 - 127.79 = 22.17 \)
- \( f(33) - f(22) = 175.10 - 149.96 = 25.14 \)
- \( f(44) - f(33) = 204.46 - 175.10 = 29.36 \)
The differences between successive \( f(x) \) values are not constant (22.17, 25.14, 29.36), indicating that the function is not linear.
Next, let's look for a constant ratio by calculating the ratios of successive values:
- \( \frac{f(22)}{f(11)} = \frac{149.96}{127.79} \approx 1.17 \)
- \( \frac{f(33)}{f(22)} = \frac{175.10}{149.96} \approx 1.17 \)
- \( \frac{f(44)}{f(33)} = \frac{204.46}{175.10} \approx 1.17 \)
The ratios are approximately constant (all around 1.17), which is a characteristic of an exponential function.
Thus, we conclude that an exponential function would better model the data because as \( x \) increases, the \( y \) values change multiplicatively. The **common ratio/multiplier** of this function is approximately **1.17**.
Linear functions have a constant rate of change (constant difference), while exponential functions have a variable rate of change (constant ratio or multiplier) that increases or decreases multiplicatively.
To examine the data, let's look at the values provided for \( f(x) \) (the money in the account) at each \( x \):
1. \( f(11) = 127.79 \)
2. \( f(22) = 149.96 \)
3. \( f(33) = 175.10 \)
4. \( f(44) = 204.46 \)
Now, let's calculate the differences between successive values to see if we have a constant change:
- \( f(22) - f(11) = 149.96 - 127.79 = 22.17 \)
- \( f(33) - f(22) = 175.10 - 149.96 = 25.14 \)
- \( f(44) - f(33) = 204.46 - 175.10 = 29.36 \)
The differences between successive \( f(x) \) values are not constant (22.17, 25.14, 29.36), indicating that the function is not linear.
Next, let's look for a constant ratio by calculating the ratios of successive values:
- \( \frac{f(22)}{f(11)} = \frac{149.96}{127.79} \approx 1.17 \)
- \( \frac{f(33)}{f(22)} = \frac{175.10}{149.96} \approx 1.17 \)
- \( \frac{f(44)}{f(33)} = \frac{204.46}{175.10} \approx 1.17 \)
The ratios are approximately constant (all around 1.17), which is a characteristic of an exponential function.
Thus, we conclude that an exponential function would better model the data because as \( x \) increases, the \( y \) values change multiplicatively. The **common ratio/multiplier** of this function is approximately **1.17**.