Question
The value of a certain investment over time is given in the table below. Answer the questions below to determine what kind of function would best fit the data, linear or exponential.
Number of Years Since Investment Made, x
1
1
2
2
3
3
4
4
Value of Investment ($), f(x)
16, comma, 510, point, 3, 4
16,510.34
13, comma, 266, point, 8, 1
13,266.81
10, comma, 778, point, 4, 9
10,778.49
8, comma, 774, point, 2, 7
8,774.27
A linear or An exponential function would best fit the data because as x increases, the y values change Additively or multiplicatively. The Common differences/slope or Common ratio/multiplier/base of this function is approximately___
Number of Years Since Investment Made, x
1
1
2
2
3
3
4
4
Value of Investment ($), f(x)
16, comma, 510, point, 3, 4
16,510.34
13, comma, 266, point, 8, 1
13,266.81
10, comma, 778, point, 4, 9
10,778.49
8, comma, 774, point, 2, 7
8,774.27
A linear or An exponential function would best fit the data because as x increases, the y values change Additively or multiplicatively. The Common differences/slope or Common ratio/multiplier/base of this function is approximately___
Answers
GPT-4o mini
To determine whether a linear or an exponential function fits the given investment data, we can analyze how the values change as \( x \) (the number of years) increases.
### Given Data:
- Years Since Investment Made (x): 1, 2, 3, 4
- Value of Investment ($), f(x):
- f(1) = 16,510.34
- f(2) = 13,266.81
- f(3) = 10,778.49
- f(4) = 8,774.27
### 1. Check Additive Changes for Linear Function:
To check for linearity, we look for a common difference:
- Difference between f(2) and f(1): 13,266.81 - 16,510.34 = -3,243.53
- Difference between f(3) and f(2): 10,778.49 - 13,266.81 = -2,488.32
- Difference between f(4) and f(3): 8,774.27 - 10,778.49 = -2,004.22
The differences are not constant; they are decreasing, which suggests that a linear model is not a good fit.
### 2. Check Multiplicative Changes for Exponential Function:
To check for exponential growth or decay, we calculate the ratios:
- Ratio between f(2) and f(1): \( \frac{13,266.81}{16,510.34} \approx 0.804 \)
- Ratio between f(3) and f(2): \( \frac{10,778.49}{13,266.81} \approx 0.811 \)
- Ratio between f(4) and f(3): \( \frac{8,774.27}{10,778.49} \approx 0.815 \)
While the ratios are not exactly constant, they are relatively close to each other and suggest a multiplicative change.
### Conclusion:
An exponential function would best fit the data because as \( x \) increases, the \( y \) values change multiplicatively.
#### Common Ratio:
The common ratio is approximately around 0.81.
So, to fill in the blank:
"The common ratio/multiplier/base of this function is approximately **0.81**."
### Given Data:
- Years Since Investment Made (x): 1, 2, 3, 4
- Value of Investment ($), f(x):
- f(1) = 16,510.34
- f(2) = 13,266.81
- f(3) = 10,778.49
- f(4) = 8,774.27
### 1. Check Additive Changes for Linear Function:
To check for linearity, we look for a common difference:
- Difference between f(2) and f(1): 13,266.81 - 16,510.34 = -3,243.53
- Difference between f(3) and f(2): 10,778.49 - 13,266.81 = -2,488.32
- Difference between f(4) and f(3): 8,774.27 - 10,778.49 = -2,004.22
The differences are not constant; they are decreasing, which suggests that a linear model is not a good fit.
### 2. Check Multiplicative Changes for Exponential Function:
To check for exponential growth or decay, we calculate the ratios:
- Ratio between f(2) and f(1): \( \frac{13,266.81}{16,510.34} \approx 0.804 \)
- Ratio between f(3) and f(2): \( \frac{10,778.49}{13,266.81} \approx 0.811 \)
- Ratio between f(4) and f(3): \( \frac{8,774.27}{10,778.49} \approx 0.815 \)
While the ratios are not exactly constant, they are relatively close to each other and suggest a multiplicative change.
### Conclusion:
An exponential function would best fit the data because as \( x \) increases, the \( y \) values change multiplicatively.
#### Common Ratio:
The common ratio is approximately around 0.81.
So, to fill in the blank:
"The common ratio/multiplier/base of this function is approximately **0.81**."