Question
The value of a certain investment 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 Investment Made, x
1
1
2
2
3
3
4
4
Value of Investment ($), f(x)
23, comma, 752, point, 9, 5
23,752.95
28, comma, 225, point, 6, 6
28,225.66
33, comma, 372, point, 7, 6
33,372.76
39, comma, 301, point, 4, 2
39,301.42
A linear or An exponential function would better model the data because as x increases, the y values change Additively or multiplicatively. The Common differences/slope or Common ratio/multiplier/base of this function is approximately ___
Number of Years Since Investment Made, x
1
1
2
2
3
3
4
4
Value of Investment ($), f(x)
23, comma, 752, point, 9, 5
23,752.95
28, comma, 225, point, 6, 6
28,225.66
33, comma, 372, point, 7, 6
33,372.76
39, comma, 301, point, 4, 2
39,301.42
A linear or An exponential function would better model 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 the investment value data is better modeled by a linear function or an exponential function, we need to look at how the values change as the number of years increases.
From the data provided:
- At year 1: $23,752.95
- At year 2: $28,225.66
- At year 3: $33,372.76
- At year 4: $39,301.42
### Step 1: Check for Linear Changes
For a linear function, the change in value (the difference between successive years) should remain constant.
Calculating the differences:
- From year 1 to year 2: $28,225.66 - $23,752.95 = $4,472.71
- From year 2 to year 3: $33,372.76 - $28,225.66 = $5,147.10
- From year 3 to year 4: $39,301.42 - $33,372.76 = $5,928.66
The differences are not constant, indicating that the data is not linear.
### Step 2: Check for Exponential Changes
For an exponential function, the ratio (the value in each year divided by the value in the previous year) should remain relatively constant.
Calculating the ratios:
- From year 1 to year 2: $28,225.66 / $23,752.95 ≈ 1.187
- From year 2 to year 3: $33,372.76 / $28,225.66 ≈ 1.182
- From year 3 to year 4: $39,301.42 / $33,372.76 ≈ 1.177
The ratios are relatively consistent, which suggests an exponential relationship.
### Conclusion
An exponential function would better model the data because as \( x \) (the number of years) increases, the investment value changes **multiplicatively** rather than additively.
The common ratio (multiplier) of this function is approximately:
\[
\text{Approximately } 1.18
\]
Thus, we can conclude:
An exponential function would better model the data because as \( x \) increases, the \( y \) values change multiplicatively. The common ratio/multiplier/base of this function is approximately **1.18**.
From the data provided:
- At year 1: $23,752.95
- At year 2: $28,225.66
- At year 3: $33,372.76
- At year 4: $39,301.42
### Step 1: Check for Linear Changes
For a linear function, the change in value (the difference between successive years) should remain constant.
Calculating the differences:
- From year 1 to year 2: $28,225.66 - $23,752.95 = $4,472.71
- From year 2 to year 3: $33,372.76 - $28,225.66 = $5,147.10
- From year 3 to year 4: $39,301.42 - $33,372.76 = $5,928.66
The differences are not constant, indicating that the data is not linear.
### Step 2: Check for Exponential Changes
For an exponential function, the ratio (the value in each year divided by the value in the previous year) should remain relatively constant.
Calculating the ratios:
- From year 1 to year 2: $28,225.66 / $23,752.95 ≈ 1.187
- From year 2 to year 3: $33,372.76 / $28,225.66 ≈ 1.182
- From year 3 to year 4: $39,301.42 / $33,372.76 ≈ 1.177
The ratios are relatively consistent, which suggests an exponential relationship.
### Conclusion
An exponential function would better model the data because as \( x \) (the number of years) increases, the investment value changes **multiplicatively** rather than additively.
The common ratio (multiplier) of this function is approximately:
\[
\text{Approximately } 1.18
\]
Thus, we can conclude:
An exponential function would better model the data because as \( x \) increases, the \( y \) values change multiplicatively. The common ratio/multiplier/base of this function is approximately **1.18**.