Question
Which function best models the dataset in the table?
x
−5
−2
−1
0 1 2 5 8 12
y
6.2 8.3 9.1 10 11 12.1 16.1 21.4 31.4
Option #1: f(x)=1.467x+10.696
Option #2: f(x)=10.004(1.100)x
Option #3: f(x)=0.066x2+0.981x+9.810
x
−5
−2
−1
0 1 2 5 8 12
y
6.2 8.3 9.1 10 11 12.1 16.1 21.4 31.4
Option #1: f(x)=1.467x+10.696
Option #2: f(x)=10.004(1.100)x
Option #3: f(x)=0.066x2+0.981x+9.810
Answers
GPT-4o mini
To determine which function best models the given dataset, we need to analyze the data and the provided options.
The dataset consists of specific \( x \) and \( y \) values:
| \( x \) | \( y \) |
|----------|----------|
| -5 | 6.2 |
| -2 | 8.3 |
| -1 | 9.1 |
| 0 | 10 |
| 1 | 11 |
| 2 | 12.1 |
| 5 | 16.1 |
| 8 | 21.4 |
| 12 | 31.4 |
### Analyzing the Options
1. **Option #1: \( f(x) = 1.467x + 10.696 \)**
- This is a linear function. We can immediately see that while the \( y \) values do increase, the rate of increase seems more than linear as \( x \) increases.
2. **Option #2: \( f(x) = 10.004(1.100)^x \)**
- This is an exponential function. The form \( a(1 + r)^x \) implies that \( y \) will increase rapidly with increasing \( x \). We can compare its values to the \( y \) values in the dataset.
3. **Option #3: \( f(x) = 0.066x^2 + 0.981x + 9.810 \)**
- This is a quadratic function. Quadratics can sometimes model datasets effectively, especially if there is a change in acceleration in the \( y \) values.
### Evaluation of Each Function
It would be ideal to calculate the expected \( y \) values for each choice at the given \( x \) values and then see how closely they match the actual \( y \) data. However, typically, the best approach is to look for the type of growth observed in the data.
#### Observations:
- The \( y \) values are not increasing linearly; they appear to accelerate as \( x \) increases, suggesting a polynomial or exponential nature.
- The quadratic function has a potential inflection point and may be able to accommodate both linear and accelerating growth.
### Best Decision
Without the exact calculations of each function for every data point, intuitively, we can suggest:
Given the characteristic shapes:
- The linear function \( (f(x) = 1.467x + 10.696) \) is likely too simple for this dataset.
- The exponential function \( f(x) = 10.004(1.100)^x \) could fit for rapidly increasing datasets, and should be checked if it tracks well as \( x \) increases significantly.
- The quadratic function \( (f(x) = 0.066x^2 + 0.981x + 9.810) \) can accommodate non-linear increases but may also introduce unnecessary curvature that could misfit lower points.
### Conclusion
Based on this consideration, **Option #2 (the exponential function)** seems likely to model the dataset best, given the rapid increase in the \( y \) values as \( x \) increases. A rigorous analysis comparing predicted \( y \) values to observed \( y \) values for each function would provide a firmer conclusion, but Option #2 aligns well with the observable trends based on description.
The dataset consists of specific \( x \) and \( y \) values:
| \( x \) | \( y \) |
|----------|----------|
| -5 | 6.2 |
| -2 | 8.3 |
| -1 | 9.1 |
| 0 | 10 |
| 1 | 11 |
| 2 | 12.1 |
| 5 | 16.1 |
| 8 | 21.4 |
| 12 | 31.4 |
### Analyzing the Options
1. **Option #1: \( f(x) = 1.467x + 10.696 \)**
- This is a linear function. We can immediately see that while the \( y \) values do increase, the rate of increase seems more than linear as \( x \) increases.
2. **Option #2: \( f(x) = 10.004(1.100)^x \)**
- This is an exponential function. The form \( a(1 + r)^x \) implies that \( y \) will increase rapidly with increasing \( x \). We can compare its values to the \( y \) values in the dataset.
3. **Option #3: \( f(x) = 0.066x^2 + 0.981x + 9.810 \)**
- This is a quadratic function. Quadratics can sometimes model datasets effectively, especially if there is a change in acceleration in the \( y \) values.
### Evaluation of Each Function
It would be ideal to calculate the expected \( y \) values for each choice at the given \( x \) values and then see how closely they match the actual \( y \) data. However, typically, the best approach is to look for the type of growth observed in the data.
#### Observations:
- The \( y \) values are not increasing linearly; they appear to accelerate as \( x \) increases, suggesting a polynomial or exponential nature.
- The quadratic function has a potential inflection point and may be able to accommodate both linear and accelerating growth.
### Best Decision
Without the exact calculations of each function for every data point, intuitively, we can suggest:
Given the characteristic shapes:
- The linear function \( (f(x) = 1.467x + 10.696) \) is likely too simple for this dataset.
- The exponential function \( f(x) = 10.004(1.100)^x \) could fit for rapidly increasing datasets, and should be checked if it tracks well as \( x \) increases significantly.
- The quadratic function \( (f(x) = 0.066x^2 + 0.981x + 9.810) \) can accommodate non-linear increases but may also introduce unnecessary curvature that could misfit lower points.
### Conclusion
Based on this consideration, **Option #2 (the exponential function)** seems likely to model the dataset best, given the rapid increase in the \( y \) values as \( x \) increases. A rigorous analysis comparing predicted \( y \) values to observed \( y \) values for each function would provide a firmer conclusion, but Option #2 aligns well with the observable trends based on description.