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Sagot :
Let's analyze the given data set using finite differences to determine the polynomial function that best models the data.
First, we'll list the given values for initial speed (X) in MPH and distance (Y) in feet:
| Initial Speed (X) | 80 | 85 | 90 | 95 | 100 | 105 | 110 | 115 |
|--------------------|-----|-----|-----|-----|-----|-----|-----|-----|
| Distance (Y) | 194 | 220 | 247 | 275 | 304 | 334 | 365 | 397 |
### Step 1: Calculate the first differences
The first differences are the differences between consecutive Y values.
[tex]\[ \Delta Y_1 = Y[i + 1] - Y[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_1(80, 85) &= 220 - 194 = 26 \\ \Delta Y_1(85, 90) &= 247 - 220 = 27 \\ \Delta Y_1(90, 95) &= 275 - 247 = 28 \\ \Delta Y_1(95, 100) &= 304 - 275 = 29 \\ \Delta Y_1(100, 105) &= 334 - 304 = 30 \\ \Delta Y_1(105, 110) &= 365 - 334 = 31 \\ \Delta Y_1(110, 115) &= 397 - 365 = 32 \\ \end{align*} \][/tex]
The first differences are:
[tex]\[ \{26, 27, 28, 29, 30, 31, 32\} \][/tex]
### Step 2: Calculate the second differences
The second differences are the differences between consecutive first differences.
[tex]\[ \Delta Y_2 = \Delta Y_1[i + 1] - \Delta Y_1[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_2(80, 90) &= 27 - 26 = 1 \\ \Delta Y_2(85, 95) &= 28 - 27 = 1 \\ \Delta Y_2(90, 100) &= 29 - 28 = 1 \\ \Delta Y_2(95, 105) &= 30 - 29 = 1 \\ \Delta Y_2(100, 110) &= 31 - 30 = 1 \\ \Delta Y_2(105, 115) &= 32 - 31 = 1 \\ \end{align*} \][/tex]
The second differences are:
[tex]\[ \{1, 1, 1, 1, 1, 1\} \][/tex]
### Step 3: Calculate the third differences
The third differences are the differences between consecutive second differences.
[tex]\[ \Delta Y_3 = \Delta Y_2[i + 1] - \Delta Y_2[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_3(80, 95) &= 1 - 1 = 0 \\ \Delta Y_3(85, 100) &= 1 - 1 = 0 \\ \Delta Y_3(90, 105) &= 1 - 1 = 0 \\ \Delta Y_3(95, 110) &= 1 - 1 = 0 \\ \Delta Y_3(100, 115) &= 1 - 1 = 0 \\ \end{align*} \][/tex]
The third differences are:
[tex]\[ \{0, 0, 0, 0, 0\} \][/tex]
### Conclusion
Since the third differences are all zero and the second differences are constant and non-zero, the second differences being constant indicates that the data can best be modeled by a quadratic function. Therefore, the polynomial function that should be used to model the data set is:
Quadratic
First, we'll list the given values for initial speed (X) in MPH and distance (Y) in feet:
| Initial Speed (X) | 80 | 85 | 90 | 95 | 100 | 105 | 110 | 115 |
|--------------------|-----|-----|-----|-----|-----|-----|-----|-----|
| Distance (Y) | 194 | 220 | 247 | 275 | 304 | 334 | 365 | 397 |
### Step 1: Calculate the first differences
The first differences are the differences between consecutive Y values.
[tex]\[ \Delta Y_1 = Y[i + 1] - Y[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_1(80, 85) &= 220 - 194 = 26 \\ \Delta Y_1(85, 90) &= 247 - 220 = 27 \\ \Delta Y_1(90, 95) &= 275 - 247 = 28 \\ \Delta Y_1(95, 100) &= 304 - 275 = 29 \\ \Delta Y_1(100, 105) &= 334 - 304 = 30 \\ \Delta Y_1(105, 110) &= 365 - 334 = 31 \\ \Delta Y_1(110, 115) &= 397 - 365 = 32 \\ \end{align*} \][/tex]
The first differences are:
[tex]\[ \{26, 27, 28, 29, 30, 31, 32\} \][/tex]
### Step 2: Calculate the second differences
The second differences are the differences between consecutive first differences.
[tex]\[ \Delta Y_2 = \Delta Y_1[i + 1] - \Delta Y_1[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_2(80, 90) &= 27 - 26 = 1 \\ \Delta Y_2(85, 95) &= 28 - 27 = 1 \\ \Delta Y_2(90, 100) &= 29 - 28 = 1 \\ \Delta Y_2(95, 105) &= 30 - 29 = 1 \\ \Delta Y_2(100, 110) &= 31 - 30 = 1 \\ \Delta Y_2(105, 115) &= 32 - 31 = 1 \\ \end{align*} \][/tex]
The second differences are:
[tex]\[ \{1, 1, 1, 1, 1, 1\} \][/tex]
### Step 3: Calculate the third differences
The third differences are the differences between consecutive second differences.
[tex]\[ \Delta Y_3 = \Delta Y_2[i + 1] - \Delta Y_2[i] \][/tex]
Calculating these:
[tex]\[ \begin{align*} \Delta Y_3(80, 95) &= 1 - 1 = 0 \\ \Delta Y_3(85, 100) &= 1 - 1 = 0 \\ \Delta Y_3(90, 105) &= 1 - 1 = 0 \\ \Delta Y_3(95, 110) &= 1 - 1 = 0 \\ \Delta Y_3(100, 115) &= 1 - 1 = 0 \\ \end{align*} \][/tex]
The third differences are:
[tex]\[ \{0, 0, 0, 0, 0\} \][/tex]
### Conclusion
Since the third differences are all zero and the second differences are constant and non-zero, the second differences being constant indicates that the data can best be modeled by a quadratic function. Therefore, the polynomial function that should be used to model the data set is:
Quadratic
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