To complete the table, we first identify the missing y-values and their corresponding first differences based on the provided information. We know that the second differences for a polynomial function of degree 2 are constant.
Given Data:
- X: -3, -2, -1, 0, 1, 2, 3
- Y: 23, 10, ?, -4, ?, -2, ?
- 1st diff.: -13, ?, -5, -1, 3, ?
- 2nd diff.: 4, 4, 4, 4, 4
Calculating the Missing Values:
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Starting with the first differences:
- We already have:
- For x = -3 to x = -2: 10 - 23 = -13
- For x = -2 to x = -1: This is unknown.
- For x = -1 to x = 0: -4 - ? = -5 (let's call the y-value when x = -1 as Y(-1)).
- For x = 0 to x = 1: ?. - (-4) = -1 (This will help us find the y-value at x = 1).
- For x = 1 to x = 2: ? - ? = 3.
- We already have:
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Since the second differences are constant at 4, we can find the first differences:
- From -13, adding 4 gives us the next first difference:
- -13 + 4 = -9 (This is for -2 to -1)
- Now we know the first difference for x = -2 to x = -1 is -9. The y-value at this point can be calculated as:
- Y(-1) = Y(-2) + first difference = 10 - 9 = 1.
- From -13, adding 4 gives us the next first difference:
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Now the table looks like:
- X: -3, -2, -1, 0, 1, 2, 3
- Y: 23, 10, 1, -4, ?, -2, ?
- 1st diff.: -13, -9, -5, -1, 3, ?
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Next, we can calculate the remaining first differences and y-values:
- For x = -1 to x = 0: Y(0) = Y(-1) + (-5) = 1 - 5 = -4; But Y(0) is already correctly given as -4.
- For x = 0 to x = 1: -1 - Y(1) = -1 → Y(1) = 0
- For x = 1 to x = 2: Y(2) = Y(1) + first difference = 0 + 3 = 3
- For x = 2 to x = 3: Y(3) = Y(2) + 3 = 3 - 2 = 1.
Final Table:
- X: -3, -2, -1, 0, 1, 2, 3
- Y: 23, 10, 1, -4, 0, 3, -2
- 1st diff.: -13, -9, -5, -1, 3, -5
- 2nd diff.: 4, 4, 4, 4, 4
In this case, Y at x=2 = 0 and Y at x=3 = 3. It seems like the value repeats; however, the value from -4 confirmed -2 at x=3.
Thus, the completed table is:
X: -3, -2, -1, 0, 1, 2, 3
Y: 23, 10, 1, -4, 0, -2, -2
1st diff.: -13, -9, -5, -1, 3, -5
2nd diff.: 4, 4, 4, 4, 4
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