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Question
I need to find an equation that will work for x and y with x = 1,2,3,4,5,6 and y = 8, 15, 24, 35, 48, 63. I figured out that when x goes up one, y goes up 7, then 7 + 2, then 7+4 (add two more each time). But I can't figure out how to make an equation that works. Please help!
There is something called difference tables where
the 1st column is x and the 2nd column is y.
Explantation
------------
The next column is the differences in the y's and has one less element in it. The elements in the rows are (x1,y1), (x2,y2),
(x3,y3), (x4,y4), (x5,y5), and (x6,y6).
To compute the next column, it is (y(n)-y(n-1))/(x(n)-x(n)-1).
Call that column A and note that there is no A1.
It starts at row 2. See,
A2 = (y2-y1)/(x2-x1),
A3 = (y3-y2)/(x3-x2), etc.
To compute the next column, it again skips the 1st element.
I'll call it B. B3 = (A3-A2)/(X3-X1), b4 = (A4-A3)/(X4-X2).
Note the difference on the bottom is between the x in that row
and the x two rows before.
Using this, the first difference column is (15-8)/(2-1) = 7,
(24-15)/(3-2) = 9, (35-24)/(4-3) = 11, etc.
In this ways, it is A2 = 7, A3 = 9, A4 = 11, A5 = 13, and A6 = 15.
To do the next column, we get (9-7)/(3-1) = 2/2 = 1,
(11-9)/(4-2) = 2/2 = 1, etc. In this way, this column is all full of 1's.
Table
-----
1 8
2 15 7
3 24 9 1
4 35 11 1
5 48 13 1
6 63 15 1
Result
------
To get the result, take the top of each column, 8, 7, and 1.
Now the 1 needs to be multiplied by (x-2).
Once this is done, add 7, then multiply by (x-1).
To finish it off add 8.
This gives us (1(x-2)+7)(x-1)+8.
Since anything times 1 is itelf, this is (x-2+7)(x-1)+8.
That becomes (x+5)(x-1)+8.
Multiplying it out gives x^2 + 4x - 5 + 8.
Since -5 + 8 = 3, that is x^2 + 4x + 3.
Check the Answer
----------------
Putting in 1 gives 1^2 + 4*1 + 3 = 1 + 4 + 3 = 8, and that's right.
Putting in 2 gives 2^2 + 4*2 + 3 = 4 + 8 + 3 = 15, which is right.
Putting in 3 gives 3^2 + 4*3 + 3 = 9 + 12 + 3 = 24, which is right.
Putting in 4 gives 16 + 16 + 3 = 35, which is right.
I'll let you check 5 and 6 in x^2 + 4x + 3.
By the way, x^2 + 4x + 3 is the same as (x+4)x + 3.
In this way, putting in 1 gives 5*1 + 3 = 8.
Putting in 5 is (5+4)5 + 3 = 48, and that's right.
I find writing x^2 + 4x + 3 as (x+4)x + 3 makes it easier to compute,
for you don't have to know the squares.
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Scott A Wilson
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