SOLUTION: I am in need of assistance in creating an equation that describes the chart as follows X Y 0 0 1 .5 2 2.0 3

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Question 189964: I am in need of assistance in creating an equation that describes the chart as follows
X Y
0 0
1 .5
2 2.0
3 4.5
4 8
5 12.5
I realize the progression of .5,1.5,2.5,4,4.5. I can't figure out the equatio

Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Looks like y = (x^2)/2

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Edwin's solution:
Here's how to get it:

First list the numbers like this skipping 
a line between them:


X    Y
0   0.0

1   0.5

2   2.0

3   4.5

4   8.0

5  12.5

Next add a column labeled "1st differences" and
on each line that was skipped write the difference
between the two y-values the skipped line is
in between.  

That is, we subtract the 1st value of
x which is 0.0 from the 2nd value of x which is 0.5
and write that in the line between the first two 
values.

we subtract the 2nd value of
x which is 0.5 from the 3rd value of x which is 2.0,
and that gives us 1.5 which we write in the line 
between the 2nd and 3rd values.  And so forth:


          1st 
X    Y   diff.   
0   0.0
          0.5
1   0.5
          1.5
2   2.0
          2.5
3   4.5
          3.5
4   8.0
          4.5
5  12.5

Next we add another column and label it "2nd
differences", subtracting each value in the
1st difference column from the one just before
it and writing it in that column between the 
two values subtracted, like this:

            1st     2nd
X    Y     diffs.  diffs.  
0   0.0
            0.5
1   0.5             1.0
            1.5
2   2.0             1.0 
            2.5
3   4.5             1.0 
            3.5
4   8.0             1.0
            4.5
5  12.5

Now we do not need to make a column for the
3rd differences, because all the 2nd differences
are the same.

Since the second differences came out all the
same number, we know to assume the equation is a
second degree polynomial,

    y = Ax˛ + Bx + C

That has 3 values A,B, and C which we must find,
so we substitute 3 of the six given points.

First we substitute x = 0, y = 0, and get

    0 = A(0)˛ + B(0) + C
    0 = 0 + 0 + C
    0 = C

Now we substitute x = 1, y = 0.5, and get

  0.5 = A(1)˛ + B(1) + C
  0.5 = A + B + C
  
Now we substitute x = 2, y = 2.0, and get

  2.0 = A(2)˛ + B(2) + C
  2.0 = 4A + 2B + C
      
So we have three equations in three unknowns:

system%280=C%2C0.5=A%2BB%2BC%2C2.0=4A%2B2B%2BC%29

or

system%28C=0%2CA%2BB%2BC=0.5%2C4A%2B2B%2BC=2.0%29

Do you know how to solve that for A, B and C?
If not post again asking how.  The answers here
are 

A=0.5, B=0, C=0

So we substitute those values in

y+=+Ax%5E2+%2B+Bx+%2B+C

and get

y+=+0.5x%5E2+%2B+0x+%2B+0

or

y=+0.5x%5E2

That's the equation you were looking for. 

Now check by substituting x=0,1,2,3,4,5

and you will get the corresponding y-values. 

Edwin