SOLUTION: Write the equation of the polynomial function that passes through these points: (1,12), (2,-10), (3,-18), (4,0), (5,56), (6,162).

To find the degree of the polynomial required, we 
make a difference table.  To do that we begin by
making a list of the y-values in order of successive
values of x

x   y
1  12
2 -10
3 -18
4   0
5  56
6 162


Then out to the right of that we make a list by
subtracting each number in the y-list from the
number just under it, except the last one.  That
is we subtract 
(-10)-(12) = -22, then 
(-18)-(-10) = -18+10 = -8, etc:

1  12  -22    
2 -10   -8  
3 -18   18  
4   0   56  
5  56  106  
6 162 
 
They are not all the same number.  So,
out to the right of that we do the same
thing again.  That is, we subtract 
(-8)-(-22) = -8+22 = 14, then 
(18)-(-8) = 18+8 = 26, etc:

1  12  -22  14    
2 -10   -8  26  
3 -18   18  38  
4   0   56  50
5  56  106  
6 162 

They are not all the same number.  So,
we do the same thing again.  That is, 
we subtract 
(26)-(14) = 12, then 
(38)-(26) = 12, then
(50)-(38) = 12 and we have

1  12  -22  14  12  
2 -10   -8  26  12
3 -18   18  38  12
4   0   56  50
5  56  106  
6 162 

Now those are all the same number, 12.

It took 3 columns to get to a column that had 
all the same numbers in it.

That tells us we must use a 3rd degree polynomial
equation, So we assume a 3rd degree polynomial equation:

y = Ax³ + Bx² + Cx + D

We substitute the first 4 points since
we have 4 unknowns A,B,C, and D

y = Ax³ + Bx² + Cx + D

Substituting (1,12):

12 = A(1)³ + B(1)² + C(1) + D
12 = A + B + C + D
A + B + C + D = 12

Substituting (2,-10):

-10 = A(2)³ + B(2)² + C(2) + D
-10 = 8A + 4B + 2C + D
8A + 4B + 2C + D = -10

Substituting (3,-18):

-18 = A(3)³ + B(3)² + C(3) + D
-18 = 27A + 9B + 3C + D
27A + 9B + 3C + D = -18

Substituting (4,0):

0 = A(4)³ + B(4)² + C(4) + D
0 = 64A + 16B + 4C + D
64A + 16B + 4C + D = 0

So we have the system of 4 equations in 4 unknowns:

(1)    A +   B +  C + D =  12
(2)   8A +  4B + 2C + D = -10
(3)  27A +  9B + 3C + D = -18
(4)  64A + 16B + 4C + D =   0

Subtract (1) from (2), (2) from (3) and (3) from (4)
to eliminate D, and have 3 equations in 3 unknowns:

(5)  7A + 3B + C = -22
(6) 19A + 5B + C =  -8
(7) 37A + 7B + C =  18

Subtract (5) from (6) and (6) from (7)
to eliminate C, and have 2 equations in 2 unknowns:

(8) 12A + 2B = 14
(9) 18A + 2B = 26

Subtract (8) from (9)
to eliminate B, and have 1 equation in 1 unknown:

(10) 6A = 12
      A = 2

Substitute A = 2 in (8)

(8) 12(2) + 2B = 14
       24 + 2B = 14
            2B = -10
             B = -5

Substitute A = 2 and B = -5 in (5)

(5)  7(2) + 3(-5) + C = -22
         14 - 15 + C = -22
              -1 + C = -22
                   C = -21 
 
Substitute A = 2 and B = -5 and C = -21 in (1)

(1)    (2) + (-5) + (-21) + D = 12
               2 - 5 - 21 + D = 12
                      -24 + D = 12
                            D = 36

So A = 2, B = -5, C = -21, D = 36

The desired polynomial

y = Ax³ + Bx² + Cx + D

becomes:

y = 2x³ - 5x² - 21x + 36

The graph is 



Edwin