There are 4 points, so assume a general polynomial of
degree which is one less, or a polynomial of degree 3.

Substitute each point (x,y) in Ax³ + Bx² + Cx + D = y:

A(-1)³ + B(-1)² + C(-1) + D = -9
 A(0)³ +  B(0)² +  C(0) + D = -2
 A(1)³ +  B(1)² +  C(1) + D =  1
 A(2)³ +  B(2)² +  C(2) + D = 12

Simplify and solve that system of 4 equations.

The second equation gives us D = -2, then the 
system simplifies to:

-A +  B -  C = -7
 A +  B +  C =  3
8A + 4B + 2C = 14

Adding the first two equations gives

     2B      = -4
      B      = -2

Substituting B = -2 in the 2nd and 3rd equations:

 A +  (-2) +  C =  3
8A + 4(-2) + 2C = 14

which simplifies to

 A +  C =  5
8A + 2C = 22

Multiplying the 1st by -2

-2A - 2C = -10
 8A + 2C =  22
--------------
 6A      =  12
  A      =   2

Substitute in 

 A +  C =  5
 2 +  C =  5
      C =  3   

So A=2, B=-2, C=3, D=-2 and

Ax³ + Bx² + Cx + D = y  becomes
2x³ - 2x² + 3x - 2 = y

or use P(x) for y and write it on the left:

P(x) = 2x³ - 2x² + 3x - 2 = y

Here's the graph.  Notice that it passes through
all four given points: 
 


Edwin