There may be more than one answer.
Here's one, assuming that you're looking for a formula that takes inputs 1, 2, 3 and produces outputs 6, 9, 13.
Let's assume that y=ax2+bx+c where y is the output and x the input, and a, b and c are constant coefficients.
So we get:
(1) For x=1, y=6: a+b+c=6
(2) For x=2, y=9: 4a+2b+c=9
(3) For x=3, y=13: 9a+3b=13
(4) Subtract eqn (1) from (2): 3a+b=3
(5) Subtract eqn (2) from (3): 5a+b=4
Subtract (4) from (5): 2a=1, so a=½. Since 3a+b=3, 3/2+b=3, so b=3-3/2=3/2.
From (1) a+b+c=6, 1/2+3/2+c=6, 2+c=6, c=6-2=4.
Therefore y=x2/2+3x/2+4, or y=½(x2+3x+8).