Question 1063017
Four points could go well for degree-three polynomial function.
Unknown constant coefficients, a, b, c, d;
{{{ax^3+bx^2+cx+d=y}}}


Use the given points to form four specific degree-three equations for a system.
{{{system((-2)^3*a+(-2)^2*b+(-2)c+d=10,(-1)^3*a+(-1)^2*b+(-1)c+d=-4,0^3*a+0^2*b+0*c+d=-2,1^3*a+1^2*b+1*c+d=-2)}}}




{{{system(-8a+4b-2c+d=10,-a+b-c+d=-4,d=-2,a+b+c+d=-2)}}}
This is the system to solve.



Substituting for d makes for a simpler three-variable system
{{{system(-8a+4b-2c=12,-a+b-c=-2,a+b+c=0)}}}


(-1/2)*E1 gives
{{{system(4a-2b+c=-6,a-b+c=2,a+b+c=0)}}}


E1-E3 and E2-E3 give
{{{system(3a-3b=-6,-2b=2)}}}


{{{system(a-b=-2,b=-1)}}}


This very simple system gives two of the variables.
{{{highlight(system(b=-1,a=-3))}}}


d was already found early, and with a, b, d, all now found, c can be evaluated, whichever equation you want.


{{{highlight(system(a=-3,b=-1,c=4,d=-2))}}}