Since the third row is constant, we know that the function h is quadratic. Say h(x) = ax^2 + bx + c. Using h(0) = 2 = a(0)^2 + b(0) + c, h(1) = 5 = a(1)^2 + b(1) + c, and h(2) = 14 = a(2)^2 + b(2) + c, we find a = 3, b = 0, and c = 2, so h(x) = 3x^2 + 2.same as computing a bunch of slopes only the denomintor is always so just the difference in the numerators
−10+29=19
−3+10=7
−2+3=1
−1+2=1
6+1=7
25−6=19
the recessive differences are quadratic, to the original function is "other" namely cubic
y=x^3−2
will work if you try the pairs
