text: request for equation? i'm trying to produce an equation that will provide the results, below. the first column is "x". the second column, "y", is the result i'm looking to produce with the equation.  so, for example, when x = 23, the equation should produce an answer of 92. when x = 12, the equation should produce a result of 840. when x = 4, the equation should produce 1080... and so on. here is the full list of values (the columns are only separated by a space because yahoo! answers won't let me create columns):  0 1104 1 1104 2 1100 3 1092 4 1080 5 1064 6 1044 7 1020 8 992 9 960 10 924 11 884 12 840 13 792 14 740 15 684 16 624 17 560 18 492 19 420 20 344 21 264 22 180 23 92 24 0  ..can anyone help me with this one?  thanks a whole whole whole bunch!  jamie checking by method of differences  . x ............. f(x) ....... δ1 ...... δ2 . 0 .......... 1104 . 1 .......... 1104 ........ 0 . 2 .......... 1100 ........ 4 ....... -4 . 3 .......... 1092 ........ 8 ....... -4 . 4 .......... 1080 ...... 12 ....... -4 . 5 .......... 1064 ...... 16 ....... -4 . 6 .......... 1044 ...... 20 ....... -4 . 7 .......... 1020 ...... 24 ....... -4 . 8 ............ 992 ...... 28 ....... -4 . 9 ............ 960 ...... 32 ....... -4 10 ............ 924 ...... 36 ....... -4 11 ............ 884 ...... 40 ....... -4 12 ............ 840 ...... 44 ....... -4 13 ............ 792 ...... 48 ....... -4 14 ............ 740 ...... 52 ....... -4 15 ............ 684 ...... 56 ....... -4 16 ............ 624 ...... 60 ....... -4 17 ............ 560 ...... 64 ....... -4 18 ............ 492 ...... 68 ....... -4 19 ............ 420 ...... 72 ....... -4 20 ............ 344 ...... 76 ....... -4 21 ............ 264 ...... 80 ....... -4 22 ............ 180 ...... 84 ....... -4 23 .............. 92 ...... 88 ....... -4 24 ................ 0 ...... 92 ....... -4  so the relation is quadratic (as δ2 is constant which imples the 2nd derivative is constant)  so the relation is of the form  f(x) = ax^2 + bx + c  now f(0) = c = 1104 f(1) = a + b + c = 1104 so a + b + 1104 = 1104 ie a + b = 0  f(2) = 4a + 2b + c = 1100 4a + 2b + 1104 = 1100 so 4a + 2b = -4  ie 2a + 2a + 2b = 2a + 2(a + b) = -4 so 2a + 0 = -4 ie a = -2 and b = 2  thus f(x) = -2x^2 + 2x + 1104 = -2(x^2 - x - 552) = -2(x - 24)(x + 23)  test f(10) = -2*(10 - 24)(10 + 23) = -2 * (-14) * (33) = 28 * 33 = 924 yesssssssssss!!