Request for Equation?

Question

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

Answer 1

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!!

Answer 2

If each pair is an exact solution to the quadratic (rather than scatter values from measurement), then any three should be enough to solve for the constants in the quadratic equation:

a*x^2 + b^x + c

Of particular value are the pairs (0,1104) and (1,1104). From the first pair, a*0 + b*0 +c = 1104, therefor c = 1104.

1104 = a + b + 1104; this means a + b = 0, and a = - b;

Then the equation is a^x^2 - a*x + 1104 = y

Now use the pair (0,24) to get

a*576 - a*24 + 1104 = 0

552*a = -1104, a = -2

since b = -a, b = 2 and the equation is

y = -2^x^2 + 2*x + 1104

You can use any other pair to check the result.

Answer 3

y = -2x^2 + 2x + 1104,
what means
y = -2xx + 2x + 1104