With the given set of points, determine the equation of the polynomial function using the algebraic method.
(1,1), (2,10), (3,42), (4,111), (5,231), (6,416)
x y 1st 2nd 3rd
1 1 - - -
2 10 9 - -
3 42 32 23 -
4 111 69 37 14
5 231 120 51 14
6 416 185 65 14
1 1=a+b+c+d
2 10=8a+4b+2c+d
3 42=27a+9b=3c+d
4 111=64a+16b+4c+d
2-1 9=7a+3b+c
3-2 32=19a+5b+c
4-3 69=37a+7b+c
6-5 23=12a+2b
7-6 37=18a+2b
9-8 14=6a
What I have so far doesn't seem right....
All of this is exactly right. Good Job. All you need to do now is solve for a and substitute to solve for b. Then use a and b to solve for c and use a similar process to get d.With the given set of points, determine the equation of the polynomial function using the algebraic method.
(1,1), (2,10), (3,42), (4,111), (5,231), (6,416)
x y 1st 2nd 3rd
1 1 - - -
2 10 9 - -
3 42 32 23 -
4 111 69 37 14
5 231 120 51 14
6 416 185 65 14
1 1=a+b+c+d
2 10=8a+4b+2c+d
3 42=27a+9b=3c+d
4 111=64a+16b+4c+d
2-1 9=7a+3b+c
3-2 32=19a+5b+c
4-3 69=37a+7b+c
6-5 23=12a+2b
7-6 37=18a+2b
9-8 14=6a
What I have so far doesn't seem right....
Johett has shown that a cubic passes through all points here:you kind of have the idea. but what you're doing will not guarantee that you will hit all the points. you have 6 points here, you should try fitting a 5th degree polynomial through them, not a cubic. it will make sure you hit all the points you're supposed to
The columns labelled 1st, 2nd and 3rd are the differences between the pairs of values in the column to the left. Since all the values in the 3rd column are the same, only a 3rd degree polynomial is needed.x y 1st 2nd 3rd
1 1 - - -
2 10 9 - -
3 42 32 23 -
4 111 69 37 14
5 231 120 51 14
6 416 185 65 14