Algebra 2!!!!!!!!!!!!?

Question

Show the nth-order difference for x^2-3x+7
I don't know how to this...Please Help!
Thanks

Answer 1

If f(x)=x^2-3x+7 then the first order difference would be: f(x+1)-f(x); the second order difference would be f(x+2)-f(x+1)-f(x+1)+f(x)...and so on.

Answer 2

Shows that the difference between y-value(s) wll be a constant for the degree of the exponent. (nth order) So this means you have a quadratic, or degree 2. Make a table of x and y values...

0 , 7
1 , 5
2 , 5
3 , 7
4 , 11
5 , 17
6 , 25
7 , 35

Now we will look at the differences(subtract until we have a constant)
7
subtract -2
5
subtract = 0
5
subtract = 2
7
subtract = 4
11
subtract = 6
17
subtract = 8
25
subtract = 10
35


Since we have different answers, we repeat with that list:

-2
........subtract = 2
0
........subtract = 2
2
............subtract = 2
4
............subtract = 2
6
............subtract = 2
8
............subtract = 2
10

Since all of our 2nd differences are the same, it is a secind degree equation