Asked by Rupert
A quadratic pattern has a third term equal to 2,a fourth term equal to -2 and the sixth term equal to-16 . calculate the second difference of this quadratic pattern show all the working
Answers
                    Answered by
            mathhelper
            
    make a chart
x y ∆y ∆(∆y)
1
2
3 2
4 -2 -4
5 k (k+2)
6 -16 (-16-k)
but we know that the first differences for a quadratic
form an arithmetic sequence, and the 2nd difference is a constant
so (-16-k) - (k+2) = k+2 + 4
-3k = 24
k = - 8
I can now complete the chart:
x y ∆y ∆(∆y)
1
2
3 2
4 -2 -4 --- -2
5 -8 -6 --- -2
6 -16 -8 -- -2
and I would be able to complete the entire chart without
actually finding the quadratic.
x y ∆y ∆(∆y)
1 4 2 ----- -2
2 4 0 ----- -2
3 2 -2 ---- -2
4 -2 -4 --- -2
5 -8 -6 --- -2
6 -16 -8 -- -2
or:
you could find the actual quadratic ax^2 + bx + c, by
setting up 3 equations in 3 variables, then filling in the missing parts.
    
x y ∆y ∆(∆y)
1
2
3 2
4 -2 -4
5 k (k+2)
6 -16 (-16-k)
but we know that the first differences for a quadratic
form an arithmetic sequence, and the 2nd difference is a constant
so (-16-k) - (k+2) = k+2 + 4
-3k = 24
k = - 8
I can now complete the chart:
x y ∆y ∆(∆y)
1
2
3 2
4 -2 -4 --- -2
5 -8 -6 --- -2
6 -16 -8 -- -2
and I would be able to complete the entire chart without
actually finding the quadratic.
x y ∆y ∆(∆y)
1 4 2 ----- -2
2 4 0 ----- -2
3 2 -2 ---- -2
4 -2 -4 --- -2
5 -8 -6 --- -2
6 -16 -8 -- -2
or:
you could find the actual quadratic ax^2 + bx + c, by
setting up 3 equations in 3 variables, then filling in the missing parts.
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