First show that this is a quadratic relationship by taking differences and second differences:
y dy d2y
3 10 8
13 18 8
31 26 8
57 34 8
91 42 8
133
If the second difference is constant, it is a quadratic relationship.
Assume the quadratic model as:
f(x)=ax2 + bx + c
then
f(x+1)=a(x+1)^2+b(x+1)+c
and
f(x+1)-f(x)=2ax+b+a
Similarly,
f(x+2)-f(x+1)=2ax+b+3a
f(x+3)-f(x+2)=2ax+b+5a
....
Thus, the second difference = 2a
2a=8
a=4
By substituting values of x and a into one of the formulas, b can be found
for x=1,
dx=10
2ax+b+a=10
b=-2
Substitute a and b in f(1), we get
2(1)2-2(1)+c=3
c=1
Thus
f(x)=4x2-2x+1
check:
f(1)=3
f(3)=31
f(6)=133
I'm having a hard time figuring out a quadratic model for this data.
x y
----------
1 3
2 13
3 31
4 57
5 91
6 133
How do I go about doing this?
Is there a website that could help me?
1 answer