Write an equation to model the data.

Question

x: 0,1,2,3,4
y: 0,2.8,11.2,25.2,44.8

How do you do this?

Answer 1

I took first differences and got
2.8 , 8.4 , 14, and 19.6
second differences yielded:
5.6 , 5.6 , 5.6

which tells me that the relationship is a quadratic function.

so let y = ax^2 + bx + c, giving us
0 = c
2.8 = a + b + c
11.2 = 4a + 2b + c
we know c = 0
so if we double the 2nd

2a + 2b = 5.6
and the third:
4a + 2b = 11.2
subtract them
2a = 5.6
a = 2.8
then in 2a+ 2b = 5.6
5.6 + 2b = 5.6
b = 0

so y = 2.8x^2

Now that I see the answer...
we could have taken out a factor of 2.8 from all the y's
2.8( 0 , 1, 4 , 9 , 16 )
and you can see the perfect squares, so
y = 2.8 x^2

Answer 2

I understand now. Thanks! How can I do it for linear data, such as the data below?

x: 1,2,3,4,5
y: 248,307,366,425,484

Answer 3

If you have consecutive values of x, like in both your cases,
in general
if the first differences in the y's are constant, you have a linear function
if the 2nd differences in the y's are constant, you have a quadratic function, (as your case above)
if the 3rd differences in the y's are constant , you have a cubic function,
etc.

So, in your new question, the first difference is 59, this in effect becomes your slope
so you could say
y = mx + b = 59x + b
sub in the first relation, x = 1, y = 248
248 = 59(1) + b
b = 189

y = 59x + 189

or, you could generalize the method I used in your first problem:
since the relationship is linear, let
y = ax + b, plug in the first two "point"
248 = a(1) + b
307 = a(2) + b
subtract them:
59 = a

back into the first:
248 = 59 + b
189 = b

y = 59x + 189 , just like before