Write an equation to model the data.
x: 0,1,2,3,4
y: 0,2.8,11.2,25.2,44.8
How do you do this?
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
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
To model the data, we need to find an equation that relates the values of x and y. In this case, we can notice that the y values are increasing quadratically as the x values increase. This suggests that a quadratic equation might be a good fit for the data.
Let's start by assuming that the equation is of the form: y = ax^2 + bx + c, where a, b, and c are constants that we need to find.
To determine the values of a, b, and c, we can plug in the given x and y values into the equation and create a system of equations. Let's start with the first data point:
When x = 0, y = 0
Substituting these values into the equation, we get:
0 = a(0)^2 + b(0) + c
Simplifying, we find: c = 0
Now let's move on to the second data point:
When x = 1, y = 2.8
Plugging these values into the equation, we have:
2.8 = a(1)^2 + b(1) + c
Simplifying, we get: a + b + c = 2.8
Similarly, we can substitute the remaining x and y values into the equation and create more equations. The resulting system of equations will allow us to solve for the values of a and b.
Using the third data point, we get:
11.2 = a(2)^2 + b(2) + c
Simplifying, we obtain: 4a + 2b + c = 11.2
Using the fourth data point, we get:
25.2 = a(3)^2 + b(3) + c
Simplifying, we have: 9a + 3b + c = 25.2
And finally, using the fifth data point, we get:
44.8 = a(4)^2 + b(4) + c
Simplifying, we obtain: 16a + 4b + c = 44.8
We now have a system of equations with three unknowns (a, b, and c). We can solve this system using various methods, such as substitution or elimination, to find the values of a, b, and c. Once we have these values, we can substitute them back into the equation y = ax^2 + bx + c to obtain the final equation that models the given data.
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