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What does it mean when the second differences of the y-values in the data set are constant?
Model: Quadratic
Equation: y=0.2x^2
We will start by determining the type of function that best models the data. Then we will be able to write an exact equation that models the data.
We want to determine the most appropriate model for the given data set. Note that the x-values have a common difference of 1. Therefore, we can check if the y-values have a common difference, a common ratio, or constant second differences. It will tell us which model is most appropriate for the data set.
| The y-values have: | The model is: |
|---|---|
| A common difference | Linear |
| A common ratio | Exponential |
| Constant second differences | Quadratic |
We can exclude the possibility of an exponential model, since the y-value in the second pair is 0. Let's check for a common difference!
| x | y | First differences |
|---|---|---|
| - 1 | 0.2 | |
| 0 | 0 | -0.2 ↩ |
| 1 | 0.2 | +0.2 ↩ |
| 2 | 0.8 | +0.6 ↩ |
| 3 | 1.8 | +1 ↩ |
| 4 | 3.2 | +1.4 ↩ |
As we can see, the differences between the consecutive y-values are not constant. Therefore, we have to check whether or not the second differences are constant.
| x | y | First differences | Second differences |
|---|---|---|---|
| - 1 | 0.2 | ||
| 0 | 0 | -0.2 ↩ | |
| 1 | 0.2 | +0.2 ↩ | +0.4 ↩ |
| 2 | 0.8 | +0.6 ↩ | +0.4 ↩ |
| 3 | 1.8 | +1 ↩ | +0.4 ↩ |
| 4 | 3.2 | +1.4 ↩ | +0.4 ↩ |
The second differences of the y-values are all 0.4, so a quadratic model fits the data.
We know that a quadratic function best models the given data. y=ax^2+bx+c To write an equation to model the data, we have to determine the values of a, b, and c. Recall that (0,c) is the y-intercept of the quadratic function. Since the pair (0,0) is included in the data set, c= . y=ax^2+bx+ ⇔ y=ax^2+bx In order to determine the values of a and b, we will use two (x,y) pairs from the data set, other than (0,0), to write a system of equations. We will use (- 1,0.2) and (1,0.2).
x= - 1, y= 0.2
(- a)^2=a^2
a(- b)=- a * b
Rearrange equation
This will be the first equation of our system. We will write the second equation using the point (1,0.2).
Let's write our system of equations. a-b=0.2 & (I) a+b=0.2 & (II) Since the coefficients of b have opposite signs, we will use the Elimination Method.The variable b will be eliminated by adding the two equations.
(I):Add (II)
(I):Add terms
(I):.LHS /2.=.RHS /2.
(II):a= 0.2
(II):LHS-0.2=RHS-0.2
We have that a=0.2 and b= . Let's finish writing the equation to model the data! y=0.2x^2+ x ⇔ y=0.2x^2 Below we have included a graph that shows how the equation models the given data.