Chapter 5.8, Problem 1LC

Solution

To calculate : The model that best fits the given set of points.

The linear model best fits the data.

Given information :

The set of points is: $\left(-2,-1\right),\left(0,3\right)\text{ and }\left(2,7\right)$ .

Formula used:

  $\left(n+1\right)$ Principle: For any set of $n+1$ points in the coordinate plane that pass the vertical line test, there is a unique polynomial of degree at most $n$ that fits the point perfectly.

Calculation :

Here, the set of points in the data are $3$ .

From the $\left(n+1\right)$ principle, there is a unique polynomial of degree at most $2$ that fits the points perfectly.

Hence, there is a quadratic model that fits the data.

Let $y=a{x}^2+bx+c$ be the model that fits the data.

Substitute the points to get the system of equations.

Substitute $\left(-2,-1\right)$ .

  $\begin{array}{c}y=a{x}^2+bx+c\\ -1=a{\left(-2\right)}^2+b\left(-2\right)+c\\ -1=4a-2b+c\mathrm{............}\left(1\right)\end{array}$

Substitute $\left(0,3\right)$ .

  $\begin{array}{c}y=a{x}^2+bx+c\\ 3=a{\left(0\right)}^2+b\left(0\right)+c\\ c=3\mathrm{............}\left(2\right)\end{array}$

Substitute $\left(2,7\right)$ .

  $\begin{array}{c}y=a{x}^2+bx+c\\ 7=a{\left(2\right)}^2+b\left(2\right)+3\left[\text{from }\left(2\right)\right]\\ 7=4a+2b+3\\ 4=4a+2b\\ 2=4a+b\left[\text{taking }2\text{ as common from both sides}\right]\\ b=2-4a\mathrm{.............}\left(3\right)\end{array}$

Substitute the values of $b\text{ and }c$ in (1)

  $\begin{array}{c}-1=4a-2b+c\\ -1=4a-2\left(2-4a\right)+3\\ -4=4a-4+4a\\ 0=8a\\ a=0\end{array}$

Substitute the value of $a$ in (3)

  $\begin{array}{c}b=2-4a\\ =2-4\left(0\right)\\ =2\end{array}$

From here, $a=0,b=2,c=3$ .

So, the model that fits the data will be:

  $\begin{array}{l}y=0x^2+2x+3\\ y=2x+3\end{array}$

Therefore, the linear model best fits the data.