Chapter 5.8, Problem 11PPSE

Solution

To calculate : A polynomial function whose graph passes through the given set of points.

The required polynomial function is $y=-0.93x^2+7.79x+4$ .

Given information :

The set of points is: $\left(7,13\right),\left(10,-11\right)\text{ and }\left(0,4\right)$ .

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 cubic 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 $\left(7,13\right)$ .

  $\begin{array}{l}y=a{x}^2+bx+c\\ 13=a{\left(7\right)}^2+b\left(7\right)+c\\ 13=49a+7b+c\mathrm{..................}\left(1\right)\end{array}$

Substitute $\left(10,-11\right)$ .

  $\begin{array}{l}y=a{x}^2+bx+c\\ -11=a{\left(10\right)}^2+b\left(10\right)+c\\ -11=100a+10b+c\mathrm{..................}\left(2\right)\end{array}$

Substitute $\left(0,4\right)$ .

  $\begin{array}{l}y=a{x}^2+bx+c\\ 4=a{\left(0\right)}^2+b\left(0\right)+c\\ 4=0+0+c\mathrm{..................}\left(3\right)\end{array}$

Write the system in the augmented matrix form:

  $\left[\begin{array}{ccc}49& 7& \begin{array}{cc}1& 13\end{array}\\ 100& 10& \begin{array}{cc}1& -11\end{array}\\ 0& 0& \begin{array}{cc}1& 4\end{array}\end{array}\right]$

Use calculator to solve the matrix.

From here, $a=-0.93,b=7.79,c=4$ .

Therefore, the required polynomial function is $y=-0.93x^2+7.79x+4$ .