Chapter 5.6, Problem 59PPSE

Solution

To calculate : The quadratic model for the given set of values, if possible.

The required quadratic model for the given set of values is $f\left(x\right)=2x^2+24x+75$ .

Given information :

The values are: $f\left(-4\right)=11,f\left(-5\right)=5,f\left(-6\right)=3$ .

Formula used:

Quadratic model is: $f\left(x\right)=a{x}^2+bx+c$

Calculation :

Let the quadratic function be $f\left(x\right)=a{x}^2+bx+c$ .

To determine whether the quadratic function exists or not, find the values of $a,b\text{ and }c$ for the given values of the function.

Consider $f\left(-4\right)=11$

Then,

  $\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(-4\right)=a{\left(-4\right)}^2+b\left(-4\right)+c\\ 11=16a-4b+c\mathrm{............}\left(1\right)\end{array}$

Consider $f\left(-5\right)=5$

Then,

  $\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(-5\right)=a{\left(-5\right)}^2+b\left(-5\right)+c\\ 5=25a-5b+c\mathrm{............}\left(2\right)\end{array}$

Consider $f\left(-6\right)=3$

Then,

  $\begin{array}{l}f\left(x\right)=a{x}^2+bx+c\\ f\left(-6\right)=a{\left(-6\right)}^2+b\left(-6\right)+c\\ 3=36a-6b+c\mathrm{............}\left(3\right)\end{array}$

Subtract $\left(3\right)\text{ from }\left(1\right)$

  $\begin{array}{l}\underset{¯}{\begin{array}{l}11=16a-4b+c\\ 3=36a-6b+c\end{array}}\\ 8=-20a+2b\\ 4=-10a+b\mathrm{.........}\left(4\right)\end{array}$

Similarly,

Subtract

  $\left(3\right)$ from $\left(2\right)$ ,

  $\begin{array}{l}\underset{¯}{\begin{array}{l}5=25a-5b+c\\ 3=36a-6b+c\end{array}}\\ 2=36a-6b+c\\ 2=-11a+b\mathrm{..........}\left(5\right)\end{array}$

Subtract equation $\left(5\right)$ from $\left(4\right)$

  $\begin{array}{l}\underset{¯}{\begin{array}{l}4=-10a+b\\ 2=-11a+b\end{array}}\\ 2=a\\ \therefore a=2\end{array}$

Substitute $a=2$ in equation $\left(5\right)$ to get the value of $b$ .

  $\begin{array}{l}2=-11\left(2\right)+b\\ b=2+22\\ \therefore b=24\end{array}$

Put $a=2\text{ and }b=24$ in equation $\left(1\right)$

  $\begin{array}{l}11=16a-4b+c\\ 11=16\left(2\right)-4\left(24\right)+c\\ 11=32-96+c\\ c=11+64\\ \therefore c=75\end{array}$

Therefore, the required quadratic model for the given set of values is $f\left(x\right)=2x^2+24x+75$ .