Chapter 4.4, Problem 98PPSE

Solution

To find: the quadratic model for the given table of values.

The quadratic model is $y=-0.14x^2+5.1x+17$ .

Given:

x	0	5	10	15	20
y	17	39	54	61	61

Concept used:

The quadratic model is in the form $y=a{x}^2+bx+c$ , where $a,b,c$ are any real numbers and $a\ne 0$ .

Calculation:

Let the required quadratic model be,

  $y=a{x}^2+bx+c\to \left(1\right)$

Substituting the point $\left(0,17\right)$ in equation (1), it follows that

  $\begin{array}{c}17=a{\left(0\right)}^2+b\left(0\right)+c\\ c=17\end{array}$

Now put $c=17$ in equation (1), it follows that

  $y=a{x}^2+bx+17\to \left(2\right)$

Now substituting the point $\left(5,39\right)$ in equation (1), it follows that

  $\begin{array}{c}39=a{\left(5\right)}^2+b\left(5\right)+17\\ 25a+5b=22\to \left(3\right)\end{array}$

Now substituting the point $\left(10,54\right)$ in equation (1), it follows that

  $\begin{array}{c}54=a{\left(10\right)}^2+b\left(10\right)+17\\ 100a+10b=37\to \left(4\right)\end{array}$

Now multiply 4 to equation (3) and then subtract equation (4) from the resulting equation.

That is,

  $\left[4\times \text{Equation}\left(3\right)\right]-\left[\text{Equation}\left(4\right)\right]$

  $\begin{array}{c}4\left(25a+5b\right)-\left(100a+10b\right)=4\left(22\right)-37\\ 100a+20b-100a-10b=88-37\\ 10b=51\\ b=5.1\end{array}$

Now put $b=5.1$ in equation (3), it follows that,

  $\begin{array}{c}25a+5\left(5.1\right)=22\\ 25a+25.5=22\\ 25a=-3.5\\ a=-0.14\end{array}$

Now put $a=-0.14,b=5.1$ , and $c=17$ in equation (1), it follows that,

  $y=-0.14x^2+5.1x+17$ .