Chapter 4.3, Problem 25PPSE

Solution

a.

To Find: The quadratic model for the data.

The quadratic formula obtained is $-0.01x^2+1.02x+18.39$ .

Given: $$

Year	$1981$	$1991$	$1995$	$1999$	$2001$	$2006$	$2007$	$2008$
Price	$18$	$29$	$32$	$33$	$34$	$39$	$41$	$42$

Explanation:

Using the given data, determine the quadratic model.

Consider $1981$ to be year $0$, $1991$ to be year $10$ , etc.

To determine the coordinate system for the given data, arrange the data as the coordinates of the system.

  $\begin{array}{c}A\left(x_1,y_1\right)=\left(0,18\right)\\ B\left(x_2,y_2\right)=\left(10,29\right)\\ C\left(x_3,y_3\right)=\left(14,32\right)\\ D\left(x_4,y_4\right)=\left(18,33\right)\\ E\left(x_5,y_5\right)=\left(25,39\right)\\ F\left(x_6,y_6\right)=\left(26,41\right)\\ G\left(x_7,y_7\right)=\left(27,42\right)\end{array}$

Sketch the above coordinates in the coordinate system:

  

Calculate the quadratic regression with the above points and the degree $2$:

  

Thus, the quadratic formula obtained is: $-0.01x^2+1.02x+18.39$

b.

To Describe: The reasonable domain and range for your model.

The reasonable domain and range for the model is $D_f=\left(0,27\right)R_f=\left(14,44.4\right)$

Given: $$

Year	$1981$	$1991$	$1995$	$1999$	$2001$	$2006$	$2007$	$2008$
Price	$18$	$29$	$32$	$33$	$34$	$39$	$41$	$42$

Explanation:

Describe the reasonable domain and range for your model:

In general, $D_f=ℝ$ is the domain of any quadratic function.

The domain of the function is more appropriately defined as the period from $1981$ to

  $2008$ , or $0$ to $27$ , using only the integers between that represent subsequent years between

  $1981$ and $2008$ .

If the function has the negative parameter $a$ , then the range of any quadratic function is $\left(-\infty ,y_0\right)$ . If the function has the positive parameter $a$ ,then the range of any quadratic function is $\left(y_0,\infty \right)$ . Here $y_0$ represents the $y$ value of the vertex.

Calculate the value $y_0$ of the equation $-0.01x^2+1.02x+18.39$ from part $\left(a\right)$ :

  $\begin{array}{l}{y}_0=\frac{4ac-b^2}{4a}\\ y_0=\frac{4\cdot \left(-0.01\right)\cdot 18.39-{1.02}^2}{4\cdot \left(-0.01\right)}\\ y_0=\frac{-0.7356-1.0404}{-0.04}\\ y_0=44.4\end{array}$

The price of the stamps never fell below $18$ cents, so it is unreasonable to assume that the range of the quadratic model is $\left(-\infty ,y_0\right)$ . As a result, this function can be interpreted within an interval of $\left(18,44.4\right).$

Thus, the reasonable domain and range for the model is $D_f=\left(0,27\right)R_f=\left(14,44.4\right)$

c.

To Estimate: The time when first-class postage was $37$ cents

By the end of $2004$ , the stamps were estimated to be worth $37$ cents.

Given: $$

Year	$1981$	$1991$	$1995$	$1999$	$2001$	$2006$	$2007$	$2008$
Price	$18$	$29$	$32$	$33$	$34$	$39$	$41$	$42$

Explanation:

Estimate the year the price of the stamps was $37$ cents:

The quadratic model obtained from part $\left(a\right)$ is $y=-0.01x^2+1.02x+18.39$ .

The given value of $37$ cents denotes $y$ .

Find the value of $x$ for the corresponding $y$ :

  $\begin{array}{c}-0.01x^2+1.02x+18.39=y\\ -0.01x^2+1.02x+18.39=37\\ -0.01x^2+1.02x+18.39-37=0\\ -0.01x^2+1.02x-18.61=0\end{array}$

Solve the quadratic equation using the Formula:

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=-0.01,b=1.02,c=-18.61$

  $\begin{array}{c}{x}_{1,2}=\frac{-1.02\pm \sqrt{\left({1.02}^2-4\left(-0.01\right)\left(-18.61\right)\right.}}{2\left(-0.01\right)}\\ x_{1,2}=\frac{-1.02\pm \sqrt{1.0404-0.7444}}{-0.02}\\ x_{1,2}=\frac{-1.02\pm \sqrt{0.296}}{-0.02}\\ x_{1,2}=\frac{-1.02\pm 0.544}{-0.02}\\ x_1=23.8\\ x_2=78.2\end{array}$

The results of the analysis of $x$ can now be analysed. Part $\left(a\right)$ and Part $\left(b\right)$ used the integers $0$ to $27$ as the domain of the function. As a consequence, the solution $x_2=78.2$ is meaningless here.

Initially, $1981$ was denoted as $0$ . So, the value of $x_1=23.8$ corresponds to the year: $1981+23.8=2004.8$ . By the end of $2004$ , the stamps were estimated to be worth $37$ cents.

d.

To Predict: When first-class postage will be $50$ cents. And Explain the prediction may not be valid.

Based on this quadratic model, the stamp cannot cost $50$ cents.

Given: $$

Year	$1981$	$1991$	$1995$	$1999$	$2001$	$2006$	$2007$	$2008$
Price	$18$	$29$	$32$	$33$	$34$	$39$	$41$	$42$

Explanation:

Predict when first-class postage will be $50$ cents:

The given value of $50$ cents denotes $y$ .

Find the value of $x$ for the corresponding $y$ :

  $\begin{array}{c}-0.01x^2+1.02x+18.39=y\\ -0.01x^2+1.02x+18.39=50\\ -0.01x^2+1.02x+18.39-50=0\\ -0.01x^2+1.02x-31.61=0\end{array}$

Solve the quadratic equation using the Formula:

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=-0.01,b=1.02,c=-18.61$

  $\begin{array}{c}{x}_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x_{1,2}=\frac{-1.02\pm \sqrt{\left({1.02}^2-4\left(-0.01\right)\left(-31.61\right)\right.}}{2\left(-0.01\right)}\\ x_{1,2}=\frac{-1.02\pm \sqrt{1.0404-1.2644}}{-0.02}\\ x_{1,2}=\frac{-1.02\pm \sqrt{-0.224}}{-0.02}\end{array}$

There is no real solution to this equation because the number under the root is negative.

In Part $b$ , it was determined that the maximum theoretical value of the stamp price is $44.4$ cents based on the range of this model. Based on this quadratic model, the stamp cannot cost $50$ cents. Since we do not have information about stamp prices after $2008$ , this may not be accurate in the real world.