Chapter 7.1, Problem 90E

(a)

To determine

To Find: To find a quadratic model and a linear model for yearly revenues of the given travel companies.

Explanation of Solution

Given information:Yearly revenues of both the online travel companies.

Since x represent the year with $x=2$ corresponding to 2012

Let $y_1$ and $y_2$ be the yearly revenues of the first and second company, then

$x$	$y_1$	$y_2$
2	4.03	5.26
3	4.77	6.79
4	5.76	8.44
5	6.67	9.22
6	8.77	10.74
7	10.06	12.68

Using the graphing utility, with regression feature the quadratic model for the first company came to be $y_1=0.128x^2+0.075x+3.366$ and the linear model for the second company came to be $y_2=1.421x+2.46$ .

(b)

To determine

To Graph: Tograph the given model functions

Explanation of Solution

Given information:Given model functions are $y_1=0.128x^2+0.075x+3.366$ and $y_2=1.421x+2.46$

Graph:

  

Interpretation:

From the graph, the x-axis shows the year x and y-axis shows yearly revenue.

The red line represents the given linear model whereas the blue line represents the given quadratic model with original scatter points.

(c)

To determine

To Find: Tofind the year in which the revenue of second company is greater then first company.

Answer to Problem 90E

The answer is from 2011.

Explanation of Solution

Given information:Given model functions are $y_1=0.128x^2+0.075x+3.366$ and $y_2=1.421x+2.46$ with graph

From the graph, it can be seen that in the year 2011 the revenue of second company exceeds the first company and it continues till 2019.

(d)

To determine

To Find: To find the solutions of the given model functions.

Answer to Problem 90E

The answer is $(4.378,6760.056)$

Explanation of Solution

Given information:Given model functions are $y_1=0.128x^2+0.075x+3.366$ and $y_2=1.421x+2.46$ with graph

Formula used: Substitution method

Calculation:

The point of intersection find out when

  $\begin{array}{l}{y}_1=y_2\\ 0.128x^2+0.075x+3.366=1.421x+2.46\\ 0.128x^2-1.346x+0.906=0\end{array}$

Using quadratic formula,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ x=\frac{-(-1.346)\pm \sqrt{{(-1.346)}^2-4\times 0.128\times 0.906}}{2\times 0.128}\\ x=\frac{1.346\pm \sqrt{1.812-0.464}}{0.256}\\ x=\frac{1.346\pm \sqrt{1.348}}{0.256}\\ x=\frac{1.346\pm 1.161}{0.256}\\ x=\frac{1.346+1.161}{0.256}\text{or}x=\frac{1.346-1.161}{0.256}\\ x=9.793\text{or}x=0.723\end{array}$

Thus, the value of x is approximately 10 and 1.

Since $x=2$ corresponding to 2012

Therefore $x=1\text{and}x=10$ corresponding to 2011 and 2020

Hence, the revenue exceeds first in the year of 2011.

(e)

To determine

To Compare: To compare the above results.

Explanation of Solution

Given information:Given model functions are $y_1=0.128x^2+0.075x+3.366$ and $y_2=1.421x+2.46$ with graph and table

By comparing the above results, both methods gives the same results