Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.6 - Transformations of Functions - Exercise Set - Page 243: 127

Answer

a. See explanations. b. $40.2\ in$, fits well. c. $0.9$ inches per month. d. $0.2$ inches per month. See explanations.

Work Step by Step

a. To obtain the graph of function $f(x)=2.9\sqrt x+20.1$ from the curve of $y=\sqrt x$, stretch the curve vertically by a factor of 2.9, then shift the result vertically $20.1\ in$ up. b. For $x=48$ month, we have $f(x)=2.9\sqrt {48}+20.1\approx40.2\ in$. Compared to the actual median height of $40.8\ in$, we can see that the model describes the actual height well. c. We can obtain the average rate of change from $0$ to $10$ month as $R_1=\frac{f(10)-f(0)}{10-0}=\frac{2.9\sqrt {10}}{10}\approx0.9$ inches per month. d. We can obtain the average rate of change from $50$ to $60$ month as $R_2=\frac{f(60)-f(50)}{60-50}=\frac{2.9(\sqrt {60}-\sqrt {50})}{10}\approx0.2$ inches per month. This growth rate is much smaller than the result from part-c, and this is shown in the graph by the slow increase of the curve when the x-value gets bigger, that is the height growth slows down when the age gets bigger.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.