Trigonometry (10th Edition)

Published by Pearson
ISBN 10: 0321671775
ISBN 13: 978-0-32167-177-6

Appendix D - Graphing Techniques - Exercises - Page 447: 61

Answer

Refer to the curve in Black in Graph I

Work Step by Step

For $f(x) = 2\sqrt{x} + 1$ when $x = 9, y = 7;$ $x = 4, y = 5;$ $x = 1, y = 3;$ $x = 0, y = 1$ As seen, $f(x) = 2\sqrt{x} + 1$ is the vertical stretching twice and the vertical translation up by 1 unit of $f(x) = \sqrt{x}$. Domain = $[0, \infty)$ Range = $[1, \infty)$
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