College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Summary, Review, and Test - Test - Page 328: 28

Answer

$Rate_{Change} = 48$

Work Step by Step

Average rate of change is expressed by the following equation: $$Rate_{Change}=\frac{f(x_{2}) - f(x_{1})}{x_{2} - x_{1}}$$ Therefore, having the values of $x_{2}$ and $x_{1}$, all we need is to find the values of $f(x_{2})$ and $f(x_{1})$: $f(x_{2}) = 3(10)^{2} - 5 = 3(100) - 5 = 300 - 5 = 295$ $f(x_{1}) = 3(6)^{2} - 5 = 3(36) - 5 = 108 - 5 = 103$ $$Rate_{Change} = \frac{295 - 103}{10 - 6} = \frac{192}{4} = 48$$
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