Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 1: Functions - Section 1.2 - Combining Functions; Shifting and Scaling Graphs - Exercises 1.2: 60

Answer

$y=1+\dfrac{9}{x^2}$

Work Step by Step

Given the graph of $y=f(x)$ and a real number $c>1$, we obtain the graph of $y=f(x/c)$ by horizontally stretching the graph of $y=f(x)$ by a factor of $c$. Hence, to horizontally stretch the graph of $y=1+\dfrac{1}{x^2}$ by a factor of 3, we replace each occurrence of $x$ in this equation by $x/3$, and, thus, obtain the equation: $$y=1+\dfrac{1}{(x/3)^2}$$ $$y=1+\dfrac{9}{x^2}.$$
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