## Thomas' Calculus 13th Edition

$y=1+\dfrac{9}{x^2}$
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: $$y=1+\dfrac{1}{(x/3)^2}$$ $$y=1+\dfrac{9}{x^2}.$$