## Thomas' Calculus 13th Edition

$y=\dfrac{1}{2}+\dfrac{1}{2x^2}$
Given the graph of $y=f(x)$ and a real number $c>1$, we obtain the graph of $y=\dfrac{1}{c}f(x)$ by vertically compressing the graph of $y=f(x)$ by a factor of $c$. Hence, to vertically compress the graph of $y=1+\dfrac{1}{x^2}$ by a factor of 2, we multiply the right hand side of this equation by $\dfrac{1}{2}$ to obtain $$y=\dfrac{1}{2}\left(1+\dfrac{1}{x^2}\right),$$ $$y=\dfrac{1}{2}+\dfrac{1}{2x^2}.$$