## Thomas' Calculus 13th Edition

$$y=\dfrac{1}{3}\sqrt{4-x^2},$$ or equivalently, $$y=\sqrt{\dfrac{4}{9}-\dfrac{x^2}{9}}$$
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 $y=\sqrt{4-x^2}$ by a factor of 3, we multiply the right hand side of this equation by $\dfrac{1}{3}$ and, thus, obtain $$y=\dfrac{1}{3}\sqrt{4-x^2},$$ or equivalently, $$y=\sqrt{\dfrac{4}{9}-\dfrac{x^2}{9}}.$$