Trigonometry (11th Edition) Clone

For $f(x) = 2\sqrt{x} + 1$ when $x = 9, y = 7;$ $x = 4, y = 5;$ $x = 1, y = 3;$ $x = 0, y = 1$ As seen, $f(x) = 2\sqrt{x} + 1$ is the vertical stretching twice and the vertical translation up by 1 unit of $f(x) = \sqrt{x}$. Domain = $[0, \infty)$ Range = $[1, \infty)$