## Trigonometry (11th Edition) Clone

$y = \frac{1}{x-4}$, where $x\geq 5$
$x = t^2+5$ $t = \sqrt{x-5}$ We can replace this value in the equation with $y$: $y = \frac{1}{t^2+1}$ $y = \frac{1}{(\sqrt{x-5})^2+1}$ $y = \frac{1}{(x-5)+1}$ $y = \frac{1}{x-4}$ Since $x = t^2+5$, then $x\geq 5$ $y = \frac{1}{x-4}$, where $x\geq 5$