Question

Find the range of $y=\frac{x^2-2}{x^2+5}$

Answer 1

$y{x}^2+5y=x^2-2$

$\Rightarrow 2+5y=(1-y)x^2$

$\Rightarrow x^2=\frac{2+5y}{1-y}$

$\Rightarrow x=\sqrt{\frac{2+5y}{1-y}}$

$\Rightarrow \frac{2+5y}{1-y}\ge 0$

$\Rightarrow \frac{5y+2}{y-1}\le 0$

$\therefore y\in [\frac{-2}{5},1)$