Question

Find the range of $y=\frac{x+2}{2x^2+3x+6},$ if $x$ is real.

Answer 1

$y=\frac{x+2}{2x^2+3x+6}$
$\Rightarrow$ $2y{x}^2+3yx+6y=x+2$
$\Rightarrow$ $2y{x}^2+(3y-1)x+6y-2=0$ (i)
case I: if $y\ne 0,$ then equation (i) is quadratic in $x$
$\because$ $x$ is real
$\therefore$ $D\ge 0$
$\Rightarrow$ ${(3y-1)}^2-8y(6y-2)\ge 0$
$\Rightarrow$ $(3y-1)(13y+1)\le 0$
$yϵ[-\frac{1}{13},\frac{1}{13}]-\left\{0\right\}$
case II: if $y=0,$ then equation becomes
$x=-2$ which is possible as $x$ is real
$\therefore$ Range $yϵ[-\frac{1}{13},\frac{1}{3}]$