Question

Find the domain of definition of the following function.
$y=\frac{\sqrt{4-3x-x^2}}{x+4}.$

Answer 1

$y=\frac{\sqrt{4-3x-x^2}}{x+4}$

$x+4\ne 0$

$x\ne -4$

$(4-3x-x^2)\ge 0$

$(-x^2-3x+4)\ge 0$

$-x^2-4x+x+4\ge 0$

$(-x+1)(x+4)\ge 0$

$x\ge -4$

$-x+1\ge 0$

$-x\ge -1$

$x\le 1$

So, the Domain of $y=(-4,1]$