Question

In the given problem, solve and graph. Express answers in both inequality and interval notation.

$\frac{x^2-x-12}{x^2+4}\le 0$

Answer 1

$\frac{x^2-x-12}{x^2+4}\le 0$

$\frac{(x+3)(x-4)}{x^2+4}\le 0$

	x<-3	x=-3	-3<x<4	x=4	x>4
x-4	-	-	-	0	+
x+3	-	0	+	+	+
$x^2+4$	+	+	+	+	+
$\frac{(x+3)(x-4)}{x^2+4}$	+	0	-	0	+

$\left[\begin{array}{cc}Solution:& -3\le x\le 4\\ IntervalNotation:& [-3,4]\end{array}\right]$