[tex]Domain:\ x\neq-2\\\\\dfrac{x+2}{x-4}<0\iff(x+2)(x-4) < 0\\\\x+2=0\to x=-2\\x-4=0\to x=4\\\\(x+2)(x-4)=x^2-4x+2x-8\\\\\text{parabola op}\text{en up}\\\\\text{Look at the picture}\\\\Answer:\ \dfrac{x+2}{x-4}<0\ for\ x\in(-2,\ 4)\to\boxed{B)\ -2 < x < 4}[/tex]
Other method:
[tex]\dfrac{x+2}{x-4}<0[/tex]
zeros of numerator and denominator are x = -2 and x = 4.
Look at the second picture.
for x < -2 → x + 2 < 0 and x - 4 < 0
therefore [tex]\dfrac{x + 2}{x - 4}=\dfrac{(-)}{(-)}>0[/tex]
for -2 < x < 4 → x + 2 > 0 and x - 4 < 0
therefore [tex]\dfrac{x+2}{x-4}=\dfrac{(+)}{(-)} < 0[/tex]
for x > 4 → x + 2 > 0 and x - 4 > 0
therefore [tex]\dfrac{x+2}{x-4}=\dfrac{(+)}{(+)} > 0[/tex]
Answer: B) -2 < x < 4
