Question

which graph shows the solution set of $\frac{x^{2}+4x - 12}{x}>0$?

Explanation:

Step1: Factor the numerator

Factor \(x^{2}+4x - 12=(x + 6)(x-2)\). So the inequality becomes \(\frac{(x + 6)(x - 2)}{x}>0\)

Step2: Find the critical points

Set the numerator and denominator equal to zero. \(x+6=0\) gives \(x=-6\), \(x - 2=0\) gives \(x=2\), and \(x = 0\) (from the denominator). The critical points are \(x=-6\), \(x=0\), and \(x = 2\)

Step3: Test the intervals

• For \(x<-6\) (e.g., \(x=-7\)), \(\frac{(-7 + 6)(-7-2)}{-7}=\frac{(-1)(-9)}{-7}=\frac{9}{-7}<0\)
• For \(-6
• For \(0
• For \(x>2\) (e.g., \(x=3\)), \(\frac{(3 + 6)(3-2)}{3}=\frac{(9)(1)}{3}=3>0\)

The inequality \(\frac{(x + 6)(x - 2)}{x}>0\) is satisfied when \(-62\). But we note that \(x
eq-6\), \(x
eq0\) and \(x
eq2\) (since the original function is undefined at \(x = 0\) and the inequality is strict). So the solution set is \(x\in(-6,0)\cup(2,\infty)\) which corresponds to the fourth graph (open - circles at \(x=-6\), \(x = 0\) and \(x=2\) and the shaded regions for \(-62\))

Answer:

The fourth graph (the one with open - circle at \(x=-6\), open - circle at \(x = 0\) and open - circle at \(x=2\) and the shaded regions for \(x<-6\) and \(0 < x<2\))