Question

which graph shows the solution set of (\frac{x^{2}+9x + 20}{x^{2}-x - 20}geq0)?

Explanation:

Step1: Factor the numerator and denominator

First, factor \(x^{2}+9x + 20=(x + 4)(x+5)\) and \(x^{2}-x - 20=(x - 5)(x + 4)\). So the inequality \(\frac{x^{2}+9x + 20}{x^{2}-x - 20}\geq0\) becomes \(\frac{(x + 4)(x + 5)}{(x - 5)(x + 4)}\geq0\), with \(x
eq5\) and \(x
eq - 4\) (to avoid division - by - zero). After canceling out the common factor \((x + 4)\) (for \(x
eq - 4\)), we get \(\frac{x + 5}{x - 5}\geq0\) with \(x
eq - 4\).

Step2: Find the critical points

The critical points of the rational function \(y=\frac{x + 5}{x - 5}\) are the values of \(x\) that make the numerator or the denominator equal to zero. The numerator \(x+5 = 0\) gives \(x=-5\) and the denominator \(x - 5=0\) gives \(x = 5\). Also, we have the excluded value \(x=-4\) from the original non - simplified rational function.

Step3: Test the intervals

We consider the intervals \((-\infty,-5)\), \((-5,-4)\), \((-4,5)\) and \((5,\infty)\).
For \(x<-5\), let \(x=-6\), then \(\frac{-6 + 5}{-6 - 5}=\frac{-1}{-11}=\frac{1}{11}\geq0\).
For \(-5For \(-4For \(x>5\), let \(x = 6\), then \(\frac{6 + 5}{6 - 5}=11\geq0\).
The solution of the inequality \(\frac{x^{2}+9x + 20}{x^{2}-x - 20}\geq0\) is \(x\leq - 5\) or \(x>5\).
On a number line, we have a closed circle at \(x=-5\) (because the inequality is \(\geq\)) and an open circle at \(x = 5\) (because the original function is undefined at \(x = 5\)) and the line extends to the left of \(x=-5\) and to the right of \(x = 5\).

Answer:

The graph with a closed circle at \(x=-5\), an open circle at \(x = 5\), and shading to the left of \(x=-5\) and to the right of \(x = 5\) is the correct one.