Question

graph the solution to the following inequality on the number line.

$$ - x ^ { 2 } - 8 x leq 15 $$

note that you can use the aleks graphing calculator to help get your answer.

Explanation:

Step1: Rewrite the inequality

Multiply both sides by - 1 (and reverse the inequality sign) to get \(x^{2}+8x\geq - 15\). Then rewrite it as \(x^{2}+8x + 15\geq0\).

Step2: Factor the quadratic

Factor \(x^{2}+8x + 15\) as \((x + 3)(x+5)\geq0\).

Step3: Find the critical points

Set \((x + 3)(x + 5)=0\). Using the zero - product property \(x+3 = 0\) gives \(x=-3\) and \(x + 5=0\) gives \(x=-5\).

Step4: Test intervals

• For \(x<-5\) (e.g., \(x=-6\)), \((-6 + 3)(-6 + 5)=(-3)\times(-1)=3>0\).
• For \(-5
• For \(x>-3\) (e.g., \(x=0\)), \((0 + 3)(0 + 5)=3\times5 = 15>0\).

Answer:

The solution of the inequality \(-x^{2}-8x\leq15\) is \(x\leq - 5\) or \(x\geq - 3\). On the number line, we place a closed circle at \(x=-5\) (since the inequality includes equality when \(x=-5\) as \((-5 + 3)(-5 + 5)=0\)) and shade to the left of \(x = - 5\), place a closed circle at \(x=-3\) (since the inequality includes equality when \(x=-3\) as \((-3+3)(-3 + 5)=0\)) and shade to the right of \(x=-3\).