Question

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

$-x^{2}+4x\leq -5$

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

Explanation:

Step1: Rewrite the inequality

First, rewrite \(-x^{2}+4x\leq - 5\) as \(x^{2}-4x - 5\geq0\).

Step2: Factor the quadratic

Factor \(x^{2}-4x - 5\). We have \(x^{2}-4x - 5=(x + 1)(x - 5)\). So the inequality becomes \((x + 1)(x - 5)\geq0\).

Step3: Find the critical points

Set \((x + 1)(x - 5)=0\). Using the zero - product property \(a\times b = 0\) implies \(a = 0\) or \(b = 0\). So \(x+1 = 0\) gives \(x=-1\), and \(x - 5=0\) gives \(x = 5\).

Step4: Test intervals

We have three intervals to test: \((-\infty,-1)\), \((-1,5)\), and \((5,\infty)\).

• For the interval \((-\infty,-1)\), let \(x=-2\). Then \((-2 + 1)(-2 - 5)=(-1)\times(-7)=7>0\).
• For the interval \((-1,5)\), let \(x = 0\). Then \((0 + 1)(0 - 5)=(1)\times(-5)=-5<0\).
• For the interval \((5,\infty)\), let \(x = 6\). Then \((6 + 1)(6 - 5)=(7)\times(1)=7>0\).

Since the inequality is \((x + 1)(x - 5)\geq0\), the solution is \(x\leq - 1\) or \(x\geq5\). On the number - line, we use closed circles at \(x=-1\) and \(x = 5\) (because the inequality includes equality) and shade the regions \(x\leq - 1\) and \(x\geq5\).

Answer:

The solution to the inequality is \(x\leq - 1\) or \(x\geq5\). On the number - line, we place a closed circle at \(x = - 1\) and shade to the left, and place a closed circle at \(x = 5\) and shade to the right.