Question

graph the solution to the following inequality on the number line.
$x^{2}-4x<12$

Explanation:

Step1: Solve the quadratic inequality

First, rewrite \(x^{2}-4x < 12\) as \(x^{2}-4x - 12<0\).
Factor the quadratic expression: \(x^{2}-4x - 12=(x - 6)(x+ 2)\).
So, the inequality becomes \((x - 6)(x + 2)<0\).

Step2: Find the critical points

Set \((x - 6)(x + 2)=0\).
Using the zero - product property \(x-6 = 0\) gives \(x = 6\), and \(x + 2=0\) gives \(x=-2\).

Step3: Determine the sign of the quadratic function

We consider the intervals \((-\infty,-2)\), \((-2,6)\) and \((6,\infty)\).
Take a test point from each interval:

• For \(x=-3\) (interval \((-\infty,-2)\)): \((-3 - 6)(-3 + 2)=(-9)\times(-1)=9>0\).
• For \(x = 0\) (interval \((-2,6)\)): \((0 - 6)(0 + 2)=(-6)\times2=-12<0\).
• For \(x = 7\) (interval \((6,\infty)\)): \((7 - 6)(7 + 2)=1\times9 = 9>0\).

Since we want \((x - 6)(x + 2)<0\), the solution of the inequality \(x^{2}-4x<12\) is the interval \(-2

Answer:

On the number - line, we use an open circle at \(x=-2\) and \(x = 6\) (because the inequality is strict, i.e., \(<\) not \(\leq\)) and shade the region between \(x=-2\) and \(x = 6\).