What is the domain of the function$\sqrt{x^2-16}$?

The given function is$\sqrt{x^2-16}$.

• The square root function $\sqrt{x^2-16}$is defined when localid="1646193121230" $x^2-16\ge 0$.
• Consider localid="1646193292700">g(x)=x2-16and factor the expression.

localid="1646193298150" $g\left(x\right)=(x-4)(x+4)$

• localid="1646193302889" $g\left(x\right)=0$implies $(x-4)=0$or$(x+4)=0$.
• So, the x -intercepts of the curve are $(4,0),(-4,0)$.
• The value of localid="1646193316302" $g\left(0\right)=-16$, so, the y-intercept is $(0,-16)$.
• Find the vertex of the curve as follows:

$\begin{array}{rcl}x& =& \frac{-b}{2a}\\ & =& 0\end{array}$

• So, the vertex is also at the y-intercept$(0,-16)$.

• Plot the graph of the function by using the obtained intercepts and vertex.

• From the graph, the region on and above the horizontal axis is $x\le -4\mathrm{or}x\ge 4$.
• The inequality is of the greater than or equal to type, so the solution set is $x\le -4\mathrm{or}x\ge 4$.