# How do you find the domain and range of y =sqrt (x^2 - 9)?

Jul 30, 2017

The domain is $\left(- \infty , - 3\right] \cup \left[3 , \infty\right)$
The range is $\left[0 , \infty\right)$

#### Explanation:

One way to find it is by graphing it, and then looking for the $y$ axis (which shows the range) and the $x$ axis (which shows the domain)

graph{sqrt(x^2-9) [-10, 10, -5, 5]}

It looks like the range is $\left[0 , \infty\right)$ (the graph isn't defined for negative numbers)

The domain is $\left(- \infty , - 3\right] \cup \left[3 , \infty\right)$ (For $x = - 3$ to $x = 3$, it is not defined).

But let's solve this algebraically.

The square root function is only defined when the number or expression under the radical sign is greater than or equal to $0$.

Therefore, $y = \sqrt{{x}^{2} - 9} \rightarrow {x}^{2} - 9 \ge 0$
Now, we set it equal to $0$
${x}^{2} - 9 = 0$
${x}^{2} = 9$
$x = - 3 , x = 3$

Now we have these points. Let us call them boundary points of the real number line. So we have 3 intervals - $\left(- \infty , - 3\right) , \left(- 3 , 3\right) , \mathmr{and} \left(3 , \infty\right)$
Just choose a point in each interval and substitute it into the original equation $\left({x}^{2} - 9 \ge 0\right)$to check if it is valid or not, and each boundary point to see if it is in the domain or not. We find that $- 3$ and $3$ is defined, $\left(- \infty , - 3\right)$ and $\left(3 , \infty\right)$ is also defined but $\left(- 3 , 3\right)$ is not. So since $\left[\right]$ means including, we get the domain is $\left(- \infty , - 3\right] \cup \left[3 , \infty\right)$. We put $\left(\right)$ around the infinity symbols because infinity is never reached, so it is not an included value.

For the range, just think about this - the lowest value a square root function can give is $0$, because the lowest point where the square root function is defined is when it is $0$. $\sqrt{0}$ IS defined, whereas $\sqrt{- 0.000001}$ is not. So it is simply $\left[0 , \infty\right)$ - and including $0$.