How do you find the domain and range of $\sqrt{x - 8}$ $sqrt(x-8)$ ?

Question

6383 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-29T16:23:05

Domain of $x$ $x$ is $[8, \infty)$ $[8, infty)$
Range of $y$ $y$ is $[0, \infty)$ $[0,infty)$

The square root is real only when the radicand is positive, or at least equal to zero. So the domain is going to be whenever

$x - 8 \geq 0$ $x-8 >= 0$

$x \geq 8$ $x >= 8$

Using interval notation, we say the domain of $x$ $x$ is $[8, \infty)$ $[8, infty)$

The range is all the the $y$ $y$ values that result from this domain. So the range starts at $x = 8$ $x=8$

$y = \sqrt{8 - 8} = \sqrt{0} = 0$ $y=sqrt(8-8)=sqrt(0)=0$

and goes up to infinity. Using interval notation the range of $y$ $y$ is $[0, \infty)$ $[0,infty)$

You can also see this by inspection in the graph

graph{sqrt(x-8)[-1,17,-2,5]}

How do you find the domain and range of √x−8sqrt(x-8)?