Parker Bird

2022-07-20

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

Killaninl2

Beginner2022-07-21Added 20 answers

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

$x-8\ge 0$

$x\ge 8$

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

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

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

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

You can also see this by inspection in the graph

graph{$\sqrt{x-8}[-1,17,-2,5]$}

$x-8\ge 0$

$x\ge 8$

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

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

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

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

You can also see this by inspection in the graph

graph{$\sqrt{x-8}[-1,17,-2,5]$}

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