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# Graphing Radical Functions

## Graphing Radical Functions 1.5 - Solution

From the diagram, we see that $f(x)$ passes through $(\text{-} 2,\text{-} 2)$ and $(2,0).$ By substituting the $x$-coordinates of these points into the function, we can determine if this is the case.
$f(x)=\sqrt{x-2}-2$
$f({\color{#0000FF}{\text{-} 2}})=\sqrt{{\color{#0000FF}{\text{-} 2}}-2}-2$
$f(\text{-} 2)=\sqrt{\text{-} 4}-2$
We can't calculate the square root of a negative number. Therefore, the function does not pass through this point. Let's do the same thing for the second point.
$f(x)=\sqrt{x-2}-2$
$f({\color{#0000FF}{2}})=\sqrt{{\color{#0000FF}{2}}-2}-2$
Simplify right-hand side
$f(2)=\sqrt{0}-2$
$f(2)=\text{-} 2$
When $x=0,$ the function's $y$-value is $\text{-} 2.$ Therefore, the function does not go through $(2,0)$ either.

### Drawing the correct graph

To draw the graph correctly, we should first find the lower limit of the domain. As already explained, the radicand has to be nonnegative, so by setting it equal to zero, we can find the domain's lower limit. \begin{aligned} x-2=0 \quad \Leftrightarrow \quad x=2 \end{aligned} When we know the lower limit of the domain, we can find it's corresponding $y$-value.
$f(x)=\sqrt{x-2}-2$
$f({\color{#0000FF}{2}})=\sqrt{{\color{#0000FF}{2}}-2}-2$
Simplify right-hand side
$f(2)=\sqrt{0}-2$
$f(2)=\text{-} 2$
The graph passes through $(2,\text{-} 2).$ By setting the function equal to $0,$ we can find where it crosses the $x$-axis.
$f(x)=\sqrt{x-2}-2$
${\color{#0000FF}{0}}=\sqrt{x-2}-2$
Solve for $x$
$2=\sqrt{x-2}$
$4=x-2$
$6=x$
$x=6$
The graph also passes through $(6,0).$ With this, we can graph the correct equation.