Answer
See below:
Work Step by Step
Consider the function,
$ f\left( x \right)=\sqrt{x}$ and $ g\left( x \right)=\sqrt{x-2}+1$
The parent function for $ g\left( x \right)=\sqrt{x-2}+1$ is $ f\left( x \right)=\sqrt{x}$.
If h is a positive real number, the graph of $ f\left( x \right)-h $ is the graph of $ y=f\left( x \right)$ shifted downward h units.
If c is a positive real number, then the graph of $ f\left( x+c \right)$ is the graph of $ y=f\left( x \right)$ shifted to the left c units.
Therefore, the graph of $ g\left( x \right)=\sqrt{x-2}+1$ is the graph of $ f\left( x \right)=\sqrt{x}$ shifted to the right $2$ units and upward $1$ unit.
The point $\left( 2,\sqrt{2} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 2,1 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$.
The point $\left( 6,\sqrt{6} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 6,3 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$.
The point $\left( 3,\sqrt{3} \right)$ on the graph of $ f\left( x \right)=\sqrt{x}$ corresponds to $\left( 3,2 \right)$ on the graph of $ g\left( x \right)=\sqrt{x-2}+1$.
