The preimage (Red) has the form of the square root function, but you need to translate it to 3 units to the left and 3 units down.
The square root function is:
[tex]f(x)=\sqrt[]{x}[/tex]
In order to do the translation 3 units left, you need to add 3 inside of the parent function:
[tex]f(x)=\sqrt[]{x+3}[/tex]
Now, to translate it vertically, you need to subtract 3 on the outside of the square root function:
[tex]f1(x)=\sqrt[]{x+3}-3[/tex]
If you graph this function, you will prove that this is the pre-image (red) function on your question:
Now, to find the image (green function), you need to translate the pre-image 6 units up (you can do this by subtracting 6 from the inside of the parent function), and 6 units to the right (you have to add 6 outside of the square root), thus, the function will look like:
[tex]\begin{gathered} f2(x)=\sqrt[]{x+3-6}-3+6 \\ f2(x)=\sqrt[]{x-3}+3 \end{gathered}[/tex]
And this is the graph of the pre-image and image:
