We shift the graph of $f(x)=\sqrt{x}$ upwards by $\green q$.

The new function becomes $$f(x)=\sqrt{x}+\green q$$

Then the origin also shifts upwards by $\green q$ and becomes equal $\rv{0, \green q}$.

Therefore the range of the function becomes equal to $\ivco{\green q}{\infty}$. The domain does not change.

We shift the graph of $f(x)=\sqrt{x}$ to the right by $\blue p$.

The new function becomes $$f(x)=\sqrt{x-\blue p}$$

Then the origin also shifts to the right by $\blue p$ and becomes equal to $\rv{\blue p, 0}$.

Therefore the domain of the function becomes equal to $\ivco{\blue p}{\infty}$. The range does not change.

We multiply the graph of $f(x)=\sqrt{x}$ by $\purple a$ relative to the $x$-axis.

The new function becomes $$f(x)=\purple a \sqrt{x}$$

If $\purple a \gt 0$, the origin does not change. Domain and range also remain the same as with the old function.

When multiplying by $\purple a \lt 0$ the function reverses. The origin and the domain remain the same, but the range becomes equal to $\ivoc{-\infty}{0}$.

If $\purple a =-1$, then we have a reflection across the $x$-axis of the old graph.