The formula y=f(x)=sqrt{x-2} definitely defines a function (for each x, you get one output f(x) that is the positive square root of x-2...the graph passes the vertical line test). Its domain is the set of all x such that x\geq 2 and its range is the set of all y such that y\geq 0.
Moreover, the function passes the horizontal line test and you can solve for x in the equation y=f(x)=sqrt{x-2} to find the inverse function: y=sqrt{x-2}\Rightarrow y^2=x-2\Rightarrow x=y^2+2.
It is technically find to write the inverse function as x=f^{-1}(y)=y^2+2. However, for the purposes of graphing both functions in the same picture and seeing the reflection property across the line y=x, you can now swap the variables and write y=f^{-1}(x)=x^2+2. (If the variables represent specific real-life quantities, you should not do this swapping).
You should only graph y=f^{-1}(x)=x^2+2 for x\geq 0 (restrict the domain) since the set of all non-negative real numbers was the range of the original function. The range of the inverse function is the set of all y such that y\geq 2.
This half-parabola is definitely a function and definitely passes the vertical line test (on the restricted domain) because the original function passed the horizontal line test.
Also note that:
(f \circ f^{-1})(x)=f(f^{-1}(x))=f(x^2+2)=sqrt{x^2+2-2}=sqrt{x^2}=abs(x)=x for x\geq 0 and
(f^{-1}\circ f)(x)=f^{-1}(f(x))=f^{-1}(sqrt{x-2})=(sqrt{x-2})^2+2=x-2+2=x for all x\geq 2.
