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#.
