If the function,
[tex]f(x)=(x-9)^2+3[/tex]
is invertible then it must pass the horizontal line test.
But as illustrated in the diagram above, this graph (in blue) failed to pass this test, because the horizontal line intersects the graph at more than one point.
This is because the domain is the entire real numbers.
However, if we restrict the domain to
[tex]x\ge 9[/tex]
the graph will now pass the horizontal line test as illustrated in the red graph above.
We can now find the inverse. To do that we let,
[tex]y=(x-9)^2+3[/tex]
Next, we interchange x and y.
[tex]x=(y-9)^2+3[/tex]
We make y the subject again.
[tex]x-3=(y-9)^2[/tex]
Take square root of both sides.
[tex]\sqrt{x-3}=y-9[/tex]
Solve for y.
[tex]\sqrt{x-3}+ 9=y[/tex]
Hence the inverse,
[tex]f^{-1}(x)=\sqrt{x-3}+9[/tex]
for
[tex]x\ge 9[/tex]
