# Graphing inverse of parabola, should the domain be restricted or not?

Find the inverse of $f(x)=y=x^2-2x+6$ for $x \ge 1$

The inverse of that above function is $|y-1|=\sqrt{(x-5)}$

The domain of this function requires $x>5$.

There are now two functions to graph, $1+\sqrt{x-5}$ when $y \ge 0$ and $1-\sqrt{x-5}$ when $y<0$

Am I correct? When I type this into online calculators like wolfram alpha, it shows me a sideways parabola. That is, it graphs both of these functions for $x>5$. Is my approach correct, or is wolframalpha correct?

My approach is I graph both functions $1+\sqrt{x-5}$ from $x \ge 6$ and $1-\sqrt{x-5}$ from $5 \le y < 6$

• It might be a good exercise to see the graph of the function defined piecewise by your last sentence, but see my answer below for more. – Travis Sep 7 '14 at 4:54

In the specification of the function $f$, we've restricted the domain $x \geq 1$, which by construction must be the image of the inverse. We can see that of the two branches (the "two functions to graph") only one takes values $y \geq 1$, namely, $y = 1 + \sqrt{x - 5}$, so the inverse of the function must be given by just that half.
Another way of seeing the issue at hand is this: If the graph of $f^{-1}$ comprised both halves of the parabola, it would fail the Vertical Line Test, and so it wouldn't be a function in the first place.