Why isn't $f(x)=\sqrt{2-x}$ reflected across the y-axis? If I try to graph this function, it does not appear to reflect across the y-axis when it comes time to do the reflection. Rather, it is reflected around the point where the function begins on the graph.
Here is what I tried:
$$\sqrt{2-x}=\sqrt{-(x-2)}$$
This makes it easier to graph the transformations:
Root function ($f(x)=\sqrt{x}$):

With the right horizontal shift ($f(x)=\sqrt{x-2}$): 

With a horizontal reflection ($f(x)=\sqrt{-(x-2)}$):
This is the part I'm confused about. Why doesn't it reflect across the y-axis?

I would expect the final graph to look like this:

 A: As you observed that $f(2-x) = f\big(-(x-2)\big)$. Since
$$
f(x) \quad \stackrel{(1)}{\longrightarrow} \quad f(-x) \quad \stackrel{(2)}{\longrightarrow} \quad f\big(-(x-2)\big),
$$
we can obtain $f(2-x)$ from $f(x)$ by transformations that correspond to (1) and (2), where
(1) is reflection across the $y$-axis;
(2) is translation to the right by $2$ units.
A: When it comes to combining transformations, you should think of it as a replacement procedure. Shifting two units to the right means replacing $x$ with $x - 2$. A reflection over the $y$-axis means replacing $x$ with $-x$. So if we apply these transformations in the given order, then we would get $y = f(-x - 2) = \sqrt{-x - 2}$.
On the other hand, the function $y = f(-(x - 2))$ suggests that we applied the two transformations in reverse order. That is, we first reflected the function over the $y$-axis, then shifted the resulting graph two units to the right. Since expanding yields $y = f(-x + 2)$, notice that this is equivalent to first shifting the original graph two units to the left, then reflecting over the $y$-axis.
