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 ($f(x)=\sqrt{x}$):

Graph of the real square root of x with x from -20 to 20.

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

Previous graph, shifted over two units to the right.

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?

Previous graph, reflected horizontally accross the point of origin of the line.

I would expect the final graph to look like this:

Same graph, but reflected across the y-axis.


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.

  • $\begingroup$ Aren't you supposed to follow the order of operations with transformations? Thus the transformation inside the parentheses should be done first, right? $\endgroup$ – Ian Sep 29 '14 at 21:20
  • $\begingroup$ We follow order of operations only when we try to evaluate or simply something. In this case, we try to transform $f(x)$ into $f(2-x)$ (in some way); by doing so, since we know the graph of $f(x)$, we can obtain the graph of $f(2-x)$ by following the transformations we performed. In (1), we can transform $f(x)$ into $f(-x)$ by replacing every $x$ with $-x$; by doing so, we reflect the graph across the $y$-axis. As for (2), we replace every $x$ with $x-2$; by doing so, we translate to the right by 2 units. $\endgroup$ – E W H Lee Sep 29 '14 at 21:26
  • $\begingroup$ Alright, that makes more sense. $\endgroup$ – Ian Sep 29 '14 at 21:29

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.

  • $\begingroup$ I apologize, I had mistyped in my original question. It should have been $-(x-2)$ not $-(x+2)$. It's fixed now. $\endgroup$ – Ian Sep 29 '14 at 20:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.