# Graphing $f(2-x)$

Sorry for this very trivial question, but I've become slightly confused by this question. Consider a graph $$y=f(x)$$. How would I draw the graph $$y=f(2-x)$$?

It seems to me that as this is obviously equal to $$y=f(-(x-2))$$ this should represent the graph being translated $$2$$ units in the positve $$x$$ direction and then reflected in the $$y$$ axis.

Is that true? It doesn't seem to be from the graphs I have plotted using Desmos. If not, please explain why it is incorrect.

EDIT: I have now slept over my problem and I believe that it lies in the following statement I have been led to believe in class:

The graph of $$f(\text{Blah}+a)$$ is ALWAYS a translation of $$a$$ units of the graph $$f(\text{Blah})$$ in the negative direction.

More specifically, I thought that the as graph of $$f(x+a)$$ is a translation of $$a$$ units of the graph $$f(x)$$ in the negative direction, then the graph of $$f(-x+a)$$ is a translation of $$a$$ units of the graph $$f(-x)$$ in the negative direction as well. After thinking it over logically however, I now think this is wrong.

This is my reasoning:

Consider $$y=f(x+a)$$. For a given $$y$$ value on the $$y=f(x+a)$$ graph, the $$x$$ value needed for it must be $$a$$ smaller than the $$x$$ value needed if it was just the function $$y=f(x)$$; hence the graph $$y=f(x+a)$$ must be the graph of $$y=f(x)$$ but shifted $$a$$ units to the negative $$x$$ direction.

But, if we consider $$y=f(-x+a)$$: For a given $$y$$ value on the $$y=f(-x+a)$$ graph, the $$x$$ value needed for it must be $$a$$ bigger than the $$x$$ value needed if it was just the function $$y=f(-x)$$; hence the graph $$y=f(-x+a)$$ must be the graph of $$y=f(-x)$$ but shifted $$a$$ units to the positive $$x$$ direction.

Is my reasoning correct now? Thanks again for your help.

• Replace every instance of $x$ with $2-x$ to see algebraically what $f(2-x)$ looks like Nov 3, 2020 at 23:09
• @Travis I know I could do that, but graphically I can't see how that helps. Nov 3, 2020 at 23:10

Denote $$g(x)=f(2-x)$$ and set $$x'=2-x$$. What you want is drawing the graph of $$g$$. Now the points $$x$$ and $$x'$$ are symmetric (on the $$x$$-axis) w.r.t. the point $$1$$ since $$\frac{x+x'}2=1$$, and $$g(x)=f(x')$$. Therefore the graph of $$g$$ is the symmetric of the graph of $$f\,$$ w.r.t. the line $$x=1$$.

• Please see my edit. Btw what does w.r.t stand for? Nov 4, 2020 at 13:38
• w.r.t.= ‘with respect to’. Nov 4, 2020 at 13:39
• I see , thanks. Nov 4, 2020 at 13:41
• Sorry, what exactly do you mean by symmetric? Symmetrical? Nov 4, 2020 at 13:45
• The midpoint of $x$ and $2-x$ is $1$. Nov 4, 2020 at 14:19

This is indeed equal to $$f(-(x-2))$$, but your interpretation of the latter is incorrect.

You identified the correct operations:

1. Translate 2 units in the positive $$x$$ direction (replace $$x$$ with $$x-2$$).
2. Reflect in the $$y$$ axis (replace $$x$$ with $$-x$$).

But what order do you have to do these in to get $$f(-(x-2))$$?

The reasoning added in revision 2 of the question, a few minutes before this edit to my answer, is correct.

• Oh, I think i get it now! Whenever we have a reflection in the $y$ axis, do we only make the $x$ negative and nothing else? Ie not make everything inside the brackets of $f()$ negative? Nov 3, 2020 at 23:13
• @A-LevelStudent Correct! Nov 3, 2020 at 23:15
• Now I'm confused for a different reason. If so, shouldn't it be a reflection in the $y$ axis first and then a translation by $2$ units in the NEGATIVE direction? Nov 3, 2020 at 23:17
• Please see my edit now. Nov 4, 2020 at 13:34

hint

If you know the graph of the curve whose equation is $$y=f(x)$$, the graph of $$y=f(-x)$$ is the symetric with respect to $$Oy$$ axis.

if you know the graph of $$y =g(x)$$ , the graph of $$y=g(x-a)$$ is got by the translation of vector $$(a,0)$$.

• Thanks, but my problem is really about the order of applying the different trasformations. Nov 3, 2020 at 23:18
• $f(-(x-2))$ implies translation and then symetry. Nov 3, 2020 at 23:24
• Please see edit now. Nov 4, 2020 at 13:35

Label the point $$x=0$$ as $$a$$ and $$x=2$$ as $$b$$. Now exchange $$a,b$$. 