Graphing $f(2-x)$

Question

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.

Thanks for your help.

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.

Answer 1

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 $x=1$.

Answer 2

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 $f(-(x-2))$?

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

Answer 3

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)$.

Answer 4

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