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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.

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

4 Answers 4

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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 $x=1$.

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

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  • $\begingroup$ 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? $\endgroup$ Commented Nov 3, 2020 at 23:13
  • $\begingroup$ @A-LevelStudent Correct! $\endgroup$ Commented Nov 3, 2020 at 23:15
  • $\begingroup$ 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? $\endgroup$ Commented Nov 3, 2020 at 23:17
  • $\begingroup$ @A-LevelStudent Please edit the question to show your working here. Then we can help you more. $\endgroup$ Commented Nov 3, 2020 at 23:19
  • $\begingroup$ Please see my edit now. $\endgroup$ Commented Nov 4, 2020 at 13:34
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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)$.

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  • $\begingroup$ Thanks, but my problem is really about the order of applying the different trasformations. $\endgroup$ Commented Nov 3, 2020 at 23:18
  • $\begingroup$ $ f(-(x-2))$ implies translation and then symetry. $\endgroup$ Commented Nov 3, 2020 at 23:24
  • $\begingroup$ Please see edit now. $\endgroup$ Commented Nov 4, 2020 at 13:35
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Label the point $x=0$ as $a$ and $x=2$ as $b$. Now exchange $a,b$.

enter image description here

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