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Take the function $\ln(3 - x)$. By the logic of transformations that I have been taught, the order of transformations goes: $\ln(x)$ to $\ln(-x)$ which reflects the curve across the $y$-axis, then $\ln(-x + 3)$. This additional $+3$ should then push the curve to the left, hence turn the asymptote to $x = -3$. (This is what I show in the black curve.) However, the way the graph is shown in my book is that the transformation actually makes the asymptote $x = 3$, which disagrees with me, although the direction of the graph relative to then $x$-axis stays is the same as mine.

Why? Someone please explain. I asked my teacher, and he said 'yea that's weird', and I am yet to find an answer.

This image is a drawing of what I've said below. It sums it all up.

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    $\begingroup$ $\ln(-x+3)=\ln(-(x-3))$ $\endgroup$
    – Dave
    May 21, 2017 at 14:41
  • $\begingroup$ This basically implies that you go from ln(x) to ln(x-3), which is shifting the curve to the right by 3, and then we reflect that in the y-axis. Still gives the black curve in my picture, and not the correct red curve. Please point out which part of my logic is wrong. This is how I've been taught. $\endgroup$ May 21, 2017 at 14:48
  • $\begingroup$ The process you just described does not result in the black curve, but rather the red one. If you transform $\ln(x)$ to $\ln(x-3)$, the curve indeed shifts to the right $3$ units. However, when you know transform $\ln(x-3)$ to $\ln(-(x-3))$, the curve simply reflects across the line $x=3$ (since this line acts like the new $y$ axis for $\ln(x-3)$), and does not create a new asymptote at $x=-3$ (i.e. the reflection of $\ln(x-3)$ to $\ln(-(x-3))$ does not reflect across the line $x=0$). $\endgroup$
    – Dave
    May 21, 2017 at 14:53
  • $\begingroup$ That's weird, I know I'd be taking up your time, but may I please ask that you explain why the line x=3 becomes the 'new y-axis'? $\endgroup$ May 21, 2017 at 14:56
  • $\begingroup$ What I mean by "like the new $y$ axis" is just that the line $x=3$ is the reflection line for $\ln(x-3)$ to $\ln(-(x-3))$, much like how the $y$ axis (i.e. $x=0$) is the reflection line for $\ln(x)$ to $\ln(-x)$. $\endgroup$
    – Dave
    May 21, 2017 at 14:58

2 Answers 2

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Elaborating on @Dave's comment, here's the sketch of what is going on.

  • You know that if you have a function $f(x)$, then $f(-x)$ reflects the function over the $y$-axis.

  • You know that if you have a function $f(x)$, then $f(x+3)$ shifts the function by $3$ to the left.

Now, consider your function:

  • Starting with $f(x)=\ln(x)$, you look at $f(-x)=\ln(-x)$, which reflects the function over the $y$-axis.

  • Now, let's call the new function we're dealing with $g$. In other words, $g(x)=\ln(-x)$. We would like to shift $3$ to the left, so we look at $g(x+3)=\ln(-(x+3))=\ln(-x-3)$.

  • If you want to shift to the right by $3$, we could apply the change $g(x-3)=\ln(-(x-3))=\ln(3-x)$, which is what you get above.

The trick is that when you substitute $x+3$, you can't put the "$+3$" in anywhere, you have to replace $x$ by $x+3$ wherever it appears and remember to distribute. Therefore, in your case, the negative in front of the $x$ must be distributed to the $+3$ as well.

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  • $\begingroup$ Wow, an amazing answer. Thank you very much :) I see the flaw in my logic now. Thank you! $\endgroup$ May 21, 2017 at 14:53
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@Dave more or less sums it up in the comment above. When you're changing $-x$ into $3-x$, you can also see that as changing $x$ into $x-3$. This should and does move the graph to the right.

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