I did a search on the order of transformations applied to graphs, and mostly found the following, e.g. in this post.

Given a function $f$ always perform transformations $$Af(Bx+C)+D$$ in the order $C,B,A,D$.

But after doing a little digging I'm not sure this is correct. For example, the functions $$y=\frac{1}{\frac{1}{2}x+1}+3\tag{1}$$ and $$y=\frac{2}{x+2}+3\tag{2}$$ are identical.

But suppose $f(x)=1/x$. Using the above approach on (1) transforms $(1,1)$ to $(0,4)$, whilst (2) transforms $(1,1)$ to $(-1,5)$.

Can someone see what might be going on here and perhaps explain? I must have a mental block on this one...


The first function performs the following transformations to $f(x)=\frac{1}{x}$:

  • Shift left 1 unit
  • Stretch horizontally by a factor of 2
  • Shift up 3 units

while the second performs the following:

  • Shift left 2 units
  • Stretch vertically 2 units
  • Shift up 3 units

As you have noted, these are not the same transformation. However, they both map the graph of $f(x)$ to the graph of $g(x)=\frac{1}{\frac{1}{2}x+1}+3=\frac{2}{x+2}+3$. They do not necessarily map each $(x,y)$ to the same point (in fact you showed they don't), but they both work.

This might seem strange to you, but consider an even simpler example: $$\frac{1}{\frac{1}{2}x}=\frac{2}{x}.$$ The first function says we should stretch $\frac{1}{x}$ horizontally by a factor of $2$, while the second says we should stretch it vertically by a factor of $2$. These are not the same transformations on $\mathbb{R}^2$, but they have the same effect on the function $\frac{1}{x}$.

For another example, consider $f(x)=\ln(x)$. We know that $\ln(ax)=\ln(a)+\ln(x)$. So for this choice of $f(x)$, compressing the function horizontally by a factor of $a$ is equivalent to shifting it vertically by $\ln(a)$ units. Again, these are not the same transformation on $\mathbb{R}^2$, but they have the same effect on the function $\ln(x)$.

  • $\begingroup$ So what's really going on is that $$\left\{g_1(x)\mid x\in\mathbb{R}\right\}=\left\{g_2(x)\mid x\in\mathbb{R}\right\}.$$ The functions $g_1$ and $g_2$ transform $f$ to the same graph, even though individual point transformations may not be the same. Makes sense now. $\endgroup$ – Pixel May 15 at 17:37
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    $\begingroup$ More like $\{T_1(x,y) \mid (x,y) \in \mathbb{R}^2\}=\{T_2(x,y) \mid (x,y) \in \mathbb{R}^2\}$ where $T_1$ and $T_2$ are the two transformations defined above, while it's not necessarily the case that $T_1 (x,y) \neq T_2(x,y)$. It is the case that $g_1(x)=g_2(x)$ for all $x \in \mathbb{R}$ if $g_1$ and $g_2$ are the same function. $\endgroup$ – kccu May 15 at 18:54
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    $\begingroup$ yes, that makes it clearer. I found that $T_{g_1}(1,1)=(0,4)$ and $T_{g_2}(2,1/2)=(0,4)$, and indeed the points $(1,1)$ and $(2,1/2)$ are on $f(x)=1/x$. $\endgroup$ – Pixel May 16 at 8:20

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