# Confusion over order of transformations of graphs

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 $$\ln(x)$$.

• 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. – Pixel May 15 at 17:37
• 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. – kccu May 15 at 18:54
• 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 $f(x)=1/x$. – Pixel May 16 at 8:20