What is the reasoning behind the way we describe function stretching and compressing transformations?

Question

If you have a function in the form of $f(kx)$, the graph is horizontally scaled by a factor of $k$ and the bigger the magnitude of $k$, the more compressed the graph gets, and the inverse is true.

So by definition, $k$ should be called the horizontal compression factor of the function, meaning if $k = \frac{1}{2}$, the graph is horizontally compressed by a factor of $\frac{1}{2}$, and since stretching is the inverse of compressing you could also say the graph is horizontally stretched by a factor of $2$. and using the same logic, if $k = 2$ then the graph is horizontally compressed by a factor of $2$ or horizontally stretched by a factor of $\frac{1}{2}$

But this is not the case and the accepted practice is to say the graph is compressed by a factor of $k$ if $|k|>1$ and stretched by a factor of $k$ if $0<|k|<1$

This seems extremely unintuitive and to me it doesn't use the word factor correctly. So my question is why do we describe stretching and compressing transformations like this?

Answer 1

Note that by the transformation:

$$f(x)\to f(kx)$$

1. if $|k|>1$ the graph of f is horizontally compressed

2. if $0<|k|<1$ the graph of f is horizontally stretched

Infact if for the original funtion $f(1)$ corresponds to the point $x=1$ in the new function f(1) correspond to $x=\frac1k$, thus if |k|>1 the function is compressed.

If k<0 the graph is also mirrored by the y-axis.

Let's try with $sin(kx)$:

http://www.wolframalpha.com/input/?i=plot+sin(4x),sin(2x),sinx,sin(x%2F2),sin(x%2F4)

Answer 2

If you have a function in the form of $f(kx)$, the graph is horizontally scaled by a factor of $k$ and the bigger the magnitude of $k$, the more compressed the graph gets, and the inverse is true.

So the transformation $y=f(x) \to y=f(kx)$ transforms $(x,y) \to \left(\dfrac xk, y \right)$. This point of view assumes that the graph is transformed and the $x-$axis and $y-$axis remain fixed.

You can also assume that the graph stays fixed and it is the $x-$axis and $y-$axis that are transformed. In particular, the transformations $x'=kx$ and $y'=y$ can be interpreted as laying the $x'-$axis and $y'-$ axis over the $x,y$ plane and then graphing $y'=f(x')$ in $(x',y')$ coordinates. From this point of view, the scaling is as you required.

Answer 3

There is a simple explanation behind all of this.

One thing is the transformation $T: \mathbb{R}\rightarrow \mathbb{R}$ defined by $T(x) = kx$. This is a transformation of the real line: a compression when $|k|<1$ and an expansion otherwise. Under this map, for example, the interval $[0, 1]$ is transformed into $[0, k]$.

This map induces another map on functions, which associates to each function $f(x)$ another function $g(x) = f(kx)$. Observe now that, the function $g$ behaves on the interval $[0,1]$ the same way as the function $f$ behaves on the interval $[0, k]$ (because when $x\in [0, 1]$, $kx\in [0, k]$).

This is the reason why, when $|k|>1$, the graph of $g$ looks stretched: in the interval $[0, 1]$, $g$ does the same thing that $f$ does in the larger interval $[0, k]$.