Shifty Stretch

Starting with the graph $y = f(x)$ (red graph), if we shift it to the right by 4, we get the graph of $y = f(x-4)$ (blue graph).

If we stretch the graph of $y = f(x-4)$ (blue graph) horizontally by 2, we get the graph $y = f ( \frac{x}{2} - 4 )$ (green graph).

Since $f ( \frac{x}{2} - 4) = f\left( \frac{1}{2} ( x - 8 ) \right)$ , this graph can also be obtained by a horizontal stretch of $2$ , and then a horizontal shift to the right by 8 .

We can choose any graph to represent this, so for simplicity start with

$f(x)=x$

Then apply the shift right 4:

$f(x)=x+4$

Then the horizontal stretch of 2:

$f(x)=2(x+4)=2x+8$

$f(x)=2x+8$ can be described as being stretched horizontally by 2 then shifted right 8.

The transformation is a translation followed by a horizontal expansion. The problem can be reduced by projection onto the $x$ -axis. Then we have that $x \mapsto 2(x+4) = 2x + 8$ that we can in fact be read as a horizontal expansion of factor $2$ followed by a right shift by $8$ . More formally, if $O$ is the origin, $\vec{e}$ the unit vector of $x$ -axis, and $\eta_{O,\lambda}$ the homothety of center $O$ and coefficient $\lambda$ , $\eta_{O} \circ \tau_{4\vec{e}} = \eta_{O,2} \circ \tau_{4\vec{e}} \circ (\eta^{-1}_{O,2} \circ \eta_{O,2}) = (\eta_{O,2} \circ \tau_{4\vec{e}} \circ \eta^{-1}_{O,2}) \circ \eta_{O,2}.$ In virtue of the stabilizer theorem, the conjugate of $\tau_{4\vec{e}}$ by the homothety $\eta_{O,2}$ is the tranlation whose vector is just $\eta_{O,2}(4\vec{e})=8\vec{e}$ .