A calculus of the absurd

4.6.2 X-axis transformations

To transform a function \(f(x)\) in the X-axis, we just evaluate \(f(g(x))\), where \(g(x)\) is a function which maps values from the \(x\)-\(y\) plane (i.e. the usual set of axis we plot things on) to one in the \(g(x)\)-\(y\) plane (i.e. like the usual set of axis we plot things on, except that wherever we had \(x=a\) (where \(a\) stands for any number) we now want \(g(x)=a\)).

This deserves a bit of explanation. Let’s imagine that \(g(x) = x - 2\). If we plot \(f(g(x))\) against \(g(x)\), we might get something like this (for this specific \(f(x)\))

(-tikz- diagram)

We don’t want a graph of \(f(g(x))\) against \(g(x)\), though! We want one of \(f(g(x))\) against \(x\). To do this, we need to work out how to write \(g(x)\) in terms of \(x\), and then work out where every point on the \(g(x)\)-axis should be on the \(x\)-axis.

As \(x - 2 = g(x)\) if we add two to each side, we obtain that \(x = g(x) + 2\). This means that if we shift every point on the \(g(x)\)-axis two to the right then we would have the X-axis!

Thus, the graph of \(f(g(x)) = f(x-2)\) and looks like

(-tikz- diagram)

We can transform the X-axis in many ways, another one is stretching the graph. For example, if we set \(g(x) = \frac {1}{2} x\), then to work out where every point on the \(g(x)\)-axis should be on the X-axis, we first rearrange \(g(x)\), obtaining that

\begin{equation} x = 2g(x) \end{equation}

and thus we stretch (not, as commonly misconceived, squish) the graph. I try to visualise it as the graph stretching as the infinite number of points on the axis are doubled (moved twice as far away as they once were).