Order of Transformations Proof

Transformations of Graphs

The transformations relating to the \(y\) value make intuitive sense; if I wanted to shift upward by \(k\) I would add \(k\) to each \(y\) value (i.e. \(f(x)+k\)), and if I wanted to stretch I would simply multiply each \(y\) value by \(k\) (i.e. \(kf(x)\)).

Consider the transformation of shifting right by \(k\). Essentially I want any point \((x,y)\) on the graph to move to the point \((x+k,y)\). I.e. my new function (let's call it \(g(x)\)) is such that \(g(x+k)=y=f(x)\), since we want \((x+k,y)\) to be a point but we also know that \(y=f(x)\). So we have:

\[g(x+k)=f(x)\]

now let \(x=u-k\) so that we can see what \(g(x)\) (our new function) is more clearly:

\[g(u-k+k)=f(u-k)\]

\[g(u)=f(u-k)\]

So hence we have shown that our new function, \(g(x)\) is equal to \(f(x-k)\).

We can do a similar argument for horizontal stretches:

Point \((x,y)\to (kx,y)\), meaning that \(g(kx)=y=f(x)\).

\[\therefore g(kx)=f(x)\]

Let \(x=\frac{u}{k}\):

\[g\left(k \times \frac{u}{k}\right)=f\left(\frac{u}{k}\right)\]

\[g(u)=f\left(\frac{u}{k}\right)\]

Therefore \(g(x)=f\left(\frac{x}{k}\right)\).

 

Order of Transformations

For transformations of a function \(f(x) \to Af(Bx+C)+D\), the transformations must be performed in the order \(C,B,A,D\).

To explain this, we can use the same reasoning as above. It doesn't actually matter if we perform the \(x\) transformations first or the \(y\) transformations, i.e. the order could be \(A,D,C,B\). So all we have to show is that the order of \(x\) transformations is \(C,B\) and the order of \(y\) transformations is \(A,D\). The \(y\) transformations are intuitive; you stretch first then add, otherwise you would be stretching both the \(y\) and the bit that you added first, giving \(A(y+D)=Ay+AD\) (i.e. the wrong transformation).

With the \(x\) transformations we can use the same argument as before:

We want \((x,y) \to (Bx+C,y)\), meaning \(g(Bx+C)=f(x)\).

\[\therefore g(Bx+C)=f(x)\]

Let \(x=\frac{u-C}{B}\):

\[g\left(B \times \frac{u-C}{B}-C\right)=f\left(\frac{u-C}{B}\right)\]

\[g(u)=f\left(\frac{u-C}{B}\right)\]

Therefore \(g(x)=f\left(\frac{x-C}{B}\right)\).  I.e. we must perform C first (giving \(x-C\)) and then B (giving \(\frac{x-C}{B}\)).

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