## 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 $$\frac{x-C}{B}$$).