In which order do I graph transformations of functions?
The 6 function transformations are:
Vertical Shifts
Horizontal Shifts
- Reflection about the x-axis
- Reflection about the y-axis
- Vertical shifting or stretching
- Horizontal shifting or stretching
Tell me if I'm wrong, but I believe that in any function, you have to do the stretching or the shrinking before the shifting. But where do the reflections fall in this process?
$\endgroup$ 14 Answers
$\begingroup$$$y=Af(B(x+\frac{C}{B}))+D$$
Can be thought of taking $f(x)=y$ and performing the following substitution.
$(x,y) \mapsto (Bx+C, \frac{y-D}{A})$
In order to understand what works and what doesn't work you need to understand what's going on.
Here is what is going on:
Let's say you have some function $y=f(x)$, it has some graph. This graph is a set $G$ consisting of points $(x,y)$ where $x$ is in the domain of the function.
If you consider $f(x,y)=y-f(x)=0$ then for every substitution you perform you'll witness an inverse mapping in the graph.
For example say we perform $x \mapsto x+1$, so now we have $y-f(x+1)=0$. You might expect the graph to be composed of points $(x+1,y)$ with respect to the old graph, but this is not true rather it is composed of points $(x-1,y)$, i.e. a shift left.
On the other hand say we perform $x \mapsto 2x$, now we have $y-f(2x)=0$. Now because the inverse of the mapping $x \mapsto 2x$ is $x \mapsto \frac{1}{2}x$ now the points become,
$$(\frac{1}{2}x,y)$$
Sometimes a combination of shifts, dilations, etc are needed, for example $y=x^2$ to $y=(2x+1)^2+1$ requires the substitution $(x,y) \mapsto (2x+1,y-1)$ whose inverse $(x,y) \mapsto (\frac{x-1}{2},y+1)$ tells you exactly what to do to the graph.
Computing the inverse of $(x,y) \mapsto (Bx+C, \frac{y-D}{A})$ will tell you everything you want to know.
I get $(x,y) \mapsto (\frac{x-C}{B},Ay+D)$. (You can perform this on points in your graph, one step at a time, in whichever way makes sense).
For example first shifting all $x$ coordinates to the left $C$, then scaling them by $\frac{1}{B}$, then scaling $y$ coordinates by $A$, then shifting up by $D$ makes sense.
But, doing all the same for $x$ and then shifting up $y$ by $D$ to get to $y+D$ then scaling by $A$ to get to $A(y+D)$ doesn't make sense!
$\endgroup$ 0 $\begingroup$For $Af (Bx+C)+D$ perform the operations in order: C, B , $A $, $D $. For the reflection, say $-A $, it does not matter if you stretch or shrink by $A $ and then reflect. Try an example with a simple function like $-3x^2$.
$\endgroup$ 2 $\begingroup$if you want to plot an expression like:
$y=a\cdot f[k\cdot (x-b)]+c$
then the connection to the parent function is:
$f[k\cdot (x-b)]=Y, where Y=f(X)$
then the y-coordinate of the point on the transformed function becomes:
$y=a\cdot Y+c$
and the x-coordinate is deducted from;
$Y=f(X)=f([k\cdot (x-b)]\Rightarrow X=k\cdot (x-b)\Rightarrow x=\frac{X}{k}+b$
therefore starting with the point $(X,Y)$ on the parent function, the chain of transformation is this:
$(X,Y)\rightarrow (\frac{X}{k}+b,a\cdot Y+c)$
I do the horizontal transformations first:
1.$(X,Y)\rightarrow(\frac{X}{k},Y)$: horizontal stretch/compression and reflection in Y-axis when k<0
2.$(\frac{X}{k},Y)\rightarrow(\frac{X}{k}+b,Y)$: horizontal shift
then I do the vertical transformations:
3.$(\frac{X}{k}+b,Y)\rightarrow(\frac{X}{k}+b,a\cdot Y)$: vertical stretch/compression and reflection in x-axis when a<0
4.$(\frac{X}{k}+b,a\cdot Y)\rightarrow(\frac{X}{k}+b,a\cdot Y+c)$: vertical shift
$\endgroup$ $\begingroup$- Scale (Stretch or shrink) Vertical or horizontal order do not matter.
- Reflect Vertical or horizontal order do not matter.
- Shift Vertical or horizontal order do not matter.
