Reflections and Transformations.

Off[General::"spell1"]; Off[Plot::"plnr"]; Off[Interpolation::"inhr"]; Needs["Miscellaneous`Units`"]Needs["Graphics`Colors`"]

Reflections

Basic Function Graphs

We need to be familiar with six basic functions and their graphs:

code

Clear[i, s, c, sr, av, r, x]; i[x_] := xs[x_] := (x^2)c[x_] := (x^3)sr[x_] := N[(Sqrt[x])]av[x_] := N[Abs[x]]r[x_] := N[(1/x)]

identity = 
      Plot[i[x], {x, -50, 50}, 
        PlotStyle -> {{Thickness[0.01], Green}}, 
        PlotLabel -> "identity", Ticks -> None, 
        DisplayFunction -> Identity];

absvalue = 
      Plot[av[x], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.01], Yellow}}, 
        PlotLabel -> absolute value, Ticks -> None, 
        DisplayFunction -> Identity];

square = 
      Plot[s[x], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.01], Mint}}, 
        PlotLabel -> "square", Ticks -> None, 
        DisplayFunction -> Identity];

cube = 
      Plot[c[x], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.01], Turquoise}}, 
        PlotLabel -> "cube", Ticks -> None, 
        DisplayFunction -> Identity];

Null

squareroot = 
      Plot[sr[x], {x, 0, 10}, PlotStyle -> {{Thickness[0.01], Purple}}, 
        PlotLabel -> "square root", Ticks -> None, 
        DisplayFunction -> Identity];

rational = 
      Plot[r[x], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.01], Violet}}, 
        PlotLabel -> "rational", Ticks -> None, 
        AspectRatio -> Automatic, DisplayFunction -> Identity];

Here are some basic graphs that you should recognize.

code

Show[
      GraphicsArray[{{identity, square}, {cube, rational}, {squareroot, 
            absvalue}}, GraphicsSpacing -> {0.3, 0}, 
        DisplayFunction -> $DisplayFunction]];

Reflect Across X Axis

Let's start with the absolute value function.  I
want to reflect it across the x axis.

code

negabs = 
      Plot[-av[x], {x, -10, 10}, Ticks -> None, 
        PlotStyle -> {{Thickness[0.01], Turquoise}}, 
        AspectRatio -> Automatic, DisplayFunction -> Identity];

both = 
      Show[{absvalue, negabs}, DisplayFunction -> $DisplayFunction];

Looks like a mirror image of the absolute value 
function.  Exactly what I wanted, the reflection of
av[x].

Look at the following graph, what can you tell me about
the relationships between the ordered pairs of av[x] and
its reflection?

code

abspts = 
      ListPlot[{{-5, 5}, {-5, -5}, {2, 2}, {2, -2}}, 
        PlotStyle -> {PointSize[0.03], Blue}, 
        DisplayFunction -> Identity];

Show[{both, abspts}, 
      Graphics[{Text["(-5,5)", {-5, 5}, {1, 0}], 
          Text["(-5,-5)", {-5, -5}, {1, 0}], 
          Text["(2,2)", {2, 2}, {-1, 0}], 
          Text["(2,-2)", {2, -2}, {-1, 0}]}], 
      DisplayFunction -> $DisplayFunction];

Hint: Look at the change in the value of the ordered pairs.
Type your answer here.

It may interest you to note that the way I got the reflection of av[x]
was to graph the function -av[x].

Reflect Across Y Axis

Let's try a different reflection.  How about one across the y axis.
What do you think will change with the coordinates this time?

For this graph I'll use the square root function sr[x].
sr[x] will be in purple and its reflection in red.

code

sqrootref = 
      Plot[sr[-x], {x, -10, 0}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        DisplayFunction -> Identity];

Show[{squareroot, sqrootref}, 
      DisplayFunction -> $DisplayFunction];

This time you pick an ordered pair and I'll show it and its reflection on the graph.  Just put a value in for x and activate 
the cell below.

code

Clear[x, y]; x := 5rootpts = 
      ListPlot[{{x, N[sr[x]]}, {-x, N[sr[x]]}}, 
        PlotStyle -> {PointSize[0.02], Yellow}, 
        DisplayFunction -> Identity]; allroots = 
      Show[{squareroot, sqrootref, rootpts}, 
        Graphics[{Text[{x, sr[x]}, {x, sr[x]}, {-1, 0}], 
            Text[{-x, sr[x]}, {-x, sr[x]}, {1, 0}]}], 
        DisplayFunction -> $DisplayFunction];

What do you notice about the change in these coordinates?
Here, you should notice that I graphed the reflection across the 
y axis by graphing sr[-x].

Reflect Across The Identity Function Y = X

Let's try one last reflection, across the axis y=x.
Look at the following two graphs and try to guess 
what is happening to the coordinates of the function.
We will use the function f[x]= -3 x + 6.

code

Clear[f, x]; f[x_] := -3 x + 6linegrph = 
      Plot[f[x], {x, -3, 10}, 
        PlotStyle -> {{Thickness[0.009], Green}}, 
        DisplayFunction -> Identity]; ident = 
      Show[Graphics[{Dashing[{0.02, 0.02}], 
            Line[{{-10, -10}, {10, 10}}]}], 
        DisplayFunction -> Identity]; g[x_] := -(x/3) + 2reflect = 
      Plot[g[x], {x, -50, 50}, 
        PlotStyle -> {{Thickness[0.009], Orange}}, 
        DisplayFunction -> Identity];

Show[{reflect, ident, linegrph}, 
      DisplayFunction -> $DisplayFunction];

Now pick some values for x and see what happens with
the ordered pairs.  Just enter the value and activate the cell.

code

x := 10pts = 
      ListPlot[{{x, f[x]}, {f[x], g[f[x]]}}, 
        PlotStyle -> {PointSize[0.03], Red}, 
        DisplayFunction -> Identity]; Show[{reflect, ident, linegrph, pts}, 
      Graphics[{Text[{x, f[x]}, {x, f[x]}, {1, 0}], 
          Text[{f[x], g[f[x]]}, {f[x], g[f[x]]}, {1, 0}]}], 
      DisplayFunction -> $DisplayFunction];

The graph of f[x] has a reflection that turns out to be the inverse of f[x].  You can see this by looking at the coordinates. What is the relationship between the points?

A Quick Review

List the rules you can find about these three special reflections.
		across the x axis:
		across the y axis:
		across the Identity Function:

Translations

Up and Down

Now let,s try some variations on some specific functions.
First we'll use the absolute value function.  av[x]
What happens if we add 4 to the function?
What if we subtract 3?

code

addav = 
      Plot[av[x] + 4, {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.009], Green}}, 
        DisplayFunction -> Identity]; subav = 
      Plot[av[x] - 3, {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.009], Blue}}, 
        DisplayFunction -> Identity]; Show[{addav, absvalue, subav}, 
      Graphics[{Text["av[x]-3", {5, 2}, {-1, 0}], 
          Text["av[x]+4", {2, 10}, {-1, 0}], 
          Text["av[x]", {2, 5}, {-1, 0}]}], 
      Ticks -> {{{-10, -10}, {-5, -5}, {5, 5}, {10, 
              10}}, {{-3, -3}, {2, 2}, {4, 4}}}, 
      DisplayFunction -> $DisplayFunction];

Very interesting how the graph went up 4 when we added
4 and went down 3 when we subtracted 3.

Let's try the squaring function.

code

adds = 
      Plot[s[x] + 10, {x, -5, 5}, 
        PlotStyle -> {{Thickness[0.009], Green}}, 
        DisplayFunction -> Identity]; subs = 
      Plot[s[x] - 8, {x, -5, 5}, 
        PlotStyle -> {{Thickness[0.009], Yellow}}, 
        DisplayFunction -> Identity]; Show[{adds, square, subs}, 
      Graphics[{Text["s[x]-8", {3, -2}, {-1, 0}], 
          Text["s[x]+10", {2, 15}, {-1, 0}], 
          Text["s[x]", {2, 5}, {-1, 0}]}], 
      Ticks -> {{{-5, -5}, {7, 7}}, {{-8, -8}, {4, 4}, {
              10, 10}}}, DisplayFunction -> $DisplayFunction];

Looks like we have a rule.  See if you can state it.

Left and Right

This time we will try to add and subtract within the function.

code

addav = 
      Plot[av[x + 4], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.009], Green}}, 
        DisplayFunction -> Identity]; subav = 
      Plot[av[x - 3], {x, -10, 10}, 
        PlotStyle -> {{Thickness[0.009], Blue}}, 
        DisplayFunction -> Identity]; Show[{addav, absvalue, subav}, 
      Graphics[{Text["av[x-3]", {5, 2}, {-1, 0}], 
          Text["av[x+4]", {2, 10}, {-1, 0}], 
          Text["av[x]", {2, 5}, {-1, 0}]}], 
      Ticks -> {{{-8, -8}, {-4, -4}, {3, 3}, {6, 6}}, {{
              -3, -3}, {2, 2}, {4, 4}}}, 
      DisplayFunction -> $DisplayFunction];

Try to guess the rule then verify your guess with the next graph.

code

adds = 
      Plot[s[x + 10], {x, -20, 20}, 
        PlotStyle -> {{Thickness[0.009], Green}}, 
        DisplayFunction -> Identity]; subs = 
      Plot[s[x - 8], {x, -20, 20}, 
        PlotStyle -> {{Thickness[0.009], Yellow}}, 
        DisplayFunction -> Identity]; Show[{adds, square, subs}, 
      Graphics[{Text["s[x-8]", {10, 80}, {-1, 0}], 
          Text["s[x+10]", {-19, 90}, {-1, 0}], 
          Text["s[x]", {0.5, 20}, {-1, 0}]}], 
      Ticks -> {{{-10, -10}, {0, 0}, {8, 8}}, Automatic}, 
      DisplayFunction -> $DisplayFunction];

Works pretty good doesn't it?

Results

You should have observed that the graph of a function
y= f(x-c)+d is a horizontal and vertical translation of the graph 
of y=f(x).

Horizontal |c| units         	   Vertical |d| units
If c<0,  shift is left                        If d<0,  shift is down
If c>0,  shift is right                      If d>0, shift is up

Test your work on the following graph.  On graph paper, graph  
a function g(x) =  (x+5)‹ - 30

code

Clear[g, x]; g[x_] := (((x + 5))^3) - 30Plot[g[x], {x, -10, 10}];

Your graph should look like this.

Dilations

we can have on last affect on these graphs.
We can multiply the function by a number.
Look at the graph of s(x)= x¤, 3x¤, and 0.6x¤

code

Clear[x]; threes[x_] := 3 (x^2)ptsixs[x_] := 0.6 (x^2)thrs = 
      Plot[threes[x], {x, -7, 7}, 
        PlotStyle -> {{Thickness[0.01], Blue}}, 
        DisplayFunction -> Identity]; ptsx = 
      Plot[ptsixs[x], {x, -7, 7}, 
        PlotStyle -> {{Thickness[0.01], Red}}, 
        DisplayFunction -> Identity]; Show[{thrs, ptsx, square}, 
      Graphics[{
          Text[FontForm["3 x¤", {"CalcMath", 10}], {3, 
              30}, {1, 0}], 
          Text[FontForm["0.6x¤", {"CalcMath", 10}], {6, 
              10}, {1, 0}], 
          Text[FontForm["x¤", {"CalcMath", 10}], {4.5, 
              25}, {1, 0}]}], DisplayFunction -> $DisplayFunction];

What happens to s(x) when we multiply by 3?
When we multiply by a number between 0 and 1?
What might happen if we use a number less than zero?