A Change of Functions
by Brian Zecher
btzecher@donovan.k12.il.us

This lesson illustrates three types of function transformations: reflections, translations and dilations.  By looking at the graphs of several functions before and after certain variables are added, the student will be able to see how the graphs are affected by the changes.  This should help them graph various functions in the future.

<< "Graphics`Colors`"

Section 1:
y = f{x} vs. y = af{x}

Activate only 1 of the following 5 functions.

Clear[f, x]; f[x_] = Abs[x + 1]  +  Abs[x - 1];

Clear[f, x]; f[x_] = Abs[x]/(x - 1);

Clear[f, x]; f[x_] = 2  x;

Clear[f, x];f[x_] =Sin[x];

Clear[f, x];f[x_] =((-1)/((x + 1))) ;

Now choose a value for a.  (As you repeat this section for different funtions, change the value of a.)  Suggestions...Try values like 2, 5, 1/2, 1/5.

a = 3;

Let's see what the graph of your function y = f(x) looks like.

t0 = 
      StyleForm[StringForm["y = f(x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\(graph1\  = 
      Plot[f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

So what does y = a f(x) look like for your value of a ???

t10 = 
      StyleForm[StringForm["y = ``f(x)",  a],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Now let's see the two graphs together.  Remember the original graph, y = f(x), was graphed in blue and the new graph, y = a f(x) was graphed in red.

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

So, what happened when we graphed y = a f(x) instead of y = f(x).  To get a better picture, let's look at smaller intervals between the two graphs.

StringForm[" ``       ``", t0, t10]\(	Show[g0]; 

    t2 = Table[
        graph1\  = 
          Plot[b\ f[x], {x, \(-10\), 10}, 
            PlotRange 
              -> {
                Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                    a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
                Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                    a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
            PlotStyle\  -> \ {Black}, DisplayFunction -> Identity, 
            ImageSize -> {300, 200}]; 

        p1\  = \ ListPlot[{{\(-4\), b\ f[\(-4\)]}}, 
            PlotStyle -> {PointSize[0.015], Blue}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p2\  = \ ListPlot[{{\(-2\), b\ f[\(-2\)]}}, 
            PlotStyle -> {PointSize[0.015], LawnGreen}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p3\  = \ ListPlot[{{0, b\ f[0]}}, 
            PlotStyle -> {PointSize[0.015], Magenta}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p4\  = \ ListPlot[{{2, b\ f[2]}}, 
            PlotStyle -> {PointSize[0.015], YellowOchre}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p5 = \ ListPlot[{{4, b\ f[4]}}, 
            PlotStyle -> {PointSize[0.015], Red}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        g1\  = Show[graph1, p1, p2, p3, p4, p5, 
            DisplayFunction -> $DisplayFunction], {b, 
          1 +  .1\ Abs[a - 1]/\((a - 1)\), \ 
          a -  .1\ Abs[a - 1]/\((a - 1)\),  .1\ Abs[a - 1]/\((a - 1)\)}]; 

    Show[g10]; \)

Animate the graphs by selecting all of the graphs we just produced and choosing "Animate Selected Graphics" from the Cell menu.
What do you think is happening as we change from y = f(x) to y = a f(x)

Maybe seeing all of these graphs together, will give you a better idea of what happened.  Again the y = f(x) is graphed in blue and y = a f(x) is red.

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, t2, g10, DisplayFunction -> $DisplayFunction, 
      GridLines\  -> Automatic, ImageSize -> {500, 340}]; \)\)

Now go back to the beginning of this section.  This time activate a different function and choose a new value for a.

Questions
1.   What happens to the graph of the function when the value of a is between 0 and 1?
2.   What happens to the graph of the function when the value of a is greater than 1?

O.K. So here's the graph of a new function, y = f(x).

Clear[f, x, a]\(f[x_] = \((x - 6)\) \((x + 5)\) \((x - 1)\); 
a = 2; 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Blue}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = 2 f(x) is graphed.

\(a = 2; 

    t10 = StyleForm[StringForm["y = ``f(x)", \ a], \ 
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\)\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

Were you correct???

Now, predict what will happen to the same original graph when y = .5f(x) is graphed.

\(a = 1\/2; 

    t10 = StyleForm[StringForm["y = ``f(x)", \ a], \ 
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\)\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

Well???

Section 2:
y = f{x} vs. y = -f{x}

Activate only 1 of the following 4 functions.

Clear[f, x]; f[x_] = (x^(3 )) - 17 x ;

Clear[f, x]; f[x_] = (x - 3) (x + 2);

Clear[f, x]; f[x_] = Sqrt[x + 10];

Clear[f, x];f[x_] =Sqrt[100 - (x^2)];

Let's see what the graph of your function y = f(x) looks like.

\(a = \(-1\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(		graph1\  = 
      Plot[f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

So what does y = -f(x) look like???

t10 = 
      StyleForm[StringForm["y = -f(x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Now let's see the two graphs together.  Remember the original graph, y = f(x), was graphed in blue and the new graph, y = -f(x) was graphed in red.

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

So, what happened to y = f(x) when we graphed y = -f(x)?  To get a better picture, let's again look at smaller intervals between the two graphs.

StringForm[" ``       ``", t0, t10]\(	Show[g0]; 

    t2 = Table[
        graph1\  = 
          Plot[b\ f[x], {x, \(-10\), 10}, 
            PlotRange 
              -> {
                Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                    a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
                Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                    a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
            PlotStyle\  -> \ {Black}, DisplayFunction -> Identity, 
            ImageSize -> {300, 200}]; 

        p1\  = \ ListPlot[{{\(-4\), b\ f[\(-4\)]}}, 
            PlotStyle -> {PointSize[0.015], Blue}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p2\  = \ ListPlot[{{\(-2\), b\ f[\(-2\)]}}, 
            PlotStyle -> {PointSize[0.015], LawnGreen}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p3\  = \ ListPlot[{{0, b\ f[0]}}, 
            PlotStyle -> {PointSize[0.015], Magenta}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p4\  = \ ListPlot[{{2, b\ f[2]}}, 
            PlotStyle -> {PointSize[0.015], YellowOchre}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p5 = \ ListPlot[{{4, b\ f[4]}}, 
            PlotStyle -> {PointSize[0.015], Red}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        g1\  = Show[graph1, p1, p2, p3, p4, p5, 
            DisplayFunction -> $DisplayFunction], {b, 
          1 +  .1\ Abs[a - 1]/\((a - 1)\), \ 
          a -  .1\ Abs[a - 1]/\((a - 1)\),  .1\ Abs[a - 1]/\((a - 1)\)}]; 

    Show[g10]; \)

Animate the graphs and think about whatis happening as we change from y = f(x) to y = -f(x)

Maybe seeing all of these graphs together, will give you a better idea of what happened.  Again the the original is graphed in blue and y = -f(x) is red.

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, t2, g10, DisplayFunction -> $DisplayFunction, 
      GridLines\  -> Automatic, ImageSize -> {500, 340}]; \)\)

Now go back to the beginning of this section 2.  This time activate a different function.

Questions
1.   What happens to the graph of the function when we graph y = -f(x)?

O.K. So here's the graph of a new function, y = f(x).

Clear[f, x, a]\(f[x_] = \((x - 3)\) \((x + 5)\); 
a = \(-1\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Blue}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = -f(x) is graphed.

t10 = 
      StyleForm[StringForm["y = -f(x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

Were you correct???

Here's the graph of another new function, y = f(x).

Clear[f, x, a]\(f[x_] = \(-\((x - 3)\)\^2\) + 20; 
a = \(-2\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Blue}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = -2f(x) is graphed.

t10 = 
      StyleForm[StringForm["y = ``f(x)",  a],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[a\ f[x], {x, \(-10\), 10}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], a\ f[\(-4\)], 
                a\ f[\(-2\)], a\ f[0], a\ f[2], a\ f[4]] + 5}, 
        PlotStyle\  -> \ {Red}, DisplayFunction -> Identity, 
        ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), a\ f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), a\ f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, a\ f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, a\ f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, a\ f[4]}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {500, 340}, GridLines -> Automatic]; 
    \)\)

Explain what happened to the graph.  Why did it happen?

Section 3:
y = f{x} vs. y = f{ax}

Activate only 1 of the following 4 functions.

Clear[f, x]; f[x_] = 2  (x^(2 )) - 3  x  +  1;

Clear[f, x];f[x_] =(x^3);

Clear[f, x]; f[x_] = (((x + 1))^4) - 25;

Clear[f, x];f[x_] = (((Mod[x, 3]))^2) - 2 Mod[x, 3];

Now choose a value for a.  (As you repeat this section for different funtions, change the value of a.)  Suggestions...Try values like 2, 4, 1/3, 1/4.

a = 3;

t0 = 
      StyleForm[StringForm["y = f(x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\(		graph1\  = 
      Plot[f[x], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

t10 = 
      StyleForm[StringForm["y = f(``x)",  a],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

StringForm[" ``       ``", t0, t10]\(	Show[g0]; 

    t2 = Table[
        graph1\  = 
          Plot[f[b\ x], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
            PlotRange 
              -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
                Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
            PlotStyle\  -> \ {Black}, DisplayFunction -> Identity, 
            ImageSize -> {300, 200}]; 

        p1\  = \ ListPlot[{{\(-4\)/b, f[\(-4\)]}}, 
            PlotStyle -> {PointSize[0.015], Blue}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p2\  = \ ListPlot[{{\(-2\)/b, f[\(-2\)]}}, 
            PlotStyle -> {PointSize[0.015], LawnGreen}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p3\  = \ ListPlot[{{0/b, f[0]}}, 
            PlotStyle -> {PointSize[0.015], Magenta}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p4\  = \ ListPlot[{{2/b, f[2]}}, 
            PlotStyle -> {PointSize[0.015], YellowOchre}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        p5 = \ ListPlot[{{4/b, f[4]}}, 
            PlotStyle -> {PointSize[0.015], Red}, 
            PlotJoined -> False, DisplayFunction -> Identity]; 

        g1\  = Show[graph1, p1, p2, p3, p4, p5, 
            DisplayFunction -> $DisplayFunction], {b, 1, \ 
          a,  .1\ Abs[a - 1]/\((a - 1)\)}]; 
Show[g10]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, t2, g10, DisplayFunction -> $DisplayFunction, 
      GridLines\  -> Automatic, ImageSize -> {500, 340}]; \)
\)

Now go back to the beginning of this section.  This time activate a different function and choose a new value for a.

Questions
1.   What happens to the graph of the function when the value of a is between 0 and 1?
2.   What happens to the graph of the function when the value of a is greater than 1?
3.   Is graphing y = a f(x) the same as graphing y = f(ax)?  If not, what is different?

O.K. So here's the graph of a new function, y = f(x).

Clear[f, x, a]\(f[x_] = 3  x\  + \ 2; 
a = 1\/4; 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(		graph1\  = 
      Plot[f[x], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = f(2x) is graphed.

\(a = 2; b =  .25; 

    t10 = StyleForm[StringForm["y = f(``x)", \ a], \ 
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\)\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), \(-4\)/b] - 5, Max[4, 4/b] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

Were you correct???

Now, predict what will happen to the same original graph when y =  f(.25x) is graphed.

\(a = 1\/4; 
		
    t10 = StyleForm[StringForm["y = f(``x)", \ a], \ 
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\)\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

How did you do???

Section 4:
y = f{x} vs. y = f{-x}

Activate only 1 of the following 4 functions.

Clear[f, x]; f[x_] = -(((x - 2))^2) + 6;

Clear[f, x]; f[x_] = 3 x  +  2;

Clear[f, x]; f[x_] = Abs[x - 2];

Clear[f, x]; f[x_] = (x^(3 )) - 17 x;

Now let's see the graphs.

\(a = \(-1\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(		graph1\  = 
      Plot[f[x], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

t10 = 
      StyleForm[StringForm["y = f(-x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

\(Show[g0]; 

    t2 = {Table[
          graph1\  = 
            Plot[f[x/b], {x, Min[\(-4\), \(-4\)/a] - 5, Max[4, 4/a] + 5}, 
              PlotRange 
                -> {Min[f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
                  Max[f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
              PlotStyle\  -> \ {Black}, DisplayFunction -> Identity, 
              ImageSize -> {300, 200}]; 

          p1\  = \ ListPlot[{{\(-4\)\ b, f[\(-4\)]}}, 
              PlotStyle -> {PointSize[0.015], Blue}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p2\  = \ ListPlot[{{\(-2\)\ b, f[\(-2\)]}}, 
              PlotStyle -> {PointSize[0.015], LawnGreen}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p3\  = \ ListPlot[{{0\ b, f[0]}}, 
              PlotStyle -> {PointSize[0.015], Magenta}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p4\  = \ ListPlot[{{2\ b, f[2]}}, 
              PlotStyle -> {PointSize[0.015], YellowOchre}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p5 = \ ListPlot[{{4\ b, f[4]}}, 
              PlotStyle -> {PointSize[0.015], Red}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          g1\  = Show[graph1, p1, p2, p3, p4, p5, 
              DisplayFunction -> $DisplayFunction], {b, 
            1 +  .1\ Abs[a - 1]/\((a - 1)\), 
            \  .1,  .1\ Abs[a - 1]/\((a - 1)\)}], 
        Table[graph1\  = 
            Plot[f[x/b], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
              PlotRange 
                -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
                  Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
              PlotStyle\  -> \ {Black}, DisplayFunction -> Identity, 
              ImageSize -> {300, 200}]; 

          p1\  = \ ListPlot[{{\(-4\)\ b, f[\(-4\)]}}, 
              PlotStyle -> {PointSize[0.015], Blue}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p2\  = \ ListPlot[{{\(-2\)\ b, f[\(-2\)]}}, 
              PlotStyle -> {PointSize[0.015], LawnGreen}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p3\  = \ ListPlot[{{0\ b, f[0]}}, 
              PlotStyle -> {PointSize[0.015], Magenta}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p4\  = \ ListPlot[{{2\ b, f[2]}}, 
              PlotStyle -> {PointSize[0.015], YellowOchre}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p5 = \ ListPlot[{{4\ b, f[4]}}, 
              PlotStyle -> {PointSize[0.015], Red}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          g1\  = Show[graph1, p1, p2, p3, p4, p5, 
              DisplayFunction -> $DisplayFunction], {b, \(- .1\), \ 
            a -  .1\ Abs[a - 1]/\((a - 1)\),  .1\ Abs[a - 1]/\((a - 1)\)}]}; 
    
Show[g10]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, t2, g10, DisplayFunction -> $DisplayFunction, 
      GridLines\  -> Automatic, ImageSize -> {500, 340}]; \)
\)

Now go back to the beginning of this section.  This time choose a different function.

Questions
1.   What happens to the graph of the function when y = f(-x) is graphed?
2.   Is graphing y = -f(x) the same as graphing y = f(-x)?  If not, what is different?

O.K. So here's the graph of yet another new function, y = f(x).

Clear[f, x, a]\(f[x_] = \((x - 5)\) \((x - 2)\) \((x + 3)\)\^2; 
a = \(-1\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = f(-x) is graphed.

t10 = 
      StyleForm[StringForm["y = f(-x)",  a],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {300, 200}]; \)\)

How did you do???

Here's the graph of another new function, y = f(x).

Clear[f, x, a]\(f[x_] = \(-\((x - 3)\)\^2\) + 20; 
a = \(-4\); 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = f(-4x) is graphed.

t10 = 
      StyleForm[StringForm["y = f(``x)",  a],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x\ a], {x, Min[\(-4\), 4/a] - 5, Max[4, \(-4\)/a] + 5}, 
        PlotRange 
          -> {Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4]] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\)/a, f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\)/a, f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0/a, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2/a, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4/a, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}, ImageSize -> {500, 340}, GridLines -> Automatic]; 
    \)\)

Explain what happened to the graph.  Why did it happen?

Section 5:
y = f{x} vs. y = f{x + h} + k

Activate only 1 of the following 4 functions.

Clear[f, x]; f[x_] = Abs[x];

Clear[f, x]; f[x_] = (x^(3 )) - x ;

Clear[f, x];f[x_] =(9/2) (((Mod[x, 3]))^2);

Clear[f, x];f[x_] =Sin[x];

Now choose a values for h and k.  (As you repeat this section for different funtions, change the values of h and k.)  Make sure you try positives and negatives for both.

h = 2; k = -3;

Here's your graph.

t0 = 
      StyleForm[StringForm["y = f(x)"],  FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\(		graph1\  = 
      Plot[f[x], {x, Min[\(-4\), \(-4\) - h] - 5, Max[4, 4 - h] + 5}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

And here is y = f(x + h) + k.

t10 = 
      StyleForm[StringForm["y = f(x + ``) + ``",  h, k],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x + h] + k, {x, Min[\(-4\), \(-4\) - h] - 5, Max[4, 4 - h] + 5}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\) - h, f[\(-4\)] + k}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\) - h, f[\(-2\)] + k}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0 - h, f[0] + k}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2 - h, f[2] + k}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4 - h, f[4] + k}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show\ [{g0, g10}]; \)\)

StringForm[" ``       ``", t0, t10]\(	Show[g0]; 

    t2 = {Table[
          graph1\  = 
            Plot[f[x + b], {x, Min[\(-4\), \(-4\) - h] - 5, 
                Max[4, 4 - h] + 5}, 
              PlotRange 
                -> {
                  Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], 
                      f[\(-4\)] + k, f[\(-2\)] + k, f[0] + k, f[2] + k, 
                      f[4] + k] - 5, 
                  Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], 
                      f[\(-4\)] + k, f[\(-2\)] + k, f[0] + k, f[2] + k, 
                      f[4] + k] + 5}, PlotStyle\  -> \ {Black}, 
              DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

          p1\  = \ ListPlot[{{\(-4\) - b, f[\(-4\)]}}, 
              PlotStyle -> {PointSize[0.015], Blue}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p2\  = \ ListPlot[{{\(-2\) - b, f[\(-2\)]}}, 
              PlotStyle -> {PointSize[0.015], LawnGreen}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p3\  = \ ListPlot[{{0 - b, f[0]}}, 
              PlotStyle -> {PointSize[0.015], Magenta}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p4\  = \ ListPlot[{{2 - b, f[2]}}, 
              PlotStyle -> {PointSize[0.015], YellowOchre}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p5 = \ ListPlot[{{4 - b, f[4]}}, 
              PlotStyle -> {PointSize[0.015], Red}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          g1\  = Show[graph1, p1, p2, p3, p4, p5, 
              DisplayFunction -> $DisplayFunction], {b, 0, 
            h,  .3\ Abs[h]/h}], 
        Table[graph1\  = 
            Plot[f[x + h] + b, {x, Min[\(-4\), \(-4\) - h] - 5, 
                Max[4, 4 - h] + 5}, 
              PlotRange 
                -> {
                  Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], 
                      f[\(-4\)] + k, f[\(-2\)] + k, f[0] + k, f[2] + k, 
                      f[4] + k] - 5, 
                  Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], 
                      f[\(-4\)] + k, f[\(-2\)] + k, f[0] + k, f[2] + k, 
                      f[4] + k] + 5}, PlotStyle\  -> \ {Black}, 
              DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

          p1\  = \ ListPlot[{{\(-4\) - h, f[\(-4\)] + b}}, 
              PlotStyle -> {PointSize[0.015], Blue}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p2\  = \ ListPlot[{{\(-2\) - h, f[\(-2\)] + b}}, 
              PlotStyle -> {PointSize[0.015], LawnGreen}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p3\  = \ ListPlot[{{0 - h, f[0] + b}}, 
              PlotStyle -> {PointSize[0.015], Magenta}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p4\  = \ ListPlot[{{2 - h, f[2] + b}}, 
              PlotStyle -> {PointSize[0.015], YellowOchre}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          p5 = \ ListPlot[{{4 - h, f[4] + b}}, 
              PlotStyle -> {PointSize[0.015], Red}, 
              PlotJoined -> False, DisplayFunction -> Identity]; 

          g1\  = Show[graph1, p1, p2, p3, p4, p5, 
              DisplayFunction -> $DisplayFunction], {b, 0, 
            k,  .3\ Abs[k]/k}]}; 
Show[g10]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, t2, g10, DisplayFunction -> $DisplayFunction, 
      GridLines\  -> Automatic, ImageSize -> {500, 340}]; \)
\)

Now go back to the beginning of this section.  This time activate a different function and choose a new values for h and k.

Questions
1.   What happens to the original graph when you graph f(x+2)?
2.   What happens to the original graph when you graph f(x)+ 5?
3.   What happens to the original graph when you graph f(x-3)-8?

O.K. So here's the graph of a new function, y = f(x).

Clear[f, x, a]\(h = \(-5\); k = 4; 
f[x_] = \((x + 2)\)\^2; 

    t0 = StyleForm[StringForm["y = f(x)"], \ FontWeight -> "Bold", 
        FontSize -> 17, FontColor -> RGBColor[0, 0, 1]]\)\(graph1\  = 
      Plot[f[x], {x, Min[\(-4\), \(-4\) - h] - 5, Max[4, 4 - h] + 5}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] + 5}, 
        PlotStyle\  -> \ {Blue, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\), f[\(-4\)]}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\), f[\(-2\)]}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0, f[0]}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2, f[2]}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4, f[4]}}, PlotStyle -> {PointSize[0.015], Red}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    g0\  = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

Predict what will happen when y = f(x - 5) + 4 is graphed.

t10 = 
      StyleForm[StringForm["y = f(x + ``) + ``",  h, k],  
        FontWeight -> "Bold", FontSize -> 17, 
        FontColor -> RGBColor[1, 0, 0]]\(		graph1\  = 
      Plot[f[x + h] + k, {x, Min[\(-4\), \(-4\) - h] - 5, Max[4, 4 - h] + 5}, 
        PlotRange 
          -> {
            Min[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] - 5, 
            Max[0, f[\(-4\)], f[\(-2\)], f[0], f[2], f[4], f[\(-4\)] + k, 
                f[\(-2\)] + k, f[0] + k, f[2] + k, f[4] + k] + 5}, 
        PlotStyle\  -> \ {Red, Thickness[ .006]}, 
        DisplayFunction -> Identity, ImageSize -> {300, 200}]; 

    p1\  = \ ListPlot[{{\(-4\) - h, f[\(-4\)] + k}}, 
        PlotStyle -> {PointSize[0.015], Blue}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    p2\  = \ ListPlot[{{\(-2\) - h, f[\(-2\)] + k}}, 
        PlotStyle -> {PointSize[0.015], LawnGreen}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p3\  = \ ListPlot[{{0 - h, f[0] + k}}, 
        PlotStyle -> {PointSize[0.015], Magenta}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p4\  = \ ListPlot[{{2 - h, f[2] + k}}, 
        PlotStyle -> {PointSize[0.015], YellowOchre}, 
        PlotJoined -> False, DisplayFunction -> Identity]; 

    p5 = \ ListPlot[{{4 - h, f[4] + k}}, 
        PlotStyle -> {PointSize[0.015], Red}, PlotJoined -> False, 
        DisplayFunction -> Identity]; 

    g10 = Show[graph1, p1, p2, p3, p4, p5, 
        DisplayFunction -> $DisplayFunction]; \)

StringForm[" ``       ``", t0, t10]\(	\(Show[g0, g10]; \)
\)

Section 6:
Final Summary

Questions
1.   What would happen to y=f(x) when you graph y = 2f(x)+3?
2.   What would happen to y=f(x) when you graph y = -f(x-2)?
3.   What would happen to y=f(x) when you graph y = 3f(-x)?
4.   What would happen to y=f(x) when you graph y = -f(2x)?
5.   What would happen to y=f(x) when you graph y = -3f(2x+3)-5?