(*********************************************************************** Mathematica-Compatible Notebook This notebook can be used on any computer system with Mathematica 3.0, MathReader 3.0, or any compatible application. The data for the notebook starts with the line of stars above. To get the notebook into a Mathematica-compatible application, do one of the following: * Save the data starting with the line of stars above into a file with a name ending in .nb, then open the file inside the application; * Copy the data starting with the line of stars above to the clipboard, then use the Paste menu command inside the application. Data for notebooks contains only printable 7-bit ASCII and can be sent directly in email or through ftp in text mode. Newlines can be CR, LF or CRLF (Unix, Macintosh or MS-DOS style). NOTE: If you modify the data for this notebook not in a Mathematica- compatible application, you must delete the line below containing the word CacheID, otherwise Mathematica-compatible applications may try to use invalid cache data. For more information on notebooks and Mathematica-compatible applications, contact Wolfram Research: web: http://www.wolfram.com email: info@wolfram.com phone: +1-217-398-0700 (U.S.) Notebook reader applications are available free of charge from Wolfram Research. ***********************************************************************) (*CacheID: 232*) (*NotebookFileLineBreakTest NotebookFileLineBreakTest*) (*NotebookOptionsPosition[ 326718, 10474]*) (*NotebookOutlinePosition[ 327595, 10507]*) (* CellTagsIndexPosition[ 327519, 10501]*) (*WindowFrame->Normal*) Notebook[{ Cell[CellGroupData[{ Cell["Graphing Techniques", "Title", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[CellGroupData[{ Cell["Read Me! (how to use this program)", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "This notebook is written using ", StyleBox["Mathematica", FontSlant->"Italic"], ", a highly sophisticated mathematical software program capable of doing \ intricate mathematics", ". ", "You will need to know only a few ", StyleBox["very basic", FontVariations->{"Underline"->True}], " things about ", StyleBox["Mathematica", FontSlant->"Italic"], " to be able to use this notebook; the programming has been done for you." }], "Text"], Cell[TextData[{ "\t1. ", StyleBox["Opening a section of the notebook", FontWeight->"Bold"], ": Scroll down the screen using the mouse on the scrollbar at the right of \ this screen until you see what appears to be a table of contents. Directly \ to the right of each topic is a short blue bracket sign. Some brackets have \ a small arrow or triangle at the bottom. This arrow indicates that there is \ hidden text which can be viewed by clicking with the mouse on the bracket \ containing the arrow. Try this on one of the arrowed brackets below. When \ you click on the bracket, new text should appear. When you are finished, you \ can close that section of the notebook by double-clicking on the same (now \ longer) bracket. If you are unsure which one it is, scroll up to the \ beginning of the section; it is the one that ", StyleBox["begins", FontSlant->"Italic"], " there. Try closing the section you just opened." }], "Text"], Cell[TextData[{ StyleBox["\t", FontWeight->"Bold"], "2. ", StyleBox["Activating a program command", FontWeight->"Bold"], ": The ", StyleBox["Mathematica", FontSlant->"Italic"], " programming always appears in bold type on a yellow background. To \ activate a program, use the mouse to move the \"I\" shaped cursor inside the \ yellow box and click once so that the straight line \"|\" cursor appears \ anywhere inside the yellow box. Now depress the \"shift\" and \"return\" \ keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], ". Try this with the following short program which generates a random \ number between 1 and 10. Remember to click on these two keys ", StyleBox["simultaneously", FontVariations->{"Underline"->True}], "." }], "Text"], Cell[BoxData[ \(Random[Integer, {1, 10}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\t3. ", StyleBox["Changing a program:", FontWeight->"Bold"], " Generally, you should not change any type in the yellow blocks. However, \ a few of the ", StyleBox["Mathematica", FontSlant->"Italic"], " programs have been designed so that you can change parts of the program. \ The parts that you should change are displayed in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], ". To change these parts, simply highlight the ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " type by sweeping over it with the cursor, and type in your change. Read, \ and take seriously, any instruction saying ", StyleBox["*do not change anything below this line*", FontColor->RGBColor[1, 0, 0]], ". Changing something there could radically change the commands, causing ", StyleBox["Mathematica", FontSlant->"Italic"], " to 'beep' in an error protest--or give you an answer to a totally \ different question!\n\n\t4. ", StyleBox["Quitting the program", FontWeight->"Bold"], ": When you have finished working and want to quit, click in the small box \ at the upper lefthand corner of the window to close it. Then choose `quit' \ or `exit' from the File menu. ", StyleBox["Mathematica", FontSlant->"Italic"], " will ask you if you want to save your changes. Say NO! Otherwise you \ will have a different notebook from the one you downloaded, and any \ misinformation you might have entered into the notebook will persist. If for \ any reason, you need to start with a fresh notebook in its original form, you \ can always trash the one you've been working with and download a new one off \ the Web." }], "Text"] }, Open ]], Cell[CellGroupData[{ Cell["Introduction", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell[TextData[{ "\tA ", StyleBox["function", FontWeight->"Bold"], " f is a rule that associates to each element x of a set X a unique element \ f(x) of a set Y. The set X is called the ", StyleBox["domain", FontWeight->"Bold"], " of the function. 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The ", StyleBox["range", FontWeight->"Bold"], " is the set of all values of y for which there is an x with y = f(x).\n\t\ The ", StyleBox["y-intercept", FontWeight->"Bold"], " of a function y = f(x) is the y-value for which the graph crosses the \ y-axis. Algebraically, it is the y-value corresponding to x = 0. An ", StyleBox["x-intercept", FontWeight->"Bold"], " is an x-value for which the graph crosses the x-axis, algebraically it is \ a value of x for which", " ", "y = f(x) = 0. The Vertical Line Test tells us there can be at most one \ y-intercept, but there can be many x-intercepts. \n\tThe example shown below \ has x-intercepts at -3, -1 and 2. 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Col202[o0015o`<04ol404?o00<0ool0Col302Wo0016o`<04Ol404Co00<0ool0D?l202Wo0017o`@0 3Ol403ko0`05o`030?oo053o0P0Yo`00B?l400[o1@10o`030?oo00Co00<0ool0D?l202Wo001:o`D0 1?l6043o1005o`<0D?l202Wo001;o``0@_l00`3o0006o`030?oo053o0P0Yo`00C_l604Go00<0o`00 1_l00`3oo`1@o`<0:?l009[o0P06o`030?oo057o0P0Xo`00X_l00`3oo`1Ao`80:?l00:;o00<0ool0 DOl202So002Ro`80D_l202So002Ro`030?oo057o0P0Xo`00X_l00`3oo`1Ao`80:?l00:;o00<0ool0 DOl302Oo002Ro`030?oo05;o0P0Wo`00X_l00`3oo`1Bo`809ol00001\ \>"], ImageRangeCache->{{{0, 287}, {176.938, 0}} -> {-4.28027, -10.3915, 0.0263433, 0.0869883}}], Cell["\<\ \tIn the sections that follow, you are expected to be able to find \ x and y intercepts of various functions, as well as finding the domain and \ range, so be sure these concepts are clear.\ \>", "Text", FontSize->16] }, Closed]], Cell[CellGroupData[{ Cell["A: The basic graphs", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell["\<\ \tThere are a few functions whose graphs you should know. By \ 'know', we mean when you see the equation of the function, you should \ immediately associate the SHAPE of the graph and its basic properties: \ domain, range, intercepts, asymptotes, symmetry. The following are the most \ essential basic functions. You need to know how to graph the shape and \ important points of all of them.\ \>", "Text", FontSize->14], Cell[CellGroupData[{ Cell["Section 1: Linear functions y = f(x) = mx + b", "Subsection"], Cell[TextData[{ "\tFor any constants m and b, the graph of y =mx", " + ", "b is a line with slope m and y-intercept b. The following yellow blocks \ contain instructions for ", StyleBox["Mathematica", FontSlant->"Italic"], " to graph some examples of linear functions. Enter the input in these \ yellow blocks to view these examples. Then, in the last block in this \ section, experiment with creating your own examples by changing the input." }], "Text", FontSize->14], Cell[TextData[{ StyleBox["Example", FontWeight->"Bold"], " ", StyleBox["1", FontWeight->"Bold"], ": y = 2x", " + ", "3" }], "Text", FontSize->14], Cell[BoxData[ \(\(Clear[\ g1]; \n\t\t g1 = Plot[2\ x\ + \ 3, {x, \(-4\), 4}, PlotRange -> {\(-4\), 4}, AspectRatio -> 1, PlotStyle -> {Hue[0.4], Thickness[0.011]}]; \ \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "We see the y-intercept is b = 3, and the slope is m = 2. To find the \ x-intercept (the x-value for which the graph crosses the x-axis), we set y = \ 0 and solve to get x = -", Cell[BoxData[ \(TraditionalForm\`3\/2\)]], "." }], "Text", FontSize->14], Cell[TextData[{ StyleBox["Example 2", FontWeight->"Bold"], ": y = -3x", " + ", "1" }], "Text", FontSize->14], Cell[BoxData[ \(Clear[\ g2]; \n g2 = Plot[\(-3\)\ x\ + \ 1, {x, \(-4\), 4}, PlotRange -> {\(-4\), 4}, AspectRatio -> 1, PlotStyle -> {Hue[0.4], Thickness[0.008]}]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "As in the first example, we readily note that the slope is -3 and the \ y-intercept is 1. Setting y = 0, we solve for the x-intercept, x = ", Cell[BoxData[ \(TraditionalForm\`1\/3\)]], "." }], "Text", FontSize->14], Cell[TextData[{ StyleBox["Example 3", FontWeight->"Bold"], ": y = 2" }], "Text", FontSize->14], Cell[BoxData[ \(Clear\ [g3]; \n g3 = Plot[0\ x\ + \ 2, {x, \(-4\), 4}, PlotRange -> {\(-4\), 4}, AspectRatio -> 1, PlotStyle -> {Hue[0.6], Thickness[0.008]}]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ Note that in this example the slope is 0, so the line is \ horizontal. It never meets the x-axis, so there is no x-intercept. In fact, \ if we were to set y = 0, we would have 0 = 2, which of course is never \ satisfied.\ \>", "Text", FontSize->14], Cell["\<\ \tFor a good comparison of these three graphs, let's plot them on \ the same set of axes.\ \>", "Text", FontSize->14], Cell[BoxData[ \(\(Show[g1, g2, g3]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tWhat follows is a small program to graph any function of the form y = \ mx", " + ", "b. ", StyleBox["Mathematica", FontSlant->"Italic"], " will only display the graph for values of x between the values specified \ for x' and x''. That is, for the example below, the x-axis of the graph will \ only be shown from -10 to 10. Try executing this program." }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ RowBox[{"m", "=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{"b", "=", StyleBox["5", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-10\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["10", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*do\ not\ change\ the\ next\ \(line!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Plot[m\ *x\ + \ b, {x, x', \(x'\)'}, PlotRange -> {x', \(x'\)'}, PlotStyle -> {Thickness[ .008], Hue[ .9]}]\), FontColor->GrayLevel[0.500008]], StyleBox[";", FontColor->GrayLevel[0.500008]], StyleBox[" ", FontColor->GrayLevel[0.500008]], StyleBox[ \(Print["\", m, "\", b, "\<.\>"]\), FontColor->GrayLevel[0.500008]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Question", FontWeight->"Bold"], ": What is the domain of any linear function? What is the range? (Be \ careful with the second question: there are two possible answers!)\n\n\tThere \ are some special cases of graphs of lines worth emphasizing:\n\t1. The ", StyleBox["identity function", FontVariations->{"Underline"->True}], ", y = x: so-called because the image of any value x is x itself, i.e., the \ function associates to a number x the same number ", "y = ", "x.\n\t2.", StyleBox[" Horizontal lines", FontVariations->{"Underline"->True}], ": the slope m of a horizontal line is 0 (since the rise is 0 for any \ run).The graph is then determined by its y-intercept b and is given by the \ formula y = b.\n\t3.", StyleBox[" Vertical lines", FontVariations->{"Underline"->True}], ": a vertical line does NOT represent a function at all; it fails the \ vertical line test! Further the slope of a vertical line is undefined since \ the run is 0 for any rise and we cannot divide by 0. However, we can give a \ formula for a vertical line after noticing that the x-value is a constant (no \ matter what the y-vaue): x = a is the formula for a vertical line. This is ", StyleBox["not", FontVariations->{"Underline"->True}], " a linear function since it is not a function." }], "Text", FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Section 2:", " ", "The squaring function y = f(x) = ", Cell[BoxData[ FormBox[ StyleBox[ SuperscriptBox[ StyleBox["x", FontWeight->"Plain", FontSlant->"Plain", FontTracking->"Plain", FontVariations->{"Underline"->False, "Outline"->False, "Shadow"->False}], "2"], FontWeight->"Bold"], TraditionalForm]], FontWeight->"Plain"] }], "Subsection"], Cell[TextData[{ "\tThe graph of the function y = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " is a parabola. Input the following data to see the graph." }], "Text", FontSize->14], Cell[BoxData[ \(\(Plot[x^2, {x, \(-3\), 3}, PlotRange -> {\(-1\), 10}, PlotStyle -> {Thickness[0.008]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[ "Notice that the domain of this function is all real numbers x (i.e. (-\ \[Infinity],\[Infinity])), and the range is the set of all nonnegative \ y-values (i.e., [0,\[Infinity])). The x and y intercepts are both 0."], "Text",\ FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Section 3:", " ", "The square root function y = f(x) = ", Cell[BoxData[ \(TraditionalForm\`\@x\)]] }], "Subsection"], Cell["\tThe graph of the square root function is as follows:", "Text", FontSize->14], Cell[BoxData[ \(\(Plot[Sqrt[x], {x, 0, 10}, PlotRange -> {\(-1\), 3}, PlotStyle -> {Thickness[0.008]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[ "Notice that for the square root function the domain is the set of all \ nonnegative x-values [0,\[Infinity]) and the range is the set of all \ nonnegative y-values [0,\[Infinity]). Again the x and y intercepts are both \ 0. Do you see any geometric relationship between the graphs of the squaring \ function and the square root function? We'll return to this question a little \ later."], "Text", FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Section 4:", " ", "The cube function y = f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]] }], "Subsection"], Cell["\tHere is the graph of the cube function:", "Text", FontSize->14], Cell[BoxData[ \(\(Plot[x^3, {x, \(-3\), 3}, PlotRange -> {\(-20\), 20}, PlotStyle -> {Thickness[0.008]}]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ What are the domain and range of the cube function? The x and y \ intercepts?\ \>", "Text", FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell[TextData[{ "Section 5:", " ", "The reciprocal function y = f(x) = ", Cell[BoxData[ \(TraditionalForm\`1\/x\)]] }], "Subsection"], Cell["\tWhat is its graph?", "Text", FontSize->14], Cell[BoxData[ \(\(Plot[1/x, {x, \(-5\), 5}, PlotRange -> {\(-5\), 5}, PlotStyle -> {Thickness[0.008]}, AspectRatio -> 1]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Notice that 0 is ", StyleBox["not", FontSlant->"Italic"], " included in the domain of the reciprocal function since 1/0 is not \ defined. Hence the domain is (-\[Infinity],0) \[Union] (0,\[Infinity]).The \ range is also the set (-\[Infinity],0) \[Union] (0,\[Infinity]). For this \ function there are no x-intercepts and no y-intercepts, since neither x nor y \ can ever be 0. (Why can't y = 0?)" }], "Text", FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell["Section 6: The absolute value function y = f(x) = |x| ", "Subsection"], Cell[TextData[ "\tSince we can take the absolute value of any number, the domain of |x| is \ all real numbers (-\[Infinity], \[Infinity]). Further since the absolute \ value of any number is nonnegative, the range of the absolute value function \ is all nonnegative numbers [0, \[Infinity]). Predict the x and y intercepts \ before looking at the graph below."], "Text", FontSize->14], Cell[BoxData[ \(\(Plot[Abs[x], {x, \(-4\), 4}, PlotRange -> {\(-4\), 4}, PlotStyle -> {Thickness[0.008]}, \ AspectRatio -> 1]; \)\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]] }, Open ]], Cell[CellGroupData[{ Cell["Section 7: The greatest integer function y = f(x) = [[x]] ", "Subsection"], Cell[TextData[{ "\tThis function might be new to you; it says f(x) = [[x]] is the ", StyleBox["greatest integer LESS than or equal to x", FontWeight->"Bold"], ".For example, [[1.5]] = 1, [[2 ", Cell[BoxData[ \(TraditionalForm\`1\/3\)]], "]] = 2, [[5]] = 5, [[-1 ", Cell[BoxData[ \(TraditionalForm\`1\/2\)]], "]] = -2. Since [[x]] is defined for all x, its domain is all x. But what \ is the range? Notice that no matter what x is, [[x]] is an integer, and any \ integer y is [[y]], so the range is the set of integers. This is a range \ unlike any we've seen (what is different about it?) and warns us that the \ graph might also be unusual. Let's look at the graph." }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ RowBox[{"xmin", ":=", StyleBox[\(-1\), FontColor->RGBColor[0, 0, 1]]}], ";", RowBox[{"xmax", ":=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", \(Clear[f, g]\), ";", \(g[n_] := Table[f[n], {n, xmin, xmax}]\), ";", \(f[n_] := Plot[n, {x, n, n\ + \ 1}, DisplayFunction -> Identity, \ PlotStyle -> {Hue[ .8], Thickness[ .01]\n\t\t}]\), ";", \(Show[g[t], DisplayFunction -> $DisplayFunction]\), ";"}]], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[ "\tThe graph of the greatest integer function is unlike the other basic \ functions in several ways. Besides the different type of set for the range, \ notice the intercepts. What are they? The y-intercept is simply y = 0, but \ the x-intercept? Recall we said a function could have many x-intercepts; this \ one has infinitely many since every point in the half-open interval [0,1) is \ an intercept. That is, for any x such that 0 \[LessEqual] x < 1, [[x]] = 0. \ Notice also that the graph has a point of 'discontinuity', i.e., it takes a \ 'step', at each integer value of x."], "Text", FontSize->14], Cell[TextData[{ StyleBox["WARNING", FontWeight->"Bold"], ": Before continuing to later sections, be sure you are very familiar with \ these basic graphs. We'll see them again, and again, and again." }], "Text", FontSize->14] }, Open ]] }, Closed]], Cell[CellGroupData[{ Cell["B: Techniques", "Section", FontColor->RGBColor[1, 1, 0], Background->RGBColor[0, 0, 1]], Cell["\<\ \tYou will learn basic techniques for describing or changing the \ shape and position of a graph. For example, what effect is there on the \ graph if the function is multiplied by a constant, or a constant is added to \ the function? If these ideas are already familiar to you, you might want to \ test your knowledge with Section 0: Pretest. Otherwise, open Section 1 and \ begin!\ \>", "Text", FontSize->14], Cell[CellGroupData[{ Cell["Section 0: Pretest", "Subsection"], Cell[TextData[{ "\tSome of the material below may be review. Use this pretest to determine \ in what areas you already know the material and in what areas you need \ review. Then you can skip to the appropriate section below until you feel \ that you know all the material. 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StyleBox["increasing", FontVariations->{"Underline"->True}], " on an interval I if for every choice of points ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], " in I, we have that f(", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ") < f(", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], "). Similarly we say that y = f(x) is ", StyleBox["decreasing", FontVariations->{"Underline"->True}], " on an interval I if for every choice of points ", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ", ", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], " in I, we have that f(", Cell[BoxData[ \(TraditionalForm\`x\_1\)]], ") > f(", Cell[BoxData[ \(TraditionalForm\`x\_2\)]], "). Geometrically, the function being increasing on I means that the graph \ is rising (going uphill) on I; being decreasing means the graph is falling \ (going downhill) on I. Note that a function which is ", StyleBox["constant", FontSlant->"Italic"], " on an interval I is neither increasing nor decreasing.\n\tAt a point c \ where the graph changes from decreasing on the left of c to increasing on the \ right of c (e.g., f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " at c = 0), the function value f(c) is larger than all nearby function \ values. In this case, the value f(c) is a ", StyleBox["local maximum", FontVariations->{"Underline"->True}], ". Similarly, if the graph changes from increasing to decreasing, f(c) is \ smaller than all nearby values and is called a ", StyleBox["local minimum", FontVariations->{"Underline"->True}], ".\n\n", StyleBox["Exercise1:", FontWeight->"Bold"], " On which intervals are the basic graphs increasing? decreasing? 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a function f(x) by -1. Compare the graph of f(x) (green) with the graph of \ -f(x) (blue) by executing the program below.\ \>", "Text", FontSize->14], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[f]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{\(f[x_]\), ":=", StyleBox[\(x^2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Change\ nothing\ in\ the\ next\ \(line!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Print["\", f[x], "\< (green) and \>", \(-f[x]\), "\< (blue).\>"]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox[ \(Plot[{f[x], \(-f[x]\)}, {x, x', \(x'\)'}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNow experiment: change the function in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " in the program above to any of the other basic functions described above. \ You may also need to change the x' and x'' values. To make these changes, \ highlight the ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " print and then type in your changes. The original function f(x) will \ always be in green; the function -f(x) in blue. When you are finished making \ your changes, press shift-return. Repeat this process with each of the basic \ functions until you are confident of correctly predicting the effect of \ multiplying f(x) by -1. What conclusions can you draw about the relationship \ between the graph of f(x) and -f(x)?" }], "Text", FontSize->14], Cell[TextData[{ "\tNow let's see what happens if we multiply x by -1 ", StyleBox["before", FontSlant->"Italic"], " applying the function f, i.e., let's compare the graphs of f(x) and \ f(-x). Again execute the program below and then change the parts in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " to experiment with a few different functions. As before, the original \ graph will be green; the new graph blue." }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[f]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{\(f[x_]\), ":=", StyleBox[\(x^3\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Change\ nothing\ in\ the\ next\ \(line!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Print["\", f[x], \ "\< (green) and \>", f[\(-x\)], "\< (blue).\>"]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], StyleBox[ \(Plot[{f[x], f[\(-x\)]}, {x, x', \(x'\)'}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tLet's summarize the predicted outcomes of the two experiments above. \ Looking back at the results of the first experiment, if we think of the \ x-axis as a mirror, the new graph is precisely the mirror image of the \ original. We say the graph of the function -f(x) is the ", StyleBox["reflection across the x-axis", FontVariations->{"Underline"->True}], " of the original graph. This is a ", StyleBox["vertical reflection", FontVariations->{"Underline"->True}], ".\n\tIn the second, the mirror is along the y-axis. The graph of f(-x) is \ the ", StyleBox["reflection across the y-axis", FontVariations->{"Underline"->True}], " of the original graph; this is a ", StyleBox["horizontal reflection", FontVariations->{"Underline"->True}], ". " }], "Text", FontSize->14], Cell[TextData[{ "\tBe sure to try several different functions in the above programs. \ Notice that when you use the second program on some functions, for example \ f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " or f(x) = |x|, only one graph is shown. This is because the two graphs \ are identical; that is, one is simply superimposed on the other. Since blue \ is the darker color, we only see the blue graph. (Does this ever happen on \ the first program?)" }], "Text", FontSize->14], Cell[TextData[{ "\tThe functions f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " and f(x) = |x| are just examples of a group of functions known as ", StyleBox["even", FontVariations->{"Underline"->True}], " functions, i.e., functions for which the graphs of f(x) and f(-x) are the \ same. We say a function f(x) is ", StyleBox["even", FontVariations->{"Underline"->True}], " if f(-x) = f(x) and for every point x in the domain, -x is also in the \ domain. We can, of course, test this algebraically. We can also see it \ geometrically: the graph of an ", StyleBox["even", FontVariations->{"Underline"->True}], " function is ", StyleBox["symmetric with respect to the y-axis", FontVariations->{"Underline"->True}], "; that is, the graph reflects across the y-axis to coincide precisely with \ itself. This explains the appearance of only one graph for some functions \ when using the second program above; those are exactly the even functions.\n\t\ The second program is repeated below. Try it again, this time with the \ notion of even functions in mind. The program starts with the example f(x) = \ ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], ", an even function. Algebraically we can see that it is even; that is we \ can see that f(x) = f(-x) since f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " and f(-x)= ", Cell[BoxData[ \(TraditionalForm\`\((\(-x\))\)\^2\)]], " = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], ". Geometrically we see that it is even by looking at the graph and \ noticing that f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " is symmetric about the y-axis. The program below will graph f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " for you. It will also graph f(-x) at the same time, but you will only be \ able to see the second graph when the function is NOT even. So for f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " you will see only one graph. Try other functions by replacing the \ function (and x', x'' values if necessary) in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], "." }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[f]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{\(f[x_]\), ":=", StyleBox[\(x^2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Change\ nothing\ in\ the\ next\ \(line!\)*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Plot[{f[x], f[\(-x\)]}, {x, x', \(x'\)'}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tA function f(x) is ", StyleBox["odd", FontVariations->{"Underline"->True}], " if f(-x) = -f(x) and for every point x in the domain, -x is also in the \ domain. (Notice f(-x)= -f(x) is the same as saying f(x) = -f(-x).) This \ notion also has a geometric interpretation: the graph of an odd function is ", StyleBox["symmetric with respect to the origin", FontVariations->{"Underline"->True}], ". This means that if we rotate the graph by 180 degrees about the origin \ we get the exact same graph. Equivalently it means that if we reflect the \ graph across the y-axis ", StyleBox["and", FontSlant->"Italic"], " the x-axis, we land precisely on the original graph. Consider the \ function f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], ". It is easy to see algebraically that it is odd, since ", Cell[BoxData[ \(TraditionalForm\`\((\(-x\))\)\^3\)]], "= -", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], ". It is also easy to see graphically. Rotate 180 degrees about the \ origin. ", StyleBox["Or", FontSlant->"Italic"], " reflect across the y-axis (this is the graph of -f(x)) ", StyleBox["then", FontSlant->"Italic"], " across the x-axis (this would give the graph of -f(-x)). In the \ following program we graph f(x) and -f(-x). So the program shows only ONE \ graph when f is ", StyleBox["odd", FontVariations->{"Underline"->True}], ". " }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[f]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{\(f[x_]\), ":=", StyleBox[\(x^3\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-3\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["3", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Change\ nothing\ below\ this\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Plot[{f[x], \(-f[\(-x\)]\)}, {x, x', \(x'\)'}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNow experiment by changing the function in ", StyleBox["blue", FontColor->RGBColor[0, 0, 1]], " to any of the other basic functions and adjusting the x' and x'' values \ as necessary. Which of the basic functions are odd? " }], "Text", FontSize->14], Cell[TextData[{ " \tSome functions are neither even nor odd. ", StyleBox["Question", FontWeight->"Bold"], ": Which of the basic functions are even? odd? neither? Use the program \ below to answer this question. The program tests graphically whether a \ function is even, odd, or neither by plotting the original function f(x) in \ green and the function f(-x) in blue. You can fill in the function that you \ want to test. (As usual, you may need to adjust the x' and x'' values to \ give a larger portion of the domain.) If you see only one graph, the \ function is even. If you see two graphs, one the mirror image of the other \ across the x-axis, the function is odd. If neither of these happens, the \ function is neither even or odd." }], "Text", FontSize->14], Cell[BoxData[ RowBox[{ StyleBox[\(Clear[f]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{\(f[x_]\), ":=", StyleBox[\(x^3\ + \ 1\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(x'\), "=", StyleBox[\(-1\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", RowBox[{\(\(x'\)'\), "=", StyleBox["2", FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*change\ nothing\ below\ this\ line*) \), FontColor->RGBColor[1, 0, 0]], "\n", StyleBox[ \(Plot[{f[x], f[\(-x\)], 0}, {x, x', \(x'\)'}, PlotStyle -> {{Hue[0.2], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}, {Hue[0, 0, 0]}}] If[f[\(-x\)] == f[x], Print["\", f[x], "\"], Print["\", f[x], "\< is not even.\>"], Print["\", f[x], "\< is not even.\>"]] If[f[\(-x\)] == \(-f[x]\), Print["\", f[x], "\"], Print["\", f[x], "\< is not odd.\>"], Print["\", f[x], "\< is not odd.\>"]]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]]}]], "Input", Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise", FontWeight->"Bold", FontVariations->{"Underline"->True}], ": For each of the following functions (and any others you'd like to test), \ first determine algebraically if the function is even or odd, then use the \ program above to test your answer.\n(a) ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + 1\n(b) 2", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], "- x", " + ", "5\n(c) 3|x| - 2 [Note: enter |x| as Abs[x] in the program.]\n(d) 3|x-2|" }], "Text", FontSize->14] }, Open ]], Cell[CellGroupData[{ Cell["Section 3: Shifts", "Subsection"], Cell[TextData[{ "\tLet's see what happens when we take one of the basic functions, say f(x) \ = ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], ", and graph ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + ", " c for different values of c." }], "Text", FontSize->14], Cell[BoxData[ \(Do[Print["\", c, "\<).\>"]; Plot[x^3\ + \ c, {x, \(-2\), 2}, PlotRange -> {\(-10\), 10}, \n\t\t PlotStyle -> {Hue[c/6], Thickness[0.008]}], {c, \(-2\), 2}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["\<\ \tNow let's have a bit of (instructive) fun with these graphs. We \ can animate the graphs to move through the various values of c. \"Select\" \ all the graphs above by clicking once on the second blue bar from the left \ (the one which includes just the graphs) and make sure that one graph is \ shown completely on your monitor. Select 'Cell' from the Menu bar at the top \ of the screen and select \"Animate Selected Graphics\" from the Cell menu. \ What you will see on the screen is all 6 graphs shown in quick succession on \ the set of axes in front of you. In fact, it is generally so quick that it's \ hard to follow, but you can control the speed. In the lower left corner of \ the graphics window is a double arrow pointing down; press on this arrow to \ slow the animation. \ \>", "Text", FontSize->14], Cell[TextData[{ "\tWhat do you see happening as c changes values from -2 to 2? Describe \ the graphical effect of adding the c-value to f. What happens if c > 0? if c \ < 0?\n\tNow let's examine the effect of adding c to x before applying the \ function f. For example, again starting with f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], ", we consider f(x + c) = ", Cell[BoxData[ \(TraditionalForm\`\((x + c)\)\^3\)]], ". Note that this is definitely not the same function as above, since ", Cell[BoxData[ \(TraditionalForm\`\((x + c)\)\^3\)]], "= ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " + 3", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], "c + 3x", Cell[BoxData[ \(TraditionalForm\`c\^2\)]], " + ", Cell[BoxData[ \(TraditionalForm\`c\^3\)]], "which is not the same as", Cell[BoxData[ \(TraditionalForm\`\(\ \ x\^3\)\)]], " + c!\n\tUse the following program to graph ", Cell[BoxData[ \(TraditionalForm\`\((x + c)\)\^3\)]], " for different values of c." }], "Text", FontSize->14], Cell[BoxData[ \(Do[Print["\", c, \*"\"\<\!\(\()\^3\)\).\>\""]; Plot[\((x\ + \ c)\)^3, {x, \(-3\), 3}, PlotRange -> {\(-20\), 20}, \n \t\tPlotStyle -> {Hue[c/6], Thickness[0.008]}], {c, \(-2\), 2}]\)], "Input", AnimationDisplayTime->1.06045, FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "Now, animate your graphs. What is the effect of adding c to x BEFORE \ applying f if c < 0? if c > 0?\n\tHere is what you should have noticed: If a \ number c is added to f(x), the graph of the new function f(x) + c has the \ same shape as that of f but is ", StyleBox["shifted vertically", FontVariations->{"Underline"->True}], " up by c (if c > 0) or down by c (if c < 0). (Hence in the animation, the \ graphs are moving ", StyleBox["up", FontSlant->"Italic"], " as c increases.) If a number c is added to x BEFORE applying f, the \ graph of the new function f(x + c) has the same SHAPE as that of f but is ", StyleBox["shifted horizontally", FontVariations->{"Underline"->True}], " to the left by c (if c > 0) or to the right by c (if c < 0). (So the \ animated graphs move ", StyleBox["left", FontSlant->"Italic"], " with increasing c. That is, when c = -2 the graph of ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " has been shifted to the right; as c increases and eventually becomes \ positive, the graph of f(x + c) moves to the left of the original ", Cell[BoxData[ \(TraditionalForm\`x\^3\)]], " graph.)" }], "Text", FontSize->14], Cell[TextData[{ "\tNow let's combine these. We start with a function f(x) and change it to \ a new function f(x + c) + d. The graph of this new function will be shifted \ vertically by d and horizontally by c. Here are a couple of examples using \ the basic functions. We'll graph the original function in green, the shifted \ graph in blue. Before looking at the graphics, predict what the two graphs \ will look like. The hit shift-return simultaneously at the end of the line \ to check your answer.\n\n\tFirst, ", Cell[BoxData[ \(TraditionalForm\`f(x)\ = \ \((x - 2)\)\^2\)]], " + 3:" }], "Text", FontSize->14], Cell[BoxData[ \(Print[ \*"\"\\""]; Plot[{x^2, \((x - 2)\)^2\ + \ 3}, {x, \(-3\), 5}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell["Now try f(x) = |x + 3| - 4:", "Text", FontSize->14], Cell[BoxData[ \(Print["\"]; Plot[{Abs[x], Abs[x\ + \ 3] - 4}, {x, \(-5\), 5}, PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]; \)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ StyleBox["Exercise 1:", FontWeight->"Bold"], " Sketch the graph of each of the following on a piece of paper. \n(a) ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], "- 2" }], "Text", FontSize->14], Cell[TextData[{ "(b) ", Cell[BoxData[ \(TraditionalForm\`\((x - 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examine the effect of \ multiplying by the constant k. We'll begin with the function f(x) = ", Cell[BoxData[ \(TraditionalForm\`x\^2\ - \ 2 x\)]], ", which is just a slight variation of the squaring function, and graph k \ f(x) for various values of k." }], "Text", FontSize->14], Cell[BoxData[ \(Do[Print["\", k, \*"\"\<(\!\(x\^2\)-2x).\>\""]; Plot[k\ \((x^2 - 2\ x)\), {x, \(-2\), 4}, PlotRange -> {\(-5\), 10}, PlotStyle -> {Hue[k/6], Thickness[0.008]}], {k, 1, 4}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tAgain animate the above sequence as described in the last section, and \ describe the effect you see. The original graph (of y = ", Cell[BoxData[ \(TraditionalForm\`x\^2\)]], " - 2 x) is the first one. \n\tIn the yellow box below, we repeat the \ program above only now you can choose the values of k. To do this, highlight \ the value in ", StyleBox["blue ", FontColor->RGBColor[0, 0, 1]], "and type in whatever value for k that you want to try. Change nothing \ else; then hit shift-return at the end of the program. The original graph \ will be in green, the graph of k\[CenterDot]f(x) in blue." }], "Text", FontSize->14], Cell[BoxData[{ RowBox[{ StyleBox[\(Clear[k]\), FontColor->GrayLevel[0.666667]], StyleBox[";", FontColor->GrayLevel[0.666667]], "\n", RowBox[{"k", "=", StyleBox[\(1/2\), FontColor->RGBColor[0, 0, 1]]}], ";", "\n", StyleBox[\( (*Do\ not\ change\ the\ next\ line*) \), FontColor->RGBColor[1, 0, 0]]}], StyleBox[ \(Print["\", k, "\<(\!\(x\^2\)-2x).\>"]; Plot[{x^2\ - \ 2\ x, k\ \((x^2\ - \ 2\ x)\)}, {x, \(-2\), 4}, \ PlotStyle -> {{Hue[0.4], Thickness[0.008]}, {Hue[0.6], Thickness[0.008]}}]; \), FontColor->GrayLevel[0.666667]]}], "Input", Background->RGBColor[1, 1, 0]], Cell["\<\ Continue this experiment until you can describe the effect when k > \ 1, when 0 < k < 1, when k < 0.\ \>", "Text", FontSize->14], Cell[TextData[{ "\tNow let's try multiplying x by k before applying the function f, i.e. \ let's look at f(", "k x", ") = ", Cell[BoxData[ \(TraditionalForm\`\((k\ x)\)\^2\ - \ 2 \((k\ x)\)\)]], ". You should again describe the effect. Remember k = 1 is the original \ function." }], "Text", FontSize->14], Cell[BoxData[ \(Do[Print["\", k, \*"\"\\"", k, "\"]; Plot[\((k\ x)\)^2 - 2\ k\ x, {x, \(-2\), 5}, PlotRange -> {\(-5\), 10}, PlotStyle -> {Hue[k/6], Thickness[0.008]}], {k, 1/2, 3, 1/2}]\)], "Input", FontColor->GrayLevel[0.666667], Background->RGBColor[1, 1, 0]], Cell[TextData[{ "\tNotice from the first series of examples above that when we multiply \ f(x) by k, the graph is vertically stretched for k > 1 and vertically \ compressed for k < 1. The graph is stretched/compressed by a factor of k. \ To see this, fix a value of x and look at the corresponding y-value in the \ graph of f(x). That y-value lies on a vertical line through x. Now look at \ the corresponding y-value in the graph of k f(x). That y-value lies on the \ same vertical line and can be obtained by multiplying the previous y-value by \ k. It's as if the vertical line has been stretched or compressed by a factor \ of k. Draw a vertical line through some x-values on any of the graphs above \ to convince yourself of this.\n\t Multiplying the variable x by k and then \ applying the function f may not produce this same effect. By looking at the \ function f(x) = ", Cell[BoxData[ \(TraditionalForm\`\(x\^2 - \ 2\ x\ \)\)]], "in the second set of examples, we can witness that the property described \ above does not hold when we multiply the variable x (instead of the whole \ function, f(x)) by k. To see this, choose the x-value x = 2. At that \ x-value the original function f(x) has an x-intercept; that is, the \ corresponding y-value, f(2), is 0. Now k\[CenterDot]0 = 0 no matter what k we \ choose so if the above property held, then at x = 2, corresponding y-value of \ f(k x) should be 0, i.e., the graph of f(k x) should always have an \ x-intercept x = 2. We can see, however, that this is NOT the case. Instead, \ that the value where they do have an x-intercept is ", Cell[BoxData[ \(TraditionalForm\`2\/k\)]], ". This provides a clue as to the graphical relationship between f(x) and \ f(k x): the y-value assigned to x by f(x) is exactly the y-value assigned to \ ", Cell[BoxData[ \(TraditionalForm\`\(x\/k\ \)\)]], "by f(k x). What this means is that if k > 1 for example, the graph \ proceeds through the same set of y-values k times as fast." }], "Text", FontSize->14], Cell[TextData[{ "\tHere is a formal description of the result of multiplying a function, \ f(x), by a constant, k:\n-if ", StyleBox["k >1", FontVariations->{"Underline"->True}], ", the graph of the new function, k f(x), is a ", StyleBox["vertical stretch", FontVariations->{"Underline"->True}], " of the graph of the original function, f(x), by a factor of k\n-if ", StyleBox["0 < k < 1", FontVariations->{"Underline"->True}], ", the graph of the new function, k f(x), is a ", StyleBox["vertical compression", FontVariations->{"Underline"->True}], " of the graph of the original function, f(x), by a factor of k\n-if ", StyleBox["k < 0", FontVariations->{"Underline"->True}], ", the graph of the new function, k f(x), is obtained by a ", StyleBox["vertical reflection", FontVariations->{"Underline"->True}], " followed by a ", StyleBox["vertical stretch/compression", FontVariations->{"Underline"->True}], " of the graph of the original function, f(x).\n\tThe following is a formal \ summary of the effect of multiplying the variable x by a constant k before \ applying f:\n-", StyleBox["if k > 1", FontVariations->{"Underline"->True}], " the graph of the new function f(k x) is a ", StyleBox["horizontal compression", FontVariations->{"Underline"->True}], " of the graph of the original function\n-", StyleBox["if 0 < k < 1", FontVariations->{"Underline"->True}], " the graph of the new function f(k x) is a ", StyleBox["horizontal stretch", FontVariations->{"Underline"->True}], " of the graph of the original function", StyleBox["\n", FontVariations->{"Underline"->True}], "-", StyleBox["if k < 0", FontVariations->{"Underline"->True}], ", the graph of the new function f(k x) is obtained by a ", StyleBox["horizontal reflection", FontVariations->{"Underline"->True}], " followed by a ", StyleBox["horizontal stretch", FontVariations->{"Underline"->True}], " of the graph of the original function, f(x).\n\t", StyleBox["Note", FontWeight->"Bold"], ": If this is not clear, go back to the two animations in this section and \ watch them slowly with the above definitions in mind.\n\t\n", StyleBox["Exercise1:", FontWeight->"Bold"], " Predict the shape of each graph." }], "Text", FontSize->14], Cell[TextData[{ "a) f(x) =", " ", Cell[BoxData[ FormBox[ StyleBox[\(\(|x | \)\/4\), FontSize->13], TraditionalForm]]] }], "Text", FontSize->14], Cell["b) f(x) = 3 |x| + 5", "Text", FontSize->14], Cell[TextData[{ "c) f(x) = -", Cell[BoxData[ \(TraditionalForm\`-\+4\%1\)]], Cell[BoxData[ \(TraditionalForm\`\(\ x\^2\)\)]], " + ", "1" }], "Text", FontSize->14], Cell[TextData[{ "d)", " ", "f(x) = ", Cell[BoxData[ \(TraditionalForm\`1\/\(3 \((x - 1)\)\)\)]] }], "Text", FontSize->14], Cell[TextData[{ "e) f(x) = 3 ", Cell[BoxData[ \(TraditionalForm\`\@\(x - 2\)\)]] }], "Text", FontSize->14], Cell[TextData[{ "f) f(x) ", Cell[BoxData[ \(TraditionalForm\`\@\(3 x\ - \ 2\)\)]] }], "Text", FontSize->14], Cell["g) f(x) = 2 [[x]]", "Text", FontSize->14], Cell["h) f(x) = [[2x]]", "Text", FontSize->14], Cell[TextData[{ StyleBox["Exercise2: ", FontWeight->"Bold", FontSlant->"Plain"], StyleBox[ "Shown below is the graph of a function f(x). 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Review this table to be sure you know the \ techniques. Then test yourself with the exercises at the end.\ \>", "Text", FontSize->14], Cell[TextData[{ StyleBox["REFLECTIONS", FontWeight->"Bold"], "\n\ty = - f(x)\t\t\t\tReflects the graph of f across the x-axis\n\ty = \ f(-x)\t\t\t\tReflects the graph of f across the y-axis\n", StyleBox["VERTICAL SHIFTS", FontWeight->"Bold"], "\n\ty = f(x) + c, c > 0\t\t\tRaises the graph of f by c units\n\ty = f(x) \ + c, c < o\t\t\tLowers the graph of f by c units\n", StyleBox["HORIZONTAL SHIFTS", FontWeight->"Bold"], "\n\ty = f(x + c), c > 0\t\t\tMoves the graph of f to the left c units\n\t\ y = f(x", " + ", "c), c < 0\t\t\tMoves the graph of f to the right c units\n", StyleBox["COMPRESSING/STRETCHING", FontWeight->"Bold"], "\n\ty = k f(x), k > 1\t\t\tStretches the graph vertically by a factor of \ k\n\ty = k f(x), 0 < k < 1\t\t\tCompresses the graph vertically by a factor \ of k\n\t\n\ty = f(k x), k > 1\t\t\tStretches the graph of f horizontally by \ a factor of k\n\ty = f(k x), 0 < k < 1\t\t\tCompresses the graph of f by a \ factor of k\t\t\t" }], "Text", FontSize->14], Cell[CellGroupData[{ Cell["Order of Graphing Operations", "Subsubsection"], Cell[TextData[{ "\t", StyleBox[ "Now we can combine all of these conclusions to graph a function of the \ form ", FontSize->14], StyleBox["g(x) = c\[CenterDot]f(ax", FontSize->14, FontSlant->"Italic"], " + ", StyleBox["b)", FontSize->14, FontSlant->"Italic"], " + ", StyleBox["d", FontSize->14, FontSlant->"Italic"], StyleBox[ ". We must BE CAREFUL and follow a very specific order of graphing \ operations when tranforming the original graph of ", FontSize->14], StyleBox["f(x) ", FontSize->14, FontSlant->"Italic"], StyleBox["into the transformed graph of ", FontSize->14], StyleBox["g(x). ", FontSize->14, FontSlant->"Italic"], StyleBox[ "This order is as follows:\n\t1. horizontal compression by a factor of ", FontSize->14], StyleBox["a", FontSize->14, FontSlant->"Italic"], StyleBox["; i.e., divide the x-coordinate of the points on the graph of ", FontSize->14], StyleBox["f(x)", FontSize->14, FontSlant->"Italic"], StyleBox[" by ", FontSize->14], StyleBox["a", FontSize->14, FontSlant->"Italic"], StyleBox["\n\t2. horizontal shift by ", FontSize->14], StyleBox["b/a", FontSize->14, FontSlant->"Italic"], StyleBox["; i.e., subtract ", FontSize->14], StyleBox["b/a", FontSize->14, FontSlant->"Italic"], StyleBox[" from the x-coordinate of the points on the graph of ", FontSize->14], StyleBox["f(x)", FontSize->14, FontSlant->"Italic"], StyleBox["\n\t3. vertical stretch by a factor of ", FontSize->14], StyleBox["c", FontSize->14, FontSlant->"Italic"], StyleBox[ "; i.e., multiply the y-coordinates of the points on the graph of ", FontSize->14], StyleBox["f(x)", FontSize->14, FontSlant->"Italic"], StyleBox[" by ", FontSize->14], StyleBox["c", FontSize->14, FontSlant->"Italic"], StyleBox["\n\t4. vertical shift by ", FontSize->14], StyleBox["d", FontSize->14, FontSlant->"Italic"], StyleBox["; i.e., add ", FontSize->14], StyleBox["d", FontSize->14, FontSlant->"Italic"], StyleBox[" to the y-coordinates of the points on the graph of ", FontSize->14], StyleBox["f(x)", FontSize->14, FontSlant->"Italic"], StyleBox[ "\nTo see why we must transform the graph in this order, let's suppose (", FontSize->14], Cell[BoxData[ \(TraditionalForm\`x\_0\)], FontSize->14], StyleBox[", ", FontSize->14], Cell[BoxData[ \(TraditionalForm\`y\_0\)], FontSize->14], StyleBox[ ") is a point on the graph of f(x). We need to figure out what x-value ", FontSize->14], Cell[BoxData[ \(TraditionalForm\`x\_0\)], FontSize->14], StyleBox[" corresponds to on the graph of ", FontSize->14], StyleBox["g(x) = c\[CenterDot]f(ax", FontSize->14, FontSlant->"Italic"], " + ", StyleBox["b)", FontSize->14, FontSlant->"Italic"], " + ", StyleBox["d. ", FontSize->14, FontSlant->"Italic"], StyleBox[" ", FontSlant->"Italic"], "In other words, we have that ", Cell[BoxData[ \(TraditionalForm\`x\_0\)]], StyleBox[" = ax", FontSlant->"Italic"], " + ", StyleBox["b", FontSlant->"Italic"], " and we want to solve for ", StyleBox["x", FontSlant->"Italic"], ". We obtain: ", StyleBox["x = ", FontSlant->"Italic"], Cell[BoxData[ \(TraditionalForm \`\((x\_0\ - \ b)\)\ /\ a\ \ = \ x\_0/a\ - \ b/a\)]], ". 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