#set document(title: "7.2 The General Sinusoidal Function", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 7.2#h(0.6em)The General Sinusoidal Function === Horizontal Shifts In the previous section we considered transformations of sinusoidal graphs, including vertical shifts, which change the midline of the graph, vertical stretches and compressions, which change its amplitude, and horizontal stretches and compressions, which occur when we change the period of the graph. In this section we consider one more transformation, shifting the graph horizontally. The figure below shows four different transformations of the graph of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$]. #figure(figph[translated sine], alt: "translated sine", caption: none) #figure(figph[scaled sine], alt: "scaled sine", caption: none) #figure(figph[horizontal compression of sine], alt: "horizontal compression of sine", caption: none) #figure(figph[horizontal shift of sine], alt: "horizontal shift of sine", caption: none) #examplebox("Example 1")[][ Graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sin ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals sin ⁡ open parenthesis x minus the fraction π over 4 close parenthesis")[$" " g ( x ) = sin ( x − display(frac(π, 4)) ) " "$] for #math.equation(block: false, alt: "minus 2 π less than or equal to x less than or equal to 2 π")[$− 2 π ≤ x ≤ 2 π$]. How is the graph of #math.equation(block: false, alt: "g")[$g$] different from the graph of #math.equation(block: false, alt: "f")[$f$]? #solutionbox[ The graphs are shown below. The graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals sin ⁡ open parenthesis x minus the fraction π over 4 close parenthesis")[$" " g ( x ) = sin ( x − display(frac(π, 4)) ) " "$] has the same amplitude, midline, and period as the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = sin ( x )$], but the graph of #math.equation(block: false, alt: "g")[$g$] is shifted to the right by #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] units, compared to the graph of #math.equation(block: false, alt: "f")[$f$]. #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) We can see why this shift occurs by studying a table of values for the two functions. #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction 5 π over 4")[$display(frac(5 π, 4))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "the fraction 7 π over 4")[$display(frac(7 π, 4))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 2 over 2")[$display(frac(sqrt(2), 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 2 over 2")[$display(frac(− sqrt(2), 2))$]], )) Notice that in the table, #math.equation(block: false, alt: "g")[$g$] has the same function values as #math.equation(block: false, alt: "f")[$f$], but each one is shifted #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] units to the right. The same thing happens in the graph: each #math.equation(block: false, alt: "y")[$y$]-value appears #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] units farther to the right on #math.equation(block: false, alt: "g")[$g$] than it does on #math.equation(block: false, alt: "f")[$f$]. ] ] Graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals cos ⁡ open parenthesis x close parenthesis")[$" " f ( x ) = cos ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals cos ⁡ open parenthesis x plus the fraction π over 4 close parenthesis")[$" " g ( x ) = cos ( x + display(frac(π, 4)) ) " "$] for #math.equation(block: false, alt: "minus 2 π less than or equal to x less than or equal to 2 π")[$− 2 π ≤ x ≤ 2 π$]. How is the graph of #math.equation(block: false, alt: "g")[$g$] different from the graph of #math.equation(block: false, alt: "f")[$f$]? The graph of #math.equation(block: false, alt: "g")[$g$] is shifted #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] units to the left of #math.equation(block: false, alt: "f")[$f$]. The examples above illustrate the following principle. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Shifts] The graphs of #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis x minus h close parenthesis and t equals cos ⁡ open parenthesis x minus h close parenthesis")[$y = sin ( x − h ) " " " " " " " " "and" " " " " " " " " t = cos ( x − h )$] are #emph[shifted horizontally] compared to the graphs of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$] and #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$]. - If #math.equation(block: false, alt: "h greater than 0")[$h > 0$], the graph is shifted to the right. - If #math.equation(block: false, alt: "h less than 0")[$h < 0$], the graph is shifted to the left. ] We can use a table of values to sketch graphs that involve horizontal shifts. #examplebox("Example 2")[][ Make a table of values and sketch a graph of #math.equation(block: false, alt: "y equals minus 2 cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$" " y = − 2 cos ( x + display(frac(π, 3)) )$]. #solutionbox[ First notice that we can write the equation as #math.equation(block: true, alt: "y equals minus 2 cos ⁡ open bracket x minus open parenthesis the fraction minus π over 3 close parenthesis close bracket")[$y = − 2 cos [ x − ( display(frac(− π, 3)) ) ]$] so that #math.equation(block: false, alt: "h equals minus the fraction π over 3")[$h = − display(frac(π, 3))$], and we expect the graph to be shifted #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the left compared to the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$]. We choose convenient values for the inputs of the cosine function, namely #math.equation(block: false, alt: "x plus the fraction π over 3")[$x + display(frac(π, 3))$]. #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x plus the fraction π over 3")[$x + display(frac(π, 3))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$cos ( x + display(frac(π, 3)) )$]], [#math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$y = cos ( x + display(frac(π, 3)) )$]]), [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(bold(italic(π)), 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$bold(italic(π))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 bold(italic(π)), 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 bold(italic(π))$]], [$#hide($0000$)$], [$#hide($0000$)$], )) From those values, we work backwards to #math.equation(block: false, alt: "x")[$x$] and forwards to #math.equation(block: false, alt: "y")[$y$]. (To obtain the values of #math.equation(block: false, alt: "x")[$x$], we subtract #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]from the values of #math.equation(block: false, alt: "x plus the fraction π over 3")[$x + display(frac(π, 3))$].) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x plus the fraction π over 3")[$x + display(frac(π, 3))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$cos ( x + display(frac(π, 3)) )$]], [#math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$y = cos ( x + display(frac(π, 3)) )$]]), [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], )) To make the graph, we'll scale the #math.equation(block: false, alt: "x")[$x$]-axis in multiples of #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]. We plot the guide points #math.equation(block: false, alt: "open parenthesis x , y close parenthesis")[$( x , y )$] from the table, and sketch a sinusoidal graph through the points. One cycle of the graph is shown below. As we expected,the graph is shifted #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the left compared to the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$" " y = cos ( x )$]. #figure(figph[transformed cosine], alt: "transformed cosine", caption: none) ] ] Complete the table and sketch a graph of #math.equation(block: false, alt: "y equals minus 2 plus sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$" " y = − 2 + sin ( x − display(frac(π, 6)) ) " "$] on the grid below. (Do you expect the graph to be shifted to the left or to the right, compared to the graph of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$" " y = sin ( x )$]?) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus the fraction π over 6")[$x − display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$sin ( x − display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "y equals minus 2 plus sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$y = − 2 + sin ( x − display(frac(π, 6)) )$]]), [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) #figure(figph[grid], alt: "grid", caption: none) #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus the fraction π over 6")[$x − display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$sin ( x − display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "y equals minus 2 plus sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$y = − 2 + sin ( x − display(frac(π, 6)) )$]]), [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "the fraction 13 π over 6")[$display(frac(13 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], )) #figure(figph[transformed sine], alt: "transformed sine", caption: none) === Combining Transformations The order in which we apply transformations to a function makes a difference in the graph. We'll compare the graphs of the two functions #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals sin ⁡ open bracket 2 open parenthesis x minus the fraction π over 3 close parenthesis close bracket and g open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$f ( x ) = sin [ 2 ( x − display(frac(π, 3)) ) ] " " " " " " " " "and" " " " " " " " " g ( x ) = sin ( 2 x − display(frac(π, 3)) )$] Graph the two functions for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$], along with #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis 2 x close parenthesis")[$" " y = sin ( 2 x ) " "$], as shown below. Each graph involves a horizontal shift relative to #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis 2 x close parenthesis")[$" " y = sin ( 2 x )$], but the graph of #math.equation(block: false, alt: "f")[$f$] is shifted #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the right, while the graph of #math.equation(block: false, alt: "g")[$g$] is shifted only #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] units to the right. This difference occurs because of the order of the transformations. #figure(figph[sinusoidal graphs], alt: "sinusoidal graphs", caption: none) #figure(figph[sinusoidal graphs], alt: "sinusoidal graphs", caption: none) We transform the graph of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$" " y = sin ( x ) " "$] into the graph of #math.equation(block: false, alt: "f")[$f$] in two steps: Step 1: First we replace #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "2 x")[$2 x$] to get #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis 2 x close parenthesis")[$" " y = sin ( 2 x ) " "$], which compresses the graph horizontally by a factor of 2. (See figure (a) below.) Step 2: Then we replace #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "x minus the fraction π over 3")[$x − display(frac(π, 3))$] to get #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open bracket 2 open parenthesis x minus the fraction π over 3 close parenthesis close bracket")[$" " f ( x ) = sin [ 2 ( x − display(frac(π, 3)) ) ]$], which shifts the graph #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the right. (See figure (b).) #figure(figph[horizontal compression of sin x], alt: "horizontal compression of sin x", caption: none) #figure(figph[translation of sin 2x], alt: "translation of sin 2x", caption: none) To transform the graph of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$" " y = sin ( x ) " "$] into the graph of #math.equation(block: false, alt: "g")[$g$], we perform the two steps in the opposite order: Step 1: We replace #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "x minus the fraction π over 3")[$x − display(frac(π, 3))$] which shifts the graph #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the right. (See figure (a).) Step 2: Then we replace #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "2 x")[$2 x$] to get #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$" " g ( x ) = sin ( 2 x − display(frac(π, 3)) ) " "$] which compresses the graph horizontally by a factor of 2, including the horizontal shift. This reduces the horizontal shift from #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] to #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]. (See figure (b.) #figure(figph[horizontal shift of sine], alt: "horizontal shift of sine", caption: none) #figure(figph[horizontal compression of shifted sine], alt: "horizontal compression of shifted sine", caption: none) It is easier to analyze the transformations in the function #math.equation(block: false, alt: "f")[$f$], because we can read the horizontal shift from the formula. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals sin ⁡ open bracket 2 open parenthesis x minus the fraction π over 3 close parenthesis close bracket")[$f ( x ) = sin [ 2 ( x − display(frac(π, 3)) ) ]$] where #math.equation(block: false, alt: "2")[$2$] is the compression factor and #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] is the horizontal shift. We can write the formula for #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] in the same easy-to-analyze form by factoring the input for the sine function: #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis equals sin ⁡ open bracket 2 open parenthesis x minus the fraction π over 6 close parenthesis close bracket")[$g ( x ) = sin ( 2 x − display(frac(π, 3)) ) = sin [ 2 ( x − display(frac(π, 6)) ) ]$] In general, if we write the formula for a sinusoidal function in #strong[standard form], we can read all the transformations from the constants in the formula. The standard form y = a sin(b(x − h)) + k with all four constants on sliders, against y = sin x (gray). Read each transformation from its constant, exactly as the section says: a is the amplitude (drag it negative to reflect), b is the compression factor — the period is 2π/b, so b = 2 fits two full cycles where the basic graph fits one — h is the horizontal shift of the starting point, and k is the midline. Try reproducing the section's worked example: set b = 2 first and then slide h, and notice the graph moves by h itself, not by bh — the reason the section insists on factoring the input before reading off the shift.{"functions":\[{"color":"\#6a6a6a","expression":"sin(x)","expression2":"","id":"basic-sine","inequality":"lt","lineDash":\[6,4\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#1f4e79","expression":"a\*sin(b\*(x - h)) + k","expression2":"","id":"sinusoid","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","expression":"k","expression2":"","id":"midline","inequality":"lt","lineDash":\[8,5\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Amplitude a","max":3,"min":-3,"name":"a","step":0.1,"value":2},{"id":"p-b","label":"Compression b (period 2\\u03c0/b)","max":4,"min":0.5,"name":"b","step":0.1,"value":2},{"id":"p-h","label":"Horizontal shift h","max":3.14,"min":-3.14,"name":"h","step":0.01,"value":0.79},{"id":"p-k","label":"Midline k","max":2,"min":-2,"name":"k","step":0.1,"value":1}\],"title":"y = a sin(b(x \\u2212 h)) + k, all four constants live","version":1,"viewport":{"centerX":0,"centerY":0,"scale":46}}#notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Standard Form for Sinusoidal Functions] The graphs of the functions #math.equation(block: true, alt: "y equals A sin ⁡ open parenthesis B open parenthesis x minus h close parenthesis close parenthesis plus k and y equals A cos ⁡ open parenthesis B open parenthesis x minus h close parenthesis close parenthesis plus k")[$bold(italic(y)) = bold(italic(A)) bold(sin) ( bold(italic(B)) ( bold(italic(x)) − bold(italic(h)) ) ) + bold(italic(k)) " " " " " " " " "and" " " " " " " " " bold(italic(y)) = bold(italic(A)) bold(cos) ( bold(italic(B)) ( bold(italic(x)) − bold(italic(h)) ) ) + bold(italic(k))$] are transformations of the sine and cosine graphs. + The amplitude is #math.equation(block: false, alt: "vertical bar A vertical bar")[$| A |$]. + The midline is #math.equation(block: false, alt: "y equals k")[$y = k$]. + The period is #math.equation(block: false, alt: "the fraction 2 π over vertical bar B vertical bar , B not equal to 0")[$display(frac(2 π, | B |)) , " " " " B ≠ 0$]. + The horizontal shift is #math.equation(block: false, alt: "h")[$h$] units to the right if #math.equation(block: false, alt: "h")[$h$] is positive, and #math.equation(block: false, alt: "h")[$h$] units to the left if #math.equation(block: false, alt: "h")[$h$] is negative. ] #examplebox("Example 3")[][ + State the midline, amplitude, period, and horizontal shift for the function #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 cos ⁡ open parenthesis 3 x plus π close parenthesis minus 2")[$f ( x ) = 4 cos ( 3 x + π ) − 2$] + Sketch a graph of the function for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. #solutionbox[ + First, we write the input for the cosine function in factored form: #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 4 cos ⁡ open bracket 3 open parenthesis x plus the fraction π over 3 close parenthesis close bracket minus 2")[$f ( x ) = 4 cos [ 3 ( x + display(frac(π, 3)) ) ] − 2$] Comparing the formula for #math.equation(block: false, alt: "f")[$f$] with the standard form, we see that #math.equation(block: false, alt: "A equals 4")[$A = 4$], #math.equation(block: false, alt: "B equals 3 , h equals the fraction minus π over 3")[$B = 3 , " " h = display(frac(− π, 3))$], and #math.equation(block: false, alt: "k equals 2")[$k = 2$]. Thus, the midline is #math.equation(block: false, alt: "y equals minus 2")[$y = − 2$], the amplitude is #math.equation(block: false, alt: "4")[$4$], the period is #math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$], and the horizontal shift is #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the left. + Use a table of values to locate the guidepoints for the graph. Begin by choosing convenient values for the input, #math.equation(block: false, alt: "3 x plus π")[$3 x + π$]. Then work backwards to find values for #math.equation(block: false, alt: "x")[$x$] and forwards to find values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "3 x")[$3 x$]], [#math.equation(block: false, alt: "3 x plus π")[$3 x + π$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 3 x plus π close parenthesis")[$cos ( 3 x + π )$]], [#math.equation(block: false, alt: "4 cos ⁡ open parenthesis 3 open parenthesis x plus the fraction π over 3 close parenthesis close parenthesis minus 2")[$4 cos ( 3 ( x + display(frac(π, 3)) ) ) − 2$]]), [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 6")[$− 6$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], )) #linebreak() Finally, plot the guidepoints and connect them with a sinusoidal curve. The graph is shown below.#figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) ] ] + State the midline, amplitude, period, and horizontal shift of the graph of #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 150 minus 25 cos ⁡ open parenthesis 4 x minus the fraction π over 2 close parenthesis")[$g ( x ) = 150 − 25 cos ( 4 x − display(frac(π, 2)) )$] + Make a table of values and sketch a graph of the function.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "4 x")[$4 x$]], [#math.equation(block: false, alt: "4 x minus the fraction π over 2")[$4 x − display(frac(π, 2))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 4 x minus the fraction π over 2 close parenthesis")[$cos ( 4 x − display(frac(π, 2)) )$]], [#math.equation(block: false, alt: "150 minus 25 cos ⁡ open parenthesis 4 x minus the fraction π over 2 close parenthesis")[$150 − 25 cos ( 4 x − display(frac(π, 2)) )$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) + midline: #math.equation(block: false, alt: "y equals 150")[$y = 150$], amplitude: #math.equation(block: false, alt: "25")[$25$], period: #math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$], horizontal shift: #math.equation(block: false, alt: "the fraction π over 8")[$display(frac(π, 8))$] to the right + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "4 x")[$4 x$]], [#math.equation(block: false, alt: "4 x minus the fraction π over 2")[$4 x − display(frac(π, 2))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 4 x minus the fraction π over 2 close parenthesis")[$cos ( 4 x − display(frac(π, 2)) )$]], [#math.equation(block: false, alt: "150 minus 25 cos ⁡ open parenthesis 4 x minus the fraction π over 2 close parenthesis")[$150 − 25 cos ( 4 x − display(frac(π, 2)) )$]]), [#math.equation(block: false, alt: "the fraction π over 8")[$display(frac(π, 8))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "125")[$125$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "the fraction 3 π over 8")[$display(frac(3 π, 8))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "175")[$175$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "the fraction 5 π over 8")[$display(frac(5 π, 8))$]], [#math.equation(block: false, alt: "the fraction 5 π over 2")[$display(frac(5 π, 2))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "125")[$125$]], )) #figure(figph[sinusoidal function], alt: "sinusoidal function", caption: none) === Modeling with Sinusoidal Functions In Section 7.1 we found formulas for sinusoidal functions that start on the midline, or at their maximum or minimum value. We can use horizontal transformations to write formulas for functions that start at other points on the cycle. #examplebox("Example 4")[][ Find a formula for the sinusoidal function whose graph is shown below. #figure(figph[sinusoidal function], alt: "sinusoidal function", caption: none) #solutionbox[ The midline of the graph is #math.equation(block: false, alt: "y equals 0")[$y = 0$], and its amplitude is #math.equation(block: false, alt: "4")[$4$]. It has the shape of a cosine, but its maximum value occurs at #math.equation(block: false, alt: "x equals the fraction π over 6")[$x = display(frac(π, 6))$] instead of at #math.equation(block: false, alt: "x equals 0")[$x = 0$]; so the graph is shifted #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] units to the right, compared to the graph of #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$]. The graph completes one cycle between #math.equation(block: false, alt: "x equals the fraction π over 6")[$x = display(frac(π, 6))$] and #math.equation(block: false, alt: "x equals the fraction 13 π over 6")[$x = display(frac(13 π, 6))$], so its period is #math.equation(block: false, alt: "2 π")[$2 π$]. Thus, we use the standard form #math.equation(block: false, alt: "y equals A cos ⁡ B open parenthesis x minus h close parenthesis plus k")[$y = A cos B ( x − h ) + k$] with #math.equation(block: false, alt: "A equals 4 , B equals 1 , h equals the fraction π over 6")[$A = 4 , " " B = 1 , " " h = display(frac(π, 6))$], and #math.equation(block: false, alt: "k equals 0")[$k = 0$], giving us #math.equation(block: true, alt: "y equals 4 cos ⁡ open parenthesis x minus the fraction π over 6 close parenthesis")[$y = 4 cos ( x − display(frac(π, 6)) )$] as our formula for the function. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the previous Example, our solution is not the only sinusoidal function that fits the given graph. For example, if we regard the graph as a sine function shifted #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] units to the left, we would use the formula #math.equation(block: false, alt: "y equals 4 sin ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$" " y = 4 sin ( x + display(frac(π, 3)) )$]. You can check that the two functions have identical graphs. Usually it is a good idea to choose a formula with a small horizontal shift. ] Find a formula for the sinusoidal function whose graph is shown at right. #figure(figph[sinusoidal function], alt: "sinusoidal function", caption: none) #math.equation(block: true, alt: "y equals 6 sin ⁡ open parenthesis x minus the fraction π over 4 close parenthesis")[$y = 6 sin ( x − display(frac(π, 4)) )$] Many natural phenomena can be modeled with sinusoidal functions. #examplebox("Example 5")[][ Sunspots are dark regions on the Sun first observed by Galileo in 1610. The number of sunspots is not constant, but varies with a period of approximately 11 years, called the solar cycle. Solar activity is directly related to this cycle, and can cause disturbances in the earth's upper atmosphere. Planning for satellite orbits and space missions requires knowledge of solar activity levels years in advance. The figure below shows sunspot data for the last solar cycle, from July, 1985 through June, 1996, and a sinusoidal function that models the data. #figure(figph[sunspot data with sine], alt: "sunspot data with sine", caption: none) + The period of the solar cycle shown is actually 10.8 years. Use this information and the graph to write a formula for the model. + Improve the fit of the model by adjusting its horizontal shift. #solutionbox[ + The graph has the shape of #math.equation(block: false, alt: "y equals minus cos ⁡ x")[$y = − cos x$]. It appears to have midline #math.equation(block: false, alt: "y equals 75")[$y = 75$] and amplitude #math.equation(block: false, alt: "75")[$75$], so #math.equation(block: false, alt: "k equals 75")[$k = 75$] and #math.equation(block: false, alt: "A equals minus 75")[$A = − 75$] To find #math.equation(block: false, alt: "B")[$B$], we compute #math.equation(block: true, alt: "B equals the fraction 2 π over period equals the fraction 2 π over 10.8 equals 0.58")[$B = display(frac(2 π, "period")) = display(frac(2 π, 10.8)) = 0.58$] Thus, a formula for the model is #math.equation(block: false, alt: "y equals 75 minus 75 cos ⁡ open bracket 0.58 open parenthesis x minus h close parenthesis close bracket")[$" " y = 75 − 75 cos [ 0.58 ( x − h ) ]$]. + Fitting a curve by eye is a subjective process, but it looks like we can get a better fit by shifting the curve slightly to the left. By trying several values for #math.equation(block: false, alt: "h")[$h$] in #math.equation(block: true, alt: "y equals 75 minus 75 cos ⁡ open bracket 0.58 open parenthesis x minus h close parenthesis close bracket")[$y = 75 − 75 cos [ 0.58 ( x − h ) ]$], we might settle on #math.equation(block: false, alt: "h equals 0.3")[$h = 0.3$], which results in the curve shown below.#figure(figph[sunspot data with sine], alt: "sunspot data with sine", caption: none) ] ] The figure below shows the sunspot data for the solar cycle that began in July 1996, and a curve of best fit calculated by NASA. This curve is not sinusoidal, but has a similar shape. #figure(figph[sunspot data with sine], alt: "sunspot data with sine", caption: none) + The minimum sunspot number occurred in October 1996. Use the graph to estimate the period, midline, and amplitude of a sinusoidal function that approximates the data. + Write a function that approximates the data. + Use your function to predict the sunspot number in January, 2005. + Period: 10 years, midline: #math.equation(block: false, alt: "y equals 61.5")[$y = 61.5$], amplitude: 53 + #math.equation(block: false, alt: "y equals 61.5 minus 53 cos ⁡ open parenthesis x minus 0.25 close parenthesis")[$y = 61.5 − 53 cos ( x − 0.25 )$] + 37 Review the following skills you will need for this section. #notebox("Algebra Refresher", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ + Write a formula for #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]. + Graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] on the same axes. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared , g open parenthesis x close parenthesis equals f open parenthesis x minus 2 close parenthesis")[$f ( x ) = x^(2) , " " g ( x ) = f ( x − 2 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the square root of x , g open parenthesis x close parenthesis equals f open parenthesis x plus 4 close parenthesis")[$f ( x ) = sqrt(x) , " " g ( x ) = f ( x + 4 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x , g open parenthesis x close parenthesis equals f open parenthesis x plus 1 close parenthesis")[$f ( x ) = display(frac(1, x)) , " " g ( x ) = f ( x + 1 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x squared , g open parenthesis x close parenthesis equals f open parenthesis x minus 1 close parenthesis")[$f ( x ) = display(frac(1, x^(2))) , " " g ( x ) = f ( x − 1 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals vertical bar x vertical bar , g open parenthesis x close parenthesis equals f open parenthesis x minus 3 close parenthesis")[$f ( x ) = | x | , " " g ( x ) = f ( x − 3 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 to the power x , g open parenthesis x close parenthesis equals f open parenthesis x plus 3 close parenthesis")[$f ( x ) = 2^(x) , " " g ( x ) = f ( x + 3 )$] #math.equation(block: true, alt: "bar")[$underline(#h(2em) #h(2em) #h(2em) #h(2em))$] Algebra Refresher Answers + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis x minus 2 close parenthesis squared")[$g ( x ) = ( x − 2 )^(2)$]#figure(figph[parabolas], alt: "parabolas", caption: none) + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the square root of x plus 4")[$g ( x ) = sqrt(x + 4)$]#figure(figph[translated square root function], alt: "translated square root function", caption: none) + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over x plus 1")[$g ( x ) = display(frac(1, x + 1))$]#figure(figph[translation of reciprocal function], alt: "translation of reciprocal function", caption: none) + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction 1 over open parenthesis x minus 1 close parenthesis squared")[$g ( x ) = display(frac(1, ( x − 1 )^(2)))$]#figure(figph[translation of inverse-square function], alt: "translation of inverse-square function", caption: none) + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals vertical bar x minus 3 vertical bar")[$g ( x ) = | x − 3 |$]#figure(figph[translation of inverse-square function], alt: "translation of inverse-square function", caption: none) + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 to the power x plus 3")[$g ( x ) = 2^(x + 3)$]#figure(figph[translation of exponential], alt: "translation of exponential", caption: none) ] === Section 7.2 Summary ==== Concepts + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Shifts] The graphs of #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis x minus h close parenthesis and t equals cos ⁡ open parenthesis x minus h close parenthesis")[$y = sin ( x − h ) " " " " " " " " "and" " " " " " " " " t = cos ( x − h )$]are #emph[shifted horizontally] compared to the graphs of #math.equation(block: false, alt: "y equals sin ⁡ open parenthesis x close parenthesis")[$y = sin ( x )$] and #math.equation(block: false, alt: "y equals cos ⁡ open parenthesis x close parenthesis")[$y = cos ( x )$]. - If #math.equation(block: false, alt: "h greater than 0")[$h > 0$], the graph is shifted to the right. - If #math.equation(block: false, alt: "h less than 0")[$h < 0$], the graph is shifted to the left. ] + The order in which we apply transformations to a function makes a difference in the graph. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Standard Form for Sinusoidal Functions] The graphs of the functions #math.equation(block: true, alt: "y equals A sin ⁡ B open parenthesis x minus h close parenthesis plus k and y equals A cos ⁡ B open parenthesis x minus h close parenthesis plus k")[$y = A sin B ( x − h ) + k " " " " " " " " "and" " " " " " " " " y = A cos B ( x − h ) + k$]are transformations of the sine and cosine graphs. + The amplitude is #math.equation(block: false, alt: "vertical bar A vertical bar")[$| A |$]. + The midline is #math.equation(block: false, alt: "y equals k")[$y = k$]. + The period is #math.equation(block: false, alt: "the fraction 2 π over vertical bar B vertical bar , B not equal to 0")[$display(frac(2 π, | B |)) , " " " " B ≠ 0$]. + The horizontal shift is #math.equation(block: false, alt: "h")[$h$] units to the right if #math.equation(block: false, alt: "h")[$h$] is positive, and #math.equation(block: false, alt: "h")[$h$] units to the left if #math.equation(block: false, alt: "h")[$h$] is negative. ] ==== Study Questions + Which of the following functions are the same? Explain your answer. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$f ( x ) = cos ( 2 x − display(frac(π, 3)) )$] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals open parenthesis cos ⁡ 2 x close parenthesis minus the fraction π over 3")[$g ( x ) = ( cos 2 x ) − display(frac(π, 3))$] #math.equation(block: true, alt: "h open parenthesis x close parenthesis equals cos ⁡ open parenthesis 2 x close parenthesis minus cos ⁡ open parenthesis the fraction π over 3 close parenthesis")[$h ( x ) = cos ( 2 x ) − cos ( display(frac(π, 3)) )$] + Which of the following functions are the same? Explain your answer. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$f ( x ) = sin ( 2 x − display(frac(π, 3)) )$] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals sin ⁡ open bracket 2 open parenthesis x minus the fraction π over 6 close parenthesis close bracket")[$g ( x ) = sin [ 2 ( x − display(frac(π, 6)) ) ]$] #math.equation(block: true, alt: "h open parenthesis x close parenthesis equals 2 sin ⁡ open parenthesis x minus the fraction π over 6 close parenthesis close parenthesis")[$h \( x \) = 2 sin \( x − display(frac(π, 6)) \) \)$] For questions 3 and 4, calculate the horizontal shift. + + #math.equation(block: false, alt: "y equals minus 2 sin ⁡ open parenthesis 3 x minus the fraction π over 4 close parenthesis")[$y = − 2 sin ( 3 x − display(frac(π, 4)) )$] + #math.equation(block: false, alt: "y equals 6 cos ⁡ open parenthesis 2 x plus the fraction 3 π over 4 close parenthesis")[$y = 6 cos ( 2 x + display(frac(3 π, 4)) )$] + + #math.equation(block: false, alt: "y equals 12 cos ⁡ open parenthesis the fraction t over 3 plus 0.8 close parenthesis")[$y = 12 cos ( display(frac(t, 3)) + 0.8 )$] + #math.equation(block: false, alt: "y equals minus 5 sin ⁡ open parenthesis the fraction t over 4 minus 0.6 close parenthesis")[$y = − 5 sin ( display(frac(t, 4)) − 0.6 )$] ==== Skills + Graph trigonometric functions using a table of values \#1–6, 11–16 + Find a formula for a transformation of a trigonometric function \#7–10, 17–26 + Solve trigonometric equations graphically \#1–6, 11–16 + Model periodic phenomena with trigonometric functions \#27–30 + Fit a circular function to data \#31–34 === Homework 7-2 #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis x close parenthesis")[$f ( x ) = sin ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals sin ⁡ open parenthesis x minus the fraction π over 3 close parenthesis")[$" " g ( x ) = sin ( x − display(frac(π, 3)) )$] + Fill in the table of values.#figure(table( columns: 14, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 5 π over 6")[$display(frac(− 5 π, 6))$]], [#math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] on the same axes.#figure(figph[grid], alt: "grid", caption: none) + What is the horizontal shift from #math.equation(block: false, alt: "f")[$f$] to #math.equation(block: false, alt: "g")[$g$]? + Find all values of #math.equation(block: false, alt: "x")[$x$] for which #math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction π over 3 close parenthesis equals 1")[$sin ( x − display(frac(π, 3)) ) = 1$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + Find all values of #math.equation(block: false, alt: "x")[$x$] for which #math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction π over 3 close parenthesis equals 0")[$sin ( x − display(frac(π, 3)) ) = 0$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + #figure(table( columns: 14, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 5 π over 6")[$display(frac(− 5 π, 6))$]], [#math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction minus the square root of 3 over 2")[$display(frac(− sqrt(3), 2))$]], [#math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction the square root of 3 over 2")[$display(frac(sqrt(3), 2))$]], )) + #figure(figph[sine and translated sine], alt: "sine and translated sine", caption: none) + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] to the right + #math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$] + #math.equation(block: false, alt: "the fraction minus 2 π over 3 , the fraction π over 3")[$display(frac(− 2 π, 3)) , " " display(frac(π, 3))$] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals cos ⁡ open parenthesis x close parenthesis")[$f ( x ) = cos ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis")[$" " g ( x ) = cos ( x + display(frac(π, 3)) )$] + Fill in the table of values.#figure(table( columns: 14, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 5 π over 6")[$display(frac(− 5 π, 6))$]], [#math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] on the same axes.#figure(figph[grid], alt: "grid", caption: none) + What is the horizontal shift from #math.equation(block: false, alt: "f")[$f$] to #math.equation(block: false, alt: "g")[$g$]? + Find all values of #math.equation(block: false, alt: "x")[$x$] for which #math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis equals 1")[$cos ( x + display(frac(π, 3)) ) = 1$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + Find all values of #math.equation(block: false, alt: "x")[$x$] for which #math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 3 close parenthesis equals 0")[$cos ( x + display(frac(π, 3)) ) = 0$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals tan ⁡ open parenthesis x close parenthesis")[$f ( x ) = tan ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals tan ⁡ open parenthesis x plus the fraction π over 4 close parenthesis")[$" " g ( x ) = tan ( x + display(frac(π, 4)) )$] + Fill in the table of values.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 3 π over 4")[$display(frac(− 3 π, 4))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 4")[$display(frac(− π, 4))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] on the same axes. #figure(figph[grid], alt: "grid", caption: none) + What is the horizontal shift from #math.equation(block: false, alt: "f")[$f$] to #math.equation(block: false, alt: "g")[$g$]? + Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis x plus the fraction π over 4 close parenthesis equals 1")[$tan ( x + display(frac(π, 4)) ) = 1$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis x plus the fraction π over 4 close parenthesis equals 0")[$tan ( x + display(frac(π, 4)) ) = 0$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 3 π over 4")[$display(frac(− 3 π, 4))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 4")[$display(frac(− π, 4))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [undef], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [undef], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [#math.equation(block: false, alt: "1")[$1$]], [undef], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [undef], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], )) + #figure(figph[translated tangent function], alt: "translated tangent function", caption: none) + #math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$] to the left + #math.equation(block: false, alt: "minus π , 0 , π")[$− π , " " 0 , " " π$] + #math.equation(block: false, alt: "the fraction minus π over 4 , the fraction minus 3 π over 4")[$display(frac(− π, 4)) , " " display(frac(− 3 π, 4))$] #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals tan ⁡ open parenthesis x close parenthesis")[$f ( x ) = tan ( x ) " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals tan ⁡ open parenthesis x minus the fraction π over 2 close parenthesis")[$" " g ( x ) = tan ( x − display(frac(π, 2)) )$] + Fill in the table of values.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "the fraction minus 3 π over 4")[$display(frac(− 3 π, 4))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "the fraction minus π over 4")[$display(frac(− π, 4))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction π over 4")[$display(frac(π, 4))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "the fraction 3 π over 4")[$display(frac(3 π, 4))$]], [#math.equation(block: false, alt: "π")[$π$]]), [#math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graphs of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$] on the same axes. #figure(figph[grid], alt: "grid", caption: none) + What is the horizontal shift from #math.equation(block: false, alt: "f")[$f$] to #math.equation(block: false, alt: "g")[$g$]? + Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis x minus the fraction π over 2 close parenthesis equals 1")[$tan ( x − display(frac(π, 2)) ) = 1$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. + Solve #math.equation(block: false, alt: "tan ⁡ open parenthesis x minus the fraction π over 2 close parenthesis equals 0")[$tan ( x − display(frac(π, 2)) ) = 0$], for #math.equation(block: false, alt: "minus π less than or equal to x less than or equal to π")[$− π ≤ x ≤ π$]. #math.equation(block: true, alt: "y equals minus 2 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$y = − 2 cos ( x + display(frac(π, 6)) )$] + What are the amplitude and the horizontal shift? + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x plus the fraction π over 6")[$x + display(frac(π, 6))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$cos ( x + display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "minus 2 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$− 2 cos ( x + display(frac(π, 6)) )$]]), [$#hide($0000$)$], [#math.equation(block: false, alt: "minus π")[$− π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "minus 2 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis equals 1")[$− 2 cos ( x + display(frac(π, 6)) ) = 1$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + Solve #math.equation(block: false, alt: "minus 2 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis equals 0")[$− 2 cos ( x + display(frac(π, 6)) ) = 0$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + amplitude 2, shift #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] to the left + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x plus the fraction π over 6")[$x + display(frac(π, 6))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$cos ( x + display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "minus 2 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$− 2 cos ( x + display(frac(π, 6)) )$]]), [#math.equation(block: false, alt: "the fraction minus 7 π over 6")[$display(frac(− 7 π, 6))$]], [#math.equation(block: false, alt: "minus π")[$− π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the fraction minus 2 π over 3")[$display(frac(− 2 π, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction minus π over 6")[$display(frac(− π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], )) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "the fraction π over 2 , the fraction 7 π over 6")[$display(frac(π, 2)) , " " display(frac(7 π, 6))$] + #math.equation(block: false, alt: "the fraction π over 3 , the fraction 4 π over 3")[$display(frac(π, 3)) , " " display(frac(4 π, 3))$] #math.equation(block: true, alt: "y equals minus 6 sin ⁡ open parenthesis x minus the fraction 2 π over 3 close parenthesis")[$y = − 6 sin ( x − display(frac(2 π, 3)) )$] + What are the amplitude and the horizontal shift? + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "x minus the fraction 2 π over 3")[$x − display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis x minus the fraction 2 π over 3 close parenthesis")[$sin ( x − display(frac(2 π, 3)) )$]], [#math.equation(block: false, alt: "minus 6 sin ⁡ open parenthesis x minus the fraction 2 π over 3 close parenthesis")[$− 6 sin ( x − display(frac(2 π, 3)) )$]]), [$#hide($0000$)$], [#math.equation(block: false, alt: "minus π")[$− π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction minus π over 2")[$display(frac(− π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "minus 6 sin ⁡ open parenthesis x minus the fraction 2 π over 3 close parenthesis equals 3")[$− 6 sin ( x − display(frac(2 π, 3)) ) = 3$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + Solve #math.equation(block: false, alt: "minus 6 sin ⁡ open parenthesis x minus the fraction 2 π over 3 close parenthesis equals 0")[$− 6 sin ( x − display(frac(2 π, 3)) ) = 0$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] The figure shows the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the sine function. + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the cosine function. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals sin ⁡ open parenthesis x plus the fraction π over 4 close parenthesis")[$f ( x ) = sin ( x + display(frac(π, 4)) )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals cos ⁡ open parenthesis x minus the fraction π over 4 close parenthesis")[$f ( x ) = cos ( x − display(frac(π, 4)) )$] The figure shows the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the sine function. + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the cosine function. The figure shows the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. #figure(figph[translated tangent function], alt: "translated tangent function", caption: none) + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the tangent function. + Find another formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a different shift of the tangent function. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals tan ⁡ open parenthesis x minus the fraction π over 3 close parenthesis")[$f ( x ) = tan ( x − display(frac(π, 3)) )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals tan ⁡ open parenthesis x plus the fraction 2 π over 3 close parenthesis")[$f ( x ) = tan ( x + display(frac(2 π, 3)) )$] The figure shows the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$]. #figure(figph[translated tangent function], alt: "translated tangent function", caption: none) + Find a formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a shift of the tangent function. + Find another formula for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] as a different shift of the tangent function. #math.equation(block: true, alt: "y equals cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$y = cos ( 2 x − display(frac(π, 3)) )$] + What are the period and the horizontal shift? (Hint: Factor out 2 from #math.equation(block: false, alt: "2 x minus the fraction π over 3")[$2 x − display(frac(π, 3))$].) + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2 x")[$2 x$]], [#math.equation(block: false, alt: "2 x minus the fraction π over 3")[$2 x − display(frac(π, 3))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$cos ( 2 x − display(frac(π, 3)) )$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis equals 1")[$cos ( 2 x − display(frac(π, 3)) ) = 1$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis equals 0")[$cos ( 2 x − display(frac(π, 3)) ) = 0$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + period #math.equation(block: false, alt: "π")[$π$], shift #math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$] to the right + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "2 x")[$2 x$]], [#math.equation(block: false, alt: "2 x minus the fraction π over 3")[$2 x − display(frac(π, 3))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis 2 x minus the fraction π over 3 close parenthesis")[$cos ( 2 x − display(frac(π, 3)) )$]]), [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 5 π over 12")[$display(frac(5 π, 12))$]], [#math.equation(block: false, alt: "the fraction 5 π over 6")[$display(frac(5 π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction 11 π over 12")[$display(frac(11 π, 12))$]], [#math.equation(block: false, alt: "the fraction 11 π over 6")[$display(frac(11 π, 6))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 7 π over 3")[$display(frac(7 π, 3))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "1")[$1$]], )) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "the fraction π over 6 , the fraction 7 π over 6")[$display(frac(π, 6)) , " " display(frac(7 π, 6))$] + #math.equation(block: false, alt: "the fraction 5 π over 12 , the fraction 11 π over 12 , the fraction 13 π over 6 , the fraction 23 π over 12")[$display(frac(5 π, 12)) , " " display(frac(11 π, 12)) , " " display(frac(13 π, 6)) , " " display(frac(23 π, 12))$] #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis 3 x plus the fraction π over 2 close parenthesis")[$y = sin ( 3 x + display(frac(π, 2)) )$] + What are the period and the horizontal shift? (Hint: Factor out 3 from #math.equation(block: false, alt: "3 x plus the fraction π over 2")[$3 x + display(frac(π, 2))$].) + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "3 x")[$3 x$]], [#math.equation(block: false, alt: "3 x plus the fraction π over 2")[$3 x + display(frac(π, 2))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis 3 x plus the fraction π over 2 close parenthesis")[$sin ( 3 x + display(frac(π, 2)) )$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis 3 x plus the fraction π over 2 close parenthesis equals 1")[$sin ( 3 x + display(frac(π, 2)) ) = 1$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis 3 x plus the fraction π over 2 close parenthesis equals 0")[$sin ( 3 x + display(frac(π, 2)) ) = 0$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] #math.equation(block: true, alt: "y equals sin ⁡ open parenthesis π x plus the fraction π over 3 close parenthesis")[$y = sin ( π x + display(frac(π, 3)) )$] + What are the period and the horizontal shift? (Hint: Factor out #math.equation(block: false, alt: "π")[$π$] from #math.equation(block: false, alt: "π x plus the fraction π over 3")[$π x + display(frac(π, 3))$].) + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "π x")[$π x$]], [#math.equation(block: false, alt: "π x plus the fraction π over 3")[$π x + display(frac(π, 3))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis π x plus the fraction π over 3 close parenthesis")[$sin ( π x + display(frac(π, 3)) )$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis π x plus the fraction π over 3 close parenthesis equals 1")[$sin ( π x + display(frac(π, 3)) ) = 1$], for #math.equation(block: false, alt: "minus 2 less than or equal to x less than or equal to 2")[$− 2 ≤ x ≤ 2$] + Solve #math.equation(block: false, alt: "sin ⁡ open parenthesis π x plus the fraction π over 3 close parenthesis equals 0")[$sin ( π x + display(frac(π, 3)) ) = 0$], for #math.equation(block: false, alt: "minus 2 less than or equal to x less than or equal to 2")[$− 2 ≤ x ≤ 2$] + period 2, shift #math.equation(block: false, alt: "the fraction 1 over 3")[$display(frac(1, 3))$] to the left + #figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "π x")[$π x$]], [#math.equation(block: false, alt: "π x plus the fraction π over 3")[$π x + display(frac(π, 3))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis π x plus the fraction π over 3 close parenthesis")[$sin ( π x + display(frac(π, 3)) )$]]), [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$display(frac(− 1, 3))$]], [#math.equation(block: false, alt: "the fraction minus π over 3")[$display(frac(− π, 3))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 1 over 6")[$display(frac(1, 6))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 7 over 6")[$display(frac(7, 6))$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "the fraction 5 over 3")[$display(frac(5, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "0")[$0$]], )) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "the fraction minus 11 over 6 , the fraction 1 over 6")[$display(frac(− 11, 6)) , " " display(frac(1, 6))$] + #math.equation(block: false, alt: "the fraction minus 4 over 3 , the fraction minus 1 over 3 , the fraction 2 over 3 , the fraction 5 over 3")[$display(frac(− 4, 3)) , " " display(frac(− 1, 3)) , " " display(frac(2, 3)) , " " display(frac(5, 3))$] #math.equation(block: true, alt: "y equals cos ⁡ open parenthesis π x minus the fraction π over 3 close parenthesis")[$y = cos ( π x − display(frac(π, 3)) )$] + What are the period and the horizontal shift? (Hint: Factor out #math.equation(block: false, alt: "π")[$π$] from #math.equation(block: false, alt: "π x minus the fraction π over 3")[$π x − display(frac(π, 3))$].) + Fill in the table of values.#figure(table( columns: 4, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "π x")[$π x$]], [#math.equation(block: false, alt: "π x minus the fraction π over 3")[$π x − display(frac(π, 3))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis π x minus the fraction π over 3 close parenthesis")[$cos ( π x − display(frac(π, 3)) )$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis π x minus the fraction π over 3 close parenthesis equals 1")[$cos ( π x − display(frac(π, 3)) ) = 1$], for #math.equation(block: false, alt: "minus 2 less than or equal to x less than or equal to 2")[$− 2 ≤ x ≤ 2$] + Solve #math.equation(block: false, alt: "cos ⁡ open parenthesis π x minus the fraction π over 3 close parenthesis equals 0")[$cos ( π x − display(frac(π, 3)) ) = 0$], for #math.equation(block: false, alt: "minus 2 less than or equal to x less than or equal to 2")[$− 2 ≤ x ≤ 2$] #math.equation(block: true, alt: "y equals 3 sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis plus 4")[$y = 3 sin ( display(frac(x, 2)) − display(frac(π, 6)) ) + 4$] + What are the midline, period, horizontal shift, and amplitude? + Fill in the table of values.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction x over 2")[$display(frac(x, 2))$]], [#math.equation(block: false, alt: "the fraction x over 2 minus the fraction π over 6")[$display(frac(x, 2)) − display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis")[$sin ( display(frac(x, 2)) − display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "3 sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis plus 4")[$3 sin ( display(frac(x, 2)) − display(frac(π, 6)) ) + 4$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$]: #math.equation(block: true, alt: "3 sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis plus 4 equals 1")[$3 sin ( display(frac(x, 2)) − display(frac(π, 6)) ) + 4 = 1$] + Solve for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$]: #math.equation(block: true, alt: "3 sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis plus 4 equals 4")[$3 sin ( display(frac(x, 2)) − display(frac(π, 6)) ) + 4 = 4$] + midline #math.equation(block: false, alt: "y equals 4")[$y = 4$], period #math.equation(block: false, alt: "4 π")[$4 π$], horizontal shift #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] to the right, amplitude 3 + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction x over 2")[$display(frac(x, 2))$]], [#math.equation(block: false, alt: "the fraction x over 2 minus the fraction π over 6")[$display(frac(x, 2)) − display(frac(π, 6))$]], [#math.equation(block: false, alt: "sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis")[$sin ( display(frac(x, 2)) − display(frac(π, 6)) )$]], [#math.equation(block: false, alt: "3 sin ⁡ open parenthesis the fraction x over 2 minus the fraction π over 6 close parenthesis plus 4")[$3 sin ( display(frac(x, 2)) − display(frac(π, 6)) ) + 4$]]), [#math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 6")[$display(frac(π, 6))$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "the fraction 4 π over 3")[$display(frac(4 π, 3))$]], [#math.equation(block: false, alt: "the fraction 2 π over 3")[$display(frac(2 π, 3))$]], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "the fraction 7 π over 3")[$display(frac(7 π, 3))$]], [#math.equation(block: false, alt: "the fraction 7 π over 6")[$display(frac(7 π, 6))$]], [#math.equation(block: false, alt: "π")[$π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "the fraction 10 π over 3")[$display(frac(10 π, 3))$]], [#math.equation(block: false, alt: "the fraction 5 π over 3")[$display(frac(5 π, 3))$]], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 13 π over 3")[$display(frac(13 π, 3))$]], [#math.equation(block: false, alt: "the fraction 13 π over 6")[$display(frac(13 π, 6))$]], [#math.equation(block: false, alt: "2 π")[$2 π$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "4")[$4$]], )) + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + no solution for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 2 π")[$0 ≤ x ≤ 2 π$] + #math.equation(block: false, alt: "the fraction π over 3")[$display(frac(π, 3))$] #math.equation(block: true, alt: "y equals 2 cos ⁡ open parenthesis the fraction x over 3 minus the fraction π over 4 close parenthesis minus 1")[$y = 2 cos ( display(frac(x, 3)) − display(frac(π, 4)) ) − 1$] + What are the midline, period, horizontal shift, and amplitude? + Fill in the table of values.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "the fraction x over 3")[$display(frac(x, 3))$]], [#math.equation(block: false, alt: "the fraction x over 3 minus the fraction π over 4")[$display(frac(x, 3)) − display(frac(π, 4))$]], [#math.equation(block: false, alt: "cos ⁡ open parenthesis the fraction x over 3 minus the fraction π over 4 close parenthesis")[$cos ( display(frac(x, 3)) − display(frac(π, 4)) )$]], [#math.equation(block: false, alt: "2 cos ⁡ open parenthesis the fraction x over 3 minus the fraction π over 4 close parenthesis minus 1")[$2 cos ( display(frac(x, 3)) − display(frac(π, 4)) ) − 1$]]), [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "0")[$0$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction π over 2")[$display(frac(π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "π")[$π$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "the fraction 3 π over 2")[$display(frac(3 π, 2))$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [#math.equation(block: false, alt: "2 π")[$2 π$]], [$#hide($0000$)$], [$#hide($0000$)$], )) + Sketch the graph.#figure(figph[grid], alt: "grid", caption: none) + Solve for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 6 π")[$0 ≤ x ≤ 6 π$]: #math.equation(block: true, alt: "2 cos ⁡ open parenthesis the fraction x over 3 minus the fraction π over 4 close parenthesis minus 1 equals 1")[$2 cos ( display(frac(x, 3)) − display(frac(π, 4)) ) − 1 = 1$] + Solve for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 6 π")[$0 ≤ x ≤ 6 π$]: #math.equation(block: true, alt: "2 cos ⁡ open parenthesis the fraction x over 3 minus the fraction π over 4 close parenthesis minus 1 equals minus 1")[$2 cos ( display(frac(x, 3)) − display(frac(π, 4)) ) − 1 = − 1$] Find a formula for a sinusoidal function that has an amplitude of 2, a period of 3, and is shifted 4 units to the left and 5 units upwards compared with the sine function. Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 3")[$0 ≤ x ≤ 3$]. #math.equation(block: true, alt: "y equals 2 sin ⁡ open parenthesis the fraction 2 π over 3 open parenthesis x plus 4 close parenthesis close parenthesis plus 5")[$y = 2 sin ( display(frac(2 π, 3)) ( x + 4 ) ) + 5$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) Find a formula for a sinusoidal function that has an amplitude of 3, a period of 24, and is shifted 2 units to the right and 4 units upwards compared with the cosine function. Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 24")[$0 ≤ x ≤ 24$]. Find a formula for a sinusoidal function that has an amplitude of 5, a period of 360, its midline at #math.equation(block: false, alt: "y equals 12")[$y = 12$], and passes through #math.equation(block: false, alt: "open parenthesis 0 , 7 close parenthesis")[$( 0 , 7 )$]. Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 360")[$0 ≤ x ≤ 360$]. #math.equation(block: true, alt: "y equals minus 5 cos ⁡ open parenthesis the fraction π x over 180 close parenthesis plus 12")[$y = − 5 cos ( display(frac(π x, 180)) ) + 12$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) Find a formula for a sinusoidal function that has an amplitude of 50, a period of 30, its midline at #math.equation(block: false, alt: "y equals 50")[$y = 50$], and passes through #math.equation(block: false, alt: "open parenthesis 0 , 100 close parenthesis")[$( 0 , 100 )$]. Sketch the graph for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 30")[$0 ≤ x ≤ 30$]. For Problems 21–26, find a formula for the circular function whose graph is shown. + Write the function in the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals A sin ⁡ open parenthesis B open parenthesis x minus h close parenthesis close parenthesis")[$f ( x ) = A sin ( B ( x − h ) )$]. + Write the function in the form #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals A cos ⁡ open parenthesis B open parenthesis x minus h close parenthesis close parenthesis")[$f ( x ) = A cos ( B ( x − h ) )$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 sin ⁡ open parenthesis x plus the fraction 2 π over 3 close parenthesis")[$f ( x ) = 3 sin ( x + display(frac(2 π, 3)) )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 3 cos ⁡ open parenthesis x plus the fraction π over 6 close parenthesis")[$f ( x ) = 3 cos ( x + display(frac(π, 6)) )$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 sin ⁡ open parenthesis 2 open parenthesis x minus the fraction π over 4 close parenthesis close parenthesis")[$f ( x ) = 2 sin ( 2 ( x − display(frac(π, 4)) ) )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 2 cos ⁡ open parenthesis 2 x close parenthesis")[$f ( x ) = − 2 cos ( 2 x )$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 4 sin ⁡ open bracket the fraction 1 over 4 open parenthesis x minus the fraction 7 π over 3 close parenthesis close bracket")[$f ( x ) = 4 sin [ display(frac(1, 4)) ( x − display(frac(7 π, 3)) ) ]$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 4 cos ⁡ open bracket the fraction 1 over 4 open parenthesis x minus the fraction π over 3 close parenthesis close bracket")[$f ( x ) = − 4 cos [ display(frac(1, 4)) ( x − display(frac(π, 3)) ) ]$] #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) The average daily high temperature in Fairbanks, Alaska can be approximated by a sinusoidal function with a period of 12 months. The low temperature of #math.equation(block: false, alt: "minus 1.6 degrees")[$− 1.6^(∘)$] occurs in January, and the high temperature of #math.equation(block: false, alt: "72.3 degrees")[$72.3^(∘)$] in July. + What are the midline, period, and amplitude? + Write a formula for the average daily high temperature #math.equation(block: false, alt: "T open parenthesis m close parenthesis")[$T ( m )$], where #math.equation(block: false, alt: "m")[$m$] is the number of months since January. + Graph #math.equation(block: false, alt: "T open parenthesis m close parenthesis")[$T ( m )$] for two periods, labeling the points that correspond to highest and lowest average temperature. + midline #math.equation(block: false, alt: "T equals 35.35")[$T = 35.35$], period 12, amplitude 36.95 + #math.equation(block: false, alt: "T open parenthesis m close parenthesis equals minus 36.95 cos ⁡ open parenthesis the fraction π over 6 m close parenthesis plus 35.35")[$T ( m ) = − 36.95 cos ( display(frac(π, 6)) m ) + 35.35$] + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) Depending on its phase, the moon looks like a disk that is partially visible and partially in shadow. The visible fraction ranges from 0% to 100%, and can be approximated by a sinusoidal function #math.equation(block: false, alt: "V open parenthesis t close parenthesis")[$V ( t )$], where #math.equation(block: false, alt: "t")[$t$] is the number of days since the last full moon. The time between successive full moons (a lunar month) is 29.5 days. + What are the period, midline, and amplitude of #math.equation(block: false, alt: "V open parenthesis t close parenthesis")[$V ( t )$]? + Write a formula for #math.equation(block: false, alt: "V open parenthesis t close parenthesis")[$V ( t )$]. + Graph your function over two periods, labeling the points that correspond to full moon, half moon, and new moon. The tide in Yorktown is approximated by the function #math.equation(block: true, alt: "h open parenthesis t close parenthesis equals 1.4 minus 1.4 cos ⁡ open parenthesis 0.51 t close parenthesis")[$h ( t ) = 1.4 − 1.4 cos ( 0.51 t )$] measured in feet above low tide, where #math.equation(block: false, alt: "t")[$t$] is the number of hours since the last low tide. + What are the midline, period, and amplitude? + Graph #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] for two periods, labeling the points that correspond to high tide and low tide. + If the last low tide occurred at 5:00 am, predict when the next high and low tides will occur. + midline #math.equation(block: false, alt: "h equals 1.4")[$h = 1.4$], period #math.equation(block: false, alt: "the fraction 2 π over 0.51 approximately equals 12.32")[$display(frac(2 π, 0.51)) ≈ 12.32$], amplitude 1.4 + #figure(figph[sinusoidal graph], alt: "sinusoidal graph", caption: none) + high 11:10 am, low 5:19 pm The height of a child's toy suspended at the end of a spring is approximated by a sinusoidal function. The toy's height ranges between 200 centimeters and 260 centimeters above the ground, and it completes one up-and-down cycle every 0.8 second. + What are the midline, period, and amplitude? + Let #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] be the height of the toy in centimeters, where #math.equation(block: false, alt: "t equals 0")[$t = 0$] seconds corresponds to a time when the object was at the midline and moving upwards. Graph #math.equation(block: false, alt: "h open parenthesis t close parenthesis")[$h ( t )$] for two periods, labeling the points that correspond to the high and low positions of the toy. + When does the toy reach its maximum height the second time? In Problems 31–34, + Estimate the amplitude, period, and midline of a circular function that fits the data. + Write a formula for the function. #figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.75")[$0.75$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.25")[$1.25$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "1.75")[$1.75$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.25")[$2.25$]], [#math.equation(block: false, alt: "2.5")[$2.5$]]), [#math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$]], [#math.equation(block: false, alt: "5.2")[$5.2$]], [#math.equation(block: false, alt: "4.26")[$4.26$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "minus 0.26")[$− 0.26$]], [#math.equation(block: false, alt: "minus 1.2")[$− 1.2$]], [#math.equation(block: false, alt: "minus 0.26")[$− 0.26$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4.26")[$4.26$]], [#math.equation(block: false, alt: "5.2")[$5.2$]], [#math.equation(block: false, alt: "4.26")[$4.26$]], [#math.equation(block: false, alt: "2")[$2$]], )) + amplitude 3.2, period 2, midline #math.equation(block: false, alt: "y equals 2")[$y = 2$] + #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 2 plus 3.2 cos ⁡ open parenthesis π t close parenthesis")[$f ( t ) = 2 + 3.2 cos ( π t )$] #figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "s")[$s$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "g open parenthesis s close parenthesis")[$g ( s )$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2.58")[$2.58$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], [#math.equation(block: false, alt: "2.58")[$2.58$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3.42")[$3.42$]], [#math.equation(block: false, alt: "3.6")[$3.6$]], [#math.equation(block: false, alt: "3.42")[$3.42$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "2.58")[$2.58$]], [#math.equation(block: false, alt: "2.4")[$2.4$]], )) #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.1")[$0.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.3")[$0.3$]], [#math.equation(block: false, alt: "0.4")[$0.4$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.7")[$0.7$]], [#math.equation(block: false, alt: "0.8")[$0.8$]]), [#math.equation(block: false, alt: "H open parenthesis x close parenthesis")[$H ( x )$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7.9")[$7.9$]], [#math.equation(block: false, alt: "9.8")[$9.8$]], [#math.equation(block: false, alt: "9.8")[$9.8$]], [#math.equation(block: false, alt: "7.9")[$7.9$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "2.1")[$2.1$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], [#math.equation(block: false, alt: "0.2")[$0.2$]], )) + amplitude 5, period 1, midline #math.equation(block: false, alt: "y equals 0")[$y = 0$] + #math.equation(block: false, alt: "H open parenthesis x close parenthesis equals 5 sin ⁡ open parenthesis 2 π x close parenthesis plus 5")[$H ( x ) = 5 sin ( 2 π x ) + 5$] #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.79")[$0.79$]], [#math.equation(block: false, alt: "1.57")[$1.57$]], [#math.equation(block: false, alt: "2.36")[$2.36$]], [#math.equation(block: false, alt: "3.14")[$3.14$]], [#math.equation(block: false, alt: "3.93")[$3.93$]], [#math.equation(block: false, alt: "4.71")[$4.71$]], [#math.equation(block: false, alt: "5.50")[$5.50$]], [#math.equation(block: false, alt: "6.28")[$6.28$]]), [#math.equation(block: false, alt: "V open parenthesis t close parenthesis")[$V ( t )$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 0.17")[$− 0.17$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 5.8")[$− 5.8$]], [#math.equation(block: false, alt: "minus 7")[$− 7$]], [#math.equation(block: false, alt: "minus 5.3")[$− 5.3$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 0.17")[$− 0.17$]], [#math.equation(block: false, alt: "1")[$1$]], ))