#set document(title: "1.4 Slope and Rate of Change", 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")) == 1.4#h(0.6em)Slope and Rate of Change === Using Ratios for Comparison Which is more expensive, a 64-ounce bottle of Velvolux dish soap that costs \$3.52, or a 60-ounce bottle of Rainfresh dish soap that costs \$3.36? You are probably familiar with the notion of comparison shopping. To decide which dish soap is the better buy, we compute the unit price, or price per ounce, for each bottle. The unit price for Velvolux is #math.equation(block: true, alt: "the fraction 352 cents over 64 ounces equals 5.5 cents per ounce")[$frac(352 " cents", 64 " ounces") = 5.5 " cents per ounce"$] and the unit price for Rainfresh is #math.equation(block: true, alt: "the fraction 336 cents over 60 ounces equals 5.6 cents per ounce")[$frac(336 " cents", 60 " ounces") = 5.6 " cents per ounce"$] The Velvolux costs less per ounce, so it is the better buy. By computing the price of each brand for #emph[the same amount of soap], it is easy to compare them. In many situations, a ratio, similar to a unit price, can provide a basis for comparison. Example uses a ratio to measure a rate of growth. #examplebox("Example 1")[][ Which grow faster, Hybrid A wheat seedlings, which grow 11.2 centimeters in 14 days, or Hybrid B seedlings, which grow 13.5 centimeters in 18 days? #solutionbox[ We compute the growth rate for each strain of wheat. Growth rate is expressed as a ratio, #math.equation(block: false, alt: "the fraction centimeters over days")[$display(frac("centimeters", "days"))$], or centimeters per day. The growth rate for Hybrid A is #math.equation(block: true, alt: "the fraction 11.2 centimeters over 14 days equals 0.8 centimeters per day")[$frac(11.2 " centimeters", 14 " days") = 0.8 " centimeters per day"$] and the growth rate for Hybrid B is #math.equation(block: true, alt: "the fraction 13.5 centimeters over 18 days equals 0.75 centimeters per day")[$frac(13.5 " centimeters", 18 " days") = 0.75 " centimeters per day"$] Because their rate of growth is larger, the Hybrid A seedlings grow faster. ] ] By computing the growth of each strain of wheat seedling over the same unit of time, a single day, we have a basis for comparison. In this case, the ratio #math.equation(block: false, alt: "the fraction centimeters over day")[$display(frac("centimeters", "day"))$] measures the rate of growth of the wheat seedlings. Delbert traveled #math.equation(block: false, alt: "258")[$258$] miles on #math.equation(block: false, alt: "12")[$12$] gallons of gas, and Francine traveled #math.equation(block: false, alt: "182")[$182$] miles on #math.equation(block: false, alt: "8")[$8$] gallons of gas. Compute the ratio #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] for each car. Delbert: \_\_\_\_\_ #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] Francine: \_\_\_\_\_ #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] Whose car gets the better gas mileage? \_\_\_\_\_ #solutionbox[ Delbert: #math.equation(block: false, alt: "the fraction 258 over 12 equals Delbert")[$display(frac(258, 12)) = "Delbert"$] mpg, Francine: #math.equation(block: false, alt: "the fraction 182 over 8 equals Francine")[$display(frac(182, 8)) = "Francine"$] mpg. Francine gets better mileage. ] Delbert traveled #math.equation(block: false, alt: "258")[$258$] miles on #math.equation(block: false, alt: "12")[$12$] gallons of gas, and Francine traveled #math.equation(block: false, alt: "182")[$182$] miles on #math.equation(block: false, alt: "8")[$8$] gallons of gas. Compute the ratio #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] for each car. Whose car gets the better gas mileage? #solutionbox[ Delbert's gas mileage is #math.equation(block: false, alt: "the fraction 258 over 12 equals 21.5")[$display(frac(258, 12)) = 21.5$] mpg, and Francine's gas mileage is #math.equation(block: false, alt: "the fraction 182 over 8 equals 22.75")[$display(frac(182, 8)) = 22.75$] mpg. Francine gets better mileage. ] The weight of #math.equation(block: false, alt: "12")[$12$] gallons of gasoline is #math.equation(block: false, alt: "67.5")[$67.5$] pounds, and the weight of #math.equation(block: false, alt: "5")[$5$] gallons of water is #math.equation(block: false, alt: "41.6")[$41.6$] pounds. Compute the ratio #math.equation(block: false, alt: "the fraction pounds over gallon")[$display(frac("pounds", "gallon"))$] for each liquid. gasoline: \_\_\_\_\_ #math.equation(block: false, alt: "the fraction pounds over gallon")[$display(frac("pounds", "gallon"))$] water: \_\_\_\_\_ #math.equation(block: false, alt: "the fraction pounds over gallon")[$display(frac("pounds", "gallon"))$] Which liquid weighs more per gallon\_\_\_\_\_ #solutionbox[ gasoline: #math.equation(block: false, alt: "the fraction 67.5 over 12 equals gas")[$display(frac(67.5, 12)) = "gas"$] pounds per gallon, water: #math.equation(block: false, alt: "the fraction 41.6 over 5 equals water")[$display(frac(41.6, 5)) = "water"$] pounds per gallon. Water weighs more per gallon. ] The weight of #math.equation(block: false, alt: "12")[$12$] gallons of gasoline is #math.equation(block: false, alt: "67.5")[$67.5$] pounds, and the weight of #math.equation(block: false, alt: "5")[$5$] gallons of water is #math.equation(block: false, alt: "41.6")[$41.6$] pounds. Compute the ratio #math.equation(block: false, alt: "the fraction pounds over gallon")[$display(frac("pounds", "gallon"))$] for each liquid. Which liquid weighs more per gallon? === Measuring Steepness In Practice 1, the ratio #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] measures the rate at which each car uses gasoline. By computing the mileage for each car for the same amount of gas, we have a basis for comparison. We can use this same idea, finding a common basis for comparison, to measure the steepness of an incline. In Practice 1, the ratio #math.equation(block: false, alt: "the fraction miles over gallon")[$display(frac("miles", "gallon"))$] measures the rate at which each car uses gasoline. By computing the mileage for each car for the same amount of gas, we have a basis for comparison. We can use this same idea, finding a common basis for comparison, to measure the steepness of an incline. Imagine you are an ant carrying a heavy burden along one of the two paths shown below. Which path is more difficult? Most ants would agree that the steeper path is more difficult. But what exactly is steepness? It is not merely the gain in altitude, because even a gentle incline will reach a great height eventually. Steepness measures how sharply the altitude increases. An ant finds the second path more difficult, or steeper, because it rises 5 feet while the first path rises only 2 feet over the same horizontal distance. #figure(figph[two paths up a slope], alt: "two paths up a slope", caption: none) To compare the steepness of two inclined paths, we compute the ratio of change in altitude to change in horizontal distance for each path. #examplebox("Example 2")[][ Which is steeper, Stony Point trail, which climbs 400 feet over a horizontal distance of 2500 feet, or Lone Pine trail, which climbs 360 feet over a horizontal distance of 1800 feet? #solutionbox[ For each trail, we compute the ratio of vertical gain to horizontal distance. For Stony Point trail, the ratio is #math.equation(block: true, alt: "the fraction 400 feet over 2500 feet equals 0.16")[$frac(400 " feet", 2500 " feet") = 0.16$] and for Lone Pine trail, the ratio is #math.equation(block: true, alt: "the fraction 360 feet over 1800 feet equals 0.20")[$frac(360 " feet", 1800 " feet") = 0.20$] Lone Pine trail is steeper, because it has a vertical gain of 0.20 foot for every foot traveled horizontally. Or, in more practical units, Lone Pine trail rises 20 feet for every 100 feet of horizontal distance, whereas Stony Point trail rises only 16 feet over a horizontal distance of 100 feet. ] ] Which is steeper, a staircase that rises #math.equation(block: false, alt: "10")[$10$] feet over a horizontal distance of #math.equation(block: false, alt: "4")[$4$] feet, or the steps in the football stadium, which rise #math.equation(block: false, alt: "20")[$20$] yards over a horizontal distance of #math.equation(block: false, alt: "12")[$12$] yards? Answer: \_\_\_\_\_ #solutionbox[ Staircase: #math.equation(block: false, alt: "the fraction 10 over 4 equals staircase")[$display(frac(10, 4)) = "staircase"$], Steps: #math.equation(block: false, alt: "the fraction 20 over 12 approximately equals steps")[$display(frac(20, 12)) ≈ "steps"$]. The staircase is steeper. ] Which is steeper, a staircase that rises #math.equation(block: false, alt: "10")[$10$] feet over a horizontal distance of #math.equation(block: false, alt: "4")[$4$] feet, or the steps in the football stadium, which rise #math.equation(block: false, alt: "20")[$20$] yards over a horizontal distance of #math.equation(block: false, alt: "12")[$12$] yards? #solutionbox[ Staircase: #math.equation(block: false, alt: "the fraction 10 over 4 equals 2.5")[$display(frac(10, 4)) = 2.5$]; steps: #math.equation(block: false, alt: "the fraction 20 over 12 approximately equals 1.67")[$display(frac(20, 12)) ≈ 1.67$]. The staircase is steeper. ] === Definition of Slope To compare the steepness of the two trails in Example, it is not enough to know which trail has the greater gain in elevation overall. Instead, we compare their elevation gains #emph[over the same horizontal distance.] Using the same horizontal distance provides a basis for comparison. The two trails are illustrated below as lines on a coordinate grid. #figure(figph[trails on grid], alt: "trails on grid", caption: none) The ratio we computed in Example, #math.equation(block: true, alt: "the fraction change in elevation over change in horizontal position")[$frac("change in elevation", "change in horizontal position")$] appears on the graphs as #math.equation(block: true, alt: "the fraction change in y -coordinate over change in x -coordinate")[$frac("change in " y "-coordinate", "change in " x "-coordinate")$] For example, as we travel along the line representing Stony Point trail, we move from the point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$] to the point #math.equation(block: false, alt: "open parenthesis 2500 , 400 close parenthesis")[$( 2500 , 400 )$]. The #math.equation(block: false, alt: "y")[$y$]-coordinate changes by #math.equation(block: false, alt: "400")[$400$] and the #math.equation(block: false, alt: "x")[$x$]-coordinate changes by #math.equation(block: false, alt: "2500")[$2500$], giving the ratio #math.equation(block: false, alt: "0.16")[$0.16$] that we found in Example. We call this ratio the #strong[slope] of the line. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Definition of Slope] The #strong[slope] of a line is the ratio #math.equation(block: true, alt: "the fraction change in y -coordinate over change in x -coordinate")[$frac("change in " y "-coordinate", "change in " x "-coordinate")$] as we move from one point to another on the line. ] #examplebox("Example 3")[][ Compute the slope of the line that passes through points #math.equation(block: false, alt: "A")[$A$] and #math.equation(block: false, alt: "B")[$B$] on the graph at right. #figure(figph[points on line], alt: "points on line", caption: none) #solutionbox[ As we move along the line from #math.equation(block: false, alt: "A")[$A$] to #math.equation(block: false, alt: "B")[$B$], the #math.equation(block: false, alt: "y")[$y$]-coordinate changes by #math.equation(block: false, alt: "3")[$3$] units, and the #math.equation(block: false, alt: "x")[$x$]-coordinate changes by #math.equation(block: false, alt: "4")[$4$] units. The slope of the line is thus #math.equation(block: true, alt: "the fraction change in y -coordinate over change in x -coordinate equals the fraction 3 over 4")[$frac("change in " y "-coordinate", "change in " x "-coordinate") = frac(3, 4)$] ] ] #figure(figph[line on grid], alt: "line on grid", caption: none) Compute the slope of the line through the indicated points on the graph above. On both axes, one square represents one unit. #math.equation(block: false, alt: "the fraction change in y -coordinate over change in x -coordinate equals")[$display(frac("change in " y "-coordinate", "change in " x "-coordinate")) =$]\_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "the fraction 2 over 8 equals the fraction 1 over 4")[$display(frac(2, 8)) = display(frac(1, 4))$] ] Compute the slope of the line through the indicated points on the graph at right. On both axes, one square represents one unit. #math.equation(block: true, alt: "the fraction change in y -coordinate over change in x -coordinate equals")[$display(frac("change in " y "-coordinate", "change in " x "-coordinate")) =$] #figure(figph[line on grid], alt: "line on grid", caption: none) #solutionbox[ The slope is #math.equation(block: false, alt: "the fraction 2 over 8 equals the fraction 1 over 4")[$" " display(frac(2, 8)) = display(frac(1, 4))$]. ] What is wrong with the following reasoning? The point #math.equation(block: false, alt: "open parenthesis 10 , 7 close parenthesis")[$" " ( 10 , 7 ) " "$] lies on the graph of a line, so the slope of the line is #math.equation(block: false, alt: "the fraction 7 over 10")[$" " display(frac(7, 10))$]. \_\_\_\_\_ #solutionbox[ We need two points to compute the slope. ] What is wrong with the following reasoning? "The point #math.equation(block: false, alt: "open parenthesis 10 , 7 close parenthesis")[$" " ( 10 , 7 ) " "$] lies on the graph of a line, so the slope of the line is #math.equation(block: false, alt: "the fraction 7 over 10")[$" " display(frac(7, 10))$]." Choose the best answer. + The slope is #math.equation(block: false, alt: "the fraction 10 over 7")[$display(frac(10, 7))$]. + #math.equation(block: false, alt: "7")[$7$] is not an intercept. + We need two points to compute the slope. + The line is decreasing. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The slope of a line is a #emph[number]. It tells us how much the #math.equation(block: false, alt: "y")[$y$]-coordinates of points on the line increase when we increase their #math.equation(block: false, alt: "x")[$x$]-coordinates by 1 unit. For instance, the slope #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] in Example means that the #math.equation(block: false, alt: "y")[$y$]-coordinate increases by #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] unit when the #math.equation(block: false, alt: "x")[$x$]-coordinate increases by 1 unit. For increasing graphs, a larger slope indicates a greater increase in altitude, and hence a steeper line. ] === Notation for Slope We use a shorthand notation for the ratio that defines slope, #math.equation(block: true, alt: "the fraction change in y -coordinate over change in x -coordinate")[$frac("change in " y "-coordinate", "change in " x "-coordinate")$] The symbol #math.equation(block: false, alt: "Δ")[$upright(Δ)$] (the Greek letter delta) is used in mathematics to denote #emph[change in]. In particular, #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] means #emph[change in #math.equation(block: false, alt: "y")[$y$]-coordinate], and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] means #emph[change in #math.equation(block: false, alt: "x")[$x$]-coordinate]. We also use the letter #math.equation(block: false, alt: "m")[$m$] to stand for slope. With these symbols, we can write the definition of slope as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Notation for Slope] The #strong[slope] of a line is given by #math.equation(block: true, alt: "m equals, the fraction Δ y over Δ x equals the fraction change in y -coordinate over change in x -coordinate; if Δ x not equal to 0")[$m = & frac(upright(Δ) y, upright(Δ) x) = frac("change in " y "-coordinate", "change in " x "-coordinate") \ & "if" " " upright(Δ) x ≠ 0$] #figure(figph[slope of line], alt: "slope of line", caption: none) ] #examplebox("Example 4")[][ The Great Pyramid of Khufu in Egypt was built around 2550 B.C. It is 147 meters tall and has a square base 229 meters on each side. Calculate the slope of the sides of the pyramid, rounded to two decimal places. #solutionbox[ From the figure, we see that #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] is only half the base of the Great Pyramid, so #math.equation(block: true, alt: "Δ x equals 0.5 open parenthesis 229 close parenthesis equals 114.5")[$upright(Δ) x = 0.5 ( 229 ) = 114.5$] and the slope of the side is #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction 147 over 114.5 equals 1.28")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(147, 114.5) = 1.28$] #figure(figph[Great Pyramid], alt: "Great Pyramid", caption: none) ] ] The Kukulcan Pyramid at Chichen Itza in Mexico was built around 800 A.D. It is #math.equation(block: false, alt: "24")[$24$] meters high, with a temple built on its top platform, as shown at right. #figure(figph[Kukulcan Pyramid], alt: "Kukulcan Pyramid", caption: none) The square base is 55 meters on each side, and the top platform is 19.5 meters on each side. Calculate the slope of the sides of the pyramid. Which pyramid is steeper, Kukulcan or the Great Pyramid (whose slope, rounded to two decimals, is 1.28)? Answer: \_\_\_\_\_ The horizontal change #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] is half the difference between the sides of the base and of the top platform. #solutionbox[ The horizontal change #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] is half the difference between the sides of the base and of the top platform: #math.equation(block: false, alt: "Δ x equals the fraction 1 over 2 open parenthesis 55 minus 19.5 close parenthesis equals Delx")[$upright(Δ) x = frac(1, 2) ( 55 − 19.5 ) = "Delx"$], so the slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction 24 over Delx approximately equals m")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(24, "Delx")) ≈ "m"$], which is greater than 1.28. Kukulcan is steeper. ] The Kukulcan Pyramid at Chichen Itza in Mexico was built around 800 A.D. It is #math.equation(block: false, alt: "24")[$24$] meters high, with a temple built on its top platform, as shown below. #figure(figph[Kukulcan Pyramid], alt: "Kukulcan Pyramid", caption: none) The square base is 55 meters on each side, and the top platform is 19.5 meters on each side. Calculate the slope of the sides of the pyramid. Which pyramid is steeper, Kukulcan or the Great Pyramid in the previous Example? #solutionbox[ The horizontal change #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] is half the difference between the sides of the base and of the top platform: #math.equation(block: false, alt: "Δ x equals the fraction 1 over 2 open parenthesis 55 minus 19.5 close parenthesis equals 17.75")[$upright(Δ) x = frac(1, 2) ( 55 − 19.5 ) = 17.75$], so the slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction 24 over 17.75 approximately equals 1.35")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(24, 17.75)) ≈ 1.35$], which is greater than 1.28. Kukulcan is steeper. ] So far, we have only considered examples in which #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] and #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] are positive numbers, but they can also be negative. #math.equation(block: true, alt: "Δ x equals open brace positive if x increases (move to the right); negative if x decreases (move to the left)")[$upright(Δ) x = \{ "positive if " x " increases (move to the right)" \ "negative if " x " decreases (move to the left)"$] #math.equation(block: true, alt: "Δ y equals open brace positive if y increases (move up); negative if y decreases (move down)")[$upright(Δ) y = \{ "positive if " y " increases (move up)" \ "negative if " y " decreases (move down)" " " " " " " " " " " " " " " " " " "$] #examplebox("Example 5")[][ Compute the slope of the line that passes through the points #math.equation(block: false, alt: "P open parenthesis minus 4 , 2 close parenthesis")[$P ( − 4 , 2 )$] and #math.equation(block: false, alt: "Q open parenthesis 5 , minus 1 close parenthesis")[$Q ( 5 , − 1 )$] shown in at right. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. #figure(figph[line with negative slope], alt: "line with negative slope", caption: none) #solutionbox[ As we move from the point #math.equation(block: false, alt: "P open parenthesis minus 4 , 2 close parenthesis")[$P ( − 4 , 2 )$] to the point #math.equation(block: false, alt: "Q open parenthesis 5 , minus 1 close parenthesis")[$Q ( 5 , − 1 )$], we move #math.equation(block: false, alt: "3")[$3$] units #emph[down], so #math.equation(block: false, alt: "Δ y equals minus 3")[$upright(Δ) y = − 3$]. We then move #math.equation(block: false, alt: "9")[$9$] units to the right, so #math.equation(block: false, alt: "Δ x equals 9")[$upright(Δ) x = 9$]\.Thus, the slope is #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction minus 3 over 9 equals the fraction minus 1 over 3")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(− 3, 9) = frac(− 1, 3)$] #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] are labeled on the graph. ] ] Suppose #math.equation(block: false, alt: "x")[$x$] represents your distance from home, in miles. Compare the meanings of the two statements #math.equation(block: false, alt: "x equals 30")[$x = 30$] and #math.equation(block: false, alt: "Δ x equals 30")[$upright(Δ) x = 30$]. \_\_\_\_\_ Suppose #math.equation(block: false, alt: "x")[$x$] represents your distance from home, in miles. Compare the meanings of the two statements #math.equation(block: false, alt: "x equals 30")[$x = 30$] and #math.equation(block: false, alt: "Δ x equals 30")[$upright(Δ) x = 30$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We can move from point to point in either direction to compute the slope. The line graphed in Example #emph[decreases] as we move from left to right and hence has a negative slope. The slope is the same if we move from point #math.equation(block: false, alt: "Q")[$Q$] to point #math.equation(block: false, alt: "P")[$P$] instead of from #math.equation(block: false, alt: "P")[$P$] to #math.equation(block: false, alt: "Q")[$Q$], as shown at right. In that case, our computation looks like this: #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction 3 over minus 9 equals the fraction minus 1 over 3")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(3, − 9) = frac(− 1, 3)$] #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] are labeled on the graph. #figure(figph[line with negative slope], alt: "line with negative slope", caption: none) ] === Lines Have Constant Slope How do we know which two points to choose when we want to compute the slope of a line? It turns out that any two points on the line will do. + Graph the line #math.equation(block: false, alt: "4 x minus 2 y equals 8")[$4 x − 2 y = 8$] by finding the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts #linebreak() #math.equation(block: false, alt: "x")[$x$]-intercept: \_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "y")[$y$]-intercept: \_\_\_\_\_ #linebreak() Note: Enter each intercept as an ordered pair. + Compute the slope of the line using the #math.equation(block: false, alt: "x")[$x$]-intercept and #math.equation(block: false, alt: "y")[$y$]-intercept. #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ + Compute the slope of the line using the points #math.equation(block: false, alt: "open parenthesis 4 , 4 close parenthesis")[$( 4 , 4 )$] and #math.equation(block: false, alt: "open parenthesis 1 , minus 2 close parenthesis")[$( 1 , − 2 )$]. #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ #solutionbox[ + A graph of the line is shown below this exercise. + The #math.equation(block: false, alt: "x")[$x$]-intercept is #math.equation(block: false, alt: "xi")[$"xi"$] and the #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "yi")[$"yi"$]. #linebreak() The slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction minus y over x equals m")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(− "y", "x")) = "m"$] + The slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction minus 2 minus 4 over 1 minus 4 equals the fraction minus 6 over minus 3 equals m")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(− 2 − 4, 1 − 4)) = display(frac(− 6, − 3)) = "m"$] ] The graph for part (a): #figure(figph[line], alt: "line", caption: none) + Graph the line #math.equation(block: false, alt: "4 x minus 2 y equals 8")[$" " 4 x − 2 y = 8 " "$] by finding the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts. + Compute the slope of the line using the #math.equation(block: false, alt: "x")[$x$]-intercept and #math.equation(block: false, alt: "y")[$y$]-intercept. + Compute the slope of the line using the points #math.equation(block: false, alt: "open parenthesis 4 , 4 close parenthesis")[$( 4 , 4 )$] and #math.equation(block: false, alt: "open parenthesis 1 , minus 2 close parenthesis")[$( 1 , − 2 )$]. #solutionbox[ + #figure(figph[line], alt: "line", caption: none) + The #math.equation(block: false, alt: "x")[$x$]-intercept is #math.equation(block: false, alt: "open parenthesis 2 , 0 close parenthesis")[$( 2 , 0 )$] and the #math.equation(block: false, alt: "y")[$y$]-intercept is #math.equation(block: false, alt: "open parenthesis 0 , minus 4 close parenthesis")[$( 0 , − 4 )$]. #linebreak() The slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction 4 over 2 equals 2")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(4, 2)) = 2$] + The slope is #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction minus 2 minus 4 over 1 minus 4 equals the fraction minus 6 over minus 3 equals 2")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(− 2 − 4, 1 − 4)) = display(frac(− 6, − 3)) = 2$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Practice 5 illustrates an important property of lines: They have constant slope. No matter which two points we use to calculate the slope, we will always get the same result. We will see later that lines are the only graphs that have this property. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Practice 5 illustrates an important property of lines: They have constant slope. No matter which two points we use to calculate the slope, we will always get the same result. We will see later that lines are the only graphs that have this property. ] We can think of the slope as a #strong[scale factor] that tells us how many units #math.equation(block: false, alt: "y")[$y$] increases (or decreases) for each unit of increase in #math.equation(block: false, alt: "x")[$x$]. Compare the lines shown below. The line y = mx + b with both constants on sliders. The slope m is the scale factor this paragraph describes: every 1 unit you move right, the line climbs m units — drag m from 2 down through 0 to −2 and watch the line tip from rising, to horizontal, to falling. The slider b moves the y-intercept without changing the tilt at all: slope is a property of the line's direction, not its position. One square on the grid is one unit, so you can read the rise over any one-unit run straight off the graph, the same counting-squares method the examples use.{"functions":\[{"color":"\#1f4e79","expression":"m\*x + b","expression2":"","id":"line","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-m","label":"Slope m","max":3,"min":-3,"name":"m","step":0.1,"value":2},{"id":"p-b","label":"Intercept b","max":4,"min":-4,"name":"b","step":0.5,"value":1}\],"title":"y = mx + b: slope m, intercept b","version":1,"viewport":{"centerX":0,"centerY":0,"scale":46}}#figure(figph[three lines show slope], alt: "three lines show slope", caption: none) Observe that a line with positive slope increases from left to right, and one with negative slope decreases. What sort of line has slope #math.equation(block: false, alt: "m equals 0")[$m = 0$]? How can you tell from the equation of a line whether its graph is decreasing? \_\_\_\_\_ #solutionbox[ The graph is decreasing when its slope is negative. ] How can you tell from the equation of a line whether its graph is decreasing? + Its slope is positive. + Its #math.equation(block: false, alt: "x")[$x$]-intercept is positive. + Its slope is negative. + Its #math.equation(block: false, alt: "x")[$x$]-intercept is negative. === Meaning of Slope In Example, we graphed the equation #math.equation(block: false, alt: "C equals 5 plus 3 t")[$" " C = 5 + 3 t " "$] showing the cost of a bicycle rental in terms of the length of the rental. The graph is reproduced at right. We can choose any two points on the line to compute its slope. Using points #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$] as shown, we find that #math.equation(block: true, alt: "m equals the fraction Δ C over Δ t equals the fraction 9 over 3 equals 3")[$m = frac(upright(Δ) C, upright(Δ) t) = frac(9, 3) = 3$] The slope of the line is #math.equation(block: false, alt: "3")[$3$]. #figure(figph[graph f cost versus time with slope], alt: "graph f cost versus time with slope", caption: none) What does this value mean for the cost of renting a bicycle? The expression #math.equation(block: true, alt: "the fraction Δ C over Δ t equals the fraction 9 over 3")[$frac(upright(Δ) C, upright(Δ) t) = frac(9, 3)$] stands for #math.equation(block: true, alt: "the fraction change in cost over change in time equals the fraction 9 dollars over 3 hours")[$frac("change in cost", "change in time") = frac(9 " dollars ", 3 " hours")$] If we increase the length of the rental by 3 hours, the cost of the rental increases by 9 dollars. The slope gives the #emph[rate of increase] in the rental fee, 3 dollars per hour. In general, we can make the following statement. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Rate of Change] The slope of a line measures the #strong[rate of change] of the output variable with respect to the input variable. ] Depending on the variables involved, this rate might be interpreted as a rate of growth or a rate of speed. A negative slope might represent a rate of decrease or a rate of consumption. The slope of a graph can give us valuable information about the variables. #examplebox("Example 6")[][ The graph at right shows the distance in miles traveled by a big-rig truck driver after #math.equation(block: false, alt: "t")[$t$] hours on the road. + Compute the slope of the graph. + What does the slope tell us about the problem? #figure(figph[graph of miles versus time], alt: "graph of miles versus time", caption: none) #solutionbox[ + We can choose any two points on the line, say #math.equation(block: false, alt: "G open parenthesis 2 , 100 close parenthesis")[$G ( 2 , 100 )$] and #math.equation(block: false, alt: "H open parenthesis 4 , 200 close parenthesis")[$H ( 4 , 200 )$], as shown. As we move from #math.equation(block: false, alt: "G")[$G$] to #math.equation(block: false, alt: "H")[$H$], we find #math.equation(block: true, alt: "m equals the fraction Δ D over Δ t equals the fraction 100 over 2 equals 50")[$m = frac(upright(Δ) D, upright(Δ) t) = frac(100, 2) = 50$] The slope of the line is #math.equation(block: false, alt: "50")[$50$]. + The best way to understand the slope is to include units in the calculation. For our example, #math.equation(block: true, alt: "the fraction Δ D over Δ t means the fraction change in distance over change in time")[$frac(upright(Δ) D, upright(Δ) t) " " " " " " " " " means " " " " " " " " " frac("change in distance", "change in time")$] or #math.equation(block: true, alt: "the fraction Δ D over Δ t equals the fraction 100 miles over 2 hours equals 50 miles per hour")[$frac(upright(Δ) D, upright(Δ) t) = frac(100 " miles", 2 " hours") = 50 " miles per hour"$] The slope represents the trucker’s average speed or velocity. ] ] The graph shows the altitude, #math.equation(block: false, alt: "a")[$a$] (in feet), of a skier #math.equation(block: false, alt: "t")[$t$] minutes after getting on a ski lift. #figure(figph[ski lift graph], alt: "ski lift graph", caption: none) + Choose two points and compute the slope (including units). (Note: Use "ft" for feet and use "min" for minutes.) #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ + What does the slope tell us about the problem? #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction y1 ft minus y0 ft over x1 min minus x0 min equals m")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac("y1" " ft" − "y0" " ft", "x1" " min" − "x0" " min")) = "m"$] + The skier's altitude increases by #math.equation(block: false, alt: "150")[$150$] feet per minute. ] The graph shows the altitude, #math.equation(block: false, alt: "a")[$a$] (in feet), of a skier #math.equation(block: false, alt: "t")[$t$] minutes after getting on a ski lift. + Choose two points and compute the slope (including units). + What does the slope tell us about the problem? #figure(figph[ski lift graph], alt: "ski lift graph", caption: none) #solutionbox[ + #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction 700 ft minus 100 ft over 4 min minus 0 min equals 150")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(700 " ft" − 100 " ft", 4 " min" − 0 " min")) = 150$] ft/min + The skier's altitude increases by #math.equation(block: false, alt: "150")[$150$] feet per minute. ] How does slope measure rate of change? \_\_\_\_\_ How does slope measure rate of change? === A Formula for Slope We have defined the slope of a line to be the ratio #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x")[$m = display(frac(upright(Δ) y, upright(Δ) x))$] as we move from one point to another on the line. So far, we have computed #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] by counting squares on the graph, but this method is not always practical. All we really need are the coordinates of two points on the graph. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Subscripts] We will use #strong[subscripts] to distinguish the two points: #math.equation(block: true, alt: "P sub 1 means \"first point\" and P sub 2 means \"second point.\"")[$P_(1) " " " " "means \"first point\" and " " " P_(2) " " " " "means \"second point.\""$] We denote the coordinates of #math.equation(block: false, alt: "P sub 1")[$P_(1)$] by #math.equation(block: false, alt: "open parenthesis x sub 1 , y sub 1 close parenthesis")[$( x_(1) , y_(1) )$] and the coordinates of #math.equation(block: false, alt: "P sub 2")[$P_(2)$] by #math.equation(block: false, alt: "open parenthesis x sub 2 , y sub 2 close parenthesis")[$( x_(2) , y_(2) )$]. ] #figure(figph[line with points labeled], alt: "line with points labeled", caption: none) Now consider a specific example. The line through the two points #math.equation(block: false, alt: "P sub 1 open parenthesis 2 , 9 close parenthesis")[$P_(1) ( 2 , 9 )$] and #math.equation(block: false, alt: "P sub 2 open parenthesis 7 , minus 6 close parenthesis")[$P_(2) ( 7 , − 6 )$] is shown at left. We can find #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] by subtracting the #math.equation(block: false, alt: "x")[$x$]-coordinates of the points: #math.equation(block: true, alt: "Δ x equals 7 minus 2 equals 5")[$upright(Δ) x = 7 − 2 = 5$] In general, we have #math.equation(block: true, alt: "Δ x equals x sub 2 minus x sub 1")[$upright(Δ) x = x_(2) − x_(1)$] and similarly #math.equation(block: true, alt: "Δ y equals y sub 2 minus y sub 1")[$upright(Δ) y = y_(2) − y_(1)$] These formulas work even if some of the coordinates are negative; in our example #math.equation(block: true, alt: "Δ y equals y sub 2 minus y sub 1 equals minus 6 minus 9 equals minus 15")[$upright(Δ) y = y_(2) − y_(1) = − 6 − 9 = − 15$] By counting squares #emph[down] from #math.equation(block: false, alt: "P sub 1")[$P_(1)$] to #math.equation(block: false, alt: "P sub 2")[$P_(2)$], we see that #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] is indeed #math.equation(block: false, alt: "minus 15")[$− 15$]. The slope of the line is #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction minus 15 over 5 equals minus 3")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(y_(2) − y_(1), x_(2) − x_(1)) = frac(− 15, 5) = − 3$] We now have a formula for the slope of a line that works even if we do not have a graph. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Two-Point Slope Formula] The slope of the line passing through the points #math.equation(block: false, alt: "P sub 1 open parenthesis x sub 1 , y sub 1 close parenthesis")[$P_(1) ( x_(1) , y_(1) )$] and #math.equation(block: false, alt: "P sub 2 open parenthesis x sub 2 , y sub 2 close parenthesis")[$P_(2) ( x_(2) , y_(2) )$] is given by #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 , if x sub 2 not equal to x sub 1")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(y_(2) − y_(1), x_(2) − x_(1)) ", if" " " x_(2) ≠ x_(1)$] ] #examplebox("Example 7")[][ Compute the slope of the line above using the points #math.equation(block: false, alt: "Q sub 1 open parenthesis 6 , minus 3 close parenthesis")[$Q_(1) ( 6 , − 3 )$] and #math.equation(block: false, alt: "Q sub 2 open parenthesis 4 , 3 close parenthesis")[$Q_(2) ( 4 , 3 )$]. #solutionbox[ We substitute the coordinates of #math.equation(block: false, alt: "Q sub 1")[$Q_(1)$] and #math.equation(block: false, alt: "Q sub 2")[$Q_(2)$] into the slope formula to find #math.equation(block: true, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction 3 minus open parenthesis minus 3 close parenthesis over 4 minus 6 equals the fraction 6 over minus 2 equals minus 3")[$m = frac(y_(2) − y_(1), x_(2) − x_(1)) = frac(3 − ( − 3 ), 4 − 6) = frac(6, − 2) = − 3$] This value for the slope, #math.equation(block: false, alt: "minus 3")[$− 3$], is the same value we found above. ] ] + Find the slope of the line passing through the points #math.equation(block: false, alt: "open parenthesis 2 , minus 3 close parenthesis")[$( 2 , − 3 )$] and #math.equation(block: false, alt: "open parenthesis minus 2 , minus 1 close parenthesis")[$( − 2 , − 1 )$]. #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ + Sketch a graph of the line by hand. #solutionbox[ + #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x equals the fraction y1 minus y0 over x1 minus x0 equals m")[$m = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac("y1" − "y0", "x1" − "x0")) = "m"$] + A graph is shown below. ] A graph for part (b): #figure(figph[line], alt: "line", caption: none) + Find the slope of the line passing through the points #math.equation(block: false, alt: "open parenthesis 2 , minus 3 close parenthesis")[$( 2 , − 3 )$] and #math.equation(block: false, alt: "open parenthesis minus 2 , minus 1 close parenthesis")[$( − 2 , − 1 )$]. + Sketch a graph of the line by hand. #solutionbox[ + #math.equation(block: true, alt: "m, equals the fraction Δ y over Δ x equals the fraction minus 1 minus open parenthesis minus 3 close parenthesis over minus 2 minus 2; equals the fraction 2 over minus 4 equals the fraction minus 1 over 2")[$m & = display(frac(upright(Δ) y, upright(Δ) x)) = display(frac(− 1 − ( − 3 ), − 2 − 2)) \ & = display(frac(2, − 4)) = display(frac(− 1, 2))$] + #figure(figph[line], alt: "line", caption: none) ] Comment on the following calculation: The intercepts of a line are #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$], so the slope of the line is #math.equation(block: false, alt: "the fraction 5 over 3")[$display(frac(5, 3))$]. \_\_\_\_\_ #solutionbox[ It is false because #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] and #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] have opposite signs. (The slope of the line is #math.equation(block: false, alt: "the fraction minus 5 over 3")[$display(frac(− 5, 3))$].) ] Which comment on the following calculation is true? "The intercepts of a line are #math.equation(block: false, alt: "open parenthesis 0 , 3 close parenthesis")[$( 0 , 3 )$] and #math.equation(block: false, alt: "open parenthesis 5 , 0 close parenthesis")[$( 5 , 0 )$], so the slope of the line is #math.equation(block: false, alt: "the fraction 5 over 3")[$display(frac(5, 3))$]." + It is true because we used two points. + It is false because #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] and #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] have opposite signs. + It is true because the values are increasing. + It is false because the slope is #math.equation(block: false, alt: "the fraction 3 over 5")[$display(frac(3, 5))$]. It will also be useful to write the slope formula with function notation. Recall that #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is another symbol for #math.equation(block: false, alt: "y")[$y$], and, in particular, that #math.equation(block: false, alt: "y sub 1 equals f open parenthesis x sub 1 close parenthesis")[$y_(1) = f ( x_(1) )$] and #math.equation(block: false, alt: "y sub 2 equals f open parenthesis x sub 2 close parenthesis")[$y_(2) = f ( x_(2) )$]. Thus, if #math.equation(block: false, alt: "x sub 2 not equal to x sub 1")[$x_(2) ≠ x_(1)$], we have this formula. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Slope Formula in Function Notation] #math.equation(block: true, alt: "m equals the fraction Δ y over Δ x equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1 equals the fraction f open parenthesis x sub 2 close parenthesis minus f open parenthesis x sub 1 close parenthesis over x sub 2 minus x sub 1 , x sub 2 not equal to x sub 1")[$m = frac(upright(Δ) y, upright(Δ) x) = frac(y_(2) − y_(1), x_(2) − x_(1)) = frac(f ( x_(2) ) − f ( x_(1) ), x_(2) − x_(1)) ", " " " " " " " " " x_(2) ≠ x_(1)$] ] #examplebox("Example 8")[][ The figure shows a graph of #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 6 x")[$f ( x ) = x^(2) − 6 x$] + Compute the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals 4")[$x = 4$]. + Compute the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals 2")[$x = 2$] and #math.equation(block: false, alt: "x equals 5")[$x = 5$]. #figure(figph[slopes on parabola], alt: "slopes on parabola", caption: none) #solutionbox[ + We set #math.equation(block: false, alt: "x sub 1 equals 1")[$x_(1) = 1$] and #math.equation(block: false, alt: "x sub 2 equals 4")[$x_(2) = 4$] and find the function values at each point. #math.equation(block: true, alt: "f open parenthesis x sub 1 close parenthesis equals f open parenthesis 1 close parenthesis equals 1 squared minus 6 open parenthesis 1 close parenthesis equals minus 5")[$f ( x_(1) ) = f ( 1 ) = 1^(2) − 6 ( 1 ) = − 5$] #math.equation(block: true, alt: "f open parenthesis x sub 2 close parenthesis equals f open parenthesis 4 close parenthesis equals 4 squared minus 6 open parenthesis 4 close parenthesis equals minus 8")[$f ( x_(2) ) = f ( 4 ) = 4^(2) − 6 ( 4 ) = − 8$] Then #math.equation(block: true, alt: "m equals the fraction f open parenthesis x sub 2 close parenthesis minus f open parenthesis x sub 1 close parenthesis over x sub 2 minus x sub 1 equals the fraction minus 8 minus open parenthesis minus 5 close parenthesis over 4 minus 1 equals the fraction minus 3 over 3 equals minus 1")[$m = frac(f ( x_(2) ) − f ( x_(1) ), x_(2) − x_(1)) = frac(− 8 − ( − 5 ), 4 − 1) = frac(− 3, 3) = − 1$] + We set #math.equation(block: false, alt: "x sub 1 equals 2")[$x_(1) = 2$] and #math.equation(block: false, alt: "x sub 2 equals 5")[$x_(2) = 5$] and find the function values at each point. #math.equation(block: true, alt: "f open parenthesis x sub 1 close parenthesis equals f open parenthesis 2 close parenthesis equals 2 squared minus 6 open parenthesis 2 close parenthesis equals minus 8")[$f ( x_(1) ) = f ( 2 ) = 2^(2) − 6 ( 2 ) = − 8$] #math.equation(block: true, alt: "f open parenthesis x sub 2 close parenthesis equals f open parenthesis 5 close parenthesis equals 5 squared minus 6 open parenthesis 5 close parenthesis equals minus 5")[$f ( x_(2) ) = f ( 5 ) = 5^(2) − 6 ( 5 ) = − 5$] Then #math.equation(block: true, alt: "m equals the fraction f open parenthesis x sub 2 close parenthesis minus f open parenthesis x sub 1 close parenthesis over x sub 2 minus x sub 1 equals the fraction minus 5 minus open parenthesis minus 8 close parenthesis over 5 minus 2 equals the fraction 3 over 3 equals 1")[$m = frac(f ( x_(2) ) − f ( x_(1) ), x_(2) − x_(1)) = frac(− 5 − ( − 8 ), 5 − 2) = frac(3, 3) = 1$] Note that the graph of #math.equation(block: false, alt: "f")[$f$] is not a straight line and that the slope is not constant. ] ] The figure shows the graph of a function #math.equation(block: false, alt: "f")[$f$]. #figure(figph[increasing curve], alt: "increasing curve", caption: none) + Find #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis")[$f ( 3 )$] and #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis")[$f ( 5 )$]. #linebreak() #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals")[$f ( 3 ) =$]\_\_\_\_\_ #linebreak() #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis equals")[$f ( 5 ) =$]\_\_\_\_\_ + Compute the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals 3")[$x = 3$] and #math.equation(block: false, alt: "x equals 5")[$x = 5$]. #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ + Write an expression for the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals a")[$x = a$] and #math.equation(block: false, alt: "x equals b")[$x = b$]. #linebreak() #math.equation(block: false, alt: "m equals")[$m =$]\_\_\_\_\_ #linebreak() Note: Be careful to use appropriate parentheses. Press the "Preview My Answers" button below to see if you are entering an appropriate expression. #solutionbox[ + #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals 2 , f open parenthesis 5 close parenthesis equals 8")[$f ( 3 ) = 2 , " " f ( 5 ) = 8$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "the fraction f open parenthesis b close parenthesis minus f open parenthesis a close parenthesis over b minus a")[$display(frac(f ( b ) − f ( a ), b − a))$] ] The figure shows the graph of a function #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$" " y = f ( x )$]. + Find #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis")[$f ( 3 )$] and #math.equation(block: false, alt: "f open parenthesis 5 close parenthesis")[$f ( 5 )$]. + Compute the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals 3")[$x = 3$] and #math.equation(block: false, alt: "x equals 5")[$x = 5$]. + Write an expression for the slope of the line segment joining the points at #math.equation(block: false, alt: "x equals a")[$x = a$] and #math.equation(block: false, alt: "x equals b")[$x = b$]. #figure(figph[increasing curve], alt: "increasing curve", caption: none) #solutionbox[ + #math.equation(block: false, alt: "f open parenthesis 3 close parenthesis equals 2 , f open parenthesis 5 close parenthesis equals 8")[$f ( 3 ) = 2 , " " f ( 5 ) = 8$] + #math.equation(block: false, alt: "3")[$3$] + #math.equation(block: false, alt: "the fraction f open parenthesis b close parenthesis minus f open parenthesis a close parenthesis over b minus a")[$display(frac(f ( b ) − f ( a ), b − a))$] ] Summarize what you learned about slope in this lesson. \_\_\_\_\_ Summarize what you learned about slope in this lesson. === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Ratio - Rate of change - Slope - Scale factor ==== CONCEPTS + We can use ratios to compare quantities. + The slope ratio, #math.equation(block: false, alt: "the fraction change in y-coordinate over change in x -coordinate")[$display(frac("change in y-coordinate", "change in " x "-coordinate"))$], measures the steepness of a line. + Notation for slope: #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x")[$m = display(frac(upright(Δ) y, upright(Δ) x))$], #math.equation(block: false, alt: "Δ x not equal to 0")[$upright(Δ) x ≠ 0$]. + Formula for slope: #math.equation(block: false, alt: "m equals the fraction y sub 2 minus y sub 1 over x sub 2 minus x sub 1")[$m = display(frac(y_(2) − y_(1), x_(2) − x_(1)))$], #math.equation(block: false, alt: "x sub 2 not equal to x sub 1")[$" " " " x_(2) ≠ x_(1)$] + Formula for slope: #math.equation(block: false, alt: "m equals the fraction f open parenthesis x sub 2 close parenthesis minus f open parenthesis x sub 1 close parenthesis over x sub 2 minus x sub 1")[$m = display(frac(f ( x_(2) ) − f ( x_(1) ), x_(2) − x_(1)))$], #math.equation(block: false, alt: "x sub 2 not equal to x sub 1")[$" " " " x_(2) ≠ x_(1)$] + Lines have constant slope. + Slope is a scale factor that tells us how many units #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] increases for each unit increase in #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] as we move along the line. + The slope gives us the rate of change. ==== STUDY QUESTIONS + Explain how to compare prices with unit pricing. + Why is #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] the numerator of the slope ratio and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] the denominator? + Which line is steeper, one with #math.equation(block: false, alt: "m equals minus 2")[$m = − 2$] or one with #math.equation(block: false, alt: "m equals minus 5")[$m = − 5$]? + A classmate says that you must always use the intercepts to calculate the slope of a line. Do you agree? Explain. + In an application, what does the slope of the graph tell you about the situation? ==== SKILLS Practice each skill in the Homework problems listed. + Use ratios for comparison: \#1–4 + Compute slope from a graph: \#5–16, 23–26 + Use slope to find #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] or #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$]: \#17–20, 27–30 + Use slope to compare steepness: \#21 and 22 + Decide whether data points lie on a straight line: \#41–46 + Interpret slope as a rate of change: \#31–40 + Use function notation to discuss graphs and slope: \#53–62 === Homework 1.4 Compute ratios to answer the questions in Problems 1–4. Carl runs #math.equation(block: false, alt: "100")[$100$] meters in #math.equation(block: false, alt: "10")[$10$] seconds. Anthony runs #math.equation(block: false, alt: "200")[$200$] meters in #math.equation(block: false, alt: "19.6")[$19.6$] seconds. Who has the faster average speed? Anthony On his #math.equation(block: false, alt: "512")[$512$]-mile round trip to Las Vegas and back, Corey needed #math.equation(block: false, alt: "16")[$16$] gallons of gasoline. He used #math.equation(block: false, alt: "13")[$13$] gallons of gasoline on a #math.equation(block: false, alt: "429")[$429$]-mile trip to Los Angeles. On which trip did he get better fuel economy? Grimy Gulch Pass rises #math.equation(block: false, alt: "0.6")[$0.6$] miles over a horizontal distance of #math.equation(block: false, alt: "26")[$26$] miles. Bob's driveway rises #math.equation(block: false, alt: "12")[$12$] feet over a horizontal distance of #math.equation(block: false, alt: "150")[$150$] feet. Which is steeper? Bob's driveway Which is steeper, the truck ramp for Acme Movers, which rises #math.equation(block: false, alt: "4")[$4$] feet over a horizontal distance of #math.equation(block: false, alt: "9")[$9$] feet, or a toy truck ramp, which rises #math.equation(block: false, alt: "3")[$3$] centimeters over a horizontal distance of #math.equation(block: false, alt: "7")[$7$] centimeters? In Problems 5–8, compute the slope of the line through the indicated points. On both axes, one square represents one unit. #figure(figph[line on grid], alt: "line on grid", caption: none) #math.equation(block: true, alt: "minus 1")[$− 1$] #figure(figph[line on grid], alt: "line on grid", caption: none) #figure(figph[line on grid], alt: "line on grid", caption: none) #math.equation(block: true, alt: "the fraction minus 2 over 3")[$display(frac(− 2, 3))$] #figure(figph[line on grid], alt: "line on grid", caption: none) For Problems 9–14, + Graph each line by the intercept method. + Use the intercepts to compute the slope. #math.equation(block: true, alt: "3 x minus 4 y equals 12")[$3 x − 4 y = 12$] + #figure(figph[3x-4y=12], alt: "3x-4y=12", caption: none) + #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] #math.equation(block: true, alt: "2 y minus 5 x equals 10")[$2 y − 5 x = 10$] #math.equation(block: true, alt: "2 y plus 6 x equals minus 18")[$2 y + 6 x = − 18$] + #figure(figph[2x+6y=-18], alt: "2x+6y=-18", caption: none) + #math.equation(block: false, alt: "minus 3")[$− 3$] #math.equation(block: true, alt: "9 x plus 12 y equals 36")[$9 x + 12 y = 36$] #math.equation(block: true, alt: "the fraction x over 5 minus the fraction y over 8 equals 1")[$display(frac(x, 5)) − display(frac(y, 8)) = 1$] + #figure(figph[x/5-y/8=1], alt: "x/5-y/8=1", caption: none) + #math.equation(block: false, alt: "the fraction 8 over 5")[$display(frac(8, 5))$] #math.equation(block: true, alt: "the fraction x over 7 minus the fraction y over 4 equals 1")[$display(frac(x, 7)) − display(frac(y, 4)) = 1$] + Use the points #math.equation(block: false, alt: "open parenthesis 0 , 2 close parenthesis")[$( 0 , 2 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 8 close parenthesis")[$( 4 , 8 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. + Use the points #math.equation(block: false, alt: "open parenthesis minus 4 , minus 4 close parenthesis")[$( − 4 , − 4 )$] and #math.equation(block: false, alt: "open parenthesis 4 , 8 close parenthesis")[$( 4 , 8 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. + Use the points #math.equation(block: false, alt: "open parenthesis 0 , 2 close parenthesis")[$( 0 , 2 )$] and #math.equation(block: false, alt: "open parenthesis minus 6 , minus 7 close parenthesis")[$( − 6 , − 7 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. #figure(figph[line through (0,2) and (4,8)], alt: "line through (0,2) and (4,8)", caption: none) + #figure(figph[vertical and horizontal changes between two points on line], alt: "vertical and horizontal changes between two points on line", caption: none) + #figure(figph[vertical and horizontal changes between two points on line], alt: "vertical and horizontal changes between two points on line", caption: none) + #figure(figph[vertical and horizontal changes between two points on line], alt: "vertical and horizontal changes between two points on line", caption: none) + Use the points #math.equation(block: false, alt: "open parenthesis 0 , minus 6 close parenthesis")[$( 0 , − 6 )$] and #math.equation(block: false, alt: "open parenthesis 8 , minus 12 close parenthesis")[$( 8 , − 12 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. + Use the points #math.equation(block: false, alt: "open parenthesis minus 8 , 0 close parenthesis")[$( − 8 , 0 )$] and #math.equation(block: false, alt: "open parenthesis 4 , minus 9 close parenthesis")[$( 4 , − 9 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. + Use the points #math.equation(block: false, alt: "open parenthesis 4 , minus 9 close parenthesis")[$( 4 , − 9 )$] and #math.equation(block: false, alt: "open parenthesis 0 , minus 6 close parenthesis")[$( 0 , − 6 )$] to compute the slope of the line. Illustrate #math.equation(block: false, alt: "Δ y")[$upright(Δ) y$] and #math.equation(block: false, alt: "Δ x")[$upright(Δ) x$] on the graph. #figure(figph[line through (4,-9) and (8,-12)], alt: "line through (4,-9) and (8,-12)", caption: none) For Problems 17–20, use the formula #math.equation(block: false, alt: "m equals the fraction Δ y over Δ x")[$m = display(frac(upright(Δ) y, upright(Δ) x))$] A line has slope #math.equation(block: false, alt: "the fraction minus 3 over 4")[$display(frac(− 3, 4))$]. + Find the vertical change associated with each horizontal change along the line. + #math.equation(block: false, alt: "Δ x equals 4")[$upright(Δ) x = 4$] + #math.equation(block: false, alt: "Δ x equals minus 8")[$upright(Δ) x = − 8$] + #math.equation(block: false, alt: "Δ x equals 2")[$upright(Δ) x = 2$] + #math.equation(block: false, alt: "Δ x equals minus 6")[$upright(Δ) x = − 6$] + Find the horizontal change associated with each vertical change along the line. + #math.equation(block: false, alt: "Δ y equals 3")[$upright(Δ) y = 3$] + #math.equation(block: false, alt: "Δ y equals minus 6")[$upright(Δ) y = − 6$] + #math.equation(block: false, alt: "Δ y equals minus 2")[$upright(Δ) y = − 2$] + #math.equation(block: false, alt: "Δ y equals 1")[$upright(Δ) y = 1$] + + #math.equation(block: false, alt: "minus 3")[$− 3$] + #math.equation(block: false, alt: "6")[$6$] + #math.equation(block: false, alt: "the fraction minus 3 over 2")[$display(frac(− 3, 2))$] + #math.equation(block: false, alt: "the fraction 9 over 2")[$display(frac(9, 2))$] + + #math.equation(block: false, alt: "minus 4")[$− 4$] + #math.equation(block: false, alt: "8")[$8$] + #math.equation(block: false, alt: "the fraction 8 over 3")[$display(frac(8, 3))$] + #math.equation(block: false, alt: "the fraction 4 over 3")[$display(frac(4, 3))$] A line has slope #math.equation(block: false, alt: "the fraction 5 over 3")[$display(frac(5, 3))$]. + Find the vertical change associated with each horizontal change along the line. + #math.equation(block: false, alt: "Δ x equals 3")[$upright(Δ) x = 3$] + #math.equation(block: false, alt: "Δ x equals minus 6")[$upright(Δ) x = − 6$] + #math.equation(block: false, alt: "Δ x equals 1")[$upright(Δ) x = 1$] + #math.equation(block: false, alt: "Δ x equals minus 24")[$upright(Δ) x = − 24$] + Find the horizontal change associated with each vertical change along the line. + #math.equation(block: false, alt: "Δ y equals minus 5")[$upright(Δ) y = − 5$] + #math.equation(block: false, alt: "Δ y equals minus 2.5")[$upright(Δ) y = − 2.5$] + #math.equation(block: false, alt: "Δ y equals minus 1")[$upright(Δ) y = − 1$] + #math.equation(block: false, alt: "Δ y equals 3")[$upright(Δ) y = 3$] Residential staircases are usually built with a slope of #math.equation(block: false, alt: "70 %")[$70 upright(%)$], or #math.equation(block: false, alt: "the fraction 7 over 10")[$display(frac(7, 10))$]. If the vertical distance between stories is #math.equation(block: false, alt: "10")[$10$] feet, how much horizontal space does the staircase require? #math.equation(block: false, alt: "the fraction 100 over 7")[$display(frac(100, 7))$] ft #math.equation(block: false, alt: "approximately equals 14.286")[$≈ 14.286$] ft #math.equation(block: false, alt: "approximately equals 14")[$≈ 14$] ft #math.equation(block: false, alt: "3.4")[$" " 3.4$] in A straight section of highway in the Midwest maintains a grade (slope) of #math.equation(block: false, alt: "4 %")[$4 upright(%)$], or #math.equation(block: false, alt: "the fraction 1 over 25")[$display(frac(1, 25))$], for #math.equation(block: false, alt: "12")[$12$] miles. How much does your elevation change as you travel the road? Choose the line with the correct slope. The scales are the same on both axes. + #math.equation(block: false, alt: "m equals 2")[$m = 2$] + #math.equation(block: false, alt: "m equals minus the fraction 1 over 2")[$m = − display(frac(1, 2))$] + #math.equation(block: false, alt: "m equals the fraction 2 over 3")[$m = display(frac(2, 3))$] + #math.equation(block: false, alt: "m equals minus the fraction 5 over 3")[$m = − display(frac(5, 3))$] #figure(figph[four lines], alt: "four lines", caption: none) + IV + III + II + I Choose the line with the correct slope. The scales are the same on both axes. + #math.equation(block: false, alt: "0 less than m less than 1")[$0 < m < 1$] + #math.equation(block: false, alt: "m less than minus 1")[$m < − 1$] + #math.equation(block: false, alt: "m greater than 1")[$m > 1$] + #math.equation(block: false, alt: "m equals 0")[$m = 0$] #figure(figph[four lines], alt: "four lines", caption: none) Compute the slope of the line in Problems 23–26. Note the scales on the axes. #figure(figph[four lines], alt: "four lines", caption: none) #math.equation(block: true, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] #figure(figph[four lines], alt: "four lines", caption: none) #figure(figph[four lines], alt: "four lines", caption: none) #math.equation(block: true, alt: "minus 4000")[$− 4000$] #figure(figph[four lines], alt: "four lines", caption: none) Each table in Problems 27–30 gives the coordinates of points on a line. + Find the slope of the line. + Fill in the missing table entries. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "minus 4")[$− 4$]], [#math.equation(block: false, alt: "minus 14")[$− 14$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [#math.equation(block: false, alt: "minus 9")[$− 9$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]], [], [], [#math.equation(block: false, alt: "11")[$11$]], )) + #math.equation(block: false, alt: "the fraction 5 over 2")[$display(frac(5, 2))$] + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "the fraction 7 over 2")[$frac(7, 2)$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "11")[$11$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "minus 5")[$− 5$]], [#math.equation(block: false, alt: "minus 3.8")[$− 3.8$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "minus 0.6")[$− 0.6$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "1.8")[$1.8$]], [], [#math.equation(block: false, alt: "4.2")[$4.2$]], [#math.equation(block: false, alt: "7")[$7$]], [], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [], [], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], )) + #math.equation(block: false, alt: "minus 3")[$− 3$] + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "12")[$12$]], )) #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "minus 10")[$− 10$]], [#math.equation(block: false, alt: "800")[$800$]], [#math.equation(block: false, alt: "minus 2")[$− 2$]], [], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "440")[$440$]], [], [#math.equation(block: false, alt: "368")[$368$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "176")[$176$]], )) A temporary typist's paycheck (before deductions) is given, in dollars, by #math.equation(block: false, alt: "S equals 8 t")[$S = 8 t$], where #math.equation(block: false, alt: "t")[$t$] is the number of hours she worked. + Make a table of values for the function.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "40")[$40$]]), [#math.equation(block: false, alt: "S")[$S$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Graph the function. + Using two points on the graph, compute the slope #math.equation(block: false, alt: "the fraction Δ S over Δ t")[$display(frac(upright(Δ) S, upright(Δ) t))$], including units. + What does the slope tell us about the typist's paycheck? + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "40")[$40$]]), [#math.equation(block: false, alt: "S")[$S$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "160")[$160$]], [#math.equation(block: false, alt: "320")[$320$]], )) + #figure(figph[salary graph], alt: "salary graph", caption: none) + 8 dollars/hour + The typist is paid \$#math.equation(block: false, alt: "8")[$8$] per hour. The distance (in miles) covered by a cross-country competitor is given by #math.equation(block: false, alt: "d equals 6 t")[$d = 6 t$], where #math.equation(block: false, alt: "t")[$t$] is the number of hours she runs. + Make a table of values for the function.#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "d")[$d$]], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], [$#hide($0000$)$], )) + Graph the function. + Using two points on the graph, compute the slope #math.equation(block: false, alt: "the fraction Δ d over Δ t")[$display(frac(upright(Δ) d, upright(Δ) t))$], including units. + What does the slope tell us about the cross-country runner? In Problems 33–40, + Choose two points and compute the slope of the graph (including units). + Explain what the slope measures in the context of the problem. The graph shows the number of barrels of oil, #math.equation(block: false, alt: "B")[$B$], that has been pumped at a drill site #math.equation(block: false, alt: "t")[$t$] days after a new drill is installed. #figure(figph[barrels vs time], alt: "barrels vs time", caption: none) + #math.equation(block: false, alt: "1250")[$1250$] barrels/day + The slope indicates that oil is pumped at a rate of #math.equation(block: false, alt: "1250")[$1250$] barrels per day. The graph shows the amount of garbage, #math.equation(block: false, alt: "G")[$G$] (in tons), that has been deposited at a dump site #math.equation(block: false, alt: "t")[$t$] years after new regulations go into effect. #figure(figph[tons of garbage vs time], alt: "tons of garbage vs time", caption: none) The graph shows the amount of emergency water, #math.equation(block: false, alt: "W")[$W$] (in liters), remaining in a southern California household #math.equation(block: false, alt: "t")[$t$] days after an earthquake. #figure(figph[water vs time], alt: "water vs time", caption: none) + #math.equation(block: false, alt: "minus 6")[$− 6$] liters/day + The slope indicates that the water is diminishing at a rate of #math.equation(block: false, alt: "6")[$6$] liters per day. The graph shows the amount of money, #math.equation(block: false, alt: "M")[$M$] (in dollars), in Tammy's bank account #math.equation(block: false, alt: "w")[$w$] weeks after she loses all sources of income. #figure(figph[dollars vs time], alt: "dollars vs time", caption: none) The graph shows the length in inches, #math.equation(block: false, alt: "i")[$i$], corresponding to various lengths in feet #math.equation(block: false, alt: "f")[$f$]. #figure(figph[inches vs ft], alt: "inches vs ft", caption: none) + #math.equation(block: false, alt: "12")[$12$] inches/foot + The slope gives the conversion rate of 12 inches per foot. The graph shows the number of ounces, #math.equation(block: false, alt: "z")[$z$], that correspond to various weights measured in pounds, #math.equation(block: false, alt: "p")[$p$]. #figure(figph[ounces vs pounds], alt: "ounces vs pounds", caption: none) The graph shows the cost, #math.equation(block: false, alt: "C")[$C$] (in dollars), of coffee beans in terms of the amount of coffee, #math.equation(block: false, alt: "b")[$b$] (in kilograms). #figure(figph[dollars vs kg], alt: "dollars vs kg", caption: none) + #math.equation(block: false, alt: "4")[$4$] dollars/kilogram + The slope gives the unit price of #math.equation(block: false, alt: "$ 4")[$upright(\$) 4$] per kilogram The graph shows Tracey's earnings, #math.equation(block: false, alt: "E")[$E$] (in dollars), in terms of the number of hours, #math.equation(block: false, alt: "h")[$h$], that she babysits. #figure(figph[dollars vs hours], alt: "dollars vs hours", caption: none) Which of the tables in Problems 41 and 42 represent variables that are related by a linear function? (Hint: Which relationships have constant slope?) + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "y")[$y$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "17")[$17$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "22")[$22$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "27")[$27$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "32")[$32$]], )) + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "P")[$P$]]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "36")[$36$]], )) (a) + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "h")[$h$]], [#math.equation(block: false, alt: "w")[$w$]]), [#math.equation(block: false, alt: "minus 6")[$− 6$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "14")[$14$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "12")[$12$]], )) + #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "d")[$d$]]), [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "24")[$24$]], )) The table shows the amount of ammonium chloride salt, in grams, that can be dissolved in 100 grams of water at different temperatures. #figure(table( columns: 8, align: left, inset: 6pt, table.header([Temperature,#math.equation(block: false, alt: "∘")[$∘$]C], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "21")[$21$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "52")[$52$]]), [Grams of salt], [#math.equation(block: false, alt: "33")[$33$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "35.5")[$35.5$]], [#math.equation(block: false, alt: "38.5")[$38.5$]], [#math.equation(block: false, alt: "40.5")[$40.5$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "54")[$54$]], )) + If you plot the data, will the points lie on a straight line? Why or why not? + Calculate the rate of change of salt dissolved with respect to temperature. + Yes, the slope between any two points is #math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]. + #math.equation(block: false, alt: "0.5")[$0.5$] grams of salt per degree Celsius A spring is suspended from the ceiling. The table shows the length of the spring, in centimeters, as it is stretched by hanging various weights from it. #figure(table( columns: 8, align: left, inset: 6pt, table.header([Weight, kg], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "20")[$20$]]), [Length, cm], [#math.equation(block: false, alt: "25.72")[$25.72$]], [#math.equation(block: false, alt: "25.88")[$25.88$]], [#math.equation(block: false, alt: "26.2")[$26.2$]], [#math.equation(block: false, alt: "26.36")[$26.36$]], [#math.equation(block: false, alt: "26.52")[$26.52$]], [#math.equation(block: false, alt: "26.84")[$26.84$]], [#math.equation(block: false, alt: "27.16")[$27.16$]], )) + If you plot the data, will the points lie on a straight line? Why or why not? + Calculate the rate of change of length with respect to weight. The table gives the radius and circumference of various circles, rounded to three decimal places. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "r")[$r$]], [#math.equation(block: false, alt: "C")[$C$]]), [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "25.133")[$25.133$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "37.699")[$37.699$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "62.832")[$62.832$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "94.248")[$94.248$]], )) + If we plot the data, will the points lie on a straight line? + What familiar number does the slope turn out to be? (Hint: Recall a formula from geometry.) + Yes + #math.equation(block: false, alt: "2 π")[$2 π$] The table gives the side and the diagonal of various squares, rounded to three decimal places. #figure(table( columns: 2, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "s")[$s$]], [#math.equation(block: false, alt: "d")[$d$]]), [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4.243")[$4.243$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "8.485")[$8.485$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "11.314")[$11.314$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "14.142")[$14.142$]], )) + If we plot the data, will the points lie on a straight line? + What familiar number does the slope turn out to be? (Hint: Draw a picture of one of the squares and use the Pythagorean theorem to compute its diagonal.) Geologists can measure the depth of the ocean at different points using a technique called echo-sounding. Scientists on board a ship send a pulse of sound toward the ocean floor and measure the time interval until the echo returns to the ship. The speed of sound in seawater is about #math.equation(block: false, alt: "1500")[$1500$] meters per second. + Write the speed of sound as a ratio. + If the echo returns in #math.equation(block: false, alt: "4.5")[$4.5$] seconds, what is the depth of the ocean at that point? + #math.equation(block: false, alt: "the fraction 1500 meters over 1 second")[$display(frac(1500 " meters", 1 " second"))$] + #math.equation(block: false, alt: "3375")[$3375$] meters Niagara Falls was discovered by Father Louis Hennepin in 1682. In 1952, much of the water of the Niagara River was diverted for hydroelectric power, but until that time erosion caused the Falls to recede upstream by #math.equation(block: false, alt: "3")[$3$] feet per year. + How far did the Falls recede from 1682 to 1952? + The Falls were formed about #math.equation(block: false, alt: "12 , 000")[$12 , 000$] years ago during the end of the last ice age. How far downstream from their current position were they then? (Give your answer in miles.) Geologists calculate the speed of seismic waves by plotting the travel times for waves to reach seismometers at known distances from the epicenter. The speed of the wave can help them determine the nature of the material it passes through. The graph shows a travel-time graph for P-waves from a shallow earthquake. #figure(figph[seismic wave: time vs distance], alt: "seismic wave: time vs distance", caption: none) + Why do you think the graph is plotted with distance as the input variable? + Use the graph to calculate the speed of the wave. + The distances are known. + #math.equation(block: false, alt: "5.7")[$5.7$] km per second Energy (supplied by heat) is required to raise the temperature of a substance, and it is also needed to melt a solid substance to a liquid. The table shows data from heating a solid sample of stearic acid. Heat was applied at a constant rate throughout the experiment. (Source: J. A. Hunt and A. Sykes, 1984) #figure(table( columns: 16, align: left, inset: 6pt, table.header([Time (minutes], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#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: "8.5")[$8.5$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "9.5")[$9.5$]], [#math.equation(block: false, alt: "10")[$10$]]), [Temperature, #math.equation(block: false, alt: "∘")[$∘$]C], [#math.equation(block: false, alt: "19")[$19$]], [#math.equation(block: false, alt: "29")[$29$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "53")[$53$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "73")[$73$]], [#math.equation(block: false, alt: "74")[$74$]], )) + Did the temperature rise at a constant rate? Describe the temperature as a function of time. + Graph temperature as a function of time. + What is the melting point of stearic acid? How long did it take the sample to melt? The graph shows the temperature of the ocean as a function of depth. #figure(figph[ocean temperature], alt: "ocean temperature", caption: none) + What is the difference in temperature between the surface of the ocean and the deepest level shown? + Over what depths does the temperature change most rapidly? + What is the average rate of change of temperature with respect to depth in the region called the thermocline? + About #math.equation(block: false, alt: "18 degrees")[$18^(∘)$]C + 0.3 km to 0.4 km + About #math.equation(block: false, alt: "minus 28 degrees")[$− 28^(∘)$]C per kilometer The graph shows the average air temperature as a function of altitude. (Figure (b) is an enlargement of the indicated region of Figure (a).) (Source: Ahrens, 1998) #figure(figph[temperature vs altitude], alt: "temperature vs altitude", caption: none) + Is temperature a decreasing function of altitude? + The #strong[lapse rate] is the rate at which the temperature changes with altitude. In which regions of the atmosphere is the lapse rate positive? + The region where the lapse rate is zero is called the isothermal zone. Give an interval of altitudes that describes the isothermal zone. + What is the lapse rate in the mesosphere? + Describe the temperature for altitudes greater than 90 kilometers. In Problems 53–56, evaluate the function at #math.equation(block: false, alt: "x equals a")[$x = a$] and #math.equation(block: false, alt: "x equals b")[$x = b$], and then find the slope of the line segment joining the two corresponding points on the graph. Illustrate the line segment on a graph of the function. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals x squared minus 2 x minus 8")[$f ( x ) = x^(2) − 2 x − 8$] + #math.equation(block: false, alt: "a equals minus 2 , b equals 1")[$a = − 2 , b = 1$] + #math.equation(block: false, alt: "a equals minus 1 , b equals 5")[$a = − 1 , b = 5$] + #math.equation(block: false, alt: "minus 3")[$− 3$]#figure(figph[segment joining two points on curve], alt: "segment joining two points on curve", caption: none) + #math.equation(block: false, alt: "2")[$2$]#figure(figph[segment joining two points on curve], alt: "segment joining two points on curve", caption: none) #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals the square root of x plus 4")[$g ( x ) = sqrt(x + 4)$] + #math.equation(block: false, alt: "a equals minus 2 , b equals 0")[$a = − 2 , b = 0$] + #math.equation(block: false, alt: "a equals 0 , b equals 5")[$a = 0 , b = 5$] #math.equation(block: true, alt: "h open parenthesis x close parenthesis equals the fraction 4 over x plus 2")[$h ( x ) = display(frac(4, x + 2))$] + #math.equation(block: false, alt: "a equals 0 , b equals 6")[$a = 0 , b = 6$] + #math.equation(block: false, alt: "a equals minus 1 , b equals 2")[$a = − 1 , b = 2$] + #math.equation(block: false, alt: "the fraction minus 1 over 4")[$display(frac(− 1, 4))$]#figure(figph[segment joining two points on curve], alt: "segment joining two points on curve", caption: none) + #math.equation(block: false, alt: "minus 1")[$− 1$]#figure(figph[segment joining two points on curve], alt: "segment joining two points on curve", caption: none) #math.equation(block: true, alt: "q open parenthesis x close parenthesis equals x cubed minus 4 x")[$q ( x ) = x^(3) − 4 x$] + #math.equation(block: false, alt: "a equals minus 1 , b equals 2")[$a = − 1 , b = 2$] + #math.equation(block: false, alt: "a equals minus 1 , b equals 3")[$a = − 1 , b = 3$] In Problems 57–62, find the coordinates of the indicated points, then write an algebraic expression using function notation for the indicated quantity. The length of the vertical line segment on the #math.equation(block: false, alt: "y")[$y$]-axis + #figure(figph[increasing graph], alt: "increasing graph", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "open parenthesis 1 , F open parenthesis 1 close parenthesis close parenthesis , open parenthesis 4 , F open parenthesis 4 close parenthesis close parenthesis")[$( 1 , F ( 1 ) ) , ( 4 , F ( 4 ) )$]; #math.equation(block: false, alt: "F open parenthesis 4 close parenthesis minus F open parenthesis 1 close parenthesis")[$" " " " " " " " F ( 4 ) − F ( 1 )$] + #math.equation(block: false, alt: "open parenthesis r , f open parenthesis r close parenthesis close parenthesis , open parenthesis s , f open parenthesis s close parenthesis close parenthesis")[$( r , f ( r ) ) , ( s , f ( s ) )$]; #math.equation(block: false, alt: "f open parenthesis s close parenthesis minus f open parenthesis r close parenthesis")[$" " " " " " " " f ( s ) − f ( r )$] The length of the vertical line segment on the #math.equation(block: false, alt: "y")[$y$]-axis + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + The increase in #math.equation(block: false, alt: "y")[$y$] as #math.equation(block: false, alt: "x")[$x$] increases from #math.equation(block: false, alt: "2")[$2$] to #math.equation(block: false, alt: "3")[$3$]#figure(figph[increasing graph], alt: "increasing graph", caption: none) + The increase in #math.equation(block: false, alt: "y")[$y$] as #math.equation(block: false, alt: "x")[$x$] increases from #math.equation(block: false, alt: "a")[$a$] to #math.equation(block: false, alt: "b")[$b$]#figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "open parenthesis 2 , H open parenthesis 2 close parenthesis close parenthesis , open parenthesis 3 , H open parenthesis 3 close parenthesis close parenthesis")[$( 2 , H ( 2 ) ) , ( 3 , H ( 3 ) )$]; #math.equation(block: false, alt: "H open parenthesis 3 close parenthesis minus H open parenthesis 2 close parenthesis")[$" " " " " " " " H ( 3 ) − H ( 2 )$] + #math.equation(block: false, alt: "open parenthesis a , g open parenthesis a close parenthesis close parenthesis , open parenthesis b , g open parenthesis b close parenthesis close parenthesis")[$( a , g ( a ) ) , ( b , g ( b ) )$]; #math.equation(block: false, alt: "g open parenthesis b close parenthesis minus g open parenthesis a close parenthesis")[$" " " " " " " " g ( b ) − g ( a )$] + The increase in #math.equation(block: false, alt: "y")[$y$] as #math.equation(block: false, alt: "x")[$x$] increases from #math.equation(block: false, alt: "minus 4")[$− 4$] to #math.equation(block: false, alt: "minus 1")[$− 1$]#figure(figph[curve], alt: "curve", caption: none) + The increase in #math.equation(block: false, alt: "y")[$y$] as #math.equation(block: false, alt: "x")[$x$] increases from #math.equation(block: false, alt: "u")[$u$] to #math.equation(block: false, alt: "v")[$v$]#figure(figph[curve], alt: "curve", caption: none) The shaded area + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "open parenthesis c , s open parenthesis c close parenthesis close parenthesis , open parenthesis d , s open parenthesis d close parenthesis close parenthesis")[$( c , s ( c ) ) , ( d , s ( d ) )$]; #math.equation(block: false, alt: "s open parenthesis c close parenthesis open parenthesis d minus c close parenthesis")[$" " " " " " " " s ( c ) ( d − c )$] + #math.equation(block: false, alt: "open parenthesis x sub 1 , q open parenthesis x sub 1 close parenthesis close parenthesis , open parenthesis x sub 2 , q open parenthesis x sub 2 close parenthesis close parenthesis")[$( x_(1) , q ( x_(1) ) ) , ( x_(2) , q ( x_(2) ) )$]; #math.equation(block: false, alt: "q open parenthesis x sub 2 close parenthesis open parenthesis x sub 2 minus x sub 1 close parenthesis")[$" " " " " " " " q ( x_(2) ) ( x_(2) − x_(1) )$] The shaded area + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) In Problems 63–66, find the coordinates of the indicated points on the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] and write an algebraic expression using function notation for the slope of the line segment joining points #math.equation(block: false, alt: "P")[$P$] and #math.equation(block: false, alt: "Q")[$Q$]. + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "open parenthesis 1 , f open parenthesis 1 close parenthesis close parenthesis , open parenthesis 5 , f open parenthesis 5 close parenthesis close parenthesis")[$( 1 , f ( 1 ) ) , ( 5 , f ( 5 ) )$]; #math.equation(block: false, alt: "the fraction f open parenthesis 5 close parenthesis minus f open parenthesis 1 close parenthesis over 4")[$" " " " " " " " display(frac(f ( 5 ) − f ( 1 ), 4))$] + #math.equation(block: false, alt: "open parenthesis minus 1 , f open parenthesis minus 1 close parenthesis close parenthesis , open parenthesis 2 , f open parenthesis 2 close parenthesis close parenthesis")[$( − 1 , f ( − 1 ) ) , ( 2 , f ( 2 ) )$]; #math.equation(block: false, alt: "the fraction f open parenthesis 2 close parenthesis minus f open parenthesis minus 1 close parenthesis over 3")[$" " " " " " " " display(frac(f ( 2 ) − f ( − 1 ), 3))$] + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none) + #math.equation(block: false, alt: "open parenthesis a , f open parenthesis a close parenthesis close parenthesis , open parenthesis b , f open parenthesis b close parenthesis close parenthesis")[$( a , f ( a ) ) , ( b , f ( b ) )$]; #math.equation(block: false, alt: "the fraction f open parenthesis b close parenthesis minus f open parenthesis a close parenthesis over b minus a")[$" " " " " " " " display(frac(f ( b ) − f ( a ), b − a))$] + #math.equation(block: false, alt: "open parenthesis a , f open parenthesis a close parenthesis close parenthesis , open parenthesis a plus Δ x , f open parenthesis a plus Δ x close parenthesis close parenthesis")[$( a , f ( a ) ) , ( a + upright(Δ) x , f ( a + upright(Δ) x ) )$]; #math.equation(block: false, alt: "the fraction f open parenthesis a plus Δ x close parenthesis minus f open parenthesis a close parenthesis over Δ x")[$" " " " " " " " display(frac(f ( a + upright(Δ) x ) − f ( a ), upright(Δ) x))$] + #figure(figph[curve], alt: "curve", caption: none) + #figure(figph[curve], alt: "curve", caption: none)