#set document(title: "7.4 Graphing Rational Functions", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 7.4#h(0.6em)Graphing Rational Functions === Introduction A rational function is the quotient of two polynomials. (As with rational numbers, the word #emph[rational] refers to a ratio.) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Rational Function] A #strong[rational function] is one of the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = frac(P ( x ), Q ( x ))$] where #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] are polynomials and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] is not the zero polynomial. ] Which of the following is not a rational function? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "y equals the fraction the square root of x squared minus 4 over x squared plus 4")[$y = display(frac(sqrt(x^(2) − 4), x^(2) + 4))$] is not a rational function because the numerator is not a polynomial. ] Which of the following is not a rational function? + #math.equation(block: false, alt: "y equals the fraction 1 over x plus 2")[$y = display(frac(1, x + 2))$] + #math.equation(block: false, alt: "y equals 3 x plus 5")[$y = 3 x + 5$] + #math.equation(block: false, alt: "y equals the fraction the square root of x squared minus 4 over x squared plus 4")[$y = display(frac(sqrt(x^(2) − 4), x^(2) + 4))$] + #math.equation(block: false, alt: "y equals the fraction 2 x cubed plus 3 x minus 7 over x to the power 4 minus x squared")[$y = display(frac(2 x^(3) + 3 x − 7, x^(4) − x^(2)))$] The graphs of rational functions can be quite different from the graphs of polynomials. #examplebox("Example 1")[][ Francine is planning a 60-mile training flight through the desert on her cycle-plane, a pedal-driven aircraft. If there is no wind, she can pedal at an average speed of 15 miles per hour, so she can complete the flight in 4 hours. + If there is a headwind of #math.equation(block: false, alt: "x")[$x$] miles per hour, it will take Francine longer to fly 60 miles. Express the time it will take to complete the training flight as a function of #math.equation(block: false, alt: "x")[$x$]. + Make a table of values for the function. + Graph the function and explain what it tells you about the time Francine should allot for the flight. #solutionbox[ + If there is a headwind of #math.equation(block: false, alt: "x")[$x$] miles per hour, Francine's ground speed will be #math.equation(block: false, alt: "15 minus x")[$15 − x$] miles per hour. Using the fact that #math.equation(block: false, alt: "time equals the fraction distance over rate")[$"time" = display(frac("distance", "rate"))$], we find that the time needed for the flight will be #math.equation(block: true, alt: "t equals f open parenthesis x close parenthesis equals the fraction 60 over 15 minus x")[$t = f ( x ) = frac(60, 15 − x)$] + We evaluate the function for several values of #math.equation(block: false, alt: "x")[$x$], as shown in the table below.#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "9")[$9$]], [#math.equation(block: false, alt: "10")[$10$]]), [#math.equation(block: false, alt: "t")[$t$]], [#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: "7.5")[$7.5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "12")[$12$]], )) #linebreak() For example, if the headwind is #math.equation(block: false, alt: "5")[$5$] miles per hour, then #math.equation(block: true, alt: "t equals the fraction 60 over 15 minus 5 equals the fraction 60 over 10 equals 6")[$t = frac(60, 15 − 5) = frac(60, 10) = 6$] Francine's effective speed is only #math.equation(block: false, alt: "10")[$10$] miles per hour, and it will take her #math.equation(block: false, alt: "6")[$6$] hours to fly the #math.equation(block: false, alt: "60")[$60$] miles. The table shows that as the speed of the headwind increases, the time required for the flight increases also. + The graph of the function is shown below. You can use your calculator with the window #math.equation(block: true, alt: "Xmin, equals minus 8.5, Xmax equals 15; Ymin, equals 0, Ymax equals 30")[$"Xmin" & = − 8.5 & & "Xmax" = 15 \ "Ymin" & = 0 & & "Ymax" = 30$] to verify the graph. In particular, the point #math.equation(block: false, alt: "open parenthesis 0 , 4 close parenthesis")[$( 0 , 4 )$] lies on the graph. This point tells us that if there is no wind, Francine can fly #math.equation(block: false, alt: "60")[$60$] miles in #math.equation(block: false, alt: "4")[$4$] hours, as we calculated earlier.#figure(figph[graph showing time to complete trip vs speed of headwind], alt: "graph showing time to complete trip vs speed of headwind", caption: none) #linebreak() The graph is increasing, as indicated by the table of values. In fact, as the speed of the wind gets close to #math.equation(block: false, alt: "15")[$15$] miles per hour, Francine's flying time becomes extremely large. In theory, if the wind speed were exactly #math.equation(block: false, alt: "15")[$15$] miles per hour, Francine would never complete her flight. On the graph, the time becomes infinite at #math.equation(block: false, alt: "x equals 15")[$x = 15$]. #linebreak() What about negative values for #math.equation(block: false, alt: "x")[$x$]? If we interpret a negative headwind as a tailwind, Francine's flying time should decrease for negative #math.equation(block: false, alt: "x")[$x$]-values. For example, if #math.equation(block: false, alt: "x equals minus 5")[$x = − 5$], there is a tailwind of #math.equation(block: false, alt: "5")[$5$] miles per hour, so Francine's effective speed is #math.equation(block: false, alt: "20")[$20$] miles per hour, and she can complete the flight in #math.equation(block: false, alt: "3")[$3$] hours. As the tailwind gets stronger (that is, as we move farther to the left in the #math.equation(block: false, alt: "x")[$x$]-direction), Francine's flying time continues to decrease, and the graph approaches the #math.equation(block: false, alt: "x")[$x$]-axis. ] ] The vertical dashed line at #math.equation(block: false, alt: "x equals 15")[$x = 15$] on the graph of #math.equation(block: false, alt: "t equals the fraction 60 over 15 minus x")[$t = display(frac(60, 15 − x))$] is a #strong[vertical asymptote] for the graph. We first encountered asymptotes in Some Basic Functions when we studied the graph of #math.equation(block: false, alt: "y equals the fraction 1 over x")[$y = display(frac(1, x))$]. Locating the vertical asymptotes of a rational function is an important part of determining the shape of the graph. Queueing theory is used to predict your waiting time in a line, or queue. For example, suppose the attendant at a toll booth can process #math.equation(block: false, alt: "6")[$6$] vehicles per minute. The average total time spent by a motorist negotiating the toll booth depends on the rate, #math.equation(block: false, alt: "r")[$r$], at which vehicles arrive, according to the formula #math.equation(block: true, alt: "T equals g open parenthesis r close parenthesis equals the fraction 12 minus r over 12 open parenthesis 6 minus r close parenthesis")[$T = g ( r ) = frac(12 − r, 12 ( 6 − r ))$] + What is the average time spent at the toll booth if vehicles arrive at a rate of #math.equation(block: false, alt: "3")[$3$] vehicles per minute? #linebreak() Answer: \_\_\_\_\_ min. + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 6 close bracket")[$[ 0 , 6 ]$]. + What is the vertical asymptote of the graph? #linebreak() Equation of asymptote: \_\_\_\_\_ + What does the vertical asymptote tell you about the queue? #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "0.25")[$0.25$] min + A graph is below. + #math.equation(block: false, alt: "r equals 6")[$r = 6$] + The wait time becomes infinite as the arrival rate approaches #math.equation(block: false, alt: "6")[$6$] vehicles per minute. ] Graph for part (b): #figure(figph[rational function], alt: "rational function", caption: none) Queueing theory is used to predict your waiting time in a line, or queue. For example, suppose the attendant at a toll booth can process #math.equation(block: false, alt: "6")[$6$] vehicles per minute. The average total time spent by a motorist negotiating the toll booth depends on the rate, #math.equation(block: false, alt: "r")[$r$], at which vehicles arrive, according to the formula #math.equation(block: true, alt: "T equals g open parenthesis r close parenthesis equals the fraction 12 minus r over 12 open parenthesis 6 minus r close parenthesis")[$T = g ( r ) = frac(12 − r, 12 ( 6 − r ))$] + What is the average time spent at the toll booth if vehicles arrive at a rate of #math.equation(block: false, alt: "3")[$3$] vehicles per minute? + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 6 close bracket")[$[ 0 , 6 ]$]. + What is the vertical asymptote of the graph? + What does the vertical asymptote tell you about the queue? #solutionbox[ + #math.equation(block: false, alt: "0.25")[$0.25$] min + #figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "r equals 6")[$r = 6$] + The wait time becomes infinite as the arrival rate approaches #math.equation(block: false, alt: "6")[$6$] vehicles per minute. ] #examplebox("Example 2")[][ EarthCare decides to sell T-shirts to raise money. The company makes an initial investment of \$100 to pay for the design of the T-shirt and to set up the printing process. After that, the T-shirts cost \$5 each for labor and materials. + Express the average cost per T-shirt as a function of the number of T-shirts EarthCare produces. + Make a table of values for the function. + Graph the function and explain what it tells you about the cost of the T-shirts. #solutionbox[ + If EarthCare produces #math.equation(block: false, alt: "x")[$x$] T-shirts, the total costs will be #math.equation(block: false, alt: "100 plus 5 x")[$100 + 5 x$] dollars. To find the average cost per T-shirt, we divide the total cost by the number of T-shirts produced, to get #math.equation(block: true, alt: "C equals g open parenthesis x close parenthesis equals the fraction 100 plus 5 x over x")[$C = g ( x ) = frac(100 + 5 x, x)$] + We evaluate the function for several values of #math.equation(block: false, alt: "x")[$x$], as shown in the table#figure(table( columns: 7, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "105")[$105$]], [#math.equation(block: false, alt: "55")[$55$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "10")[$10$]], )) #linebreak() If EarthCare makes only one T-shirt, its cost is \$105. But if more than one T-shirt is made, the cost of the original \$100 investment is distributed among them. For example, the average cost per T-shirt for #math.equation(block: false, alt: "2")[$2$] T-shirts is #math.equation(block: true, alt: "the fraction 100 plus 5 open parenthesis 2 close parenthesis over 2 equals 55")[$frac(100 + 5 ( 2 ), 2) = 55$] and the average cost for #math.equation(block: false, alt: "5")[$5$] T-shirts is #math.equation(block: true, alt: "the fraction 100 plus 5 open parenthesis 5 close parenthesis over 5 equals 25")[$frac(100 + 5 ( 5 ), 5) = 25$] + The graph is shown below. You can use your calculator with the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 470; Ymin, equals 0, Ymax equals 30")[$"Xmin" & = 0 & & "Xmax" = 470 \ "Ymin" & = 0 & & "Ymax" = 30$] to verify the graph. Use the #emph[Trace] to locate on the graph several points from the table of values. For example, the point #math.equation(block: false, alt: "open parenthesis 5 , 25 close parenthesis")[$( 5 , 25 )$] indicates that if EarthCare makes #math.equation(block: false, alt: "5")[$5$] T-shirts, the cost per shirt is \$#math.equation(block: false, alt: "25")[$25$].#figure(figph[graph of average cost of production], alt: "graph of average cost of production", caption: none) #linebreak() The graph shows that as the number of T-shirts increases, the average cost per shirt continues to decrease, but not as rapidly as at first. Eventually the average cost levels off and approaches \$5 per T-shirt. For example, if EarthCare produces #math.equation(block: false, alt: "400")[$400$] T-shirts, the average cost per shirt is #math.equation(block: true, alt: "the fraction 100 plus 5 open parenthesis 400 close parenthesis over 400 equals 5.25")[$frac(100 + 5 ( 400 ), 400) = 5.25$] ] ] The horizontal line #math.equation(block: false, alt: "C equals 5")[$C = 5$] on the graph of #math.equation(block: false, alt: "C equals the fraction 100 plus 5 x over x")[$C = display(frac(100 + 5 x, x))$] is a #strong[horizontal asymptote]. As #math.equation(block: false, alt: "x")[$x$] increases, the graph approaches the line #math.equation(block: false, alt: "C equals 5")[$C = 5$] but never actually meets it. The average price per T-shirt will always be slightly more than \$5. Horizontal asymptotes are also important in sketching the graphs of rational functions. Delbert prepares a 20% glucose solution of by mixing 2 mL of glucose with 8 mL of water. If he adds #math.equation(block: false, alt: "x")[$x$] ml of glucose to the solution, its concentration is given by #math.equation(block: true, alt: "C open parenthesis x close parenthesis equals the fraction 2 plus x over 8 plus x")[$C ( x ) = frac(2 + x, 8 + x)$] + How many milliliters of glucose should Delbert add to increase the concentration to 50%? #linebreak() Answer: \_\_\_\_\_ ml + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 100 close bracket")[$[ 0 , 100 ]$]. + What is the horizontal asymptote of the graph? #linebreak() Equation of asymptote: \_\_\_\_\_ Use #math.equation(block: false, alt: "x")[$x$] for the independent variable, and use #math.equation(block: false, alt: "C")[$C$] for the dependent variable. + What does the horizontal asymptote tell you about the solution? #linebreak() \_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "4")[$4$] ml + A graph is below. + #math.equation(block: false, alt: "C equals 1")[$C = 1$] + As Delbert adds more glucose to the mixture, its concentration increases toward #math.equation(block: false, alt: "100 %")[$100 upright(%)$]. ] Graph for part (b): #figure(figph[rational function], alt: "rational function", caption: none) Delbert prepares a 20% glucose solution of by mixing 2 mL of glucose with 8 mL of water. If he adds #math.equation(block: false, alt: "x")[$x$] ml of glucose to the solution, its concentration is given by #math.equation(block: true, alt: "C open parenthesis x close parenthesis equals the fraction 2 plus x over 8 plus x")[$C ( x ) = frac(2 + x, 8 + x)$] + How many milliliters of glucose should Delbert add to increase the concentration to 50%? + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 100 close bracket")[$[ 0 , 100 ]$]. + What is the horizontal asymptote of the graph? + What does the horizontal asymptote tell you about the solution? #solutionbox[ + #math.equation(block: false, alt: "4")[$4$] ml + #figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "C equals 1")[$C = 1$] + As Delbert adds more glucose to the mixture, its concentration increases toward #math.equation(block: false, alt: "100 %")[$100 upright(%)$]. ] === Domain of a Rational Function Most applications of rational functions have restricted domains, that is, they make sense for only a subset of the real numbers on the #math.equation(block: false, alt: "x")[$x$]-axis. Consequently, only a portion of the graph is useful for analyzing the application. However, a knowledge of the general shape and properties of the whole graph can be very helpful in understanding a rational function. As we stated earlier, a rational function is a quotient of two polynomials. Some examples of rational functions are shown below. #math.equation(block: true, alt: "f open parenthesis x close parenthesis, equals the fraction 2 over open parenthesis x minus 3 close parenthesis squared, g open parenthesis x close parenthesis, equals the fraction x over x plus 1; h open parenthesis x close parenthesis, equals the fraction 2 x squared over x squared plus 4, k open parenthesis x close parenthesis, equals the fraction x squared minus 1 over x squared minus 9")[$f ( x ) & = frac(2, ( x − 3 )^(2)) & & & g ( x ) & = frac(x, x + 1) \ h ( x ) & = frac(2 x^(2), x^(2) + 4) & & & k ( x ) & = frac(x^(2) − 1, x^(2) − 9)$] Because we cannot divide by zero, a rational function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$" " f ( x ) = display(frac(P ( x ), Q ( x )) " ")$] is undefined for any value #math.equation(block: false, alt: "x equals a")[$x = a$] where #math.equation(block: false, alt: "Q open parenthesis a close parenthesis equals 0")[$Q ( a ) = 0$]. These #math.equation(block: false, alt: "x")[$x$]-values are not in the domain of the function. #examplebox("Example 3")[][ Find the domains of the rational functions #math.equation(block: false, alt: "f")[$f$], #math.equation(block: false, alt: "g")[$g$], #math.equation(block: false, alt: "h")[$h$], and #math.equation(block: false, alt: "k")[$k$] defined above. #solutionbox[ The domain of #math.equation(block: false, alt: "f")[$f$] is the set of all real numbers except #math.equation(block: false, alt: "3")[$3$], because the denominator, #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared")[$( x − 3 )^(2)$], equals #math.equation(block: false, alt: "0")[$0$] when #math.equation(block: false, alt: "x equals 3")[$x = 3$]. The domain of #math.equation(block: false, alt: "g")[$g$] is the set of all real numbers except #math.equation(block: false, alt: "minus 1")[$− 1$], because #math.equation(block: false, alt: "x plus 1")[$x + 1$] equals zero when #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. The denominator of the function #math.equation(block: false, alt: "h")[$h$], #math.equation(block: false, alt: "x squared plus 4")[$x^(2) + 4$], is never equal to zero, so the domain of #math.equation(block: false, alt: "h")[$h$] is all the real numbers. The domain of #math.equation(block: false, alt: "k")[$k$] is the set of all real numbers except #math.equation(block: false, alt: "3")[$3$] and #math.equation(block: false, alt: "minus 3")[$− 3$], because #math.equation(block: false, alt: "x squared minus 9")[$x^(2) − 9$] equals #math.equation(block: false, alt: "0")[$0$] when #math.equation(block: false, alt: "x equals 3")[$x = 3$] or #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$]. ] ] The fraction #math.equation(block: false, alt: "the fraction x squared minus 1 over x squared minus 4")[$display(frac(x^(2) − 1, x^(2) − 4))$] \_\_\_\_\_ #solutionbox[ is equal to zero at #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$] because the numerator (and not the denominator) is zero there. ] The fraction #math.equation(block: false, alt: "the fraction x squared minus 1 over x squared minus 4")[$display(frac(x^(2) − 1, x^(2) − 4))$] + can be reduced to #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$]. + is undefined at #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. + is equivalent to #math.equation(block: false, alt: "the fraction open parenthesis x minus 1 close parenthesis squared over open parenthesis x minus 2 close parenthesis squared")[$display(frac(( x − 1 )^(2), ( x − 2 )^(2)))$]. + is equal to zero at #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ We only need to exclude the zeros of the #emph[denominator] from the domain of a rational function. We do not exclude the zeros of the numerator. In fact, the zeros of the numerator include the zeros of the rational function itself, because a fraction is equal to #math.equation(block: false, alt: "0")[$0$] when its numerator is #math.equation(block: false, alt: "0")[$0$] but its denominator is not #math.equation(block: false, alt: "0")[$0$]. ] + Find the domain of #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals the fraction x minus 2 over x plus 4")[$F ( x ) = display(frac(x − 2, x + 4))$]. #linebreak() \_\_\_\_\_ You may use the inequality symbols (#math.equation(block: false, alt: "less than")[$<$], #math.equation(block: false, alt: "less than or equal to")[$≤$], #math.equation(block: false, alt: "greater than")[$>$], #math.equation(block: false, alt: "greater than or equal to")[$≥$]) or enter your answers using interval notation. When using interval notation, use "inf" for #math.equation(block: false, alt: "∞")[$upright(∞)$] and use "U" (an upper case letter u) for the union symbol #math.equation(block: false, alt: "union")[$∪$]. + Find the zeros of #math.equation(block: false, alt: "F open parenthesis x close parenthesis")[$F ( x )$]. #linebreak() \_\_\_\_\_ Separate different zeros with a comma. #solutionbox[ + #math.equation(block: false, alt: "x not equal to minus 4")[$x ≠ − 4$], which is #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 4 close parenthesis union open parenthesis minus 4 , ∞ close parenthesis")[$( − upright(∞) , − 4 ) ∪ ( − 4 , upright(∞) )$] + #math.equation(block: false, alt: "x equals 2")[$x = 2$] ] + Find the domain of #math.equation(block: false, alt: "F open parenthesis x close parenthesis equals the fraction x minus 2 over x plus 4")[$F ( x ) = display(frac(x − 2, x + 4))$]. + Find the zeros of #math.equation(block: false, alt: "F open parenthesis x close parenthesis")[$F ( x )$]. #solutionbox[ + #math.equation(block: false, alt: "x not equal to minus 4")[$x ≠ − 4$], which is #math.equation(block: false, alt: "open parenthesis minus ∞ , minus 4 close parenthesis union open parenthesis minus 4 , ∞ close parenthesis")[$( − upright(∞) , − 4 ) ∪ ( − 4 , upright(∞) )$] + #math.equation(block: false, alt: "x equals 2")[$x = 2$] ] === Vertical Asymptotes As we saw in Graphing Polynomial Functions, a polynomial function is defined for all values of #math.equation(block: false, alt: "x")[$x$], and its graph is a smooth curve without any breaks or holes. The graph of a rational function, on the other hand, will have breaks or holes at those #math.equation(block: false, alt: "x")[$x$]-values where it is undefined. #examplebox("Example 4")[][ Investigate the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 2 over open parenthesis x minus 3 close parenthesis squared")[$" " f ( x ) = display(frac(2, ( x − 3 )^(2)))$] near #math.equation(block: false, alt: "x equals 3")[$x = 3$]. #solutionbox[ This function is undefined for #math.equation(block: false, alt: "x equals 3")[$x = 3$], so there is no point on the graph with #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "3")[$3$]. However, we can make a table of values for other values of #math.equation(block: false, alt: "x")[$x$]. Plotting the ordered pairs in the table results in the points shown below. #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: "0")[$0$]], [#math.equation(block: false, alt: "the fraction 2 over 9")[$frac(2, 9)$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [undefined], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "the fraction 1 over 2")[$frac(1, 2)$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "the fraction 2 over 9")[$frac(2, 9)$]], )) #figure(figph[graph showing 6 points about a vertical asymptote], alt: "graph showing 6 points about a vertical asymptote", caption: none) Next, we make a table showing #math.equation(block: false, alt: "x")[$x$]-values close to #math.equation(block: false, alt: "3")[$3$], as shown below. As we choose #math.equation(block: false, alt: "x")[$x$]-values closer and closer to #math.equation(block: false, alt: "3")[$3$], #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis squared")[$( x − 3 )^(2)$] gets closer to #math.equation(block: false, alt: "0")[$0$], so the fraction #math.equation(block: false, alt: "the fraction 2 over open parenthesis x minus 3 close parenthesis squared")[$display(frac(2, ( x − 3 )^(2)))$] gets very large. This means that the graph approaches, but never touches, the vertical line #math.equation(block: false, alt: "x equals 3")[$x = 3$]. In other words, the graph has a vertical asymptote at #math.equation(block: false, alt: "x equals 3")[$x = 3$]. We indicate the vertical asymptote by a dashed line, as shown in the figure. #figure(figph[table of values and graph of rational function], alt: "table of values and graph of rational function", caption: none) ] ] In general, we have the following result. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertical Asymptotes] If #math.equation(block: false, alt: "Q open parenthesis a close parenthesis equals 0")[$Q ( a ) = 0$] but #math.equation(block: false, alt: "P open parenthesis a close parenthesis not equal to 0")[$P ( a ) ≠ 0$], then the graph of the rational function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] has a #strong[vertical asymptote] at #math.equation(block: false, alt: "x equals a")[$x = a$]. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If #math.equation(block: false, alt: "P open parenthesis a close parenthesis")[$P ( a )$] and #math.equation(block: false, alt: "Q open parenthesis a close parenthesis")[$Q ( a )$] are both zero, then the graph of the rational function #math.equation(block: false, alt: "the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$display(frac(P ( x ), Q ( x )))$] may have a hole at #math.equation(block: false, alt: "x equals a")[$x = a$] rather than an asymptote. (This possibility is considered in the homework exercises.) ] Find the vertical asymptotes of #math.equation(block: false, alt: "G open parenthesis x close parenthesis equals the fraction 4 x squared over x squared minus 4")[$G ( x ) = display(frac(4 x^(2), x^(2) − 4))$]. Equations of vertical asymptotes: \_\_\_\_\_ Separate different equations with a comma. #solutionbox[ #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] and #math.equation(block: false, alt: "x equals 2")[$x = 2$] ] Find the vertical asymptotes of #math.equation(block: false, alt: "G open parenthesis x close parenthesis equals the fraction 4 x squared over x squared minus 4")[$G ( x ) = display(frac(4 x^(2), x^(2) − 4))$]. #solutionbox[ #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] and #math.equation(block: false, alt: "x equals 2")[$x = 2$] ] If the numerator and denominator of a rational function are both equal to zero at #math.equation(block: false, alt: "x equals a")[$x = a$], then \_\_\_\_\_ #solutionbox[ there is a hole in the graph at #math.equation(block: false, alt: "x equals a")[$x = a$]. ] If the numerator and denominator of a rational function are both equal to zero at #math.equation(block: false, alt: "x equals a")[$x = a$], then + #math.equation(block: false, alt: "f open parenthesis a close parenthesis equals 0")[$f ( a ) = 0$] + #math.equation(block: false, alt: "f open parenthesis a close parenthesis equals 1")[$f ( a ) = 1$] + there is a vertical asymptote at #math.equation(block: false, alt: "x equals a")[$x = a$]. + there is a hole in the graph at #math.equation(block: false, alt: "x equals a")[$x = a$]. Near a vertical asymptote, the graph of a rational function has one of the four characteristic shapes, illustrated below. Locating the vertical asymptotes can help us make a quick sketch of a rational function. A rational function y = (ax + b)/(x − c) whose asymptotes you control. The vertical dashed line x = c is where the denominator is zero: slide c and the break in the graph follows it, with the two branches showing the characteristic shapes this paragraph catalogs. The horizontal dashed line y = a is the horizontal asymptote: numerator and denominator have the same degree, so for large |x| the function levels off at the ratio of leading coefficients — slide a and both the line and the far ends of the graph move together. One special position to find: when b = −ac the “asymptote” disappears entirely, because the numerator shares the factor (x − c) and the graph collapses to a horizontal line with a hole — the exceptional case the section sends to the homework.{"functions":\[{"color":"\#1f4e79","expression":"(a\*x + b)/(x - c)","expression2":"","id":"rational","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","expression":"a + 0\*x","expression2":"","id":"horizontal-asymptote","inequality":"lt","lineDash":\[8,5\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"x","visible":true},{"color":"\#b8442a","expression":"c + 0\*t","expression2":"t","id":"vertical-asymptote","inequality":"lt","lineDash":\[8,5\],"lineWidth":1.5,"mode":"parametric","tMax":8,"tMin":-8,"variable":"x","visible":true}\],"grid":{"majorSpacing":1,"showAxes":true,"showLabels":true},"parameters":\[{"id":"p-a","label":"Leading ratio a","max":2,"min":-2,"name":"a","step":0.1,"value":1},{"id":"p-b","label":"Numerator constant b","max":3,"min":-3,"name":"b","step":0.1,"value":2},{"id":"p-c","label":"Break at x = c","max":3,"min":-3,"name":"c","step":0.1,"value":1}\],"title":"y = (ax + b)/(x \\u2212 c) with its two asymptotes","version":1,"viewport":{"centerX":0,"centerY":0,"scale":46}}#figure(figph[4 cases of behavior near vertical asymptote], alt: "4 cases of behavior near vertical asymptote", caption: none) #examplebox("Example 5")[][ Locate the vertical asymptotes and sketch the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x over x plus 1")[$g ( x ) = display(frac(x, x + 1))$]. #solutionbox[ The denominator, #math.equation(block: false, alt: "x plus 1")[$x + 1$], equals zero when #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. Because the numerator does not equal zero when #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$], there is a vertical asymptote at #math.equation(block: false, alt: "x equals minus 1")[$x = − 1$]. The asymptote separates the graph into two pieces. We can use the #strong[Table] feature of a calculator to evaluate #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] for several values of #math.equation(block: false, alt: "x")[$x$] on either side of the asymptote, as shown in figure (a). We plot the points found in this way; then connect the points on either side of the asymptote to obtain the graph shown in figure (b). #figure(figph[table and graph of rational funciton], alt: "table and graph of rational funciton", caption: none) ] ] + Find the vertical asymptotes of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x squared minus 4")[$f ( x ) = display(frac(1, x^(2) − 4))$] . #linebreak() Equations of vertical asymptotes: \_\_\_\_\_ Separate different equations with a comma. + Locate any #math.equation(block: false, alt: "x")[$x$]-intercepts. #linebreak() \_\_\_\_\_ + Evaluate the function at #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$], #math.equation(block: false, alt: "minus 1")[$− 1$], #math.equation(block: false, alt: "1")[$1$], and #math.equation(block: false, alt: "3")[$3$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "y")[$y$]], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], [\_\_\_\_\_], )) + Sketch a graph of the function. #solutionbox[ + #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] and #math.equation(block: false, alt: "x equals 2")[$x = 2$] + no #math.equation(block: false, alt: "x")[$x$]-intercepts + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "the fraction 1 over 5")[$frac(1, 5)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$]], [#math.equation(block: false, alt: "the fraction 1 over 5")[$frac(1, 5)$]], )) + A graph is below. ] Graph for part (b): #figure(figph[rational function], alt: "rational function", caption: none) + Find the vertical asymptotes of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction 1 over x squared minus 4")[$f ( x ) = display(frac(1, x^(2) − 4))$]. + Locate any #math.equation(block: false, alt: "x")[$x$]-intercepts. + Evaluate the function at #math.equation(block: false, alt: "x equals minus 3")[$x = − 3$], #math.equation(block: false, alt: "minus 1")[$− 1$], #math.equation(block: false, alt: "1")[$1$], and #math.equation(block: false, alt: "3")[$3$].#figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "y")[$y$]], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], )) + Sketch a graph of the function. #solutionbox[ + #math.equation(block: false, alt: "x equals minus 2")[$x = − 2$] and #math.equation(block: false, alt: "x equals 2")[$x = 2$] + There are no #math.equation(block: false, alt: "x")[$x$]-intercepts. + #figure(table( columns: 5, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "minus 3")[$− 3$]], [#math.equation(block: false, alt: "minus 1")[$− 1$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "3")[$3$]]), [#math.equation(block: false, alt: "y")[$y$]], [#math.equation(block: false, alt: "the fraction 1 over 5")[$frac(1, 5)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$]], [#math.equation(block: false, alt: "the fraction minus 1 over 3")[$frac(− 1, 3)$]], [#math.equation(block: false, alt: "the fraction 1 over 5")[$frac(1, 5)$]], )) + #figure(figph[rational function], alt: "rational function", caption: none) ] === Horizontal Asymptotes Look again at the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x over x plus 1")[$g ( x ) = display(frac(x, x + 1))$] in Example. As #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$] gets large—that is, as we move away from the origin along the #math.equation(block: false, alt: "x")[$x$]-axis in either direction—the corresponding #math.equation(block: false, alt: "y")[$y$]-values get closer and closer to #math.equation(block: false, alt: "1")[$1$]. The graph approaches, but never coincides with, the line #math.equation(block: false, alt: "y equals 1")[$y = 1$]. We say that the graph has a #strong[horizontal asymptote] at #math.equation(block: false, alt: "y equals 1")[$y = 1$]. When does a rational function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] have a horizontal asymptote? It depends on the degrees of the two polynomials #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]. The degree of the numerator of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals the fraction x over x plus 1")[$g ( x ) = display(frac(x, x + 1))$] is equal to the degree of the denominator. In other words, the highest power of #math.equation(block: false, alt: "x")[$x$] in the numerator (1, in this case) is the same as the highest power in the denominator. Consider the three rational functions whose graphs are shown below. #figure(figph[3 cases of end behavior], alt: "3 cases of end behavior", caption: none) - The graph of #math.equation(block: false, alt: "r open parenthesis x close parenthesis equals the fraction x plus 1 over x squared")[$r ( x ) = display(frac(x + 1, x^(2)))$] in figure (a) has a horizontal asymptote at #math.equation(block: false, alt: "y equals 0")[$y = 0$], the #math.equation(block: false, alt: "x")[$x$]-axis, because the degree of the denominator is larger than the degree of the numerator. Higher powers of #math.equation(block: false, alt: "x")[$x$] grow much more rapidly than smaller powers. Thus, for large values of #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$], the denominator is much larger in absolute value than the numerator of #math.equation(block: false, alt: "r open parenthesis x close parenthesis")[$r ( x )$], so the function values approach #math.equation(block: false, alt: "0")[$0$]. - The graph of #math.equation(block: false, alt: "s open parenthesis x close parenthesis equals the fraction x squared minus 4 over 2 x squared")[$s ( x ) = display(frac(x^(2) − 4, 2 x^(2)))$] in figure (b) has a horizontal asymptote at #math.equation(block: false, alt: "y equals the fraction 1 over 2")[$y = display(frac(1, 2))$], because the numerator and denominator of the fraction have the same degree. For large values of #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$], the terms of lower degree are negligible compared to the squared terms. As #math.equation(block: false, alt: "x")[$x$] increases, #math.equation(block: false, alt: "s open parenthesis x close parenthesis")[$s ( x )$] is approximately equal to #math.equation(block: false, alt: "the fraction x squared over 2 x squared")[$display(frac(x^(2), 2 x^(2)))$], or #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. Thus, the function values approach a constant value of #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. - The graph of #math.equation(block: false, alt: "t open parenthesis x close parenthesis equals the fraction x squared plus 1 over x minus 1")[$t ( x ) = display(frac(x^(2) + 1, x − 1))$] in figure (c) does not have a horizontal asymptote, because the degree of the numerator is larger than the degree of the denominator. As #math.equation(block: false, alt: "vertical bar x vertical bar")[$| x |$] increases, #math.equation(block: false, alt: "x squared plus 1")[$x^(2) + 1$] grows much faster than #math.equation(block: false, alt: "x minus 1")[$x − 1$], so their ratio does not approach a constant value. The function values increase without bound. We summarize our discussion as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Asymptotes] Suppose #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] is a rational function, where the degree of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] is #math.equation(block: false, alt: "m")[$m$] and the degree of #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] is #math.equation(block: false, alt: "n")[$n$]. + If #math.equation(block: false, alt: "m less than n")[$m < n$], the graph of #math.equation(block: false, alt: "f")[$f$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals 0")[$y = 0$]. + If #math.equation(block: false, alt: "m equals n")[$m = n$], the graph of #math.equation(block: false, alt: "f")[$f$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals the fraction a over b")[$y = display(frac(a, b))$], where #math.equation(block: false, alt: "a")[$a$] is the lead coefficient of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "b")[$b$] is the lead coefficient of #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]. + If #math.equation(block: false, alt: "m greater than n")[$m > n$], the graph of #math.equation(block: false, alt: "f")[$f$] does not have a horizontal asymptote. ] Which of these functions has a horizontal asymptote at #math.equation(block: false, alt: "y equals 2")[$y = 2$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "y equals the fraction 2 x squared plus 4 over x squared plus 1")[$y = display(frac(2 x^(2) + 4, x^(2) + 1))$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals 2")[$y = 2$] because numerator and denominator have the same degree, and the ratio of the lead coefficients is 2. ] Which of these functions has a horizontal asymptote at #math.equation(block: false, alt: "y equals 2")[$y = 2$] ? + #math.equation(block: false, alt: "y equals the fraction 2 x plus 1 over x squared plus 1")[$y = display(frac(2 x + 1, x^(2) + 1))$] + #math.equation(block: false, alt: "y equals the fraction 2 x squared minus 4 over x minus 2")[$y = display(frac(2 x^(2) − 4, x − 2))$] + #math.equation(block: false, alt: "y equals the fraction 2 x squared plus 4 over x squared plus 1")[$y = display(frac(2 x^(2) + 4, x^(2) + 1))$] + #math.equation(block: false, alt: "y equals the fraction 2 x squared plus 4 over x plus 2")[$y = display(frac(2 x^(2) + 4, x + 2))$] #examplebox("Example 6")[][ Locate the horizontal asymptotes and sketch the graph of #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals the fraction 2 x squared over x squared plus 4")[$" " h ( x ) = display(frac(2 x^(2), x^(2) + 4))$]. #solutionbox[ The numerator and denominator of the fraction are both second-degree polynomials, so the graph does have a horizontal asymptote. The lead coefficients of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] are #math.equation(block: false, alt: "2")[$2$] and #math.equation(block: false, alt: "1")[$1$], respectively, so the horizontal asymptote is #math.equation(block: false, alt: "y equals the fraction 2 over 1")[$y = display(frac(2, 1))$], or #math.equation(block: false, alt: "y equals 2")[$y = 2$]. The function #math.equation(block: false, alt: "h")[$h$] does not have a vertical asymptote because the denominator, #math.equation(block: false, alt: "x squared plus 4")[$x^(2) + 4$], is never equal to zero. The #math.equation(block: false, alt: "y")[$y$]-intercept of the graph is the point #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. We can plot several points by evaluating the function at convenient #math.equation(block: false, alt: "x")[$x$]-values, and use the asymptote to help us sketch the graph, as shown below. #figure(figph[rational function with horizontal asymptote at y=2], alt: "rational function with horizontal asymptote at y=2", caption: none) ] ] Locate the horizontal and vertical asymptotes and sketch the graph of #math.equation(block: false, alt: "k open parenthesis x close parenthesis equals the fraction x squared minus 1 over x squared minus 9")[$k ( x ) = display(frac(x^(2) − 1, x^(2) − 9))$]. Label the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts with their coordinates. Equations of asymptotes: \_\_\_\_\_ Use #math.equation(block: false, alt: "x")[$x$] for the independant variable and #math.equation(block: false, alt: "y")[$y$] for the dependant variable. Separate different equations with a comma. #solutionbox[ #math.equation(block: true, alt: "y equals 1")[$y = 1$]; #math.equation(block: true, alt: "x equals minus 3")[$x = − 3$], #math.equation(block: true, alt: "x equals 3")[$x = 3$] A graph is below ] #figure(figph[rational function], alt: "rational function", caption: none) Locate the horizontal and vertical asymptotes and sketch the graph of #math.equation(block: false, alt: "k open parenthesis x close parenthesis equals the fraction x squared minus 1 over x squared minus 9")[$" " k ( x ) = display(frac(x^(2) − 1, x^(2) − 9))$]. Label the #math.equation(block: false, alt: "x")[$x$]- and #math.equation(block: false, alt: "y")[$y$]-intercepts with their coordinates. #solutionbox[ #math.equation(block: true, alt: "y equals 1")[$y = 1$]; #math.equation(block: true, alt: "x equals minus 3")[$x = − 3$], #math.equation(block: true, alt: "x equals 3")[$x = 3$] #figure(figph[rational function], alt: "rational function", caption: none) ] Explain how to reduce an algebraic fraction, and describe some mistakes to avoid. (See Appendix A.9 Working with Algebraic Fractions to review reducing algebraic fractions.) \_\_\_\_\_ Explain how to reduce an algebraic fraction, and describe some mistakes to avoid. (See Appendix A.9 Working with Algebraic Fractions to review reducing algebraic fractions.) === Applications It is often useful to simplify the formula for a rational function before using it. (See Algebra Skills Refresher Working with Algebraic Fractions to review operations on algebraic fractions.) #examplebox("Example 7")[][ When estimating their travel time, pilots must take into account the prevailing winds. A tailwind adds to the plane's ground speed, while a headwind decreases the ground speed. Skyhigh Airlines is setting up a shuttle service from Dallas to Phoenix, a distance of #math.equation(block: false, alt: "800")[$800$] miles. + Express the time needed for a one-way trip, without wind, as a function of the speed of the plane. + Suppose there is a prevailing wind of #math.equation(block: false, alt: "30")[$30$] miles per hour blowing from the west. Write expressions for the flying time from Dallas to Phoenix and from Phoenix to Dallas. + Write an expression for the round-trip flying time, excluding stops, with a #math.equation(block: false, alt: "30")[$30$]-mile-per-hour wind from the west, as a function of the plane's speed. Simplify your expression. #solutionbox[ + Recall that #math.equation(block: false, alt: "time equals the fraction distance over rate")[$"time" = display(frac("distance", "rate"))$]. If we let #math.equation(block: false, alt: "r")[$r$] represent the speed of the plane in still air, then the time required for a one-way trip is #math.equation(block: true, alt: "f open parenthesis r close parenthesis equals the fraction 800 over r")[$f ( r ) = frac(800, r)$] + On the trip from Dallas to Phoenix, the plane encounters a headwind of #math.equation(block: false, alt: "30")[$30$] miles per hour, so its actual ground speed is #math.equation(block: false, alt: "r minus 30")[$r − 30$]. On the return trip, the plane enjoys a tailwind of #math.equation(block: false, alt: "30")[$30$] miles per hour, so its actual ground speed is #math.equation(block: false, alt: "r plus 30")[$r + 30$]. Therefore, the flying times are #math.equation(block: true, alt: "Dallas to Phoenix: the fraction 800 over r minus 30")[$"Dallas to Phoenix:" #hide($b l a n k$) frac(800, r − 30)$] and #math.equation(block: true, alt: "Phoenix to Dallas: the fraction 800 over r plus 30")[$"Phoenix to Dallas:" #hide($b l a n k$) frac(800, r + 30)$] + The round-trip flying time from Dallas to Phoenix and back is #math.equation(block: true, alt: "F open parenthesis r close parenthesis equals the fraction 800 over r minus 30 plus the fraction 800 over r plus 30")[$F ( r ) = frac(800, r − 30) + frac(800, r + 30)$] The LCD for these fractions is #math.equation(block: false, alt: "open parenthesis r minus 30 close parenthesis open parenthesis r plus 30 close parenthesis")[$( r − 30 ) ( r + 30 )$]. Thus, #math.equation(block: true, alt: "the fraction 800 over r minus 30 plus the fraction 800 over r plus 30, equals the fraction 800 open parenthesis r plus 30 close parenthesis over open parenthesis r minus 30 close parenthesis open parenthesis r plus 30 close parenthesis plus the fraction 800 open parenthesis r minus 30 close parenthesis over open parenthesis r plus 30 close parenthesis open parenthesis r minus 30 close parenthesis; equals the fraction open parenthesis 800 r plus 24000 close parenthesis plus open parenthesis 800 r minus 24000 close parenthesis over open parenthesis r plus 30 close parenthesis open parenthesis r minus 30 close parenthesis; equals the fraction 1600 r over r squared minus 900")[$frac(800, r − 30) + frac(800, r + 30) & = frac(800 ( bold(italic(r)) + 30 ), ( r − 30 ) ( bold(italic(r)) + 30 )) + frac(800 ( bold(italic(r)) − 30 ), ( r + 30 ) ( bold(italic(r)) − 30 )) \ & = frac(( 800 r + 24000 ) + ( 800 r − 24000 ), ( r + 30 ) ( r − 30 )) \ & = frac(1600 r, r^(2) − 900)$] (See Algebra Skills Refresher Working with Algebraic Fractions to review adding fractions.) ] ] Navid took his outboard motorboat 20 miles upstream to a fishing site, returning downstream later that day. His boat travels 10 miles per hour in still water. Write an expression for the time Navid spent traveling, as a function of the speed of the current. Let #math.equation(block: false, alt: "x")[$x$] represent the speed of the current in miles per hour. \_\_\_\_\_ hrs Add the time for the upstream trip to the time of the downstream trip. #solutionbox[ #math.equation(block: false, alt: "the fraction 20 over 10 minus x plus the fraction 20 over 10 plus x equals the fraction 400 over 100 minus x squared")[$display(frac(20, 10 − x)) + display(frac(20, 10 + x)) = display(frac(400, 100 − x^(2)))$] hrs ] Navid took his outboard motorboat 20 miles upstream to a fishing site, returning downstream later that day. His boat travels 10 miles per hour in still water. Write an expression for the time Navid spent traveling, as a function of the speed of the current. Let #math.equation(block: false, alt: "x")[$x$] represent the speed of the current in miles per hour. #solutionbox[ #math.equation(block: false, alt: "the fraction 20 over 10 minus x plus the fraction 20 over 10 plus x equals the fraction 400 over 100 minus x squared")[$display(frac(20, 10 − x)) + display(frac(20, 10 + x)) = display(frac(400, 100 − x^(2)))$] hrs ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Rational function - Vertical asymptote - Horizontal asymptote ==== CONCEPTS + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Rational Function] A #strong[rational function] is one of the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = frac(P ( x ), Q ( x ))$]where #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] are polynomials and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] is not the zero polynomial. ] + A rational function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] is undefined for any value #math.equation(block: false, alt: "x equals 1")[$x = 1$] where #math.equation(block: false, alt: "Q open parenthesis a close parenthesis equals 0")[$Q ( a ) = 0$]. These #math.equation(block: false, alt: "x")[$x$]-values are not in the domain of the function. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertical Asymptotes] If #math.equation(block: false, alt: "Q open parenthesis a close parenthesis equals 0")[$Q ( a ) = 0$] but #math.equation(block: false, alt: "P open parenthesis a close parenthesis not equal to 0")[$P ( a ) ≠ 0$], then the graph of the rational function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] has a #strong[vertical asymptote] at #math.equation(block: false, alt: "x equals a")[$x = a$]. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Horizontal Asymptotes] Suppose #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals the fraction P open parenthesis x close parenthesis over Q open parenthesis x close parenthesis")[$f ( x ) = display(frac(P ( x ), Q ( x )))$] is a rational function, where the degree of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] is #math.equation(block: false, alt: "m")[$m$] and the degree of #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] is #math.equation(block: false, alt: "n")[$n$]. + If #math.equation(block: false, alt: "m less than n")[$m < n$], the graph of #math.equation(block: false, alt: "f")[$f$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals 0")[$y = 0$]. + If #math.equation(block: false, alt: "m equals n")[$m = n$], the graph of #math.equation(block: false, alt: "f")[$f$] has a horizontal asymptote at #math.equation(block: false, alt: "y equals the fraction a over b")[$y = display(frac(a, b))$], where #math.equation(block: false, alt: "a")[$a$] is the lead coefficient of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] and #math.equation(block: false, alt: "b")[$b$] is the lead coefficient of #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]. + If #math.equation(block: false, alt: "m greater than n")[$m > n$], the graph of #math.equation(block: false, alt: "f")[$f$] does not have a horizontal asymptote. ] ==== STUDY QUESTIONS + Why does the word #strong[rational] refer to a quotient? + How are the graphs of rational functions different from the graphs of polynomials? + What do the zeros of the numerator of a rational function tell you? What about the zeros of the denominator? + Under what circumstances can the graph of a rational function have a horizontal asymptote? ==== SKILLS Practice each skill in the Homework problems listed. + Find the vertical asymptotes of a rational function: \#13–32 + Find the horizontal asymptotes of a rational function: \#13–32 + Interpret the significance of horizontal and vertical asymptotes in context: \#1–10 + Sketch the graph of a rational function: \#13–36, 51–54 + Write a rational function to model a situation: \#37–42 === Homework 7.4 The eider duck, one of the world's fastest flying birds, can exceed an airspeed of #math.equation(block: false, alt: "65")[$65$] miles per hour. A flock of eider ducks is migrating south at an average airspeed of #math.equation(block: false, alt: "50")[$50$] miles per hour against a moderate headwind. Their next feeding grounds are #math.equation(block: false, alt: "150")[$150$] miles away. + Express the ducks' travel time, #math.equation(block: false, alt: "t")[$t$], as a function of the windspeed, #math.equation(block: false, alt: "v")[$v$]. + Complete the table showing the travel time for various windspeeds.#figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "35")[$35$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "50")[$50$]]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($00$)$], [$#hide($00$)$], [], [], [], [], [], [], [], [], [], )) #linebreak() What happens to the travel time as the headwind increases? + Use the table to choose an appropriate window and graph your function #math.equation(block: false, alt: "t open parenthesis v close parenthesis")[$t ( v )$]. Give the equations of any horizontal or vertical asymptotes. What does the vertical asymptote signify in the context of the problem? + #math.equation(block: false, alt: "t equals the fraction 150 over 50 minus v")[$t = display(frac(150, 50 − v))$] + #figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "35")[$35$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "50")[$50$]]), [#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3.33")[$3.33$]], [#math.equation(block: false, alt: "3.75")[$3.75$]], [#math.equation(block: false, alt: "4.29")[$4.29$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7.5")[$7.5$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "30")[$30$]], [—], )) #linebreak() The travel time increases as the headwind speed increases. + #figure(figph[rational function], alt: "rational function", caption: none) The fastest fish in the sea may be the bluefin tuna, which has been clocked at #math.equation(block: false, alt: "43")[$43$] miles per hour in short sprints. A school of tuna is migrating a distance of #math.equation(block: false, alt: "200")[$200$] miles at an average speed of #math.equation(block: false, alt: "36")[$36$] miles per hour in still water, but they have run into a current flowing against their direction of travel. + Express the tuna's travel time, #math.equation(block: false, alt: "t")[$t$], as a function of the current speed, #math.equation(block: false, alt: "v")[$v$]. + Complete the table showing the travel time for various current speeds.#figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "36")[$36$]]), [#math.equation(block: false, alt: "t")[$t$]], [$#hide($00$)$], [$#hide($00$)$], [$#hide($00$)$], [], [], [], [], [], [], [], )) #linebreak() What happens to the travel time as the current increases? + Use the table to choose an appropriate window and graph your function #math.equation(block: false, alt: "t open parenthesis v close parenthesis")[$t ( v )$]. Give the equations of any horizontal or vertical asymptotes. What does the vertical asymptote signify in the context of the problem? The cost, in thousands of dollars, for immunizing #math.equation(block: false, alt: "p")[$p$] percent of the residents of Emporia against a dangerous new disease is given by the function #math.equation(block: true, alt: "C open parenthesis p close parenthesis equals the fraction 72 p over 100 minus p")[$C ( p ) = frac(72 p, 100 − p)$] + What is the domain of #math.equation(block: false, alt: "C")[$C$]? + Complete the table showing the cost of immunizing various percentages of the population.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "75")[$75$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "C")[$C$]], [$#hide($00$)$], [$#hide($00$)$], [], [], [], [], [], [], [], )) + Graph the function #math.equation(block: false, alt: "C")[$C$]. (Use #math.equation(block: false, alt: "Xmin equals 6")[$"Xmin" = 6$], #math.equation(block: false, alt: "Xmax equals 100")[$"Xmax" = 100$], and appropriate values of #math.equation(block: false, alt: "Ymin")[$"Ymin"$] and #math.equation(block: false, alt: "Ymax .")[$"Ymax" .$]) What percentage of the population can be immunized if the city is able to spend \$#math.equation(block: false, alt: "108 , 000")[$108 , 000$]? + For what values of #math.equation(block: false, alt: "p")[$p$] is the total cost more than \$#math.equation(block: false, alt: "1 , 728 , 000")[$1 , 728 , 000$]? + The graph has a vertical asymptote. What is it? What is its significance in the context of this problem? + #math.equation(block: false, alt: "0 less than or equal to p less than 100")[$0 ≤ p < 100$] + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "75")[$75$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "12.7")[$12.7$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "72")[$72$]], [#math.equation(block: false, alt: "216")[$216$]], [#math.equation(block: false, alt: "288")[$288$]], [#math.equation(block: false, alt: "648")[$648$]], [—], )) + #math.equation(block: false, alt: "60 %")[$60 upright(%)$]#figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "p greater than 96 %")[$p > 96 upright(%)$] + #math.equation(block: false, alt: "p equals 100")[$p = 100$]; As the percentage immunized approaches #math.equation(block: false, alt: "100")[$100$], the cost grows without bound. The cost, in thousands of dollars, for extracting #math.equation(block: false, alt: "p")[$p$] percent of a precious ore from a mine is given by the equation #math.equation(block: true, alt: "C open parenthesis p close parenthesis equals the fraction 360 p over 100 minus p")[$C ( p ) = frac(360 p, 100 − p)$] + What is the domain of #math.equation(block: false, alt: "C")[$C$]? + Complete the table showing the cost of extracting various percentages of the ore.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "15")[$15$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "75")[$75$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "C")[$C$]], [$#hide($00$)$], [$#hide($00$)$], [], [], [], [], [], [], [], )) + Graph the function #math.equation(block: false, alt: "C")[$C$]. (Use #math.equation(block: false, alt: "Xmin equals 6")[$"Xmin" = 6$], #math.equation(block: false, alt: "Xmax equals 100")[$"Xmax" = 100$], and appropriate values of #math.equation(block: false, alt: "Ymin")[$"Ymin"$] and #math.equation(block: false, alt: "Ymax .")[$"Ymax" .$]) What percentage of the ore can be extracted if #math.equation(block: false, alt: "$ 540 , 000")[$\$ 540 , 000$] can be spent on the extraction? + For what values of #math.equation(block: false, alt: "p")[$p$] is the total cost less than #math.equation(block: false, alt: "$ 1 , 440 , 000")[$\$ 1 , 440 , 000$]? + The graph has a vertical asymptote. What is it? What does it tell us about the cost of extracting ore from the mine? The total cost in dollars of producing #math.equation(block: false, alt: "n")[$n$] calculators is approximately #math.equation(block: false, alt: "20 , 000 plus 8 n")[$20 , 000 + 8 n$]. + Express the cost per calculator, #math.equation(block: false, alt: "C")[$C$], as a function of the number #math.equation(block: false, alt: "n")[$n$] of calculators produced. + Complete the table showing the cost per calculator for various production levels.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "2000")[$2000$]], [#math.equation(block: false, alt: "4000")[$4000$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "8000")[$8000$]]), [#math.equation(block: false, alt: "C")[$C$]], [], [], [], [], [], [], [], [], [], )) + Graph the function #math.equation(block: false, alt: "C open parenthesis n close parenthesis")[$C ( n )$] for the cost per calculator. Use the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 9400; Ymin, equals 0, Ymax equals 50")[$"Xmin" & = 0 & & "Xmax" = 9400 \ "Ymin" & = 0 & & "Ymax" = 50$] + How many calculators should be produced so that the cost per calculator is \$#math.equation(block: false, alt: "18")[$18$]? + For what values of #math.equation(block: false, alt: "n")[$n$] is the cost less than \$#math.equation(block: false, alt: "12")[$12$] per calculator? + Find the horizontal asymptote of the graph. What does it tell us about the cost of extracting the precious ore? + #math.equation(block: false, alt: "C equals 8 plus the fraction 20 , 000 over n")[$C = 8 + display(frac(20 "," 000, n))$] + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "n")[$n$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "2000")[$2000$]], [#math.equation(block: false, alt: "4000")[$4000$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "8000")[$8000$]]), [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "208")[$208$]], [#math.equation(block: false, alt: "108")[$108$]], [#math.equation(block: false, alt: "58")[$58$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "13")[$13$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "10.5")[$10.5$]], )) + #figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "2000")[$2000$] + #math.equation(block: false, alt: "n greater than 5000")[$n > 5000$] + #math.equation(block: false, alt: "C equals 8")[$C = 8$]; As #math.equation(block: false, alt: "n")[$n$] increases, the average cost per calculator approaches \$#math.equation(block: false, alt: "8")[$8$]. The number of loaves of Mom’s Bread sold each day is approximated by the demand function #math.equation(block: true, alt: "D open parenthesis p close parenthesis equals the fraction 100 over 1 plus open parenthesis p minus 1.10 close parenthesis to the power 4")[$D ( p ) = frac(100, 1 + ( p − 1.10 )^(4))$] where #math.equation(block: false, alt: "p")[$p$] is the price per loaf in dollars. + Complete the table showing the demand for Mom's Bread at various prices per loaf. Round the values of #math.equation(block: false, alt: "D open parenthesis p close parenthesis")[$D ( p )$] to the nearest whole number.#figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.50")[$0.50$]], [#math.equation(block: false, alt: "1.00")[$1.00$]], [#math.equation(block: false, alt: "1.25")[$1.25$]], [#math.equation(block: false, alt: "1.50")[$1.50$]], [#math.equation(block: false, alt: "1.75")[$1.75$]], [#math.equation(block: false, alt: "2.00")[$2.00$]], [#math.equation(block: false, alt: "2.25")[$2.25$]], [#math.equation(block: false, alt: "2.50")[$2.50$]], [#math.equation(block: false, alt: "2.75")[$2.75$]], [#math.equation(block: false, alt: "3.00")[$3.00$]]), [Demand], [], [], [], [], [], [], [], [], [], [], [], )) + Graph the demand function #math.equation(block: false, alt: "C open parenthesis n close parenthesis")[$C ( n )$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 3.74; Ymin, equals 0, Ymax equals 170")[$"Xmin" & = 0 & & "Xmax" = 3.74 \ "Ymin" & = 0 & & "Ymax" = 170$] What happens to the demand for Mom's Bread as the price increases? + Add a row to your table to show the daily revenue from Mom’s Bread at various prices.#figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "p")[$p$]], [#math.equation(block: false, alt: "0.25")[$0.25$]], [#math.equation(block: false, alt: "0.50")[$0.50$]], [#math.equation(block: false, alt: "1.00")[$1.00$]], [#math.equation(block: false, alt: "1.25")[$1.25$]], [#math.equation(block: false, alt: "1.50")[$1.50$]], [#math.equation(block: false, alt: "1.75")[$1.75$]], [#math.equation(block: false, alt: "2.00")[$2.00$]], [#math.equation(block: false, alt: "2.25")[$2.25$]], [#math.equation(block: false, alt: "2.50")[$2.50$]], [#math.equation(block: false, alt: "2.75")[$2.75$]], [#math.equation(block: false, alt: "3.00")[$3.00$]]), [Demand], [], [], [], [], [], [], [], [], [], [], [], [Revenue], [], [], [], [], [], [], [], [], [], [], [], )) + Using the formula for #math.equation(block: false, alt: "D open parenthesis p close parenthesis")[$D ( p )$], write an expression #math.equation(block: false, alt: "R open parenthesis p close parenthesis")[$R ( p )$] that approximates the total daily revenue as a function of the price, #math.equation(block: false, alt: "p")[$p$]. + Graph the revenue function #math.equation(block: false, alt: "R open parenthesis p close parenthesis")[$R ( p )$] in the same window with #math.equation(block: false, alt: "D open parenthesis p close parenthesis")[$D ( p )$]. Estimate the maximum possible revenue. Does the maximum for #math.equation(block: false, alt: "D open parenthesis p close parenthesis")[$D ( p )$] occur at the same value of #math.equation(block: false, alt: "p")[$p$] as the maximum for #math.equation(block: false, alt: "R open parenthesis p close parenthesis")[$R ( p )$]? + Find the horizontal asymptote of the graphs. What does it represent in this context? A computer store sells approximately #math.equation(block: false, alt: "300")[$300$] of its most popular model per year. The manager would like to minimize her annual inventory cost by ordering the optimal number of computers, #math.equation(block: false, alt: "x")[$x$], at regular intervals. If she orders #math.equation(block: false, alt: "x")[$x$] computers in each shipment, the cost of storage will be #math.equation(block: false, alt: "6 x")[$6 x$] dollars, and the cost of reordering will be #math.equation(block: false, alt: "the fraction 300 over x open parenthesis 15 x plus 10 close parenthesis")[$display(frac(300, x)) ( 15 x + 10 )$] dollars. The inventory cost is the sum of the storage cost and the reordering cost. + Use the distributive law to simplify the expression for the reordering cost. Then express the inventory cost, #math.equation(block: false, alt: "C")[$C$], as a function of #math.equation(block: false, alt: "x")[$x$]. + Complete the table of values for the inventory cost for various reorder sizes.#figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "C")[$C$]], [], [], [], [], [], [], [], [], [], [], )) + Graph the function #math.equation(block: false, alt: "C")[$C$] for the cost per calculator. Use the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 150; Ymin, equals 4500, Ymax equals 5500")[$"Xmin" & = 0 & & "Xmax" = 150 \ "Ymin" & = 4500 & & "Ymax" = 5500$] Estimate the minimum possible value for #math.equation(block: false, alt: "C")[$C$]. + How many computers should the manager order in each shipment so as to minimize the inventory cost? How many orders will she make during the year? + Graph the function #math.equation(block: false, alt: "y equals 6 x plus 4500")[$y = 6 x + 4500$] in the same window with the function #math.equation(block: false, alt: "C")[$C$]. What do you observe? + #math.equation(block: false, alt: "4500 plus the fraction 3000 over x")[$4500 + display(frac(3000, x))$]; #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 6 x plus 4500 plus the fraction 3000 over x")[$C ( x ) = 6 x + 4500 + display(frac(3000, x))$] + #figure(table( columns: 11, align: left, inset: 6pt, [#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "C")[$C$]], [#math.equation(block: false, alt: "4860")[$4860$]], [#math.equation(block: false, alt: "4770")[$4770$]], [#math.equation(block: false, alt: "4780")[$4780$]], [#math.equation(block: false, alt: "4815")[$4815$]], [#math.equation(block: false, alt: "4860")[$4860$]], [#math.equation(block: false, alt: "4910")[$4910$]], [#math.equation(block: false, alt: "5018")[$5018$]], [#math.equation(block: false, alt: "5073")[$5073$]], [#math.equation(block: false, alt: "5130")[$5130$]], [], )) + \$#math.equation(block: false, alt: "4768.33")[$4768.33$]#figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "22")[$22$]; #math.equation(block: false, alt: "14")[$14$] + #figure(figph[rational function with slant asymptote], alt: "rational function with slant asymptote", caption: none) #linebreak() The graph of #math.equation(block: false, alt: "C")[$C$] approaches the line as an asymptote. A chain of electronics stores sells approximately #math.equation(block: false, alt: "500")[$500$] cellphones every year. The owner would like to minimize his annual inventory cost by ordering the optimal number of phones, #math.equation(block: false, alt: "x")[$x$], at regular intervals. The cost of storing the phones will then be #math.equation(block: false, alt: "2 x")[$2 x$] dollars, and the cost of reordering will be #math.equation(block: false, alt: "the fraction 500 over x open parenthesis 4 x plus 10 close parenthesis")[$display(frac(500, x)) ( 4 x + 10 )$]. The total annual inventory cost is the sum of the storage cost and the reordering cost. + Use the distributive law to simplify the expression for the reordering cost. Then express the inventory cost, #math.equation(block: false, alt: "C")[$C$], as a function of #math.equation(block: false, alt: "x")[$x$]. + Complete the table of values for the inventory cost for various reorder sizes.#figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "90")[$90$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "C")[$C$]], [], [], [], [], [], [], [], [], [], [], )) + Graph the function #math.equation(block: false, alt: "C")[$C$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 150; Ymin, equals 2000, Ymax equals 2500")[$"Xmin" & = 0 & & "Xmax" = 150 \ "Ymin" & = 2000 & & "Ymax" = 2500$] Estimate the minimum possible value for #math.equation(block: false, alt: "C")[$C$]. + How many cellphones should the manager order in each shipment so as to minimize the inventory cost? How many orders will he make during the year? + Graph the function #math.equation(block: false, alt: "y equals 2 x plus 2000")[$y = 2 x + 2000$] in the same window with the function #math.equation(block: false, alt: "C")[$C$]. What do you observe? Francine wants to make a rectangular box. In order to simplify construction and keep her costs down, she plans for the box to have a square base and a total surface area of #math.equation(block: false, alt: "96")[$96$] square centimeters. She would like to know the largest volume that such a box can have. + If the square base has length #math.equation(block: false, alt: "x")[$x$] centimeters, show that the height of the box is #math.equation(block: false, alt: "h equals the fraction 24 over x minus the fraction x over 2")[$h = display(frac(24, x)) − display(frac(x, 2))$] centimeters. (#emph[Hint]: The surface area of the box is the sum of the areas of the six sides of the box.) + Write an expression for the volume, #math.equation(block: false, alt: "V")[$V$], of the box as a function of the length, #math.equation(block: false, alt: "x")[$x$], of its base. + Complete the table showing the heights and volumes of the box for various base lengths.#figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]]), [#math.equation(block: false, alt: "h")[$h$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "V")[$V$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) #linebreak() Explain why the values of #math.equation(block: false, alt: "h")[$h$] and #math.equation(block: false, alt: "V")[$V$] are negative when #math.equation(block: false, alt: "x equals 7")[$x = 7$]. + Graph your expression for volume #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$] in an appropriate window. Approximate the maximum possible volume for a box of surface area #math.equation(block: false, alt: "96")[$96$] square centimeters. + What value of #math.equation(block: false, alt: "x")[$x$] gives the maximum volume? + Graph the height, #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$], in the same window with #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$]. What is the height of the box with greatest volume? (Find the height directly from your graph and verify by using the formula given for #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$].) + The surface area is #math.equation(block: false, alt: "2 x squared plus 4 x h equals 96")[$2 x^(2) + 4 x h = 96$]. Solving for #math.equation(block: false, alt: "h")[$h$], #math.equation(block: false, alt: "h equals the fraction 96 minus 2 x squared over 4 x equals the fraction 24 over x minus the fraction x over 2")[$h = display(frac(96 − 2 x^(2), 4 x)) = display(frac(24, x)) − display(frac(x, 2))$]. + #math.equation(block: false, alt: "V equals 24 x minus the fraction 1 over 2 x cubed")[$V = 24 x − display(frac(1, 2)) x^(3)$] + #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]]), [#math.equation(block: false, alt: "h")[$h$]], [#math.equation(block: false, alt: "23.5")[$23.5$]], [#math.equation(block: false, alt: "11")[$11$]], [#math.equation(block: false, alt: "6.5")[$6.5$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "2.3")[$2.3$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "minus 0.07")[$− 0.07$]], [#math.equation(block: false, alt: "V")[$V$]], [#math.equation(block: false, alt: "23.5")[$23.5$]], [#math.equation(block: false, alt: "44")[$44$]], [#math.equation(block: false, alt: "58.5")[$58.5$]], [#math.equation(block: false, alt: "64")[$64$]], [#math.equation(block: false, alt: "57.5")[$57.5$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "minus 3.5")[$− 3.5$]], )) #linebreak() If the base is more than #math.equation(block: false, alt: "7")[$7$] cm, the top and bottom alone exceed the total area allowed. + #figure(figph[cubic function], alt: "cubic function", caption: none) #linebreak() Maximum of #math.equation(block: false, alt: "64")[$64$] cu. cm + #math.equation(block: false, alt: "4")[$4$] cm + #math.equation(block: false, alt: "h equals 4")[$h = 4$] cm#figure(figph[cubic and rational], alt: "cubic and rational", caption: none) Delbert wants to make a box with a square base and a volume of #math.equation(block: false, alt: "64")[$64$] cubic centimeters. He would like to know the smallest surface area that such a box can have. + If the square base has length #math.equation(block: false, alt: "x")[$x$] centimeters, show that the height of the box is #math.equation(block: false, alt: "h equals the fraction 64 over x squared")[$h = display(frac(64, x^(2)))$] centimeters. + Write an expression for the surfacd area, #math.equation(block: false, alt: "S")[$S$], of the box as a function of the length, #math.equation(block: false, alt: "x")[$x$], of its base. (#emph[Hint]: The surface area of the box is the sum of the areas of the six sides of the box.) + Complete the table showing the heights and surface areas of the box for various base lengths.#figure(table( columns: 9, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "8")[$8$]]), [#math.equation(block: false, alt: "h")[$h$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [#math.equation(block: false, alt: "S")[$S$]], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], )) + Graph your expression for surface area #math.equation(block: false, alt: "S open parenthesis x close parenthesis")[$S ( x )$] in an appropriate window. Approximate the minimum possible surface area for Delbert's box. + What value of #math.equation(block: false, alt: "x")[$x$] gives the minimum surface area? + Graph the height, #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$], in the same window with #math.equation(block: false, alt: "S open parenthesis x close parenthesis")[$S ( x )$]. What is the height of the box with smallest surface area? (Find the height directly from your graph and verify by using the formula given for #math.equation(block: false, alt: "h open parenthesis x close parenthesis")[$h ( x )$].) A train whistle sounds higher when the train is approaching you than when it is moving away from you. This phenomenon is known as the Doppler effect. If the actual pitch of the whistle is #math.equation(block: false, alt: "440")[$440$] hertz (this is the A note below middle C), then the note you hear will have the pitch #math.equation(block: true, alt: "P open parenthesis v close parenthesis equals the fraction 440 open parenthesis 332 close parenthesis over 332 minus v")[$P ( v ) = frac(440 ( 332 ), 332 − v)$] where the velocity, #math.equation(block: false, alt: "v")[$v$], in meters per second is positive as the train approaches and is negative when the train is moving away. (The number #math.equation(block: false, alt: "332")[$332$] that appears in this expression is the speed of sound in meters per second.) + Complete the table of values showing the pitch of the whistle at various train velocities.#figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "minus 100")[$− 100$]], [#math.equation(block: false, alt: "minus 75")[$− 75$]], [#math.equation(block: false, alt: "minus 50")[$− 50$]], [#math.equation(block: false, alt: "minus 25")[$− 25$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "75")[$75$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "P")[$P$]], [], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [], )) + Graph the function #math.equation(block: false, alt: "P")[$P$]. (Use the window #math.equation(block: false, alt: "Xmin equals minus 94")[$"Xmin" = − 94$], #math.equation(block: false, alt: "Xmax equals 94")[$"Xmax" = 94$], and appropriate values of #math.equation(block: false, alt: "Ymin")[$"Ymin"$] and #math.equation(block: false, alt: "Ymax .")[$"Ymax" .$]) + What is the velocity of the train if the note you hear has a pitch of #math.equation(block: false, alt: "415")[$415$] hertz (corresponding to the note A-flat)? A pitch of #math.equation(block: false, alt: "553.3 bar")[$553. limits(3)^(―)$] hertz (C-sharp)? + For what velocities will the pitch you hear be greater than #math.equation(block: false, alt: "456.5")[$456.5$] hertz? + The graph has a vertical asymptote (although it is not visible in the suggested window). Where is it and what is its significance in this context? + #figure(table( columns: 10, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "minus 100")[$− 100$]], [#math.equation(block: false, alt: "minus 75")[$− 75$]], [#math.equation(block: false, alt: "minus 50")[$− 50$]], [#math.equation(block: false, alt: "minus 25")[$− 25$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "25")[$25$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "75")[$75$]], [#math.equation(block: false, alt: "100")[$100$]]), [#math.equation(block: false, alt: "P")[$P$]], [#math.equation(block: false, alt: "338.15")[$338.15$]], [#math.equation(block: false, alt: "358.92")[$358.92$]], [#math.equation(block: false, alt: "382.41")[$382.41$]], [#math.equation(block: false, alt: "409.19")[$409.19$]], [#math.equation(block: false, alt: "440")[$440$]], [#math.equation(block: false, alt: "475.83")[$475.83$]], [#math.equation(block: false, alt: "518.01")[$518.01$]], [#math.equation(block: false, alt: "568.4")[$568.4$]], [#math.equation(block: false, alt: "629.66")[$629.66$]], )) + #figure(figph[rational function], alt: "rational function", caption: none) + #math.equation(block: false, alt: "minus 20")[$− 20$] m/sec; #math.equation(block: false, alt: "68")[$68$] m/sec + #math.equation(block: false, alt: "v greater than 12")[$v > 12$] m/sec + #math.equation(block: false, alt: "v equals 332")[$v = 332$]; As #math.equation(block: false, alt: "v")[$v$] approaches #math.equation(block: false, alt: "332")[$332$] m per sec, the pitch increases without bound. The maximum altitude (in meters) attained by a projectile shot from the surface of the Earth is #math.equation(block: true, alt: "h open parenthesis v close parenthesis equals the fraction 6.4 times 10 to the power 6 v squared over 19.6 times 6.4 times 10 to the power 6 minus v squared")[$h ( v ) = frac(6.4 × 10^(6) v^(2), 19.6 ⋅ 6.4 × 10^(6) − v^(2))$] where #math.equation(block: false, alt: "v")[$v$] is the speed (in meters per second) at which the projectile was launched. (The radius of the Earth is #math.equation(block: false, alt: "6.4 times 10 to the power 6")[$6.4 × 10^(6)$] meters, and the constant #math.equation(block: false, alt: "19.6")[$19.6$] is related to the Earth’s gravitational constant.) + Complete the table of values showing the maximum altitude for various launch velocities.#figure(table( columns: 11, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "200")[$200$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "400")[$400$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "600")[$600$]], [#math.equation(block: false, alt: "700")[$700$]], [#math.equation(block: false, alt: "800")[$800$]], [#math.equation(block: false, alt: "900")[$900$]], [#math.equation(block: false, alt: "1000")[$1000$]]), [#math.equation(block: false, alt: "h")[$h$]], [], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [$#hide($000$)$], [], [], )) + Graph the function #math.equation(block: false, alt: "h")[$h$]. (Use the window #math.equation(block: false, alt: "Xmin equals 0")[$"Xmin" = 0$], #math.equation(block: false, alt: "Xmax equals 940")[$"Xmax" = 940$], and appropriate values of #math.equation(block: false, alt: "Ymin")[$"Ymin"$] and #math.equation(block: false, alt: "Ymax .")[$"Ymax" .$]) + Approximately what speed is needed to attain an altitude of #math.equation(block: false, alt: "4000")[$4000$] meters? An altitude of #math.equation(block: false, alt: "16")[$16$] kilometers? + For what velocities will the projectile attain an altitude exceeding #math.equation(block: false, alt: "32")[$32$] kilometers? + The graph has a vertical asymptote (although it is not visible in the suggested window). Where is it and what is its significance in this context? For Problems 13–30, + Sketch the horizontal and vertical asymptotes for each function. + Use the asymptotes to help you sketch the rest of the graph. #math.equation(block: true, alt: "y equals the fraction 1 over x plus 3")[$y = display(frac(1, x + 3))$] #figure(figph[translate of reciprocal], alt: "translate of reciprocal", caption: none) #math.equation(block: true, alt: "y equals the fraction 1 over x minus 3")[$y = display(frac(1, x − 3))$] #math.equation(block: true, alt: "y equals the fraction 2 over x squared minus 5 x plus 4")[$y = display(frac(2, x^(2) − 5 x + 4))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction 4 over x squared minus x minus 6")[$y = display(frac(4, x^(2) − x − 6))$] #math.equation(block: true, alt: "y equals the fraction x over x plus 3")[$y = display(frac(x, x + 3))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction x over x minus 2")[$y = display(frac(x, x − 2))$] #math.equation(block: true, alt: "y equals the fraction x plus 1 over x plus 2")[$y = display(frac(x + 1, x + 2))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction x minus 1 over x minus 3")[$y = display(frac(x − 1, x − 3))$] #math.equation(block: true, alt: "y equals the fraction 2 x over x squared minus 4")[$y = display(frac(2 x, x^(2) − 4))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction x over x squared minus 9")[$y = display(frac(x, x^(2) − 9))$] #math.equation(block: true, alt: "y equals the fraction x minus 2 over x squared plus 5 x plus 4")[$y = display(frac(x − 2, x^(2) + 5 x + 4))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction x plus 1 over x squared minus x minus 6")[$y = display(frac(x + 1, x^(2) − x − 6))$] #math.equation(block: true, alt: "y equals the fraction x squared minus 1 over x squared minus 4")[$y = display(frac(x^(2) − 1, x^(2) − 4))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction 2 x squared over x squared minus 1")[$y = display(frac(2 x^(2), x^(2) − 1))$] #math.equation(block: true, alt: "y equals the fraction x plus 1 over open parenthesis x minus 1 close parenthesis squared")[$y = display(frac(x + 1, ( x − 1 )^(2)))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction 2 open parenthesis x squared minus 1 close parenthesis over x squared plus 4")[$y = display(frac(2 ( x^(2) − 1 ), x^(2) + 4))$] #math.equation(block: true, alt: "y equals the fraction x over x squared plus 3")[$y = display(frac(x, x^(2) + 3))$] #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction x squared plus 2 over x squared plus 4")[$y = display(frac(x^(2) + 2, x^(2) + 4))$] Graph the curve known as Newton's Serpentine: #math.equation(block: false, alt: "y equals the fraction 4 x over x squared plus 1")[$y = display(frac(4 x, x^(2) + 1))$]. #figure(figph[Newtons Serpentine], alt: "Newtons Serpentine", caption: none) Graph the curve known as the Witch of Agnesi: #math.equation(block: false, alt: "y equals the fraction 8 over x squared plus 4")[$y = display(frac(8, x^(2) + 4))$]. For Problems 33–38, + Use polynomial division to write the fraction in the form #math.equation(block: false, alt: "y equals the fraction k over p open parenthesis x close parenthesis plus c")[$y = display(frac(k, p ( x ))) + c$], where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "c")[$c$] are constants. + Use transformations to sketch the graph. #math.equation(block: true, alt: "y equals the fraction 2 x plus 2 over x")[$y = display(frac(2 x + 2, x))$] + #math.equation(block: false, alt: "y equals the fraction 2 over x plus 2")[$y = display(frac(2, x)) + 2$] + #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction 4 x squared plus 3 over x squared")[$y = display(frac(4 x^(2) + 3, x^(2)))$] #math.equation(block: true, alt: "y equals the fraction x plus 2 over x plus 1")[$y = display(frac(x + 2, x + 1))$] + #math.equation(block: false, alt: "y equals the fraction 1 over x plus 1 plus 1")[$y = display(frac(1, x + 1)) + 1$] + #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction 7 minus 2 x over x minus 3")[$y = display(frac(7 − 2 x, x − 3))$] #math.equation(block: true, alt: "y equals the fraction 3 x squared minus 12 x plus 13 over open parenthesis x minus 2 close parenthesis squared")[$y = display(frac(3 x^(2) − 12 x + 13, ( x − 2 )^(2)))$] + #math.equation(block: false, alt: "y equals the fraction 1 over open parenthesis x minus 2 close parenthesis squared plus 3")[$y = display(frac(1, ( x − 2 )^(2))) + 3$] + #figure(figph[rational], alt: "rational", caption: none) #math.equation(block: true, alt: "y equals the fraction minus 4 x squared plus 8 x minus 3 over open parenthesis x minus 1 close parenthesis squared")[$y = display(frac(− 4 x^(2) + 8 x − 3, ( x − 1 )^(2)))$] Problems 39–46 involve operations on algebraic fractions. To review operations on algebraic fractions, see Algebra Skills Refresher Working with Algebraic Fractions. River Queen Tours offers a #math.equation(block: false, alt: "50")[$50$]-mile round-trip excursion on the Mississippi River on a paddle wheel boat. The current in the Mississippi is #math.equation(block: false, alt: "8")[$8$] miles per hour. + Express the time required for the downstream journey as a function of the speed of the paddle wheel boat in still water. + Write a function for the time required for the return trip upstream. + Write and simplify an expression for the time needed for the round trip as a function of the boat's speed. + #math.equation(block: false, alt: "the fraction 25 over s plus 8")[$display(frac(25, s + 8))$] + #math.equation(block: false, alt: "the fraction 25 over s minus 8")[$display(frac(25, s − 8))$] + #math.equation(block: false, alt: "the fraction 50 s over s squared minus 64")[$display(frac(50 s, s^(2) − 64))$] A rowing team can maintain a speed of #math.equation(block: false, alt: "15")[$15$] miles per hour in still water. The team's daily training session includes a #math.equation(block: false, alt: "5")[$5$]-mile run up the Red Cedar River and the return downstream. + Express the team's time on the upstream leg as a function of the speed of the current. + Write a function for the team's time on the downstream leg. + Write and simplify an expression for the total time for the training run as a function of the current's speed. Two pilots for the Flying Express parcel service receive packages simultaneously. Orville leaves Boston for Chicago at the same time Wilbur leaves Chicago for Boston. Each selects an airspeed of #math.equation(block: false, alt: "400")[$400$] miles per hour for the #math.equation(block: false, alt: "900")[$900$]-mile trip. The prevailing winds blow from east to west. + Express Orville's flying time as a function of the windspeed. + Write a function for Wilbur's flying time. + Who reaches his destination first? By how much time (in terms of windspeed)? + #math.equation(block: false, alt: "the fraction 900 over 400 plus w")[$display(frac(900, 400 + w))$] + #math.equation(block: false, alt: "the fraction 900 over 400 minus w")[$display(frac(900, 400 − w))$] + Orville by #math.equation(block: false, alt: "the fraction 1800 w over 160 , 000 minus w squared")[$display(frac(1800 w, 160 "," 000 − w^(2)))$] hours On New Year's Day, a blimp leaves its berth in Carson, California, and heads north for the Rose Bowl, #math.equation(block: false, alt: "23")[$23$] miles away. There is a breeze from the north at #math.equation(block: false, alt: "6")[$6$] miles per hour. + Express the time required for the trip as a function of the blimp's airspeed. + Write a function for the time needed for the return trip. + Which trip takes longer? By how much time (in terms of the blimp's airspeed)? The focal length of a lens is given by the formula #math.equation(block: true, alt: "the fraction 1 over f equals the fraction 1 over p plus the fraction 1 over q")[$frac(1, f) = frac(1, p) + frac(1, q)$] where #math.equation(block: false, alt: "f")[$f$] stands for the focal length, #math.equation(block: false, alt: "p")[$p$] is the distance from the object viewed to the lens, and #math.equation(block: false, alt: "q")[$q$] is the distance from the image to the lens. Suppose you estimate that the distance from your cat (the object viewed) to your camera lens is #math.equation(block: false, alt: "60")[$60$] inches greater than the distance from the lens to the film inside the camera, where the image forms. + Express #math.equation(block: false, alt: "1 / f")[$1 / f$] as a single fraction in terms of #math.equation(block: false, alt: "q")[$q$]. + Write an expression for #math.equation(block: false, alt: "f")[$f$] as a function of #math.equation(block: false, alt: "q")[$q$]. + #math.equation(block: false, alt: "the fraction 1 over f equals the fraction 2 q plus 60 over q squared plus 60 q")[$display(frac(1, f)) = display(frac(2 q + 60, q^(2) + 60 q))$] + #math.equation(block: false, alt: "f equals the fraction q squared plus 60 q over 2 q plus 60")[$f = display(frac(q^(2) + 60 q, 2 q + 60))$] If two resistors, #math.equation(block: false, alt: "R sub 1")[$R_(1)$] and #math.equation(block: false, alt: "R sub 2")[$R_(2)$], in an electrical circuit are connected in parallel, the total resistance #math.equation(block: false, alt: "R")[$R$] in the circuit is given by #math.equation(block: true, alt: "the fraction 1 over R equals the fraction 1 over R sub 1 plus the fraction 1 over R sub 2")[$frac(1, R) = frac(1, R_(1)) + frac(1, R_(2))$] + Suppose that the second resistor, #math.equation(block: false, alt: "R sub 2")[$R_(2)$], is #math.equation(block: false, alt: "10")[$10$] ohms greater than the first. Express #math.equation(block: false, alt: "1 / R")[$1 / R$] as a single fraction in terms of #math.equation(block: false, alt: "R sub 1")[$R_(1)$]. + Write an expression for #math.equation(block: false, alt: "R")[$R$] as a function of #math.equation(block: false, alt: "R sub 1")[$R_(1)$]. + Show that the equation #math.equation(block: false, alt: "the fraction 1 over y minus the fraction 1 over x equals the fraction 1 over k")[$display(frac(1, y)) − display(frac(1, x)) = display(frac(1, k))$] is equivalent to #math.equation(block: false, alt: "y equals the fraction k x over x plus k")[$y = display(frac(k x, x + k))$] on their common domain. + Graph the functions #math.equation(block: false, alt: "y equals the fraction k x over x plus k")[$y = display(frac(k x, x + k))$] for #math.equation(block: false, alt: "k equals 1 , 2 ,")[$k = 1 , 2 ,$] and #math.equation(block: false, alt: "3")[$3$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 30; Ymin, equals 0, Ymax equals 4")[$"Xmin" & = 0 & & "Xmax" = 30 \ "Ymin" & = 0 & & "Ymax" = 4$] Describe the graphs. + #math.equation(block: false, alt: "the fraction 1 over y equals the fraction 1 over x plus the fraction 1 over k equals the fraction k plus x over x k")[$display(frac(1, y)) = display(frac(1, x)) + display(frac(1, k)) = display(frac(k + x, x k))$], so by taking reciprocals, #math.equation(block: false, alt: "y equals the fraction k x over x plus k")[$y = display(frac(k x, x + k))$]. + #figure(figph[three rational functions], alt: "three rational functions", caption: none) #linebreak() The graphs increase from the origin and approach a horizontal asymptote at #math.equation(block: false, alt: "y equals k")[$y = k$]. Consider the graph of #math.equation(block: false, alt: "y equals the fraction a x over x plus k")[$y = display(frac(a x, x + k))$], where #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "k")[$k$] are positive constants. + What is the horizontal asymptote of the graph? + Show that for #math.equation(block: false, alt: "x equals k")[$x = k$], #math.equation(block: false, alt: "y equals the fraction a over 2")[$y = display(frac(a, 2))$]. + Sketch the graph of #math.equation(block: false, alt: "y equals the fraction a x over x plus k")[$y = display(frac(a x, x + k))$]for #math.equation(block: false, alt: "a equals 4")[$a = 4$] and #math.equation(block: false, alt: "k equals 10")[$k = 10$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 60; Ymin, equals 0, Ymax equals 5")[$"Xmin" & = 0 & & "Xmax" = 60 \ "Ymin" & = 0 & & "Ymax" = 5$] Illustrate your answers to parts (a) and (b) on the graph. For Problems 47–48, + Use your answers to Problem 46 to find equations of the form #math.equation(block: false, alt: "y equals the fraction a x over x plus k")[$y = display(frac(a x, x + k))$] for the graphs shown. + Check your answer with a graphing calculator. #figure(figph[horizontal asymptote], alt: "horizontal asymptote", caption: none) #math.equation(block: true, alt: "the fraction 12 x over x plus 20")[$display(frac(12 x, x + 20))$] #figure(figph[horizontal asymptote], alt: "horizontal asymptote", caption: none) The Michaelis-Menten equation is the rate equation for chemical reactions catalyzed by enzymes. The speed of the reaction #math.equation(block: false, alt: "v")[$v$] is a function of the initial concentration of the reactant #math.equation(block: false, alt: "s")[$s$] and is given by #math.equation(block: true, alt: "v equals f open parenthesis s close parenthesis equals the fraction V s over s plus K")[$v = f ( s ) = frac(V s, s + K)$] where #math.equation(block: false, alt: "V")[$V$] is the maximum possible reaction rate and #math.equation(block: false, alt: "K")[$K$] is called the Michaelis constant. (Source: Holme and Peck, 1993) + What value does #math.equation(block: false, alt: "v")[$v$] approach as #math.equation(block: false, alt: "s")[$s$] increases? + What is the value of #math.equation(block: false, alt: "v")[$v$] when #math.equation(block: false, alt: "s equals K")[$s = K$]? + The table gives data from reactions of the enzyme D-amino acid oxidase.#figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "s")[$s$]], [#math.equation(block: false, alt: "0.33")[$0.33$]], [#math.equation(block: false, alt: "0.66")[$0.66$]], [#math.equation(block: false, alt: "1.00")[$1.00$]], [#math.equation(block: false, alt: "1.66")[$1.66$]], [#math.equation(block: false, alt: "2.50")[$2.50$]], [#math.equation(block: false, alt: "3.33")[$3.33$]], [#math.equation(block: false, alt: "6.66")[$6.66$]]), [#math.equation(block: false, alt: "v")[$v$]], [#math.equation(block: false, alt: "0.08")[$0.08$]], [#math.equation(block: false, alt: "0.14")[$0.14$]], [#math.equation(block: false, alt: "0.20")[$0.20$]], [#math.equation(block: false, alt: "0.30")[$0.30$]], [#math.equation(block: false, alt: "0.39")[$0.39$]], [#math.equation(block: false, alt: "0.46")[$0.46$]], [#math.equation(block: false, alt: "0.58")[$0.58$]], )) #linebreak() Plot the data and estimate the values of #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "K")[$K$] from your graph. + Graph the function #math.equation(block: false, alt: "v equals the fraction 0.88 s over s plus 3.34")[$v = display(frac(0.88 s, s + 3.34))$] on top of your data points. + #math.equation(block: false, alt: "V")[$V$] + #math.equation(block: false, alt: "the fraction V over 2")[$display(frac(V, 2))$] + #figure(figph[rational on data points], alt: "rational on data points", caption: none) #linebreak() #math.equation(block: false, alt: "V approximately equals 0.7 , K approximately equals 2.2")[$V ≈ 0.7 , " " K ≈ 2.2$] (many answers are possible) + (See figure.) Show that #math.equation(block: true, alt: "the fraction 1 over v equals the fraction 1 over V plus the fraction K over V s")[$frac(1, v) = frac(1, V) + frac(K, V s)$] is another form of the Michaelis-Menten equation. (See Problem 49.) + Refer to the Michaelis-Menten equation in Problem 49. Solve for #math.equation(block: false, alt: "the fraction 1 over v")[$display(frac(1, v))$], then write your new equation in the form #math.equation(block: false, alt: "the fraction 1 over v equals a times the fraction 1 over s plus b")[$display(frac(1, v)) = a ⋅ display(frac(1, s)) + b$]. Express #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] in terms of #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "K")[$K$]. + Use the data from part (c) of Problem 49 to make a table of values for #math.equation(block: false, alt: "open parenthesis the fraction 1 over s , the fraction 1 over v close parenthesis")[$( display(frac(1, s)) , display(frac(1, v)) )$]. + Plot the points #math.equation(block: false, alt: "open parenthesis the fraction 1 over s , the fraction 1 over v close parenthesis")[$( display(frac(1, s)) , display(frac(1, v)) )$], then use linear regression to find the line of best fit. + Use your values for #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] to solve for #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "K")[$K$]. + #math.equation(block: false, alt: "the fraction 1 over v equals the fraction K over V times the fraction 1 over s plus the fraction 1 over V")[$display(frac(1, v)) = display(frac(K, V)) ⋅ display(frac(1, s)) + display(frac(1, V))$]; Therefore, #math.equation(block: false, alt: "a equals the fraction K over V")[$a = display(frac(K, V))$] and #math.equation(block: false, alt: "b equals the fraction 1 over V")[$b = display(frac(1, V))$] + #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "the fraction 1 over s")[$frac(1, s)$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "0.6")[$0.6$]], [#math.equation(block: false, alt: "0.4")[$0.4$]], [#math.equation(block: false, alt: "0.3")[$0.3$]], [#math.equation(block: false, alt: "0.15")[$0.15$]]), [#math.equation(block: false, alt: "the fraction 1 over v")[$frac(1, v)$]], [#math.equation(block: false, alt: "12.5")[$12.5$]], [#math.equation(block: false, alt: "7.1")[$7.1$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "3.3")[$3.3$]], [#math.equation(block: false, alt: "2.6")[$2.6$]], [#math.equation(block: false, alt: "2.2")[$2.2$]], [#math.equation(block: false, alt: "1.7")[$1.7$]], )) + #figure(figph[linear fit on data points], alt: "linear fit on data points", caption: none) #linebreak() #math.equation(block: false, alt: "the fraction 1 over v equals 3.8 times the fraction 1 over s plus 1.1")[$display(frac(1, v)) = 3.8 ⋅ display(frac(1, s)) + 1.1$] + #math.equation(block: false, alt: "V approximately equals 0.89 , K approximately equals 3.37")[$V ≈ 0.89 , " " K ≈ 3.37$] + Refer to the Michaelis-Menten equation in Problem 49. Write an equation for #math.equation(block: false, alt: "the fraction s over v")[$display(frac(s, v))$] in the form #math.equation(block: false, alt: "the fraction s over v equals c s plus d")[$display(frac(s, v)) = c s + d$]. Express #math.equation(block: false, alt: "c")[$c$] and #math.equation(block: false, alt: "d")[$d$] in terms of #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "K")[$K$]. + Use the data from part (c) of Problem 49 to make a table of values for #math.equation(block: false, alt: "open parenthesis s , the fraction s over v close parenthesis")[$( s , display(frac(s, v)) )$]. + Plot the points #math.equation(block: false, alt: "open parenthesis s , the fraction s over v close parenthesis")[$( s , display(frac(s, v)) )$], then use linear regression to find the line of best fit. + Use your values for #math.equation(block: false, alt: "c")[$c$] and #math.equation(block: false, alt: "d")[$d$] to solve for #math.equation(block: false, alt: "V")[$V$] and #math.equation(block: false, alt: "K")[$K$]. Problems 53-56 give examples of functions whose graphs have holes. + Find the domain of the function. + Reduce the fraction to lowest terms. + Graph the function. (#emph[Hint]: The graph of the original function is identical to the graph of the function in part (b) except that certain points are excluded from the domain.) Indicate a hole in the graph by an open circle. #math.equation(block: true, alt: "y equals the fraction x squared minus 4 over x minus 2")[$y = display(frac(x^(2) − 4, x − 2))$] + #math.equation(block: false, alt: "x not equal to 2")[$x ≠ 2$] + #math.equation(block: false, alt: "x plus 2")[$x + 2$] + #figure(figph[line with hole], alt: "line with hole", caption: none) #math.equation(block: true, alt: "y equals the fraction x squared minus 1 over x plus 1")[$y = display(frac(x^(2) − 1, x + 1))$] #math.equation(block: true, alt: "y equals the fraction x plus 1 over x squared minus 1")[$y = display(frac(x + 1, x^(2) − 1))$] + #math.equation(block: false, alt: "x not equal to plus or minus 1")[$x ≠ ± 1$] + #math.equation(block: false, alt: "the fraction 1 over x minus 1")[$display(frac(1, x − 1))$] + #figure(figph[line with hole], alt: "line with hole", caption: none) #math.equation(block: true, alt: "y equals the fraction x minus 3 over x squared minus 9")[$y = display(frac(x − 3, x^(2) − 9))$]