#set document(title: "6.4 Problem Solving", 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")) == 6.4#h(0.6em)Problem Solving Many quadratic models arise as the product of two variables, one of which increases while the other decreases. For example, the area of a rectangle is the product of its length and its width, or #math.equation(block: false, alt: "A equals l w")[$A = l w$]. If we require that the rectangle have a certain perimeter, then as we increase its length, we must also decrease its width. (We analyzed this problem in Perimeter and Area of Modeling with Functions.) For Revenue from Theater Tickets, recall the formula for the revenue from sales of an item: #math.equation(block: true, alt: "Revenue equals open parenthesis price of one item close parenthesis open parenthesis number of items sold close parenthesis")[$"Revenue" = ( "price of one item" ) ( "number of items sold" )$] Usually, when the price of an item increases, the number of items sold decreases. #notebox("Investigation", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Revenue from Theater Tickets] The local theater group sold tickets to its opening night performance for #math.equation(block: false, alt: "$ 5")[$upright(\$) 5$] and drew an audience of #math.equation(block: false, alt: "100")[$100$] people. The next night, the group reduced the ticket price by #math.equation(block: false, alt: "$ 0.25")[$upright(\$) 0.25$] and #math.equation(block: false, alt: "10")[$10$] more people attended; that is, #math.equation(block: false, alt: "110")[$110$] people bought tickets at #math.equation(block: false, alt: "$ 4.75")[$upright(\$) 4.75$] apiece. In fact, for each #math.equation(block: false, alt: "$ 0.25")[$upright(\$) 0.25$] reduction in ticket price, #math.equation(block: false, alt: "10")[$10$] additional tickets can be sold. + Complete the table.#figure(table( columns: 4, align: left, inset: 6pt, table.header([Number #linebreak() of price #linebreak() reductions #linebreak()], [Price of #linebreak() ticket #linebreak()], [Number #linebreak() of tickets #linebreak() sold #linebreak()], [Total #linebreak() revenue #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "5.00")[$5.00$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "500")[$500$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "4.75")[$4.75$]], [#math.equation(block: false, alt: "110")[$110$]], [#math.equation(block: false, alt: "522.50")[$522.50$]], [#math.equation(block: false, alt: "2")[$2$]], [], [], [], [#math.equation(block: false, alt: "3")[$3$]], [], [], [], [#math.equation(block: false, alt: "4")[$4$]], [], [], [], [#math.equation(block: false, alt: "5")[$5$]], [], [], [], [#math.equation(block: false, alt: "6")[$6$]], [], [], [], [#math.equation(block: false, alt: "7")[$7$]], [], [], [], [#math.equation(block: false, alt: "8")[$8$]], [], [], [], [#math.equation(block: false, alt: "9")[$9$]], [], [], [], [#math.equation(block: false, alt: "10")[$10$]], [], [], [], [#math.equation(block: false, alt: "11")[$11$]], [], [], [], [#math.equation(block: false, alt: "12")[$12$]], [], [], [], )) + Use your table to make a graph. Plot total revenue on the vertical axis versus number of price reductions on the horizontal axis.#figure(figph[blank grid], alt: "blank grid", caption: none) + Let #math.equation(block: false, alt: "x")[$x$] represent the number of price reductions, as in the first column of the table. Write algebraic expressions in terms of #math.equation(block: false, alt: "x")[$x$] for each quantity. #linebreak() The price of a ticket after #math.equation(block: false, alt: "x")[$x$] price reductions: #math.equation(block: true, alt: "Price equals")[$"Price" =$] #linebreak() The number of tickets sold at that price: #math.equation(block: true, alt: "Number equals")[$"Number" =$] #linebreak() The total revenue from ticket sales: #math.equation(block: true, alt: "Revenue equals")[$"Revenue" =$] + Enter your expressions for the price of a ticket, the number of tickets sold, and the total revenue into the calculator as #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$], #math.equation(block: false, alt: "Y sub 2")[$Y_(2)$], and #math.equation(block: false, alt: "Y sub 3")[$Y_(3)$]. Use the Table feature to verify that your algebraic expressions agree with your table from part (1). + Use your calculator to graph your expression for total revenue in terms of #math.equation(block: false, alt: "x")[$x$]. Use your table to choose appropriate window settings that show the high point of the graph and both xintercepts. + What is the maximum revenue possible from ticket sales? What price should the theater group charge for a ticket to generate that revenue? How many tickets will the group sell at that price? ] === Maximum or Minimum Values Finding the maximum or minimum value for a variable expression is a common problem in many applications. For example, if you own a company that manufactures blue jeans, you might like to know how much to charge for your jeans in order to maximize your revenue. As you increase the price of the jeans, your revenue may increase for a while. But if you charge too much for the jeans, consumers will not buy as many pairs, and your revenue may actually start to decrease. Is there some optimum price you should charge for a pair of jeans in order to achieve the greatest revenue? #examplebox("Example 1")[][ Late Nite Blues finds that it can sell #math.equation(block: false, alt: "600 minus 15 x")[$600 − 15 x$] pairs of jeans per week if it charges #math.equation(block: false, alt: "x")[$x$] dollars per pair. (Notice that as the price increases, the number of pairs of jeans sold decreases.) + Write an equation for the revenue as a function of the price of a pair of jeans. + Graph the function. + How much should Late Nite Blues charge for a pair of jeans in order to maximize its revenue? #solutionbox[ + Using the formula for revenue stated above, we find #math.equation(block: true, alt: "Revenue, equals open parenthesis price of one item close parenthesis open parenthesis number of items sold close parenthesis; R, equals x open parenthesis 600 minus 15 x close parenthesis; R, equals 600 x minus 15 x squared")[$"Revenue" & = ( "price of one item" ) ( "number of items sold" ) \ R & = x ( 600 − 15 x ) \ R & = 600 x − 15 x^(2)$] + We recognize the function as quadratic, so the graph is a parabola. You can use your calculator to verify the graph below.#figure(figph[graph showing revenue from sales of jeans], alt: "graph showing revenue from sales of jeans", caption: none) + The maximum value of #math.equation(block: false, alt: "R")[$R$] occurs at the vertex of the parabola. Thus, #math.equation(block: true, alt: "x sub v, equals the fraction minus b over 2 a equals the fraction minus 600 over 2 open parenthesis minus 15 close parenthesis equals 20; y sub v, equals 600 open parenthesis 20 close parenthesis minus 15 open parenthesis 20 close parenthesis squared equals 6000")[$x_(v) & = frac(− b, 2 a) = frac(− 600, 2 ( − 15 )) = 20 \ y_(v) & = 600 ( 20 ) − 15 ( 20 )^(2) = 6000$] The revenue takes on its maximum value when #math.equation(block: false, alt: "x equals 20")[$x = 20$], and the maximum value is #math.equation(block: false, alt: "R equals 6000")[$R = 6000$]. This means that Late Nite Blues should charge \$#math.equation(block: false, alt: "20")[$20$] for a pair of jeans in order to maximize revenue at \$#math.equation(block: false, alt: "6000")[$6000$] a week. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ The maximum or minimum value of a quadratic function is easy to find: It is the value at the vertex. If the parabola opens downward, as in Example, there is a maximum value at the vertex. If the parabola opens upward, there is a minimum value at the vertex. ] The height of golf ball in meters is given by #math.equation(block: false, alt: "h equals minus 4.9 t squared plus 20 t")[$h = − 4.9 t^(2) + 20 t$]. What does the vertex of the graph tell us about the golf ball? \_\_\_\_\_ #solutionbox[ When the golf ball reaches its maximum height ] The height of golf ball in meters is given by #math.equation(block: false, alt: "h equals minus 4.9 t squared plus 20 t")[$h = − 4.9 t^(2) + 20 t$]. What does the vertex of the graph tell us about the golf ball? + How long the golf ball is in the air + The starting height of the golf ball + When the golf ball reaches its maximum height + The speed of the golf ball The Metro Rail service sells #math.equation(block: false, alt: "1200 minus 80 x")[$1200 − 80 x$] tickets each day when it charges #math.equation(block: false, alt: "x")[$x$] dollars per ticket. + Write an equation for the revenue, #math.equation(block: false, alt: "R")[$R$], as a function of the price of a ticket. #linebreak() #math.equation(block: false, alt: "R equals")[$R =$]\_\_\_\_\_ + What ticket price will return the maximum revenue? #linebreak() \$\_\_\_\_\_ #linebreak() What is the maximum revenue? #linebreak() \$\_\_\_\_\_ #solutionbox[ + The revenue is the product of the number of tickets times the price per ticket: #math.equation(block: false, alt: "R equals 1200 x minus 80 x squared")[$R = 1200 x − 80 x^(2)$] + \$#math.equation(block: false, alt: "7.50 ,")[$7.50 ,$] \$#math.equation(block: false, alt: "4500")[$4500$] ] The Metro Rail service sells #math.equation(block: false, alt: "1200 minus 80 x")[$1200 − 80 x$] tickets each day when it charges #math.equation(block: false, alt: "x")[$x$] dollars per ticket. + Write an equation for the revenue, #math.equation(block: false, alt: "R")[$R$], as a function of the price of a ticket. + What ticket price will return the maximum revenue? What is the maximum revenue? #solutionbox[ + The revenue is the product of the number of tickets times the price per ticket: #math.equation(block: false, alt: "R equals 1200 x minus 80 x squared")[$R = 1200 x − 80 x^(2)$] + ticket price: \$#math.equation(block: false, alt: "7.50")[$7.50$], maximum revenue: \$#math.equation(block: false, alt: "4500")[$4500$] ] Explain why revenue will probably not increase indefinitely as price increases. \_\_\_\_\_ Explain why revenue will probably not increase indefinitely as price increases. === The Vertex Form for a Parabola What is the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex of the parabola #math.equation(block: false, alt: "y equals 4 open parenthesis x minus 2 close parenthesis open parenthesis x plus 8 close parenthesis")[$y = 4 ( x − 2 ) ( x + 8 )$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "minus 3")[$− 3$] ] What is the #math.equation(block: false, alt: "x")[$x$]-coordinate of the vertex of the parabola #math.equation(block: false, alt: "y equals 4 open parenthesis x minus 2 close parenthesis open parenthesis x plus 8 close parenthesis")[$y = 4 ( x − 2 ) ( x + 8 )$] ? + #math.equation(block: false, alt: "4")[$4$] + #math.equation(block: false, alt: "minus 16")[$− 16$] + #math.equation(block: false, alt: "6")[$6$] + #math.equation(block: false, alt: "minus 3")[$− 3$] Consider the quadratic equation #math.equation(block: true, alt: "y equals 2 open parenthesis x minus 3 close parenthesis squared minus 8")[$y = 2 ( x − 3 )^(2) − 8$] By expanding the squared expression and collecting like terms, we can rewrite the equation in standard form as #math.equation(block: true, alt: "y, equals 2 open parenthesis x squared minus 6 x plus 9 close parenthesis minus 8; y, equals 2 x squared minus 12 x plus 10")[$y & = 2 ( x^(2) − 6 x + 9 ) − 8 \ y & = 2 x^(2) − 12 x + 10$] The vertex of this parabola is #math.equation(block: true, alt: "x sub v, equals the fraction minus open parenthesis minus 12 close parenthesis over 2 open parenthesis 2 close parenthesis equals 3; y sub v, equals 2 open parenthesis 3 close parenthesis squared minus 12 open parenthesis 3 close parenthesis plus 10 equals minus 8")[$x_(v) & = frac(− ( − 12 ), 2 ( 2 )) = 3 \ y_(v) & = 2 ( 3 )^(2) − 12 ( 3 ) + 10 = − 8$] and its graph is shown below. #figure(figph[parabola with vertex labeled], alt: "parabola with vertex labeled", caption: none) Notice that the coordinates of the vertex, #math.equation(block: false, alt: "open parenthesis 3 , minus 8 close parenthesis")[$( 3 , − 8 )$], are apparent in the original equation; we don’t need to do any computation to find the vertex. #math.equation(block: true, alt: "y equals 2 open parenthesis x minus, 3 close parenthesis squared, minus 8; x sub v, y sub v")[$y = 2 ( x − & 3 )^(2) " " & − 8 \ & bold(italic(x))_(bold(italic(v))) & bold(italic(y))_(bold(italic(v)))$] This equation is an example of the #strong[vertex form] for a quadratic function. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertex Form for a Quadratic Function] A quadratic function #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$], #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$], can be written in the vertex form #math.equation(block: true, alt: "y equals a open parenthesis x minus x sub v close parenthesis squared plus y sub v")[$y = a ( x − x_(v) )^(2) + y_(v)$] where the vertex of the graph is #math.equation(block: false, alt: "open parenthesis x sub v , y sub v close parenthesis")[$( x_(v) , y_(v) )$]. ] #examplebox("Example 2")[][ Find the vertex of the graph of #math.equation(block: false, alt: "y equals minus 3 open parenthesis x minus 4 close parenthesis squared plus 6")[$y = − 3 ( x − 4 )^(2) + 6$]. Is the vertex a maximum or a minimum point of the graph? #solutionbox[ Compare the equation to the vertex form to see that the coordinates of the vertex are #math.equation(block: false, alt: "open parenthesis 4 , 6 close parenthesis")[$( 4 , 6 )$]. For this equation, #math.equation(block: false, alt: "a equals minus 3 less than 0")[$a = − 3 < 0$], so the parabola opens downward. The vertex is the maximum point of the graph. ] ] To understand why the vertex form works, substitute #math.equation(block: false, alt: "x sub v equals 4")[$x_(v) = 4$] into #math.equation(block: false, alt: "y equals minus 3 open parenthesis x minus 4 close parenthesis squared plus 6")[$y = − 3 ( x − 4 )^(2) + 6$] from Example to find #math.equation(block: true, alt: "y equals minus 3 open parenthesis 4 minus 4 close parenthesis squared plus 6 equals 6")[$y = − 3 ( 4 − 4 )^(2) + 6 = 6$] which confirms that when #math.equation(block: false, alt: "x equals 4")[$x = 4$], #math.equation(block: false, alt: "y equals 6")[$y = 6$]. Next, notice that if #math.equation(block: false, alt: "x")[$x$] is any number except #math.equation(block: false, alt: "4")[$4$], the expression #math.equation(block: false, alt: "minus 3 open parenthesis x minus 4 close parenthesis squared")[$− 3 ( x − 4 )^(2)$] is negative, so #math.equation(block: false, alt: "y less than 6")[$y < 6$]. Therefore, #math.equation(block: false, alt: "6")[$6$] is the maximum value for #math.equation(block: false, alt: "y")[$y$] on the graph, so #math.equation(block: false, alt: "open parenthesis 4 , 6 close parenthesis")[$( 4 , 6 )$] is the high point or vertex. You can also rewrite #math.equation(block: false, alt: "y equals minus 3 open parenthesis x minus 4 close parenthesis squared plus 6")[$y = − 3 ( x − 4 )^(2) + 6$] in standard form and use the formula #math.equation(block: false, alt: "x sub v equals the fraction minus b over 2 a")[$x_(v) = display(frac(− b, 2 a))$] to confirm that the vertex is the point #math.equation(block: false, alt: "open parenthesis 4 , 6 close parenthesis")[$( 4 , 6 )$]. + Find the vertex of the graph of #math.equation(block: false, alt: "y equals 5 minus the fraction 1 over 2 open parenthesis x plus 2 close parenthesis squared")[$y = 5 − display(frac(1, 2)) ( x + 2 )^(2)$]. #linebreak() Vertex: \_\_\_\_\_ + Write the equation of the parabola in standard form. #linebreak() #math.equation(block: false, alt: "y equals")[$y =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "open parenthesis minus 2 , 5 close parenthesis")[$( − 2 , 5 )$] + #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared minus 2 x plus 3")[$y = − frac(1, 2) x^(2) − 2 x + 3$] ] + Find the vertex of the graph of #math.equation(block: false, alt: "y equals 5 minus the fraction 1 over 2 open parenthesis x plus 2 close parenthesis squared")[$y = 5 − display(frac(1, 2)) ( x + 2 )^(2)$]. + Write the equation of the parabola in standard form. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis minus 2 , 5 close parenthesis")[$( − 2 , 5 )$] + #math.equation(block: false, alt: "y equals minus the fraction 1 over 2 x squared minus 2 x plus 3")[$y = − frac(1, 2) x^(2) − 2 x + 3$] ] What is the smallest #math.equation(block: false, alt: "y")[$y$]-value on the graph of #math.equation(block: false, alt: "y equals 4 open parenthesis x minus 6 close parenthesis squared plus 12")[$y = 4 ( x − 6 )^(2) + 12$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "12")[$12$] ] What is the smallest #math.equation(block: false, alt: "y")[$y$]-value on the graph of #math.equation(block: false, alt: "y equals 4 open parenthesis x minus 6 close parenthesis squared plus 12")[$y = 4 ( x − 6 )^(2) + 12$] ? + #math.equation(block: false, alt: "6")[$6$] + #math.equation(block: false, alt: "4")[$4$] + #math.equation(block: false, alt: "12")[$12$] + We can't tell without graphing Any quadratic equation in vertex form can be written in standard form by expanding, and any quadratic equation in standard form can be put into vertex form by completing the square. #examplebox("Example 3")[][ Write the equation #math.equation(block: false, alt: "y equals 3 x squared minus 6 x minus 1")[$y = 3 x^(2) − 6 x − 1$] in vertex form and find the vertex of its graph. #solutionbox[ We factor the lead coefficient, #math.equation(block: false, alt: "3")[$3$], from the variable terms, leaving a space to complete the square. #math.equation(block: true, alt: "y equals 3 open parenthesis x squared minus 2 x plus bar close parenthesis minus 1")[$y = 3 ( x^(2) − 2 x + underline(#h(3em)) ) − 1$] Next, we complete the square inside parentheses. Take half the coefficient of #math.equation(block: false, alt: "x")[$x$] and square the result: #math.equation(block: true, alt: "p equals the fraction 1 over 2 open parenthesis minus 2 close parenthesis equals minus 1 , and p squared equals open parenthesis minus 1 close parenthesis squared equals 1")[$p = frac(1, 2) ( − 2 ) = − 1 , " " " " "and" " " " " p^(2) = ( − 1 )^(2) = 1$] . We must add #math.equation(block: false, alt: "1")[$1$] to complete the square. However, we are really adding #math.equation(block: false, alt: "3 open parenthesis 1 close parenthesis")[$3 ( 1 )$] to the right side of the equation, so we must also subtract #math.equation(block: false, alt: "3")[$3$] to compensate: #math.equation(block: true, alt: "y equals 3 open parenthesis x squared minus 2 x plus 1 close parenthesis minus 1 minus 3")[$y = 3 ( x^(2) − 2 x + 1 ) − 1 #h(0.167em) − #h(0.167em) 3$] The expression inside parentheses is now a perfect square, and the vertex form is #math.equation(block: true, alt: "y equals 3 open parenthesis x minus 1 close parenthesis squared minus 4")[$y = 3 ( x − 1 )^(2) − 4$] The vertex of the parabola is #math.equation(block: false, alt: "open parenthesis 1 , minus 4 close parenthesis")[$( 1 , − 4 )$]. ] ] Write the equation #math.equation(block: false, alt: "y equals 2 x squared plus 12 x plus 13")[$y = 2 x^(2) + 12 x + 13$] in vertex form, and find the vertex of its graph. The vertex form is #math.equation(block: false, alt: "y equals a open parenthesis x minus x sub v close parenthesis squared plus y sub v")[$y = a ( x − x_(v) )^(2) + y_(v)$], where #math.equation(block: false, alt: "a equals")[$a =$]\_\_\_\_\_ #math.equation(block: false, alt: "x sub v equals")[$x_(v) =$]\_\_\_\_\_ #math.equation(block: false, alt: "y sub v equals")[$y_(v) =$]\_\_\_\_\_ The vertex of the graph is \_\_\_\_\_ + Factor #math.equation(block: false, alt: "2")[$2$] from the variable terms. + Complete the square inside parentheses. + Subtract #math.equation(block: false, alt: "2 p squared")[$2 p^(2)$] outside parentheses. + Write the vertex form. #solutionbox[ #math.equation(block: true, alt: "y equals 2 open parenthesis x plus 3 close parenthesis squared minus 5")[$y = 2 ( x + 3 )^(2) − 5$]; #math.equation(block: true, alt: "open parenthesis minus 3 , minus 5 close parenthesis")[$( − 3 , − 5 )$] ] Write the equation #math.equation(block: false, alt: "y equals 2 x squared plus 12 x plus 13")[$y = 2 x^(2) + 12 x + 13$] in vertex form, and find the vertex of its graph. + Factor #math.equation(block: false, alt: "2")[$2$] from the variable terms. + Complete the square inside parentheses. + Subtract #math.equation(block: false, alt: "2 p squared")[$2 p^(2)$] outside parentheses. + Write the vertex form. #solutionbox[ #math.equation(block: false, alt: "y equals 2 open parenthesis x plus 3 close parenthesis squared minus 5")[$y = 2 ( x + 3 )^(2) − 5$]; vertex: #math.equation(block: false, alt: "open parenthesis minus 3 , minus 5 close parenthesis")[$( − 3 , − 5 )$] ] Why do we need to know a second point besides the vertex to find the equation of a parabola? \_\_\_\_\_ Why do we need to know a second point besides the vertex to find the equation of a parabola? === Graphing with the Vertex Form We can also use the vertex form to sketch a graph, using what we know about transformations. #examplebox("Example 4")[][ Use transformations to graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals minus 3 open parenthesis x minus 4 close parenthesis squared plus 6")[$f ( x ) = − 3 ( x − 4 )^(2) + 6$]. #solutionbox[ We can graph #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] by applying transformations to the basic parabola, #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$]. To identify the transformations, think of evaluating the function for a specific value of #math.equation(block: false, alt: "x")[$x$]. What operations would we perform on #math.equation(block: false, alt: "x")[$x$], besides squaring? #figure(table( columns: 3, align: left, inset: 6pt, table.header([1. Subtract 4:], [#math.equation(block: false, alt: "y equals open parenthesis x minus 4 close parenthesis squared")[$y = ( x #h(0.167em) − #h(0.167em) 4 )^(2)$]], [#math.equation(block: false, alt: "Shift 4 units right.")[$"Shift 4 units right."$]]), [2. Multiply by -3:], [#math.equation(block: false, alt: "y equals minus 3 open parenthesis x minus 4 close parenthesis squared")[$y = − 3 ( x − 4 )^(2)$]], [#math.equation(block: false, alt: "Stretch by a factor of 3,")[$"Stretch by a factor of 3,"$] #linebreak() #math.equation(block: false, alt: "and reflect about the x -axis.")[$"and reflect about the " bold(italic(x)) "-axis."$] #linebreak()], [3. Add 6:], [#math.equation(block: false, alt: "y equals minus 3 open parenthesis x minus 4 close parenthesis squared plus 6")[$y = − 3 ( x − 4 )^(2) #h(0.167em) + #h(0.167em) 6$]], [#math.equation(block: false, alt: "Shift up 6 units.")[$"Shift up 6 units."$]], )) We perform the same transformations on the graph of #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$], as shown in the figure. #figure(figph[three transformations on the basic parabola], alt: "three transformations on the basic parabola", caption: none) ] ] + List the transformations of #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] needed to graph #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 minus the fraction 1 over 2 open parenthesis x plus 2 close parenthesis squared")[$g ( x ) = 5 − display(frac(1, 2)) ( x + 2 )^(2)$].- Shift 2 units \_\_\_\_\_ - Reflect about the #math.equation(block: false, alt: "x")[$x$]-axis and \_\_\_\_\_ by a factor of #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$]. - Shift 5 units \_\_\_\_\_ + Use transformations to sketch the graph. #solutionbox[ + Shift 2 units left, reflect about #math.equation(block: false, alt: "x")[$x$]-axis and compress by a factor of 2, shift 5 units up. + A graph is below. ] #figure(figph[parabola], alt: "parabola", caption: none) + List the transformations of #math.equation(block: false, alt: "y equals x squared")[$y = x^(2)$] needed to graph #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 5 minus the fraction 1 over 2 open parenthesis x plus 2 close parenthesis squared")[$g ( x ) = 5 − display(frac(1, 2)) ( x + 2 )^(2)$]. + Use transformations to sketch the graph. #solutionbox[ + Shift 2 units left, reflect about #math.equation(block: false, alt: "x")[$x$]-axis and compress by a factor of 2, shift 5 units up. + #figure(figph[parabola], alt: "parabola", caption: none) ] === Systems Involving Quadratic Equations Recall that the solution to a #math.equation(block: false, alt: "2 times 2")[$2 × 2$] system of linear equations is the intersection point of the graphs of the equations. (See Algebra Review Refresher Linear Systems in Two Variables.) This is also true of systems in which one or both of the equations is quadratic. The figure below shows the three cases for systems of one quadratic and one linear equation. #figure(figph[graphs of 3 cases for quadratic system], alt: "graphs of 3 cases for quadratic system", caption: none) In Example, we use both graphical and algebraic techniques to solve the system. #examplebox("Example 5")[][ The Pizza Connection calculates that the cost, in dollars, of producing #math.equation(block: false, alt: "x")[$x$] pizzas per day is given by #math.equation(block: true, alt: "C equals 0.15 x squared plus 0.75 x plus 180")[$C = 0.15 x^(2) + 0.75 x + 180$] The Pizza Connection charges \$#math.equation(block: false, alt: "15")[$15$] per pizza, so the revenue from selling #math.equation(block: false, alt: "x")[$x$] pizzas is #math.equation(block: true, alt: "R equals 15 x")[$R = 15 x$] How many pizzas per day must the Pizza Connection sell in order to break even? #solutionbox[ To break even means to make zero profit. Because #math.equation(block: true, alt: "Profit equals Revenue minus Cost")[$"Profit" = "Revenue" − "Cost"$] the break-even points occur when revenue equals cost. In mathematical terms, we would like to find any values of #math.equation(block: false, alt: "x")[$x$] for which #math.equation(block: false, alt: "R equals C")[$R = C$]. If we graph the revenue and cost functions on the same axes, these values correspond to points where the two graphs intersect. Use the window settings #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 94; Ymin, equals 0, Ymax equals 1400")[$"Xmin" & = 0 & & "Xmax" = 94 \ "Ymin" & = 0 & & "Ymax" = 1400$] on your calculator to obtain the graph shown below. You can verify that the two intersection points are #math.equation(block: false, alt: "open parenthesis 15 , 225 close parenthesis")[$( 15 , 225 )$] and #math.equation(block: false, alt: "open parenthesis 80 , 1200 close parenthesis")[$( 80 , 1200 )$]. #figure(figph[GC figure for quadratic system], alt: "GC figure for quadratic system", caption: none) Thus, the Pizza Connection must sell either #math.equation(block: false, alt: "15")[$15$] or #math.equation(block: false, alt: "80")[$80$] pizzas in order to break even. On the graph we see that revenue is greater than cost for #math.equation(block: false, alt: "x")[$x$]-values between #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "80")[$80$], so the Pizza Connection will make a profit if it sells between #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "80")[$80$] pizzas. We can also solve algebraically for the break-even points. The intersection points of the two graphs correspond to the solutions of the system of equations #math.equation(block: true, alt: "y, equals 0.15 x squared plus 0.75 x plus 180; y, equals 15 x")[$y & = 0.15 x^(2) + 0.75 x + 180 \ y & = 15 x$] We equate the two expressions for #math.equation(block: false, alt: "y")[$y$] and solve for #math.equation(block: false, alt: "x")[$x$]: #math.equation(block: true, alt: "0.15 x squared plus 0.75 x plus 180, equals 15 x, Subtract 15 x from both sides.; 0.15 x squared minus 14.25 x plus 180, equals 0, Use the quadratic formula.")[$0.15 x^(2) + 0.75 x + 180 & = 15 x & & "Subtract " 15 bold(italic(x)) " from both sides." \ 0.15 x^(2) − 14.25 x + 180 & = 0 & & "Use the quadratic formula."$] #math.equation(block: true, alt: "x, equals the fraction 14.25 plus or minus the square root of 14.25 squared minus 4 open parenthesis 0.15 close parenthesis open parenthesis 180 close parenthesis over 2 open parenthesis minus 0.05 close parenthesis, Simplify.; equals the fraction 14.25 plus or minus 9.75 over 0.3")[$x & = frac(14.25 ± sqrt(14.25^(2) − 4 ( 0.15 ) ( 180 )), 2 ( − 0.05 )) & & "Simplify." \ & = frac(14.25 ± 9.75, 0.3)$] The solutions are #math.equation(block: false, alt: "15")[$15$] and #math.equation(block: false, alt: "80")[$80$], as we found from the graph. ] ] + Solve the system algebraically: #math.equation(block: true, alt: "y, equals x squared minus 6 x minus 7; y, equals 13 minus x squared")[$y & = x^(2) − 6 x − 7 \ y & = 13 − x^(2)$] #linebreak() Solutions: \_\_\_\_\_ Note: list solutions as ordered pairs, and use a comma to separate solutions. + Graph both equations, and show the solutions on the graph. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis minus 2 , 9 close parenthesis , open parenthesis 5 , minus 12 close parenthesis")[$( − 2 , 9 ) , ( 5 , − 12 )$] ] #figure(figph[two parabolas], alt: "two parabolas", caption: none) + Solve the system algebraically: #math.equation(block: true, alt: "y, equals x squared minus 6 x minus 7; y, equals 13 minus x squared")[$y & = x^(2) − 6 x − 7 \ y & = 13 − x^(2)$] + Graph both equations, and show the solutions on the graph. #solutionbox[ + #math.equation(block: false, alt: "open parenthesis minus 2 , 9 close parenthesis , open parenthesis 5 , minus 12 close parenthesis")[$( − 2 , 9 ) , ( 5 , − 12 )$] + #figure(figph[two parabolas], alt: "two parabolas", caption: none) ] Explain how to find the intersection points of two parabolas. \_\_\_\_\_ Explain how to find the intersection points of two parabolas. #notebox("Technology", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Solving Systems with the Graphing Calculator] We can use the intersect feature of the graphing calculator to solve systems of quadratic equations. Consider the system #math.equation(block: true, alt: "y, equals open parenthesis x plus 1.1 close parenthesis squared; y, equals 7.825 minus 2 x minus 2.5 x squared")[$y & = ( x + 1.1 )^(2) \ y & = 7.825 − 2 x − 2.5 x^(2)$] We will graph these two equations in the standard window. The two intersection points are visible in the window, but we do not find their exact coordinates when we trace the graphs. We can use the intersect command to locate one of the solutions, as shown below. You can check that the point #math.equation(block: false, alt: "open parenthesis 0.9 , 4 close parenthesis")[$( 0.9 , 4 )$] is an exact solution to the system by substituting #math.equation(block: false, alt: "x equals 0.9")[$x = 0.9$] and #math.equation(block: false, alt: "y equals 4")[$y = 4$] into each equation of the system. (The calculator is not always able to find the exact coordinates, but it usually gives a very good approximation.) #figure(figph[GC graph showing quadratic system], alt: "GC graph showing quadratic system", caption: none) You can find the other solution of the system by following the same steps and moving the bug close to the other intersection point. You should verify that the other solution is the point #math.equation(block: false, alt: "open parenthesis minus 2.1 , 1 close parenthesis")[$( − 2.1 , 1 )$]. ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Maximum value - Minimum value - Vertex form ==== CONCEPTS + Quadratic models may arise as the product of two variables. + The maximum or minimum of a quadratic function occurs at the vertex. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Vertex Form for a Quadratic Function] A quadratic function #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$], #math.equation(block: false, alt: "a not equal to 0")[$a ≠ 0$], can be written in the vertex form #math.equation(block: true, alt: "y equals a open parenthesis x minus x sub v close parenthesis squared plus y sub v")[$y = a ( x − x_(v) )^(2) + y_(v)$]where the vertex of the graph is #math.equation(block: false, alt: "open parenthesis x sub v , y sub v close parenthesis")[$( x_(v) , y_(v) )$]. ] + We can convert a quadratic equation to vertex form by completing the square. + We can graph a quadratic equation in vertex form using transformations. + A #math.equation(block: false, alt: "2 times 2")[$2 × 2$] system involving quadratic equations may have one, two, or no solutions. ==== STUDY QUESTIONS + How can you tell whether a variable given by a quadratic equation has a maximum value or a minimum value? + Correct the following false statement. - The maximum or minimum value given by a quadratic equation is the average of the #math.equation(block: false, alt: "x")[$x$]-intercepts. (#math.equation(block: false, alt: "False!")[$"False!"$]) + Explain why #math.equation(block: false, alt: "minus 4")[$− 4$] is the smallest function value for #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 open parenthesis x minus 3 close parenthesis squared minus 4")[$f ( x ) = 2 ( x − 3 )^(2) − 4$]. + In the equation #math.equation(block: false, alt: "y equals the fraction 1 over 3 open parenthesis x plus 5 close parenthesis squared minus 2")[$y = display(frac(1, 3)) ( x + 5 )^(2) − 2$], what does each of the constants tell you about the graph? + Francine attempts to write the equation #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 x squared minus 6 x plus 1")[$g ( x ) = 2 x^(2) − 6 x + 1$] in vertex form as follows: #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals open parenthesis 2 x squared minus 6 x plus 9 close parenthesis plus 1 minus 9")[$g ( x ) = ( 2 x^(2) − 6 x + 9 ) + 1 − 9$]. What is wrong with her work? + Without doing any calculations, solve the system #math.equation(block: false, alt: "y equals x squared plus 4 , y equals 2 minus 3 x squared")[$y = x^(2) + 4 , y = 2 − 3 x^(2)$]. (#emph[Hint]: Visualize the graphs.) ==== SKILLS Practice each skill in the Homework problems listed. + Find the maximum or minimum value of a quadratic function: \#1–14 + Convert a quadratic equation from vertex form to standard form: \#19–22 + Convert a quadratic equation from standard form to vertex form: \#23–28 + Use transformations to graph a quadratic equation: \#15–28 + USolve a system involving quadratic equations: \#31–50 === Homework 6.4 The owner of a motel has #math.equation(block: false, alt: "60")[$60$] rooms to rent. She finds that if she charges \$#math.equation(block: false, alt: "0")[$0$] per room per night, all the rooms will be rented. For every \$#math.equation(block: false, alt: "2")[$2$] that she increases the price of a room, #math.equation(block: false, alt: "3")[$3$] rooms will stand vacant. + Complete the table. The first two rows are filled in for you.#figure(table( columns: 4, align: left, inset: 6pt, table.header([No. of price #linebreak() increases #linebreak()], [Price of #linebreak() room #linebreak()], [No. of rooms #linebreak() rented #linebreak()], [Total #linebreak() revenue #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "1200")[$1200$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "22")[$22$]], [#math.equation(block: false, alt: "57")[$57$]], [#math.equation(block: false, alt: "1254")[$1254$]], [#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: "10")[$10$]], [], [], [], [#math.equation(block: false, alt: "12")[$12$]], [], [], [], [#math.equation(block: false, alt: "16")[$16$]], [], [], [], [#math.equation(block: false, alt: "20")[$20$]], [], [], [], )) + Let #math.equation(block: false, alt: "x")[$x$] stand for the number of #math.equation(block: false, alt: "$ 2")[$upright(\$) 2$] price increases the owner makes. Write algebraic expressions for the price of a room, the number of rooms that will be rented, and the total revenue earned at that price. + Use your calculator to make a table of values for your algebraic expressions. Let #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$] stand for the price of a room, #math.equation(block: false, alt: "Y sub 2")[$Y_(2)$] for the number of rooms rented, and #math.equation(block: false, alt: "Y sub 3")[$Y_(3)$] for the total revenue. Verify the values you calculated in part (a). + Use your table to find a value of #math.equation(block: false, alt: "x")[$x$] that causes the total revenue to be zero. + Use your graphing calculator to graph your formula for total revenue. + What is the lowest price that the owner can charge for a room if she wants her revenue to exceed #math.equation(block: false, alt: "$ 1296")[$upright(\$) 1296$] per night? What is the highest price she can charge to obtain this revenue? + What is the maximum revenue the owner can earn in one night? How much should she charge for a room to maximize her revenue? How many rooms will she rent at that price? + #figure(table( columns: 4, align: left, inset: 6pt, table.header([No. of price #linebreak() increases #linebreak()], [Price of #linebreak() room #linebreak()], [No. of rooms #linebreak() rented #linebreak()], [Total #linebreak() revenue #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "1200")[$1200$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "22")[$22$]], [#math.equation(block: false, alt: "57")[$57$]], [#math.equation(block: false, alt: "1254")[$1254$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "54")[$54$]], [#math.equation(block: false, alt: "1296")[$1296$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "26")[$26$]], [#math.equation(block: false, alt: "51")[$51$]], [#math.equation(block: false, alt: "1326")[$1326$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "28")[$28$]], [#math.equation(block: false, alt: "48")[$48$]], [#math.equation(block: false, alt: "1344")[$1344$]], [#math.equation(block: false, alt: "5")[$5$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "45")[$45$]], [#math.equation(block: false, alt: "1350")[$1350$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "42")[$42$]], [#math.equation(block: false, alt: "1344")[$1344$]], [#math.equation(block: false, alt: "7")[$7$]], [#math.equation(block: false, alt: "34")[$34$]], [#math.equation(block: false, alt: "39")[$39$]], [#math.equation(block: false, alt: "1326")[$1326$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "36")[$36$]], [#math.equation(block: false, alt: "1296")[$1296$]], [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "1200")[$1200$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "44")[$44$]], [#math.equation(block: false, alt: "24")[$24$]], [#math.equation(block: false, alt: "1056")[$1056$]], [#math.equation(block: false, alt: "16")[$16$]], [#math.equation(block: false, alt: "52")[$52$]], [#math.equation(block: false, alt: "12")[$12$]], [#math.equation(block: false, alt: "624")[$624$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0")[$0$]], )) + Price of a room: #math.equation(block: false, alt: "20 plus 2 x")[$20 + 2 x$]; Rooms rented: #math.equation(block: false, alt: "60 minus 3 x")[$60 − 3 x$]; Revenue: #math.equation(block: false, alt: "1200 plus 60 x minus 6 x squared")[$1200 + 60 x − 6 x^(2)$] + #math.equation(block: false, alt: "20")[$20$] + #figure(figph[parabola], alt: "parabola", caption: none) + \$#math.equation(block: false, alt: "24 ;")[$24 ; " "$] \$#math.equation(block: false, alt: "36")[$36$] + \$#math.equation(block: false, alt: "1350 ;")[$1350 ; " "$] \$#math.equation(block: false, alt: "30 ;")[$30 ;$] #math.equation(block: false, alt: "45")[$" " 45 " "$]rooms The owner of a video store sells #math.equation(block: false, alt: "96")[$96$] blank tapes per week if he charges #math.equation(block: false, alt: "$ 6")[$upright(\$) 6$] per tape. For every #math.equation(block: false, alt: "$ 0.50")[$upright(\$) 0.50$] he increases the price, he sells #math.equation(block: false, alt: "4")[$4$] fewer tapes per week. + Complete the table. The first two rows are filled in for you.#figure(table( columns: 4, align: left, inset: 6pt, table.header([No. of price #linebreak() increases #linebreak()], [Price of #linebreak() tape #linebreak()], [No. of tapes #linebreak() sold #linebreak()], [Total #linebreak() revenue #linebreak()]), [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "6")[$6$]], [#math.equation(block: false, alt: "96")[$96$]], [#math.equation(block: false, alt: "576")[$576$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "6.50")[$6.50$]], [#math.equation(block: false, alt: "92")[$92$]], [#math.equation(block: false, alt: "598")[$598$]], [#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: "12")[$12$]], [], [], [], [#math.equation(block: false, alt: "16")[$16$]], [], [], [], [#math.equation(block: false, alt: "20")[$20$]], [], [], [], [#math.equation(block: false, alt: "24")[$24$]], [], [], [], )) + Let #math.equation(block: false, alt: "x")[$x$] stand for the number of #math.equation(block: false, alt: "$ 0.50")[$upright(\$) 0.50$] price increases the owner makes. Write algebraic expressions for the price of a tape, the number of tapes sold, and the total revenue. + Use your calculator to make a table of values for your algebraic expressions. Let #math.equation(block: false, alt: "Y sub 1")[$Y_(1)$] stand for the price of a tape, #math.equation(block: false, alt: "Y sub 2")[$Y_(2)$] for the number of tapes sold, and #math.equation(block: false, alt: "Y sub 3")[$Y_(3)$] for the total revenue. Verify the values you calculated in part (a). + Use your table to find a value of #math.equation(block: false, alt: "x")[$x$] that causes the total revenue to be zero. + Use your graphing calculator to graph your formula for total revenue. + How much should the owner charge for a tape in order to bring in #math.equation(block: false, alt: "$ 630")[$upright(\$) 630$] per week from tapes? (You should have two answers.) + What is the maximum revenue the owner can earn from tapes in one week? How much should he charge for a tape to maximize his revenue? How many tapes will he sell at that price? + Give the dimensions of two different rectangles with perimeter #math.equation(block: false, alt: "60")[$60$] meters. Compute the areas of the two rectangles. + A rectangle has a perimeter of #math.equation(block: false, alt: "60")[$60$] meters. If the length of the rectangle is #math.equation(block: false, alt: "x")[$x$] meters, write an expression for its width. + Write an expression for the area of the rectangle. + (For example) #math.equation(block: false, alt: "10")[$10$] m by #math.equation(block: false, alt: "20")[$20$] m with area #math.equation(block: false, alt: "200")[$200$] sq m; or #math.equation(block: false, alt: "15")[$15$] m by #math.equation(block: false, alt: "15")[$15$] m, area #math.equation(block: false, alt: "225")[$225$] sq m + #math.equation(block: false, alt: "30 minus x")[$30 − x$] + #math.equation(block: false, alt: "30 x minus x squared")[$30 x − x^(2)$] + Give the dimensions of two different rectangles with perimeter #math.equation(block: false, alt: "48")[$48$] inches. Compute the areas of the two rectangles. + A rectangle has a perimeter of #math.equation(block: false, alt: "48")[$48$] inches. If the width of the rectangle is #math.equation(block: false, alt: "w")[$w$] inches, write an expression for its length. + Write an expression for the area of the rectangle. For Problems 5–8, + Find the maximum or minimum value algebraically. + Obtain a good graph on your calculator and verify your answer. (Use the coordinates of the vertex and the vertical intercept to help you choose an appropriate window for the graph.) Delbert launches a toy water rocket from ground level. Its distance above the ground #math.equation(block: false, alt: "t")[$t$] seconds after launch is given, in feet, by #math.equation(block: true, alt: "d equals 96 t minus 16 t squared")[$d = 96 t − 16 t^(2)$] When will the rocket reach its greatest height, and what will that height be? #math.equation(block: false, alt: "3")[$3$] sec, #math.equation(block: false, alt: "144")[$144$] ft Francine throws a wrench into the air from the bottom of a trench #math.equation(block: false, alt: "12")[$12$] feet deep. Its height #math.equation(block: false, alt: "t")[$t$] seconds later is given, in feet, by #math.equation(block: true, alt: "h equals minus 12 plus 32 t minus 16 t squared")[$h = − 12 + 32 t − 16 t^(2)$] When will the wrench reach its greatest height, and what will that height be? The owners of a small fruit orchard decide to produce gift baskets as a sideline. The cost per basket for producing #math.equation(block: false, alt: "x")[$x$] baskets is #math.equation(block: true, alt: "C equals 0.01 x squared minus 2 x plus 120")[$C = 0.01 x^(2) − 2 x + 120$] How many baskets should they produce in order to minimize the cost per basket? What will their total cost be at that production level? #math.equation(block: false, alt: "100")[$100$] baskets, #math.equation(block: false, alt: "$ 2000")[$upright(\$) 2000$] A new electronics firm is considering marketing a line of telephones. The cost per phone for producing #math.equation(block: false, alt: "x")[$x$] telephones is #math.equation(block: true, alt: "C equals 0.001 x squared minus 3 x plus 2270")[$C = 0.001 x^(2) − 3 x + 2270$] How many telephones should the firm produce in order to minimize the cost per phone? What will the firm's total cost be at that production level? As part of a collage for her art class, Sheila wants to enclose a rectangle with #math.equation(block: false, alt: "100")[$100$] inches of yarn. + Let #math.equation(block: false, alt: "w")[$w$] represent the width of the rectangle, and write an expression for its length. Then write an expression that gives the area, #math.equation(block: false, alt: "A")[$A$], of the rectangle as a function of its width, #math.equation(block: false, alt: "w")[$w$]. + What is the area of the largest rectangle that Sheila can enclose with #math.equation(block: false, alt: "100")[$100$] inches of yarn? + Length: #math.equation(block: false, alt: "50 minus w")[$50 − w$]; Area: #math.equation(block: false, alt: "50 w minus w squared")[$50 w − w^(2)$] + #math.equation(block: false, alt: "625")[$625$] sq in Gavin has rented space for a booth at the county fair. As part of his display, he wants to rope off a rectangular area with #math.equation(block: false, alt: "80")[$80$] yards of rope. + Let #math.equation(block: false, alt: "w")[$w$] represent the width of the roped-off rectangle, and write an expression for its length. Then write an expression that gives the area, #math.equation(block: false, alt: "A")[$A$], of the roped-off space as a function of its width, #math.equation(block: false, alt: "w")[$w$]. + What is the largest area that Gavin can rope off? What will the dimensions of the rectangle be? A farmer plans to fence a rectangular grazing area along a river with 300 yards of fence as shown in the figure. #figure(figph[rectangle against river], alt: "rectangle against river", caption: none) + Write an expression that gives the area, #math.equation(block: false, alt: "A")[$A$], of the grazing land as a function of the width, #math.equation(block: false, alt: "w")[$w$], of the rectangle. + What is the largest area the farmer can enclose? + #math.equation(block: false, alt: "300 w minus 2 w squared")[$300 w − 2 w^(2)$] + #math.equation(block: false, alt: "11 , 250")[$11 , 250$] sq yd A breeder of horses wants to fence two rectangular grazing areas along a river with #math.equation(block: false, alt: "600")[$600$] meters of fence as shown in the figure. #figure(figph[two adjoining rectangles against river], alt: "two adjoining rectangles against river", caption: none) + Write an expression that gives the area, #math.equation(block: false, alt: "A")[$A$], of the grazing land as a function of the width, #math.equation(block: false, alt: "w")[$w$], of the rectangles. + What is the largest area the breeder can enclose? A travel agent offers a group rate of #math.equation(block: false, alt: "$ 2400")[$upright(\$) 2400$] per person for a week in London if #math.equation(block: false, alt: "16")[$16$] people sign up for the tour. For each additional person who signs up, the price per person is reduced by #math.equation(block: false, alt: "$ 100")[$upright(\$) 100$]. + Let #math.equation(block: false, alt: "x")[$x$] represent the number of additional people who sign up. Write expressions for the total number of people signed up, the price per person, and the total revenue. + How many people must sign up for the tour in order for the travel agent to maximize her revenue? + Number of people: #math.equation(block: false, alt: "16 plus x")[$16 + x$]; Price per person: #math.equation(block: false, alt: "2400 minus 100 x")[$2400 − 100 x$]; Total revenue: #math.equation(block: false, alt: "38 , 400 plus 800 x minus 100 x squared")[$38 , 400 + 800 x − 100 x^(2)$] + #math.equation(block: false, alt: "20")[$20$] An entrepreneur buys an apartment building with #math.equation(block: false, alt: "40")[$40$] units. The previous owner charged #math.equation(block: false, alt: "$ 240")[$upright(\$) 240$] per month for a single apartment and on the average rented #math.equation(block: false, alt: "32")[$32$] apartments at that price. The entrepreneur discovers that for every #math.equation(block: false, alt: "$ 20")[$upright(\$) 20$] he raises the price, another apartment stands vacant. + Let #math.equation(block: false, alt: "x")[$x$] represent the number of #math.equation(block: false, alt: "$ 20")[$upright(\$) 20$] price increases. Write expressions for the new price, the number of rented apartments, and the total revenue. + What price should the entrepreneur charge for an apartment in order to maximize his revenue? During a statistical survey, a public interest group obtains two estimates for the average monthly income of young adults aged 18 to 25. The first estimate is #math.equation(block: false, alt: "$ 860")[$upright(\$) 860$] and the second estimate is #math.equation(block: false, alt: "$ 918")[$upright(\$) 918$]. To refine its estimate, the group will take a weighted average of these two figures: #math.equation(block: true, alt: "I equals 860 a plus 918 open parenthesis 1 minus a close parenthesis where 0 less than or equal to a less than or equal to 1")[$I = 860 a + 918 ( 1 − a ) #hide($00$) " where " #hide($00$) 0 ≤ a ≤ 1$] To get the best estimate, the group must choose #math.equation(block: false, alt: "a")[$a$] to minimize the function #math.equation(block: true, alt: "V equals 576 a squared plus 5184 open parenthesis 1 minus a close parenthesis squared")[$V = 576 a^(2) + 5184 ( 1 − a )^(2)$] (The numbers that appear in this expression reflect the #strong[variance] of the data, which measures how closely the data cluster around the mean, or average.) Find the value of #math.equation(block: false, alt: "a")[$a$] that minimizes #math.equation(block: false, alt: "V")[$V$], and use this value to get a refined estimate for the average income. #math.equation(block: true, alt: "a equals 0.9")[$a = 0.9$]; #math.equation(block: true, alt: "I equals $ 865.80")[$I = upright(\$) 865.80$] The rate at which an antigen precipitates during an antigen-antibody reaction depends upon the amount of antigen present. For a fixed quantity of antibody, the time required for a particular antigen to precipitate is given in minutes by the function #math.equation(block: true, alt: "t equals 2 w squared minus 20 w plus 54")[$t = 2 w^(2) − 20 w + 54$] where #math.equation(block: false, alt: "w")[$w$] is the quantity of antigen present, in grams. For what quantity of antigen will the reaction proceed most rapidly, and how long will the precipitation take? For Problems 17–20, use transformations to graph the parabola. What is the vertex of each graph? + #math.equation(block: false, alt: "y equals open parenthesis x minus 3 close parenthesis squared")[$y = ( x − 3 )^(2)$] + #math.equation(block: false, alt: "y equals minus open parenthesis x minus 3 close parenthesis squared")[$y = − ( x − 3 )^(2)$] + #math.equation(block: false, alt: "y equals minus open parenthesis x minus 3 close parenthesis squared plus 4")[$y = − ( x − 3 )^(2) + 4$] #figure(figph[three parabolas], alt: "three parabolas", caption: none) + #math.equation(block: false, alt: "y equals open parenthesis x plus 1 close parenthesis squared")[$y = ( x + 1 )^(2)$] + #math.equation(block: false, alt: "y equals 2 open parenthesis x plus 1 close parenthesis squared")[$y = 2 ( x + 1 )^(2)$] + #math.equation(block: false, alt: "y equals 2 open parenthesis x plus 1 close parenthesis squared minus 4")[$y = 2 ( x + 1 )^(2) − 4$] + #math.equation(block: false, alt: "y equals open parenthesis x plus 4 close parenthesis squared")[$y = ( x + 4 )^(2)$] + #math.equation(block: false, alt: "y equals the fraction 1 over 2 open parenthesis x plus 4 close parenthesis squared")[$y = display(frac(1, 2)) ( x + 4 )^(2)$] + #math.equation(block: false, alt: "y equals 3 plus the fraction 1 over 2 open parenthesis x plus 4 close parenthesis squared")[$y = 3 + display(frac(1, 2)) ( x + 4 )^(2)$] #figure(figph[three parabolas], alt: "three parabolas", caption: none) + #math.equation(block: false, alt: "y equals open parenthesis x minus 2 close parenthesis squared")[$y = ( x − 2 )^(2)$] + #math.equation(block: false, alt: "y equals minus open parenthesis x minus 2 close parenthesis squared")[$y = − ( x − 2 )^(2)$] + #math.equation(block: false, alt: "y equals minus 3 minus open parenthesis x minus 2 close parenthesis squared")[$y = − 3 − ( x − 2 )^(2)$] In Problems 21–24, + Find the vertex of the parabola. + Use transformations to sketch the graph. + Write the equation in standard form. #math.equation(block: true, alt: "y equals 2 open parenthesis x minus 3 close parenthesis squared plus 4")[$y = 2 ( x − 3 )^(2) + 4$] + #math.equation(block: false, alt: "open parenthesis 3 , 4 close parenthesis")[$( 3 , 4 )$] + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "y equals 2 x squared minus 12 x plus 22")[$y = 2 x^(2) − 12 x + 22$] #math.equation(block: true, alt: "y equals minus 3 open parenthesis x plus 1 close parenthesis squared minus 2")[$y = − 3 ( x + 1 )^(2) − 2$] #math.equation(block: true, alt: "y equals minus the fraction 1 over 2 open parenthesis x plus 4 close parenthesis squared minus 3")[$y = − display(frac(1, 2)) ( x + 4 )^(2) − 3$] + #math.equation(block: false, alt: "open parenthesis minus 4 , minus 3 close parenthesis")[$( − 4 , − 3 )$] + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "y equals the fraction minus 1 over 2 x squared minus 4 x minus 11")[$y = display(frac(− 1, 2)) x^(2) − 4 x − 11$] #math.equation(block: true, alt: "y equals 4 open parenthesis x minus 2 close parenthesis squared minus 6")[$y = 4 ( x − 2 )^(2) − 6$] For Problems 25–30, + Write each equation in the form #math.equation(block: false, alt: "y equals a open parenthesis x minus p close parenthesis squared plus q")[$y = a ( x − p )^(2) + q$] by completing the square. + Using horizontal and vertical translations, sketch the graph by hand. #math.equation(block: true, alt: "y equals x squared minus 4 x plus 7")[$y = x^(2) − 4 x + 7$] + #math.equation(block: false, alt: "y equals open parenthesis x minus 2 close parenthesis squared plus 3")[$y = ( x − 2 )^(2) + 3$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals x squared minus 2 x minus 1")[$y = x^(2) − 2 x − 1$] #math.equation(block: true, alt: "y equals 3 x squared plus 6 x minus 2")[$y = 3 x^(2) + 6 x − 2$] + #math.equation(block: false, alt: "y equals 3 open parenthesis x plus 1 close parenthesis squared minus 5")[$y = 3 ( x + 1 )^(2) − 5$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals the fraction 1 over 2 x squared plus 2 x plus 5")[$y = display(frac(1, 2)) x^(2) + 2 x + 5$] #math.equation(block: true, alt: "y equals minus 2 x squared minus 8 x plus 3")[$y = − 2 x^(2) − 8 x + 3$] + #math.equation(block: false, alt: "y equals minus 2 open parenthesis x plus 2 close parenthesis squared plus 11")[$y = − 2 ( x + 2 )^(2) + 11$] + #figure(figph[parabola], alt: "parabola", caption: none) #math.equation(block: true, alt: "y equals minus x squared plus 5 x plus 2")[$y = − x^(2) + 5 x + 2$] A system of two quadratic equations may have no solution, one solution, or two solutions. Sketch a system illustrating each case. In your sketches, one of the parabolas should open up, and the other down. No solutions: #figure(figph[two non-intersecting parabolas], alt: "two non-intersecting parabolas", caption: none) One solution: #figure(figph[two tangent parabolas], alt: "two tangent parabolas", caption: none) Two solutions: #figure(figph[two parabolas intersecting in two points], alt: "two parabolas intersecting in two points", caption: none) A system of two quadratic equations may have no solution, one solution, or two solutions. Sketch a system illustrating each case. In your sketches, both parabolas should open up. For Problems 33–44, solve the system algebraically. Use your calculator to graph both equations and verify your solutions. #math.equation(block: true, alt: "y, equals x squared minus 4 x plus 7; y, equals 11 minus x")[$y & = x^(2) − 4 x + 7 \ y & = 11 − x$] #math.equation(block: true, alt: "open parenthesis minus 1 , 12 close parenthesis , open parenthesis 4 , 7 close parenthesis")[$( − 1 , 12 ) , ( 4 , 7 )$] #math.equation(block: true, alt: "y, equals x squared plus 6 x plus 4; y, equals 3 x plus 8")[$y & = x^(2) + 6 x + 4 \ y & = 3 x + 8$] #math.equation(block: true, alt: "y, equals minus x squared minus 2 x plus 7; y, equals 2 x plus 11")[$y & = − x^(2) − 2 x + 7 \ y & = 2 x + 11$] #math.equation(block: true, alt: "open parenthesis minus 2 , 7 close parenthesis")[$( − 2 , 7 )$] #math.equation(block: true, alt: "y equals x squared minus 8 x plus 17; y plus 4 x equals 13")[$& y = x^(2) − 8 x + 17 \ & y + 4 x = 13$] #math.equation(block: true, alt: "y equals x squared plus 8 x plus 8; 3 y plus 2 x equals minus 36")[$& y = x^(2) + 8 x + 8 \ & 3 y + 2 x = − 36$] No solution #math.equation(block: true, alt: "y equals minus x squared plus 4 x plus 2; 4 y minus 3 x equals 24")[$& y = − x^(2) + 4 x + 2 \ & 4 y − 3 x = 24$] #math.equation(block: true, alt: "y, equals x squared minus 9; y, equals minus 2 x squared plus 9 x plus 21")[$y & = x^(2) − 9 \ y & = − 2 x^(2) + 9 x + 21$] #math.equation(block: true, alt: "open parenthesis minus 2 , minus 5 close parenthesis , open parenthesis 5 , 16 close parenthesis")[$( − 2 , − 5 ) , ( 5 , 16 )$] #math.equation(block: true, alt: "y, equals 4 minus x squared; y, equals 3 x squared minus 12 x minus 12")[$y & = 4 − x^(2) \ y & = 3 x^(2) − 12 x − 12$] #math.equation(block: true, alt: "y, equals x squared minus 0.5 x plus 3.5; y, equals minus x squared plus 3.5 x plus 1.5")[$y & = x^(2) − 0.5 x + 3.5 \ y & = − x^(2) + 3.5 x + 1.5$] #math.equation(block: true, alt: "open parenthesis 1 , 4 close parenthesis")[$( 1 , 4 )$] #math.equation(block: true, alt: "y, equals x squared plus 10 x plus 22; y, equals minus 0.5 x squared minus 8 x minus 32")[$y & = x^(2) + 10 x + 22 \ y & = − 0.5 x^(2) − 8 x − 32$] #math.equation(block: true, alt: "y, equals x squared minus 4 x plus 4; y, equals x squared minus 8 x plus 16")[$y & = x^(2) − 4 x + 4 \ y & = x^(2) − 8 x + 16$] #math.equation(block: true, alt: "open parenthesis 3 , 1 close parenthesis")[$( 3 , 1 )$] #math.equation(block: true, alt: "y, equals 0.5 x squared plus 3 x plus 5.5; y, equals 2 x squared plus 12 x plus 4")[$y & = 0.5 x^(2) + 3 x + 5.5 \ y & = 2 x^(2) + 12 x + 4$] Problems 45–48 deal with wildlife management and sustainable yield. In Problem of Graphing Parabolas, you graphed the annual growth rate of a population of fish, #math.equation(block: true, alt: "R equals f open parenthesis x close parenthesis equals 0.4 x minus 0.0001 x squared")[$R = f ( x ) = 0.4 x − 0.0001 x^(2)$] where #math.equation(block: false, alt: "x")[$x$] is the current biomass of the population, in tons. + Suppose that fishermen harvest #math.equation(block: false, alt: "300")[$300$] tons of fish each year. Sketch the graph of #math.equation(block: false, alt: "H equals 300")[$H = 300$] on the same axes with your graph of #math.equation(block: false, alt: "y")[$y$]. + If the biomass is currently #math.equation(block: false, alt: "2500")[$2500$] tons and #math.equation(block: false, alt: "300")[$300$] tons are harvested, will the population be larger or smaller next year? By how much? What if the biomass is currently #math.equation(block: false, alt: "3500")[$3500$] tons? + What sizes of biomass will remain stable from year to year if #math.equation(block: false, alt: "300")[$300$] tons are harvested annually? + If the biomass ever falls below #math.equation(block: false, alt: "1000")[$1000$] tons, what will happen after several years of harvesting #math.equation(block: false, alt: "300")[$300$] tons annually? + #figure(figph[parabola and horizontal line], alt: "parabola and horizontal line", caption: none) + Larger, by #math.equation(block: false, alt: "75")[$75$] tons. Smaller, by #math.equation(block: false, alt: "125")[$125$] tons. + #math.equation(block: false, alt: "1000")[$1000$] tons and #math.equation(block: false, alt: "3000")[$3000$] tons + The fish population will decrease each year until it is completely depleted. In Problem of Graphing Parabolas, you graphed the annual increase, #math.equation(block: false, alt: "I")[$I$], in the deer population in a national park, #math.equation(block: true, alt: "I equals g open parenthesis x close parenthesis equals 1.2 x minus 0.0002 x squared")[$I = g ( x ) = 1.2 x − 0.0002 x^(2)$] where #math.equation(block: false, alt: "x")[$x$] is the current population. + Suppose hunters are allowed to kill #math.equation(block: false, alt: "1000")[$1000$] deer per year. Sketch the graph of #math.equation(block: false, alt: "H equals 1000")[$H = 1000$] on the same axes with a graph of #math.equation(block: false, alt: "y")[$y$]. + What sizes of deer populations will remain stable from year to year if #math.equation(block: false, alt: "1000")[$1000$] deer are hunted annually? + Suppose #math.equation(block: false, alt: "1600")[$1600$] deer are killed annually. What sizes of deer populations will remain stable? + What is the largest annual harvest that still allows for a stable population? (This harvest is called the maximum sustainable yield.) What is the stable population? + What eventually happens if the population falls below the stable value but hunting continues at the maximum sustainable yield? The annual increase, #math.equation(block: false, alt: "N")[$N$], in a bear population of size #math.equation(block: false, alt: "x")[$x$] is given by #math.equation(block: true, alt: "N equals F open parenthesis x close parenthesis equals 0.0002 x open parenthesis 2000 minus x close parenthesis")[$N = F ( x ) = 0.0002 x ( 2000 − x )$] if the bears are not hunted. The number of bears killed each year by hunters is related to the bear population by the equation #math.equation(block: false, alt: "K equals 0.2 x")[$K = 0.2 x$]. (Notice that in this model, hunting is adjusted to the size of the bear population.) + Sketch the graphs of #math.equation(block: false, alt: "N")[$N$] and #math.equation(block: false, alt: "K")[$K$] on the same axes. + When the bear population is #math.equation(block: false, alt: "1200")[$1200$], which is greater, #math.equation(block: false, alt: "N")[$N$] or #math.equation(block: false, alt: "K")[$K$]? Will the population increase or decrease in the next year? By how many bears? + When the bear population is #math.equation(block: false, alt: "900")[$900$], will the population increase or decrease in the next year? By how many bears? + What sizes of bear population will remain stable after hunting? + What sizes of bear populations will increase despite hunting? What sizes of populations will decrease? + Toward what size will the population tend over time? + Suppose hunting limits are raised so that #math.equation(block: false, alt: "K equals 0.3 x")[$K = 0.3 x$]. Toward what size will the population tend over time? + #figure(figph[parabola and line of positive slope], alt: "parabola and line of positive slope", caption: none) + #math.equation(block: false, alt: "K greater than N")[$K > N$]. The population will decrease by #math.equation(block: false, alt: "48")[$48$] bears. + The population will increase by #math.equation(block: false, alt: "18")[$18$] bears. + #math.equation(block: false, alt: "1000")[$1000$] + Populations between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "1000")[$1000$] will increase; populations over #math.equation(block: false, alt: "1000")[$1000$] will decrease. + #math.equation(block: false, alt: "1000")[$1000$] (unless the population is #math.equation(block: false, alt: "0")[$0$]) + #math.equation(block: false, alt: "500")[$500$] (unless the population is #math.equation(block: false, alt: "0")[$0$]) The annual increase in the biomass of a whale population is given in tons by #math.equation(block: true, alt: "w equals G open parenthesis x close parenthesis equals 0.001 x open parenthesis 1000 minus x close parenthesis")[$w = G ( x ) = 0.001 x ( 1000 − x )$] where #math.equation(block: false, alt: "x")[$x$] is the current population, also in tons. + Sketch a graph of #math.equation(block: false, alt: "w")[$w$] for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 1100")[$0 ≤ x ≤ 1100$]. What size biomass remains stable? + Each year hunters are allowed to harvest a biomass given by #math.equation(block: false, alt: "H equals 0.6 x")[$H = 0.6 x$]. Sketch #math.equation(block: false, alt: "H")[$H$] on the same graph with #math.equation(block: false, alt: "w")[$w$]. What is the stable biomass with hunting? + What sizes of populations will increase despite hunting? What sizes will decrease? + What size will the population approach over time? What biomass are hunters allowed to harvest for that size population? + Find a value of #math.equation(block: false, alt: "k")[$k$] so that the graph of #math.equation(block: false, alt: "H equals k x")[$H = k x$] will pass through the vertex of #math.equation(block: false, alt: "w equals 0.001 x open parenthesis 1000 minus x close parenthesis")[$w = 0.001 x ( 1000 − x )$]. + For the value of #math.equation(block: false, alt: "k")[$k$] found in part (e), what size will the population approach over time? What biomass are hunters allowed to harvest for that size population? + Explain why the whaling industry should prefer hunting quotas of #math.equation(block: false, alt: "k x")[$k x$] rather than #math.equation(block: false, alt: "0.6 x")[$0.6 x$] for a long-term strategy, even though #math.equation(block: false, alt: "0.6 x greater than k x")[$0.6 x > k x$] for any positive value of #math.equation(block: false, alt: "x")[$x$]. For Problems 49–52, + Find the break-even points by solving a system of equations. + Graph the equations for Revenue and Cost in the same window and verify your solutions on the graph. + Use the fact that #math.equation(block: true, alt: "Profit equals Revenue minus Cost")[$"Profit" = "Revenue" − "Cost"$] to find the value of #math.equation(block: false, alt: "x")[$x$] for which profit is maximum. Writewell, Inc. makes fountain pens. It costs Writewell #math.equation(block: true, alt: "C equals 0.0075 x squared plus x plus 2100")[$C = 0.0075 x^(2) + x + 2100$] dollars to manufacture #math.equation(block: false, alt: "x")[$x$] pens, and the company receives #math.equation(block: false, alt: "R equals 13 x")[$R = 13 x$] dollars in revenue from the sale of the pens. + #math.equation(block: false, alt: "open parenthesis 200 , 2600 close parenthesis , open parenthesis 1400 , 18 , 200 close parenthesis")[$( 200 , " " 2600 ) , ( 1400 , " " 18 , 200 )$] + #figure(figph[parabola and line], alt: "parabola and line", caption: none) + #math.equation(block: false, alt: "x equals 800")[$x = 800$] It costs The Sweetshop #math.equation(block: true, alt: "C equals 0.01 x squared plus 1836")[$C = 0.01 x^(2) + 1836$] dollars to produce #math.equation(block: false, alt: "x")[$x$] pounds of chocolate creams. The company brings in #math.equation(block: false, alt: "R equals 12 x")[$R = 12 x$] dollars revenue from the sale of the chocolates. It costs an appliance manufacturer #math.equation(block: true, alt: "C equals 1.625 x squared plus 33 , 150")[$C = 1.625 x^(2) + 33 , 150$] dollars to produce #math.equation(block: false, alt: "x")[$x$] front-loading washing machines, which will then bring in revenues of #math.equation(block: false, alt: "R equals 650 x")[$R = 650 x$] dollars. + #math.equation(block: false, alt: "open parenthesis 60 , 39 , 000 close parenthesis , open parenthesis 340 , 221 , 000 close parenthesis")[$( 60 , " " 39 , 000 ) , ( 340 , " " 221 , 000 )$] + #figure(figph[parabola and line], alt: "parabola and line", caption: none) + #math.equation(block: false, alt: "x equals 200")[$x = 200$] A company can produce #math.equation(block: false, alt: "x")[$x$] lawn mowers for a cost of #math.equation(block: true, alt: "C equals 0.125 x squared plus 100 , 000")[$C = 0.125 x^(2) + 100 , 000$] dollars. The sale of the lawn mowers will generate #math.equation(block: false, alt: "R equals 300 x")[$R = 300 x$] dollars in revenue. Problems 53 and 54 prove that the vertical line #math.equation(block: false, alt: "x equals the fraction minus b over 2 a")[$x = display(frac(− b, 2 a))$] is the axis of symmetry of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$]. A graph is #strong[symmetric about the line #math.equation(block: false, alt: "x equals h")[$x = h$]] if the point #math.equation(block: false, alt: "open parenthesis h plus d , v close parenthesis")[$( h + d , v )$] lies on the graph whenever the point #math.equation(block: false, alt: "open parenthesis h minus d , v close parenthesis")[$( h − d , v )$] lies on the graph. + Sketch a parabola #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a ( x − h )^(2) + k$] and the line #math.equation(block: false, alt: "x equals h")[$x = h$]. We will show that the parabola is symmetric about the line #math.equation(block: false, alt: "x equals h")[$x = h$]. + Label a point on the parabola with #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "x equals h plus d")[$x = h + d$], where #math.equation(block: false, alt: "d greater than 0")[$d > 0$]. What is the #math.equation(block: false, alt: "y")[$y$]-coordinate of that point? + Label the point on the parabola with #math.equation(block: false, alt: "x")[$x$]-coordinate #math.equation(block: false, alt: "x equals h minus d")[$x = h − d$]. What is the #math.equation(block: false, alt: "y")[$y$]-coordinate of that point? + Explain why your answers to parts (b) and (c) prove that the line #math.equation(block: false, alt: "x equals h")[$x = h$] is the axis of symmetry for the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus h close parenthesis squared plus k")[$f ( x ) = a ( x − h )^(2) + k$]. + #figure(figph[parabola and axis of symmetry], alt: "parabola and axis of symmetry", caption: none) + See graph and (c) + #math.equation(block: false, alt: "a d squared plus k")[$a d^(2) + k$] + The two points on the parabola that are the same horizontal distance from the line #math.equation(block: false, alt: "x equals h")[$x = h$] the axis of symmetry have the same #math.equation(block: false, alt: "y")[$y$]-coordinate, so they are symmetric about that line. To find the axis of symmetry for the graph of #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals a x squared plus b x plus c")[$g ( x ) = a x^(2) + b x + c$], we will use the results of Problem 51 and the technique of completing the square. + Write the equation #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] in vertex form by completing the square. (Follow the steps in Example.) + Your answer to part (a) has the form #math.equation(block: false, alt: "y equals a open parenthesis x minus h close parenthesis squared plus k")[$y = a ( x − h )^(2) + k$]. What is your value of #math.equation(block: false, alt: "h")[$h$]? What is your value of #math.equation(block: false, alt: "k")[$k$]? + What is the axis of symmetry for the parabola #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals a x squared plus b x plus c")[$g ( x ) = a x^(2) + b x + c$]?