#set document(title: "6.1 Factors and x -Intercepts", 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")) == 6.1#h(0.6em)Factors and #math.equation(block: false, alt: "x")[$x$]-Intercepts In Height of a Baseball, perhaps you recognized the graph of the baseball's height as a parabola. In this chapter, we shall see that the graph of any quadratic function is a parabola. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quadratic Function] A #strong[quadratic function] is one that can be written in the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] where #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] are constants, and #math.equation(block: false, alt: "a")[$a$] is not equal to zero. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ In the definition above, notice that if #math.equation(block: false, alt: "a")[$a$] is zero, there is no #math.equation(block: false, alt: "x")[$x$]-squared term, so the function is not quadratic. ] In Height of a Baseball, the height of a baseball #math.equation(block: false, alt: "t")[$t$] seconds after being hit was given by #math.equation(block: true, alt: "h equals minus 16 t squared plus 64 t plus 4")[$h = − 16 t^(2) + 64 t + 4$] We used a graph to find two times when the baseball was #math.equation(block: false, alt: "64")[$64$] feet high. Can we solve the same problem algebraically? We are looking for values of #math.equation(block: false, alt: "t")[$t$] that produce #math.equation(block: false, alt: "h equals 64")[$h = 64$] in the height equation. So, if we substitute #math.equation(block: false, alt: "h equals 64")[$h = 64$] into the height equation, we would like to solve the #strong[quadratic equation] #math.equation(block: true, alt: "64 equals minus 16 t squared plus 64 t plus 4")[$64 = − 16 t^(2) + 64 t + 4$] This equation cannot be solved by extraction of roots, because there are two terms containing the variable #math.equation(block: false, alt: "t")[$t$], and they cannot be combined. To solve this equation, we will appeal to a property of our number system, called the #strong[zero-factor principle]. === Zero-Factor Principle Can you multiply two numbers together and obtain a product of zero? Only if one of the two numbers happens to be zero. This property of numbers is called the #strong[zero-factor principle]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero-Factor Principle] The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols, #math.equation(block: true, alt: "a b equals 0 if and only if a equals 0 or b equals 0")[$a b = 0 " " " " " if and only if " " " " " a = 0 " " " " " or " " " " " b = 0$] ] The principle is true even if the numbers #math.equation(block: false, alt: "a")[$a$] and #math.equation(block: false, alt: "b")[$b$] are represented by algebraic expressions, such as #math.equation(block: false, alt: "x minus 5")[$x − 5$] or #math.equation(block: false, alt: "2 x plus 1")[$2 x + 1$]. For example, if #math.equation(block: true, alt: "open parenthesis x minus 5 close parenthesis open parenthesis 2 x plus 1 close parenthesis equals 0")[$( x − 5 ) ( 2 x + 1 ) = 0$] then it must be true that either #math.equation(block: false, alt: "x minus 5 equals 0")[$x − 5 = 0$] or #math.equation(block: false, alt: "2 x plus 1 equals 0")[$2 x + 1 = 0$]. Thus, we can use the zero-factor principle to solve equations. #examplebox("Example 1")[][ + Solve the equation #math.equation(block: false, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 2 close parenthesis equals 0")[$( x − 6 ) ( x + 2 ) = 0$]. + Find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 4 x minus 12")[$f ( x ) = x^(2) − 4 x − 12$]. #solutionbox[ + We apply the zero-factor principle to the product #math.equation(block: false, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 2 close parenthesis")[$( x − 6 ) ( x + 2 )$]. #math.equation(block: true, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 2 close parenthesis, equals 0, Set each factor equal to zero.; x minus 6 equals 0, or x plus 2 equals 0, Solve each equation.; x equals 6, or x equals minus 2")[$( x − 6 ) ( x + 2 ) & = 0 & & "Set each factor equal to zero." \ x − 6 = 0 " " " " & "or" " " " " x + 2 = 0 & & "Solve each equation." \ " " " " " " x = 6 " " " " & "or" " " " " x = − 2$] There are two solutions, #math.equation(block: false, alt: "6")[$6$] and #math.equation(block: false, alt: "minus 2")[$− 2$]. (You should check that both of these values satisfy the original equation.) + To find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph, we set #math.equation(block: false, alt: "y equals 0")[$y = 0$] and solve the equation #math.equation(block: true, alt: "0 equals x squared minus 4 x minus 12")[$0 = x^(2) − 4 x − 12$] But this is the equation we solved in part (a), because #math.equation(block: false, alt: "open parenthesis x minus 6 close parenthesis open parenthesis x plus 2 close parenthesis equals x squared minus 4 x minus 12")[$( x − 6 ) ( x + 2 ) = x^(2) − 4 x − 12$]. The solutions of that equation were #math.equation(block: false, alt: "6")[$6$] and #math.equation(block: false, alt: "minus 2")[$− 2$], so the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph are #math.equation(block: false, alt: "6")[$6$] and #math.equation(block: false, alt: "minus 2")[$− 2$]. You can see this by graphing the equation on your calculator, as shown below.#figure(figph[parabola with x-intercepts], alt: "parabola with x-intercepts", caption: none) ] ] Example illustrates an important fact about the #math.equation(block: false, alt: "x")[$x$]-intercepts of a graph. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[#math.equation(block: false, alt: "x")[$x$]-Intercepts of a Graph] The #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] are the solutions of the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$]. ] What are the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals open parenthesis 2 x minus 7 close parenthesis open parenthesis x plus 2 close parenthesis")[$y = ( 2 x − 7 ) ( x + 2 )$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "the fraction 7 over 2")[$display(frac(7, 2))$] and #math.equation(block: false, alt: "minus 2")[$− 2$] ] What are the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals open parenthesis 2 x minus 7 close parenthesis open parenthesis x plus 2 close parenthesis")[$y = ( 2 x − 7 ) ( x + 2 )$] ? + #math.equation(block: false, alt: "the fraction 7 over 2")[$display(frac(7, 2))$] and #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "minus the fraction 7 over 2")[$− display(frac(7, 2))$] and #math.equation(block: false, alt: "2")[$2$] + #math.equation(block: false, alt: "the fraction 2 over 7")[$display(frac(2, 7))$] and #math.equation(block: false, alt: "minus 2")[$− 2$] + #math.equation(block: false, alt: "minus the fraction 2 over 7")[$− display(frac(2, 7))$] and #math.equation(block: false, alt: "2")[$2$] Graph the function #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x minus 3 close parenthesis open parenthesis 2 x plus 3 close parenthesis")[$f ( x ) = ( x − 3 ) ( 2 x + 3 )$] on a calculator, and use your graph to solve the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$]. (Use #math.equation(block: false, alt: "Xmin equals minus 9.4")[$"Xmin" = − 9.4$], #math.equation(block: false, alt: "Xmax equals 9.4")[$"Xmax" = 9.4$].) Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ \[Separate different values with a comma.\] Check your answer with the zero-factor principle. #solutionbox[ #math.equation(block: true, alt: "x equals minus the fraction 3 over 2")[$x = − display(frac(3, 2))$], #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] Graph the function #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals open parenthesis x minus 3 close parenthesis open parenthesis 2 x plus 3 close parenthesis")[$f ( x ) = ( x − 3 ) ( 2 x + 3 )$] with technology, and use your graph to solve the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$]. (Use #math.equation(block: false, alt: "Xmin equals minus 9.4")[$"Xmin" = − 9.4$], #math.equation(block: false, alt: "Xmax equals 9.4")[$"Xmax" = 9.4$].) Check your answer with the zero-factor principle. #solutionbox[ #math.equation(block: true, alt: "x equals the fraction minus 3 over 2")[$x = display(frac(− 3, 2))$], #math.equation(block: true, alt: "x equals 3")[$x = 3$] ] How can you use a graph to factor a quadratic expression? \_\_\_\_\_ How can you use a graph to factor a quadratic expression? === Solving Quadratic Equations by Factoring Before we apply the zero-factor principle to solve a quadratic equation, we must first write the equation so that one side of the equation is zero. Let us introduce some terminology. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Forms for Quadratic Equations] + A quadratic equation written #math.equation(block: true, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] is in #strong[standard form]. + A quadratic equation written #math.equation(block: true, alt: "a open parenthesis x minus r sub 1 close parenthesis open parenthesis x minus r sub 2 close parenthesis equals 0")[$a ( x − r_(1) ) ( x − r_(2) ) = 0$] is in #strong[factored form]. ] Once we have written the equation in standard form, we factor the left side and set each variable factor equal to zero separately. (See Appendix A.8 Factoring Quadratic Trinomials to review factoring.) #examplebox("Example 2")[][ Solve #math.equation(block: false, alt: "3 x open parenthesis x plus 1 close parenthesis equals 2 x plus 2")[$" " " " 3 x ( x + 1 ) = 2 x + 2$] #solutionbox[ First, we write the equation in standard form. #math.equation(block: true, alt: "3 x open parenthesis x plus 1 close parenthesis, equals 2 x plus 2, Apply the distributive law to the left side.; 3 x squared plus 3 x, equals 2 x plus 2, Subtract 2 x plus 2 from both sides.; 3 x squared plus x minus 2, equals 0")[$3 x ( x + 1 ) & = 2 x + 2 & & "Apply the distributive law to the left side." \ 3 x^(2) + 3 x & = 2 x + 2 & & "Subtract " 2 bold(italic(x)) + 2 " from both sides." \ 3 x^(2) + x − 2 & = 0$] Next, we factor the left side to obtain #math.equation(block: true, alt: "open parenthesis 3 x minus 2 close parenthesis open parenthesis x plus 1 close parenthesis equals 0")[$( 3 x − 2 ) ( x + 1 ) = 0$] We then apply the zero-factor principle by setting each factor equal to zero. #math.equation(block: true, alt: "3 x minus 2 equals 0 or x plus 1 equals 0")[$3 x − 2 = 0 " " " " " or " " " " " x + 1 = 0$] Finally, we solve each equation to find #math.equation(block: true, alt: "x equals the fraction 2 over 3 or x equals minus 1")[$x = frac(2, 3) " " " " " or " " " " " x = − 1$] The solutions are #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] and #math.equation(block: false, alt: "minus 1")[$− 1$]. ] ] #notebox("Caution", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ When we apply the zero-factor principle, one side of the equation #emph[must be zero]. For example, to solve the equation #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x minus 4 close parenthesis equals 15")[$( x − 2 ) ( x − 4 ) = 15$] it is incorrect to set each factor equal to #math.equation(block: false, alt: "15")[$15$]! (There are many ways that the product of two numbers can equal #math.equation(block: false, alt: "15")[$15$]; it is not necessary that one of the numbers be #math.equation(block: false, alt: "15")[$15$].) We must first simplify the left side and write the equation in standard form. (The correct solutions are #math.equation(block: false, alt: "7")[$7$] and #math.equation(block: false, alt: "minus 1")[$− 1$]; check that you can find these solutions.) ] We summarize the factoring method for solving quadratic equations as follows. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Quadratic Equation by Factoring] + Write the equation in standard form. + Factor the left side of the equation. + Apply the zero-factor principle: Set each factor equal to zero. + Solve each equation. There are two solutions (which may be equal). ] Solve by factoring: #math.equation(block: false, alt: "open parenthesis t minus 3 close parenthesis squared equals 3 open parenthesis 9 minus t close parenthesis")[$( t − 3 )^(2) = 3 ( 9 − t )$] Solutions: #math.equation(block: false, alt: "t equals")[$t =$]\_\_\_\_\_ \[Separate different values with a comma.\] #solutionbox[ After rewriting the equation in standard form and then factoring, we find #math.equation(block: false, alt: "t equals minus 3")[$t = − 3$] and #math.equation(block: false, alt: "t equals 6")[$t = 6$]. ] Solve by factoring: #math.equation(block: false, alt: "open parenthesis t minus 3 close parenthesis squared equals 3 open parenthesis 9 minus t close parenthesis")[$( t − 3 )^(2) = 3 ( 9 − t )$] #solutionbox[ After rewriting the equation in standard form and then factoring, we find #math.equation(block: false, alt: "t equals minus 3 , t equals 6")[$t = − 3 , " " " " t = 6$]. ] We can use factoring to solve the equation from Height of a Baseball. #examplebox("Example 3")[][ The height, #math.equation(block: false, alt: "h")[$h$], of a baseball #math.equation(block: false, alt: "t")[$t$] seconds after being hit is given by #math.equation(block: true, alt: "h equals minus 16 t squared plus 64 t plus 4")[$h = − 16 t^(2) + 64 t + 4$] When will the baseball reach a height of #math.equation(block: false, alt: "64")[$64$] feet? #solutionbox[ We substitute #math.equation(block: false, alt: "64")[$64$] for #math.equation(block: false, alt: "h")[$h$] in the formula, and solve for #math.equation(block: false, alt: "t")[$t$]. #math.equation(block: true, alt: "64 equals minus 16 t squared plus 64 t, plus 4, Write the equation in standard form.; 16 t squared minus 64 t plus 60, equals 0, Factor 4 from the left side.; 4 open parenthesis 4 t squared minus 16 t plus 15 close parenthesis, equals 0, Factor the quadratic expression.; 4 open parenthesis 2 t minus 3 close parenthesis open parenthesis 2 t minus 5 close parenthesis, equals 0, Set each variable factor equal to zero.; 2 t minus 3 equals 0 or 2 t minus 5, equals 0, Solve each equation.; t equals the fraction 3 over 2 or t, equals the fraction 5 over 2")[$64 = − 16 t^(2) + 64 t & + 4 & & "Write the equation in standard form." \ 16 t^(2) − 64 t + 60 & = 0 & & "Factor 4 from the left side." \ 4 ( 4 t^(2) − 16 t + 15 ) & = 0 & & "Factor the quadratic expression." \ 4 ( 2 t − 3 ) ( 2 t − 5 ) & = 0 & & "Set each variable factor equal to zero." \ 2 t − 3 = 0 " " " " " or " " " " " 2 t − 5 & = 0 & & "Solve each equation." \ t = frac(3, 2) " " " " " or " " " " " t & = frac(5, 2)$] There are two solutions to the quadratic equation. At #math.equation(block: false, alt: "t equals the fraction 3 over 2")[$t = display(frac(3, 2))$] seconds, the ball reaches a height of #math.equation(block: false, alt: "64")[$64$] feet on the way up, and at #math.equation(block: false, alt: "t equals the fraction 5 over 2")[$t = display(frac(5, 2))$] seconds, the ball is #math.equation(block: false, alt: "64")[$64$] feet high on its way down. #figure(figph[height of baseball], alt: "height of baseball", caption: none) ] ] In the solution to Example, the factor #math.equation(block: false, alt: "4")[$4$] does not affect the solutions of the equation at all. You can understand why this is true by looking at some graphs. First, check that the two equations #math.equation(block: true, alt: "x squared minus 4 x plus 3 equals 0 and 4 open parenthesis x squared minus 4 x plus 3 close parenthesis equals 0")[$x^(2) − 4 x + 3 = 0 " " " " " and " " " " " 4 ( x^(2) − 4 x + 3 ) = 0$] have the same solutions, #math.equation(block: false, alt: "x equals 1")[$x = 1$] and #math.equation(block: false, alt: "x equals 3")[$x = 3$]. Then use your graphing calculator to graph the equation #math.equation(block: true, alt: "Y sub 1 equals X squared minus 4 X plus 3")[$Y_(1) = X^(2) − 4 X + 3$] in the window #math.equation(block: true, alt: "Xmin, equals minus 2, Xmax equals 8; Ymin, equals minus 5, Ymax equals 10")[$"Xmin" & = − 2 & & "Xmax" = 8 \ "Ymin" & = − 5 & & "Ymax" = 10$] Notice that when #math.equation(block: false, alt: "y equals 0")[$y = 0$], #math.equation(block: false, alt: "x equals 3")[$x = 3$] or #math.equation(block: false, alt: "x equals 1")[$x = 1$]. These two points are the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph. In the same window, now graph #math.equation(block: true, alt: "Y sub 2 equals 4 open parenthesis X squared minus 4 X plus 3 close parenthesis")[$Y_(2) = 4 ( X^(2) − 4 X + 3 )$] This graph has the same #math.equation(block: false, alt: "x")[$x$]-values when #math.equation(block: false, alt: "y equals 0")[$y = 0$]. The factor of #math.equation(block: false, alt: "4")[$4$] stretches the graph vertically but does not change the location of the #math.equation(block: false, alt: "x")[$x$]-intercepts. #figure(figph[two parabolas with the same x-intercepts], alt: "two parabolas with the same x-intercepts", caption: none) What happens to the #math.equation(block: false, alt: "x")[$x$]-intercepts when you multiply the right side of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] by 3? \_\_\_\_\_ #solutionbox[ They are unchanged. ] What happens to the #math.equation(block: false, alt: "x")[$x$]-intercepts when you multiply the right side of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] by 3? + They are tripled. + They are divided by 3. + They move 3 units to the right. + They are unchanged. The value of the constant factor #math.equation(block: false, alt: "a")[$a$] in the factored form of a quadratic function, #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals a open parenthesis x minus r sub 1 close parenthesis open parenthesis x minus r sub 2 close parenthesis")[$f ( x ) = a ( x − r_(1) ) ( x − r_(2) )$], does not affect the location of the #math.equation(block: false, alt: "x")[$x$]-intercepts, because it does not affect the solutions of the equation #math.equation(block: false, alt: "a open parenthesis x minus r sub 1 close parenthesis open parenthesis x minus r sub 2 close parenthesis equals 0")[$a ( x − r_(1) ) ( x − r_(2) ) = 0$]. + Solve #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 4 t minus t squared equals 0")[$f ( t ) = 4 t − t^(2) = 0$] by factoring. #linebreak() Solutions: #math.equation(block: false, alt: "t equals")[$t =$]\_\_\_\_\_ \[Separate different values with a comma.\] + Solve #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals 20 t minus 5 t squared equals 0")[$g ( t ) = 20 t − 5 t^(2) = 0$] by factoring. #linebreak() Solutions: #math.equation(block: false, alt: "t equals")[$t =$]\_\_\_\_\_ \[Separate different values with a comma.\] + Graph #math.equation(block: false, alt: "y equals f open parenthesis t close parenthesis")[$y = f ( t )$] and #math.equation(block: false, alt: "y equals g open parenthesis t close parenthesis")[$y = g ( t )$] together in the window #math.equation(block: true, alt: "Xmin, equals minus 2, Xmax equals 6; Ymin, equals minus 20, Ymax equals 25")[$"Xmin" & = − 2 & & "Xmax" = 6 \ "Ymin" & = − 20 & & "Ymax" = 25$] and locate the horizontal intercepts of each graph. #linebreak() Horizontal intercepts: \_\_\_\_\_ \[Separate different ordered pairs with a comma.\] #solutionbox[ + #math.equation(block: false, alt: "t equals 0")[$t = 0$], #math.equation(block: false, alt: "t equals 4")[$t = 4$] + #math.equation(block: false, alt: "t equals 0")[$t = 0$], #math.equation(block: false, alt: "t equals 4")[$t = 4$] + #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$] ] + Solve #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals 4 t minus t squared equals 0")[$f ( t ) = 4 t − t^(2) = 0$] by factoring. + Solve #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals 20 t minus 5 t squared equals 0")[$g ( t ) = 20 t − 5 t^(2) = 0$] by factoring. + Graph #math.equation(block: false, alt: "y equals f open parenthesis t close parenthesis")[$y = f ( t )$] and #math.equation(block: false, alt: "y equals g open parenthesis t close parenthesis")[$y = g ( t )$] together in the window #math.equation(block: true, alt: "Xmin, equals minus 2, Xmax equals 6; Ymin, equals minus 20, Ymax equals 25")[$"Xmin" & = − 2 & & "Xmax" = 6 \ "Ymin" & = − 20 & & "Ymax" = 25$] and locate the horizontal intercepts of each graph. #solutionbox[ + #math.equation(block: false, alt: "t equals 0")[$t = 0$], #math.equation(block: false, alt: "t equals 4")[$t = 4$] + #math.equation(block: false, alt: "t equals 0")[$t = 0$], #math.equation(block: false, alt: "t equals 4")[$t = 4$] + #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$], #math.equation(block: false, alt: "open parenthesis 4 , 0 close parenthesis")[$( 4 , 0 )$] ] Explain why the solutions of #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x minus 6 close parenthesis equals 10")[$( x − 3 ) ( x − 6 ) = 10$] are not 3 and 6. \_\_\_\_\_ Explain why the solutions of #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x minus 6 close parenthesis equals 10")[$( x − 3 ) ( x − 6 ) = 10$] are not 3 and 6. === Applications Here is another example of how quadratic equations arise in applications. #examplebox("Example 4")[][ #figure(figph[computer screen], alt: "computer screen", caption: none) The size of a rectangular computer monitor screen is given by the length of its diagonal, as shown at left. If the length of the screen should be #math.equation(block: false, alt: "3")[$3$] inches greater than its width, what are the dimensions of a #math.equation(block: false, alt: "15")[$15$]-inch monitor? #solutionbox[ We express the two dimensions of the screen in terms of a single variable: #figure(table( columns: 1, align: left, inset: 6pt, [Width of screen: #math.equation(block: false, alt: "w")[$" " " " w$]], [Length of screen: #math.equation(block: false, alt: "w plus 3")[$" " " " w + 3$]], )) We can use the Pythagorean theorem to write an equation. #math.equation(block: true, alt: "w squared plus open parenthesis w plus 3 close parenthesis squared equals 15 squared")[$w^(2) + ( w + 3 )^(2) = 15^(2)$] Solve the equation. Begin by simplifying the left side. #math.equation(block: true, alt: "w squared plus w squared plus 6 w plus 9, equals 225, Write the equation in standard form.; 2 w squared plus 6 w minus 216, equals 0, Factor 2 from the left side.; 2 open parenthesis w squared plus 3 w minus 108 close parenthesis, equals 0, Factor the quadratic expression.; 2 open parenthesis w minus 9 close parenthesis open parenthesis w plus 12 close parenthesis, equals 0, Set each variable factor equal to zero.; w minus 9 equals 0 or w plus 12, equals 0, Solve each equation.; w equals 9 or w, equals minus 12")[$w^(2) + w^(2) + 6 w + 9 & = 225 & & "Write the equation in standard form." \ 2 w^(2) + 6 w − 216 & = 0 & & "Factor 2 from the left side." \ 2 ( w^(2) + 3 w − 108 ) & = 0 & & "Factor the quadratic expression." \ 2 ( w − 9 ) ( w + 12 ) & = 0 & & "Set each variable factor equal to zero." \ w − 9 = 0 " " " " " or " " " " " w + 12 & = 0 & & "Solve each equation." \ w = 9 " " " " " " " or " " " " " " " w & = − 12$] Because the width of the screen cannot be a negative number, we discard the solution #math.equation(block: false, alt: "w equals minus 12")[$w = − 12$]. Thus, the width is #math.equation(block: false, alt: "w equals 9")[$w = 9$] inches, and the length is #math.equation(block: false, alt: "w plus 3 equals 12")[$w + 3 = 12$] inches. ] ] #figure(figph[garden], alt: "garden", caption: none) Francine is designing the layout for a botanical garden. The plan includes a square herb garden, with a path 5 feet wide through the center of the garden, as shown above. To include all the species of herbs, the planted area must be 300 square feet. Find the dimensions of the herb garden. Answer: \_\_\_\_\_ feet by \_\_\_\_\_ feet #solutionbox[ 20 feet by 20 feet. ] Francine is designing the layout for a botanical garden. The plan includes a square herb garden, with a path 5 feet wide through the center of the garden, as shown below. To include all the species of herbs, the planted area must be 300 square feet. Find the dimensions of the herb garden. #figure(figph[garden], alt: "garden", caption: none) #solutionbox[ The dimensions of the planted area are #math.equation(block: false, alt: "x")[$x$] feet by #math.equation(block: false, alt: "x minus 5")[$x − 5$] feet. Write an equation about the area of the planted area and solve, to find that the dimensions of the garden are 20 feet by 20 feet. ] If the perimeter of a rectangle is 56 inches and its width is #math.equation(block: false, alt: "x")[$x$] inches, what is an expression for its length? \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "28 minus x")[$28 − x$] ] If the perimeter of a rectangle is 56 inches and its width is #math.equation(block: false, alt: "x")[$x$] inches, what is an expression for its length? + #math.equation(block: false, alt: "56 minus x")[$56 − x$] + #math.equation(block: false, alt: "28 minus x")[$28 − x$] + #math.equation(block: false, alt: "56 x")[$56 x$] + #math.equation(block: false, alt: "the fraction 28 over x")[$display(frac(28, x))$] === Solutions of Quadratic Equations We have seen that the solutions of the quadratic equation #math.equation(block: true, alt: "a open parenthesis x minus r sub 1 close parenthesis open parenthesis x minus r sub 2 close parenthesis equals 0")[$a ( x − r_(1) ) ( x − r_(2) ) = 0$] are #math.equation(block: false, alt: "r sub 1")[$r_(1)$] and #math.equation(block: false, alt: "r sub 2")[$r_(2)$]. Thus, if we know the two solutions of a quadratic equation, we can work backward and reconstruct the equation, starting from its factored form. We can then write the equation in standard form by multiplying together the factors. #examplebox("Example 5")[][ Find a quadratic equation whose solutions are #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] and #math.equation(block: false, alt: "minus 3")[$− 3$]. #solutionbox[ The quadratic equation is #math.equation(block: true, alt: "open parenthesis x minus the fraction 1 over 2 close parenthesis open bracket x minus open parenthesis minus 3 close parenthesis close bracket, equals 0; open parenthesis x minus the fraction 1 over 2 close parenthesis open parenthesis x plus 3 close parenthesis, equals 0")[$( x − frac(1, 2) ) [ x − ( − 3 ) ] & = 0 \ ( x − frac(1, 2) ) ( x + 3 ) & = 0$] To write the equation in standard form, we multiply the factors together. #math.equation(block: true, alt: "x squared plus the fraction 5 over 2 x minus the fraction 3 over 2 equals 0")[$x^(2) + frac(5, 2) x − frac(3, 2) = 0$] We can also find an equation with integer coefficients if we clear the equation of fractions. Multiply both sides by #math.equation(block: false, alt: "2")[$2$]: #math.equation(block: true, alt: "2 open parenthesis x squared plus the fraction 5 over 2 x minus the fraction 3 over 2 close parenthesis, equals 2 open parenthesis 0 close parenthesis; 2 x squared plus 5 x minus 3, equals 0")[$2 ( x^(2) + frac(5, 2) x − frac(3, 2) ) & = 2 ( 0 ) \ 2 x^(2) + 5 x − 3 & = 0$] You can check that the solutions of this last equation are in fact #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] and #math.equation(block: false, alt: "minus 3")[$− 3$]. Multiplying both sides of an equation by a constant factor does not change its solutions. ] ] Which statement is true? \_\_\_\_\_ #solutionbox[ If you know the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals x squared plus b x plus c")[$y = x^(2) + b x + c$], you can write it in factored form. ] Which statement is true? + All rectangles with the same perimeter have the same area. + The solutions of #math.equation(block: false, alt: "x open parenthesis 18 minus x close parenthesis equals 80")[$x ( 18 − x ) = 80$] are 18 and 80. + If the perimeter of a rectangle is 20 cm, the largest area it can have is 20 sq cm. + If you know the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals x squared plus b x plus c")[$y = x^(2) + b x + c$], you can write it in factored form. Find a quadratic equation with integer coefficients whose solutions are #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] and #math.equation(block: false, alt: "minus 5")[$− 5$]. Use the smallest possible positive coefficient for #math.equation(block: false, alt: "x squared")[$x^(2)$]. \_\_\_\_\_#math.equation(block: false, alt: "equals 0")[$= 0$] #solutionbox[ #math.equation(block: true, alt: "3 x squared plus 13 x minus 10 equals 0")[$3 x^(2) + 13 x − 10 = 0$] ] Find a quadratic equation with integer coefficients whose solutions are #math.equation(block: false, alt: "the fraction 2 over 3")[$display(frac(2, 3))$] and #math.equation(block: false, alt: "minus 5")[$− 5$]. Use the smallest possible positive coefficient for #math.equation(block: false, alt: "x squared")[$x^(2)$]. #solutionbox[ #math.equation(block: true, alt: "3 x squared plus 13 x minus 10 equals 0")[$3 x^(2) + 13 x − 10 = 0$] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A quadratic equation in one variable always has two solutions. However, in some cases, the solutions may be equal. For example, the equation #math.equation(block: false, alt: "x squared minus 2 x plus 1 equals 0")[$x^(2) − 2 x + 1 = 0$] can be solved by factoring as follows: #math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis open parenthesis x minus 1 close parenthesis, equals 0, Apply the zero-factor principle.; x minus 1 equals 0 or x minus 1, equals 0")[$( x − 1 ) ( x − 1 ) & = 0 & & "Apply the zero-factor principle." \ x − 1 = 0 " " " " " or " " " " " x − 1 & = 0$] Both of these equations have solution #math.equation(block: false, alt: "1")[$1$]. We say that #math.equation(block: false, alt: "1")[$1$] is a solution of #strong[multiplicity] two, meaning that it occurs twice as a solution of the quadratic equation. ] === Equations Quadratic in Form The equation #math.equation(block: true, alt: "x to the power 6 minus 4 x cubed minus 5 equals 0")[$x^(6) − 4 x^(3) − 5 = 0$] is not quadratic, but if we make the substitution #math.equation(block: false, alt: "u equals x cubed")[$u = x^(3)$], the equation becomes #math.equation(block: true, alt: "u squared minus 4 u minus 5 equals 0")[$u^(2) − 4 u − 5 = 0$] An equation is called #strong[quadratic in form] if we can use a substitution to write it as #math.equation(block: true, alt: "a u squared plus b u plus c equals 0")[$a u^(2) + b u + c = 0$] where #math.equation(block: false, alt: "u")[$u$] stands for an algebraic expression. Such equations can be solved by the same techniques we use to solve quadratic equations. #examplebox("Example 6")[][ Use the substitution #math.equation(block: false, alt: "u equals x cubed")[$u = x^(3)$] to solve the equation #math.equation(block: true, alt: "x to the power 6 minus 4 x cubed minus 5 equals 0")[$x^(6) − 4 x^(3) − 5 = 0$] #solutionbox[ We set #math.equation(block: false, alt: "u equals x cubed")[$u = x^(3)$], so that #math.equation(block: false, alt: "u squared equals open parenthesis x cubed close parenthesis squared equals x to the power 6")[$u^(2) = attach(( x^(3) ), t: 2) = x^(6)$]. The original equation then becomes a quadratic equation in the variable #math.equation(block: false, alt: "u")[$u$], which we can solve by factoring. #math.equation(block: true, alt: "u squared minus 4 u minus 5, equals 0, Factor the left side.; open parenthesis u plus 1 close parenthesis open parenthesis u minus 5 close parenthesis, equals 0, Apply the zero-factor principle.; u plus 1, equals 0 or u minus 5 equals 0, Solve each equation for u .; u, equals minus 1 or u equals 5")[$u^(2) − 4 u − 5 & = 0 & & "Factor the left side." \ ( u + 1 ) ( u − 5 ) & = 0 & & "Apply the zero-factor principle." \ u + 1 & = 0 " " " " " or " " " " " u − 5 = 0 & & "Solve each equation for " bold(italic(u)) . \ u & = − 1 " " " " " or " " " " " u = 5$] Finally, we replace #math.equation(block: false, alt: "u")[$u$] by #math.equation(block: false, alt: "x cubed")[$x^(3)$] and solve for #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "x cubed, equals minus 1 or, x cubed, equals 5, Take cube roots.; x, equals the cube root of minus 1 equals minus 1 or, x, equals the cube root of 5")[$x^(3) & = − 1 " " " " " " " " " " " " " " " or " & x^(3) & = 5 & & "Take cube roots." \ x & = root(3, − 1) = − 1 " " " " " " " or " & x & = root(3, 5)$] You can verify that the solutions of the original equation are #math.equation(block: false, alt: "minus 1")[$− 1$] and #math.equation(block: false, alt: "the cube root of 5")[$root(3, 5)$]. ] ] We say that the equation in Example, #math.equation(block: false, alt: "x to the power 6 minus 4 x cubed minus 5 equals 0")[$" " x^(6) − 4 x^(3) − 5 = 0$], is #strong[quadratic in] #math.equation(block: false, alt: "x cubed")[$x^(3)$]. We chose the substitution #math.equation(block: false, alt: "u equals x cubed")[$u = x^(3)$] because #math.equation(block: false, alt: "x to the power 6 equals u squared")[$x^(6) = u^(2)$]. Use the substitution #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] to solve the equation #math.equation(block: false, alt: "x to the power 4 minus 5 x squared plus 6 equals 0")[$x^(4) − 5 x^(2) + 6 = 0$]. Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ List all the values that are solutions. Use "sqrt(5)" to get #math.equation(block: false, alt: "the square root of 5")[$sqrt(5)$], and use a comma to separate different solutions. #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus the square root of 2")[$x = ± sqrt(2)$], #math.equation(block: true, alt: "x equals plus or minus the square root of 3")[$x = ± sqrt(3)$] ] Use the substitution #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] to solve the equation #math.equation(block: false, alt: "x to the power 4 minus 5 x squared plus 6 equals 0")[$x^(4) − 5 x^(2) + 6 = 0$]. #solutionbox[ #math.equation(block: true, alt: "x equals plus or minus the square root of 2")[$x = ± sqrt(2)$], #math.equation(block: true, alt: "x equals plus or minus the square root of 3")[$x = ± sqrt(3)$] ] Usually, you can choose the simpler variable term in the equation for the #math.equation(block: false, alt: "u")[$u$]-substitution. For example, in Practice 6 we chose #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] because #math.equation(block: false, alt: "u squared equals open parenthesis x squared close parenthesis squared equals x to the power 4")[$u^(2) = attach(( x^(2) ), t: 2) = x^(4)$], which is the first term of the equation. Once you have chosen the #math.equation(block: false, alt: "u")[$u$]-substitution, you should check that the other variable term is then a multiple of #math.equation(block: false, alt: "u squared")[$u^(2)$]; otherwise, the equation is not quadratic in form. Usually, you can choose the simpler variable term in the equation for the #math.equation(block: false, alt: "u")[$u$]-substitution. For example, in Practice 6 we chose #math.equation(block: false, alt: "u equals x squared")[$u = x^(2)$] because #math.equation(block: false, alt: "u squared equals open parenthesis x squared close parenthesis squared equals x to the power 4")[$u^(2) = attach(( x^(2) ), t: 2) = x^(4)$], which is the first term of the equation. Once you have chosen the #math.equation(block: false, alt: "u")[$u$]-substitution, you should check that the other variable term is then a multiple of #math.equation(block: false, alt: "u squared")[$u^(2)$]; otherwise, the equation is not quadratic in form. #examplebox("Example 7")[][ Solve the equation #math.equation(block: false, alt: "e to the power 2 x minus 7 e to the power x plus 12 equals 0")[$e^(2 x) − 7 e^(x) + 12 = 0$]. #solutionbox[ We use the substitution #math.equation(block: false, alt: "u equals e to the power x")[$u = e^(x)$], because #math.equation(block: false, alt: "u squared equals open parenthesis e to the power x close parenthesis squared equals e to the power 2 x")[$u^(2) = attach(( e^(x) ), t: 2) = e^(2 x)$]. The original equation then becomes #math.equation(block: true, alt: "u squared minus 7 u plus 12, equals 0, Factor the left side.; open parenthesis u minus 3 close parenthesis open parenthesis u minus 4 close parenthesis, equals 0, Apply the zero-factor principle.; u minus 3, equals 0 or u minus 4 equals 0, Solve each equation for u .; u, equals 3 or u equals 4")[$u^(2) − 7 u + 12 & = 0 & & "Factor the left side." \ ( u − 3 ) ( u − 4 ) & = 0 & & "Apply the zero-factor principle." \ u − 3 & = 0 " " " " " or " " " " " u − 4 = 0 & & "Solve each equation for" bold(italic(u)) . \ u & = 3 " " " " " or " " " " " u = 4$] Finally, we replace #math.equation(block: false, alt: "u")[$u$] by #math.equation(block: false, alt: "e to the power x")[$e^(x)$] and solve for #math.equation(block: false, alt: "x")[$x$]. #math.equation(block: true, alt: "e to the power x, equals 3 or, e to the power x, equals 4; x, equals ln ⁡ 3 or, x, equals ln ⁡ 4")[$e^(x) & = 3 " " " " " " " " " " " or " & e^(x) & = 4 \ x & = ln 3 " " " " " " " or " & x & = ln 4$] You should verify that the solutions of the original equation are #math.equation(block: false, alt: "ln ⁡ 3")[$ln 3$] and #math.equation(block: false, alt: "ln ⁡ 4")[$ln 4$]. ] ] Solve the equation #math.equation(block: false, alt: "10 to the power 2 x minus 3 times 10 to the power x plus 2 equals 0")[$10^(2 x) − 3 ⋅ 10^(x) + 2 = 0$], and check the solutions. Solutions: #math.equation(block: false, alt: "x equals")[$x =$]\_\_\_\_\_ List all the values that are solutions. Use a comma to separate different solutions. #solutionbox[ #math.equation(block: true, alt: "x equals 0")[$x = 0$], #math.equation(block: true, alt: "x equals log ⁡ 2")[$x = log 2$] ] Solve the equation #math.equation(block: false, alt: "10 to the power 2 x minus 3 times 10 to the power x plus 2 equals 0")[$10^(2 x) − 3 ⋅ 10^(x) + 2 = 0$], and check the solutions. #solutionbox[ #math.equation(block: true, alt: "x equals 0")[$x = 0$], #math.equation(block: true, alt: "x equals log ⁡ 2")[$x = log 2$] ] Explain why we cannot "cancel" #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis")[$( x − 5 )$] from both sides of the equation #math.equation(block: false, alt: "3 x open parenthesis x minus 5 close parenthesis equals 6 open parenthesis x minus 5 close parenthesis")[$3 x ( x − 5 ) = 6 ( x − 5 )$]. What are the solutions of the equation? \_\_\_\_\_ Explain why we cannot "cancel" #math.equation(block: false, alt: "open parenthesis x minus 5 close parenthesis")[$( x − 5 )$] from both sides of the equation #math.equation(block: false, alt: "3 x open parenthesis x minus 5 close parenthesis equals 6 open parenthesis x minus 5 close parenthesis")[$3 x ( x − 5 ) = 6 ( x − 5 )$]. What are the solutions of the equation? === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Quadratic function - Zero-factor principle - Standard form - Factored form - Multiplicity - Monotonic ==== CONCEPTS + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Quadratic Function] A #strong[quadratic function] is one that can be written in the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$]where #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$] are constants, and #math.equation(block: false, alt: "a")[$a$] is not equal to zero. ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Zero-Factor Principle] The product of two factors equals zero if and only if one or both of the factors equals zero. In symbols, #math.equation(block: true, alt: "a b equals 0 if and only if a equals 0 or b equals 0")[$a b = 0 " " " " " if and only if " " " " " a = 0 " " " " " or " " " " " b = 0$] ] + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[#math.equation(block: false, alt: "x")[$x$]-Intercepts of a Graph] The #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] are the solutions of the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 0")[$f ( x ) = 0$]. ] + A quadratic equation written as #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$] is in #strong[standard form]. #linebreak() A quadratic equation written as #math.equation(block: false, alt: "a open parenthesis x minus r sub 1 close parenthesis open parenthesis x minus r sub 2 close parenthesis equals 0")[$a ( x − r_(1) ) ( x − r_(2) ) = 0$] is in #strong[factored form]. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[To Solve a Quadratic Equation by Factoring] + Write the equation in standard form. + Factor the left side of the equation. + Apply the zero-factor principle: Set each factor equal to zero. + Solve each equation. There are two solutions (which may be equal). ] + Every quadratic equation has two solutions, which may be the same. + The value of the constant #math.equation(block: false, alt: "a")[$a$] in the factored form of a quadratic equation does not affect the solutions. + Each solution of a quadratic equation corresponds to a factor in the factored form. + An equation is called #strong[quadratic in form] if we can use a substitution to write it as #math.equation(block: false, alt: "a u squared plus b u plus c equals 0")[$a u^(2) + b u + c = 0$], where #math.equation(block: false, alt: "u")[$u$] stands for an algebraic expression. ==== STUDY QUESTIONS + + Find a pair of numbers whose product is #math.equation(block: false, alt: "6")[$6$]. Now find a different pair of numbers whose product is #math.equation(block: false, alt: "6")[$6$]. Can you find more such pairs? + Find a pair of numbers whose product is #math.equation(block: false, alt: "0")[$0$]. What is true about any such pair? + Before you begin factoring to solve a quadratic equation, what should you do? + How can you find the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph of #math.equation(block: false, alt: "y equals f open parenthesis x close parenthesis")[$y = f ( x )$] without looking at the graph? + How many solutions does a quadratic equation have? + + Write a linear equation whose only solution is #math.equation(block: false, alt: "x equals 3")[$x = 3$]. + Write a quadratic equation whose only solution is #math.equation(block: false, alt: "x equals 3")[$x = 3$]. + If you know the solutions of #math.equation(block: false, alt: "a x squared plus b x plus c equals 0")[$a x^(2) + b x + c = 0$], how can you find the solutions of #math.equation(block: false, alt: "5 open parenthesis a x squared plus b x plus c close parenthesis equals 0")[$5 ( a x^(2) + b x + c ) = 0$]? + Is the equation #math.equation(block: false, alt: "x to the power 9 minus 6 x cubed plus 8 equals 0")[$x^(9) − 6 x^(3) + 8 = 0$] quadratic in form? Why or why not? + Delbert says that he can solve the equation #math.equation(block: false, alt: "x open parenthesis x plus 5 close parenthesis equals 2 open parenthesis x plus 5 close parenthesis")[$x ( x + 5 ) = 2 ( x + 5 )$] by canceling the factor #math.equation(block: false, alt: "open parenthesis x plus 5 close parenthesis")[$( x + 5 )$] to get #math.equation(block: false, alt: "x equals 2")[$x = 2$]. Comment on his method. ==== SKILLS Practice each skill in the Homework problems listed. + Use the zero-factor principle and find #math.equation(block: false, alt: "x")[$x$]-intercepts: \#3–10 + Solve quadratic equations by factoring: \#11–24 + Use the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph to factor a quadratic equation: \#25–28, 37–40 + Write a quadratic equation with given solutions: \#29–36 + Solve applied problems involving quadratic equations: \#41–50 + Solve equations that are quadratic in form: \#51–62 === Homework 6.1 Delbert stands at the top of a #math.equation(block: false, alt: "300")[$300$]-foot cliff and throws his algebra book directly upward with a velocity of #math.equation(block: false, alt: "20")[$20$] feet per second. The height of his book above the ground #math.equation(block: false, alt: "t")[$t$] seconds later is given by the equation #math.equation(block: true, alt: "h equals minus 16 t squared plus 20 t plus 300")[$h = − 16 t^(2) + 20 t + 300$] where #math.equation(block: false, alt: "h")[$h$] is in feet. + Use your calculator to make a table of values for the height function, with increments of #math.equation(block: false, alt: "0.5")[$0.5$] second. + Graph the height function on your calculator. Use your table of values to help you choose appropriate window settings. + What is the highest altitude Delbert’s book reaches? When does it reach that height? Use the TRACE feature to find approximate answers first. Then use the #strong[Table] feature to improve your estimate. + When does Delbert's book pass him on its way down? (Delbert is standing at a height of #math.equation(block: false, alt: "300")[$300$] feet.) Use the #strong[intersect] command. + How long will it take Delbert's book to hit the ground at the bottom of the cliff? + #figure(table( columns: 12, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "t")[$t$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "0.5")[$0.5$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "1.5")[$1.5$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "2.5")[$2.5$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "3.5")[$3.5$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "4.5")[$4.5$]], [#math.equation(block: false, alt: "5")[$5$]]), [#math.equation(block: false, alt: "h")[$h$]], [#math.equation(block: false, alt: "300")[$300$]], [#math.equation(block: false, alt: "306")[$306$]], [#math.equation(block: false, alt: "304")[$304$]], [#math.equation(block: false, alt: "294")[$294$]], [#math.equation(block: false, alt: "276")[$276$]], [#math.equation(block: false, alt: "250")[$250$]], [#math.equation(block: false, alt: "216")[$216$]], [#math.equation(block: false, alt: "174")[$174$]], [#math.equation(block: false, alt: "124")[$124$]], [#math.equation(block: false, alt: "66")[$66$]], [#math.equation(block: false, alt: "0")[$0$]], )) #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "306.25")[$306.25$] ft at #math.equation(block: false, alt: "0.625")[$0.625$] sec + #math.equation(block: false, alt: "1.25")[$1.25$] sec + #math.equation(block: false, alt: "5")[$5$] sec James Bond stands on top of a #math.equation(block: false, alt: "240")[$240$]-foot building and throws a film canister upward to a fellow agent in a helicopter #math.equation(block: false, alt: "16")[$16$] feet above the building. The height of the film above the ground #math.equation(block: false, alt: "t")[$t$] seconds later is given by the formula #math.equation(block: true, alt: "h equals minus 16 t squared plus 32 t plus 240")[$h = − 16 t^(2) + 32 t + 240$] where #math.equation(block: false, alt: "h")[$h$] is in feet. + Use your calculator to make a table of values for the height function, with increments of #math.equation(block: false, alt: "0.5")[$0.5$] second. + Graph the height function on your calculator. Use your table of values to help you choose appropriate window settings. + How long will it take the film canister to reach the agent in the helicopter? (What is the agent's altitude?) Use the TRACE feature to find approximate answers first. Then use the #strong[Table] feature to improve your estimate. + If the agent misses the canister, when will it pass James Bond on the way down? Use the intersect command. + How long will it take it to hit the ground? In Problems 3–10, use a graph to solve the equation #math.equation(block: false, alt: "y equals 0")[$y = 0$]. (Use #math.equation(block: false, alt: "Xmin equals minus 9.4")[$"Xmin" = − 9.4$], #math.equation(block: false, alt: "Xmax equals 9.4")[$"Xmax" = 9.4$].) Check your answers with the zero-factor principle. #math.equation(block: true, alt: "y equals open parenthesis 2 x plus 5 close parenthesis open parenthesis x minus 2 close parenthesis")[$y = ( 2 x + 5 ) ( x − 2 )$] #math.equation(block: true, alt: "the fraction minus 5 over 2")[$display(frac(− 5, 2))$], #math.equation(block: true, alt: "2")[$" " 2$] #math.equation(block: true, alt: "y equals open parenthesis x plus 1 close parenthesis open parenthesis 4 x minus 1 close parenthesis")[$y = ( x + 1 ) ( 4 x − 1 )$] #math.equation(block: true, alt: "y equals x open parenthesis 3 x plus 10 close parenthesis")[$y = x ( 3 x + 10 )$] #math.equation(block: true, alt: "0")[$0$], #math.equation(block: true, alt: "the fraction minus 10 over 3")[$" " display(frac(− 10, 3))$] #math.equation(block: true, alt: "y equals x open parenthesis 3 x minus 7 close parenthesis")[$y = x ( 3 x − 7 )$] #math.equation(block: true, alt: "y equals open parenthesis 4 x plus 3 close parenthesis open parenthesis x plus 8 close parenthesis")[$y = ( 4 x + 3 ) ( x + 8 )$] #math.equation(block: true, alt: "the fraction minus 3 over 4")[$display(frac(− 3, 4))$], #math.equation(block: true, alt: "minus 8")[$" " − 8$] #math.equation(block: true, alt: "y equals open parenthesis x minus 2 close parenthesis open parenthesis x minus 9 close parenthesis")[$y = ( x − 2 ) ( x − 9 )$] #math.equation(block: true, alt: "y equals open parenthesis x minus 4 close parenthesis squared")[$y = ( x − 4 )^(2)$] #math.equation(block: true, alt: "4")[$4$] #math.equation(block: true, alt: "y equals open parenthesis x plus 6 close parenthesis squared")[$y = ( x + 6 )^(2)$] For Problems 11-24, solve by factoring. (See Algebra Skills Refresher Appendix to review factoring.) #math.equation(block: true, alt: "2 a squared plus 5 a minus 3 equals 0")[$2 a^(2) + 5 a − 3 = 0$] #math.equation(block: true, alt: "the fraction 1 over 2")[$display(frac(1, 2))$], #math.equation(block: true, alt: "minus 3")[$" " − 3$] #math.equation(block: true, alt: "3 b squared minus 4 b minus 4 equals 0")[$3 b^(2) − 4 b − 4 = 0$] #math.equation(block: true, alt: "2 x squared equals 6 x")[$2 x^(2) = 6 x$] #math.equation(block: true, alt: "0")[$0$], #math.equation(block: true, alt: "3")[$" " 3$] #math.equation(block: true, alt: "5 z squared equals 5 z")[$5 z^(2) = 5 z$] #math.equation(block: true, alt: "3 y squared minus 6 y equals minus 3")[$3 y^(2) − 6 y = − 3$] #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "4 y squared plus 4 y equals 8")[$4 y^(2) + 4 y = 8$] #math.equation(block: true, alt: "x open parenthesis 2 x minus 3 close parenthesis equals minus 1")[$x ( 2 x − 3 ) = − 1$] #math.equation(block: true, alt: "the fraction 1 over 2")[$display(frac(1, 2))$], #math.equation(block: true, alt: "1")[$" " 1$] #math.equation(block: true, alt: "2 x open parenthesis x minus 2 close parenthesis equals x plus 3")[$2 x ( x − 2 ) = x + 3$] #math.equation(block: true, alt: "t open parenthesis t minus 3 close parenthesis equals 2 open parenthesis t minus 3 close parenthesis")[$t ( t − 3 ) = 2 ( t − 3 )$] #math.equation(block: true, alt: "2")[$2$], #math.equation(block: true, alt: "3")[$" " 3$] #math.equation(block: true, alt: "5 open parenthesis t plus 2 close parenthesis equals t open parenthesis t plus 2 close parenthesis")[$5 ( t + 2 ) = t ( t + 2 )$] #math.equation(block: true, alt: "z open parenthesis 3 z plus 2 close parenthesis equals open parenthesis z plus 2 close parenthesis squared")[$z ( 3 z + 2 ) = ( z + 2 )^(2)$] #math.equation(block: true, alt: "minus 1")[$− 1$], #math.equation(block: true, alt: "2")[$" " 2$] #math.equation(block: true, alt: "open parenthesis z minus 1 close parenthesis squared equals 2 z squared plus 3 z minus 5")[$( z − 1 )^(2) = 2 z^(2) + 3 z − 5$] #math.equation(block: true, alt: "open parenthesis v plus 2 close parenthesis open parenthesis v minus 5 close parenthesis equals 8")[$( v + 2 ) ( v − 5 ) = 8$] #math.equation(block: true, alt: "minus 3")[$− 3$], #math.equation(block: true, alt: "6")[$" " 6$] #math.equation(block: true, alt: "open parenthesis w plus 1 close parenthesis open parenthesis 2 w minus 3 close parenthesis equals 3")[$( w + 1 ) ( 2 w − 3 ) = 3$] In Problems 25–28, graph each set of functions in the standard window.What do you notice about the #math.equation(block: false, alt: "x")[$x$]-intercepts? Generalize your observation, and test your idea with examples. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus x minus 20")[$f ( x ) = x^(2) − x − 20$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 2 open parenthesis x squared minus x minus 20 close parenthesis")[$g ( x ) = 2 ( x^(2) − x − 20 )$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals 0.5 open parenthesis x squared minus x minus 20 close parenthesis")[$h ( x ) = 0.5 ( x^(2) − x − 20 )$] The 3 graphs have the same #math.equation(block: false, alt: "x")[$x$]-intercepts. In general, the graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] has the same #math.equation(block: false, alt: "x")[$x$]-intercepts as the graph of #math.equation(block: false, alt: "y equals k open parenthesis a x squared plus b x plus c close parenthesis")[$y = k ( a x^(2) + b x + c )$]. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 2 x minus 15")[$f ( x ) = x^(2) + 2 x − 15$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals 3 open parenthesis x squared plus 2 x minus 15 close parenthesis")[$g ( x ) = 3 ( x^(2) + 2 x − 15 )$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals 0.2 open parenthesis x squared plus 2 x minus 15 close parenthesis")[$h ( x ) = 0.2 ( x^(2) + 2 x − 15 )$] + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared plus 6 x minus 16")[$f ( x ) = x^(2) + 6 x − 16$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals minus 2 open parenthesis x squared plus 6 x minus 16 close parenthesis")[$g ( x ) = − 2 ( x^(2) + 6 x − 16 )$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals minus 0.1 open parenthesis x squared plus 6 x minus 16 close parenthesis")[$h ( x ) = − 0.1 ( x^(2) + 6 x − 16 )$] The 3 graphs have the same #math.equation(block: false, alt: "x")[$x$]-intercepts. In general, the graph of #math.equation(block: false, alt: "y equals a x squared plus b x plus c")[$y = a x^(2) + b x + c$] has the same #math.equation(block: false, alt: "x")[$x$]-intercepts as the graph of #math.equation(block: false, alt: "y equals k open parenthesis a x squared plus b x plus c close parenthesis")[$y = k ( a x^(2) + b x + c )$]. + #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x squared minus 16")[$f ( x ) = x^(2) − 16$] + #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals minus 1.5 open parenthesis x squared minus 16 close parenthesis")[$g ( x ) = − 1.5 ( x^(2) − 16 )$] + #math.equation(block: false, alt: "h open parenthesis x close parenthesis equals minus 0.4 open parenthesis x squared minus 16 close parenthesis")[$h ( x ) = − 0.4 ( x^(2) − 16 )$] In Problems 29–36, write a quadratic equation whose solutions are given. The equation should be in standard form with integer coefficients. #math.equation(block: false, alt: "minus 2")[$− 2$] and #math.equation(block: false, alt: "1")[$1$] #math.equation(block: true, alt: "x squared plus x minus 2 equals 0")[$x^(2) + x − 2 = 0$] #math.equation(block: false, alt: "minus 4")[$− 4$] and #math.equation(block: false, alt: "3")[$3$] #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "minus 5")[$− 5$] #math.equation(block: true, alt: "x squared plus 5 x equals 0")[$x^(2) + 5 x = 0$] #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "5")[$5$] #math.equation(block: false, alt: "minus 3")[$− 3$] and #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] #math.equation(block: true, alt: "2 x squared plus 5 x minus 3 equals 0")[$2 x^(2) + 5 x − 3 = 0$] #math.equation(block: false, alt: "the fraction minus 2 over 3")[$display(frac(− 2, 3))$] and #math.equation(block: false, alt: "4")[$4$] #math.equation(block: false, alt: "the fraction minus 1 over 4")[$display(frac(− 1, 4))$] and #math.equation(block: false, alt: "the fraction 3 over 2")[$display(frac(3, 2))$] #math.equation(block: true, alt: "8 x squared minus 10 x minus 3 equals 0")[$8 x^(2) − 10 x − 3 = 0$] #math.equation(block: false, alt: "the fraction minus 1 over 3")[$display(frac(− 1, 3))$] and #math.equation(block: false, alt: "the fraction minus 1 over 2")[$display(frac(− 1, 2))$] For problems 37-40, graph the function in the #strong[ZInteger] window, and locate the #math.equation(block: false, alt: "x")[$x$]-intercepts of the graph. Use the #math.equation(block: false, alt: "x")[$x$]-intercepts to write the quadratic expression in factored form. #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 0.1 open parenthesis x squared minus 3 x minus 270 close parenthesis")[$f ( x ) = 0.1 ( x^(2) − 3 x − 270 )$] #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 0.1 open parenthesis x minus 18 close parenthesis open parenthesis x plus 15 close parenthesis")[$f ( x ) = 0.1 ( x − 18 ) ( x + 15 )$] #math.equation(block: true, alt: "h open parenthesis x close parenthesis equals 0.1 open parenthesis x squared plus 9 x minus 360 close parenthesis")[$h ( x ) = 0.1 ( x^(2) + 9 x − 360 )$] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals minus 0.08 open parenthesis x squared plus 14 x minus 576 close parenthesis")[$g ( x ) = − 0.08 ( x^(2) + 14 x − 576 )$] #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals minus 0.08 open parenthesis x minus 18 close parenthesis open parenthesis x plus 32 close parenthesis")[$g ( x ) = − 0.08 ( x − 18 ) ( x + 32 )$] #math.equation(block: true, alt: "F open parenthesis x close parenthesis equals minus 0.06 open parenthesis x squared minus 22 x minus 504 close parenthesis")[$F ( x ) = − 0.06 ( x^(2) − 22 x − 504 )$] Use the Pythagorean theorem to solve Problems 41 and 42. (See Algebra Skills Refresher Appendix to review the Pythagorean theorem.) One end of a ladder is #math.equation(block: false, alt: "10")[$10$] feet from the base of a wall, and the other end reaches a window in the wall. The ladder is #math.equation(block: false, alt: "2")[$2$] feet longer than the height of the window. + Write a quadratic equation about the height of the window. + Solve your equation to find the height of the window. #figure(figph[ladder leaning against wall], alt: "ladder leaning against wall", caption: none) + #math.equation(block: false, alt: "10 squared plus h squared equals open parenthesis h plus 2 close parenthesis squared")[$10^(2) + h^(2) = ( h + 2 )^(2)$] + #math.equation(block: false, alt: "24")[$24$] ft The diagonal of a rectangle is #math.equation(block: false, alt: "20")[$20$] inches. One side of the rectangle is #math.equation(block: false, alt: "4")[$4$] inches shorter than the other side. + Write a quadratic equation about the length of the rectangle. + Solve your equation to find the dimensions of the rectangle. Use the following formula to answer Problems 43 and 44. If an object is thrown into the air from a height #math.equation(block: false, alt: "s sub 0")[$s_(0)$] above the ground with an initial velocity #math.equation(block: false, alt: "v sub 0")[$v_(0)$], its height #math.equation(block: false, alt: "t")[$t$] seconds later is given by the formula #math.equation(block: true, alt: "h equals minus the fraction 1 over 2 g t squared plus v sub 0 t plus s sub 0")[$h = − frac(1, 2) g t^(2) + v_(0) t + s_(0)$] where #math.equation(block: false, alt: "g")[$g$] is a constant that measures the force of gravity. A tennis ball is thrown into the air with an initial velocity of #math.equation(block: false, alt: "16")[$16$] feet per second from a height of #math.equation(block: false, alt: "8")[$8$] feet. The value of #math.equation(block: false, alt: "g")[$g$] is #math.equation(block: false, alt: "32")[$32$]. + Write a quadratic function that gives the height of the tennis ball at time #math.equation(block: false, alt: "t")[$t$]. + Find the height of the tennis ball at #math.equation(block: false, alt: "t equals the fraction 1 over 2")[$t = display(frac(1, 2))$] second and at #math.equation(block: false, alt: "t equals 1")[$t = 1$] second. + Write and solve an equation to answer the question: At what time is the tennis ball #math.equation(block: false, alt: "11")[$11$] feet high? + Use the Table feature on your calculator to verify your answers to parts (b) and (c). (What value of #math.equation(block: false, alt: "Δ Tbl")[$upright(Δ) "Tbl"$] is useful for this problem?) + Graph your function from part (a) on your calculator. Use your table to help you choose an appropriate window. + If nobody hits the tennis ball, approximately how long will it be in the air? + #math.equation(block: false, alt: "h equals minus 16 t squared plus 16 t plus 8")[$h = − 16 t^(2) + 16 t + 8$] + #math.equation(block: false, alt: "12")[$12$] ft; #math.equation(block: false, alt: "8")[$8$] ft + #math.equation(block: false, alt: "11 equals minus 16 t squared plus 16 t plus 8")[$11 = − 16 t^(2) + 16 t + 8$]; at #math.equation(block: false, alt: "the fraction 1 over 4")[$display(frac(1, 4))$] sec and #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] sec + #math.equation(block: false, alt: "Δ Tbl equals 0.25")[$upright(Δ) "Tbl" = 0.25$] + #figure(figph[parabola], alt: "parabola", caption: none) + #math.equation(block: false, alt: "1.37")[$1.37$] sec A mountain climber stands on a ledge #math.equation(block: false, alt: "80")[$80$] feet above the ground and tosses a rope down to a companion clinging to the rock face below the ledge. The initial velocity of the rope is #math.equation(block: false, alt: "minus 8")[$− 8$] feet per second, and the value of #math.equation(block: false, alt: "g")[$g$] is #math.equation(block: false, alt: "32")[$32$]. + Write a quadratic function that gives the height of the rope at time #math.equation(block: false, alt: "t")[$t$]. + What is the height of the rope after #math.equation(block: false, alt: "the fraction 1 over 2")[$display(frac(1, 2))$] second? After #math.equation(block: false, alt: "1")[$1$] second? + Write and solve an equation to answer the question: How long does it take the rope to reach the second climber, who is #math.equation(block: false, alt: "17")[$17$] feet above the ground? + Use the Table feature on your calculator to verify your answers to parts (b) and (c). (What value of #math.equation(block: false, alt: "Δ Tbl")[$upright(Δ) "Tbl"$] is useful for this problem?) + Graph your function from part (a) on your calculator. Use your table to help you choose an appropriate window. + If the second climber misses the rope, approximately how long will the rope take to reach the ground? For Problems 45 and 46, you may want to review Investigation, Perimeter and Area, in Modeling with Functions. A rancher has #math.equation(block: false, alt: "360")[$360$] yards of fence to enclose a rectangular pasture. If the pasture should be #math.equation(block: false, alt: "8000")[$8000$] square yards in area, what should its dimensions be? We will use 3 methods to solve this problem: a table of values, a graph, and an algebraic equation. + Make a table by hand that shows the areas of pastures of various widths, as shown here.#figure(table( columns: 3, align: left, inset: 6pt, table.header([Width], [Length], [Area]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "170")[$170$]], [#math.equation(block: false, alt: "1700")[$1700$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], )) #linebreak() (To find the length of each pasture, ask yourself, What is the sum of the length plus the width if there are #math.equation(block: false, alt: "360")[$360$] yards of fence?) Continue the table until you find the pasture whose area is #math.equation(block: false, alt: "8000")[$8000$] square yards. + Write an expression for the length of the pasture if its width is #math.equation(block: false, alt: "x")[$x$]. Next, write an expression for the area, #math.equation(block: false, alt: "A")[$A$], of the pasture if its width is #math.equation(block: false, alt: "x")[$x$]. Graph the equation for #math.equation(block: false, alt: "A")[$A$] on your calculator, and use the graph to find the pasture of area #math.equation(block: false, alt: "8000")[$8000$] square yards. + Write an equation for the area, #math.equation(block: false, alt: "A")[$A$], of the pasture in terms of its width #math.equation(block: false, alt: "x")[$x$]. Solve your equation algebraically for #math.equation(block: false, alt: "A equals 8000")[$A = 8000$]. Explain why there are two solutions. + #figure(table( columns: 3, align: left, inset: 6pt, table.header([Width], [Length], [Area]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "170")[$170$]], [#math.equation(block: false, alt: "1700")[$1700$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "160")[$160$]], [#math.equation(block: false, alt: "3200")[$3200$]], [#math.equation(block: false, alt: "30")[$30$]], [#math.equation(block: false, alt: "150")[$150$]], [#math.equation(block: false, alt: "4500")[$4500$]], [#math.equation(block: false, alt: "40")[$40$]], [#math.equation(block: false, alt: "140")[$140$]], [#math.equation(block: false, alt: "5600")[$5600$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "130")[$130$]], [#math.equation(block: false, alt: "6500")[$6500$]], [#math.equation(block: false, alt: "60")[$60$]], [#math.equation(block: false, alt: "120")[$120$]], [#math.equation(block: false, alt: "7200")[$7200$]], [#math.equation(block: false, alt: "70")[$70$]], [#math.equation(block: false, alt: "110")[$110$]], [#math.equation(block: false, alt: "7700")[$7700$]], [#math.equation(block: false, alt: "80")[$80$]], [#math.equation(block: false, alt: "100")[$100$]], [#math.equation(block: false, alt: "8000")[$8000$]], )) + #math.equation(block: false, alt: "l equals 180 minus x")[$l = 180 − x$], #math.equation(block: false, alt: "A equals 180 x minus x squared")[$A = 180 x − x^(2)$]; #math.equation(block: false, alt: "80")[$80$] yd by #math.equation(block: false, alt: "100")[$100$] yd + #math.equation(block: false, alt: "180 x minus x squared equals 8000")[$180 x − x^(2) = 8000$], #math.equation(block: false, alt: "80")[$80$] yd by #math.equation(block: false, alt: "100")[$100$] yd, or #math.equation(block: false, alt: "100")[$100$] yd by #math.equation(block: false, alt: "80")[$80$] yd. There are two solutions because the pasture can be oriented in two directions. If the rancher in Problem 45 uses a riverbank to border one side of the pasture as shown in the figure, he can enclose #math.equation(block: false, alt: "16 , 000")[$16 , 000$] square yards with #math.equation(block: false, alt: "360")[$360$] yards of fence. What will the dimensions of the pasture be then? We will use three methods to solve this problem: a table of values, a graph, and an algebraic equation. #figure(figph[river and adjoining rectangle], alt: "river and adjoining rectangle", caption: none) + Make a table by hand that shows the areas of pastures of various widths, as shown here.#figure(table( columns: 3, align: left, inset: 6pt, table.header([Width], [Length], [Area]), [#math.equation(block: false, alt: "10")[$10$]], [#math.equation(block: false, alt: "340")[$340$]], [#math.equation(block: false, alt: "3400")[$3400$]], [#math.equation(block: false, alt: "20")[$20$]], [#math.equation(block: false, alt: "320")[$320$]], [#math.equation(block: false, alt: "6400")[$6400$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], [#math.equation(block: false, alt: "⋮")[$⋮$]], )) #linebreak() (Be careful computing the length of the pasture: Remember that one side of the pasture does not need any fence!) Continue the table until you find the pasture whose area is #math.equation(block: false, alt: "16 , 000")[$16 , 000$] square yards. + Write an expression for the length of the pasture if its width is #math.equation(block: false, alt: "x")[$x$]. Next, write an expression for the area, #math.equation(block: false, alt: "A")[$A$], of the pasture if its width is #math.equation(block: false, alt: "x")[$x$]. Graph the equation for #math.equation(block: false, alt: "A")[$A$], and use the graph to find the pasture of area #math.equation(block: false, alt: "16 , 000")[$16 , 000$] square yards. + Write an equation for the area, #math.equation(block: false, alt: "A")[$A$], of the pasture in terms of its width #math.equation(block: false, alt: "x")[$x$]. Solve your equation algebraically for #math.equation(block: false, alt: "A equals 16 , 000")[$A = 16 , 000$]. For Problems 47 and 48, you will need the formula for the volume of a box. A box is made from a square piece of cardboard by cutting #math.equation(block: false, alt: "2")[$2$]-inch squares from each corner and turning up the edges. #figure(figph[cardboard box], alt: "cardboard box", caption: none) + If the piece of cardboard is #math.equation(block: false, alt: "x")[$x$] inches square, write expressions for the length, width, and height of the box. Then write an expression for the volume, #math.equation(block: false, alt: "V")[$V$], of the box in terms of #math.equation(block: false, alt: "x")[$x$]. + Use your calculator to make a table of values showing the volumes of boxes made from cardboard squares of side #math.equation(block: false, alt: "4")[$4$] inches, #math.equation(block: false, alt: "5")[$5$] inches, and so on. + Graph your function for the volume on your calculator. What happens to #math.equation(block: false, alt: "V")[$V$] as #math.equation(block: false, alt: "x")[$x$] increases? + Use your table or your graph to find what size cardboard you need to make a box with volume #math.equation(block: false, alt: "50")[$50$] cubic inches. + Write and solve a quadratic equation to answer part (d). + #math.equation(block: false, alt: "l equals x minus 4")[$l = x − 4$], #math.equation(block: false, alt: "w equals x minus 4")[$" " w = x − 4$], #math.equation(block: false, alt: "h equals 2")[$" " h = 2$], #math.equation(block: false, alt: "V equals 2 open parenthesis x minus 4 close parenthesis squared")[$" " V = 2 ( x − 4 )^(2)$] + #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#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: "V")[$V$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "8")[$8$]], [#math.equation(block: false, alt: "18")[$18$]], [#math.equation(block: false, alt: "32")[$32$]], [#math.equation(block: false, alt: "50")[$50$]], [#math.equation(block: false, alt: "72")[$72$]], )) + As #math.equation(block: false, alt: "x")[$x$] increases, #math.equation(block: false, alt: "V")[$V$] increases. + #math.equation(block: false, alt: "9")[$9$] inches by #math.equation(block: false, alt: "9")[$9$] inches. + #math.equation(block: false, alt: "2 open parenthesis x minus 4 close parenthesis squared equals 50")[$2 ( x − 4 )^(2) = 50$], #math.equation(block: false, alt: "x equals 9")[$" " x = 9$] A length of rain gutter is made from a piece of aluminum #math.equation(block: false, alt: "6")[$6$] feet long and #math.equation(block: false, alt: "1")[$1$] foot wide. #figure(figph[rain gutter], alt: "rain gutter", caption: none) + If a strip of width #math.equation(block: false, alt: "x")[$x$] is turned up along each long edge, write expressions for the length, width, and height of the gutter. Then write an expression for the volume, #math.equation(block: false, alt: "V")[$V$], of the gutter in terms of #math.equation(block: false, alt: "x")[$x$]. + Use your calculator to make a table of values showing the volumes of various rain gutters formed by turning up edges of #math.equation(block: false, alt: "0.1")[$0.1$] foot, #math.equation(block: false, alt: "0.2")[$0.2$] foot, and so on. + Graph your function for the volume. What happens to #math.equation(block: false, alt: "V")[$V$] as #math.equation(block: false, alt: "x")[$x$] increases? + Use your table or your graph to discover how much metal should be turned up along each long edge so that the gutter has a capacity of #math.equation(block: false, alt: "the fraction 3 over 4")[$display(frac(3, 4))$] cubic foot of rainwater. + Write and solve a quadratic equation to answer part (d). Problems 49 and 50 deal with wildlife management. The annual increase, #math.equation(block: false, alt: "I")[$I$], in a population often depends on the size #math.equation(block: false, alt: "x")[$x$] of the population, according to the formula #math.equation(block: true, alt: "I equals k C x minus k x squared")[$I = k C x − k x^(2)$] where #math.equation(block: false, alt: "k")[$k$] and #math.equation(block: false, alt: "C")[$C$] are constants related to the fertility of the population and the availability of food. The annual increase, #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$], in the deer population in a national park is given by #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 1.2 x minus 0.0002 x squared")[$f ( x ) = 1.2 x − 0.0002 x^(2)$] where #math.equation(block: false, alt: "x")[$x$] is the size of the population that year. + Make a table of values for #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 7000")[$0 ≤ x ≤ 7000$]. Use increments of #math.equation(block: false, alt: "500")[$500$] in #math.equation(block: false, alt: "x")[$x$]. + How much will a population of #math.equation(block: false, alt: "2000")[$2000$] deer increase? A population of #math.equation(block: false, alt: "5000")[$5000$] deer? A population of #math.equation(block: false, alt: "7000")[$7000$] deer? + Use your calculator to graph the annual increase, #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$], versus the size of the population, #math.equation(block: false, alt: "x")[$x$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 7000")[$0 ≤ x ≤ 7000$]. + What do the #math.equation(block: false, alt: "x")[$x$]-intercepts tell us about the deer population? + Estimate the population size that results in the largest annual increase. What is that increase? + #figure(table( columns: 9, 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: "500")[$500$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "1500")[$1500$]], [#math.equation(block: false, alt: "2000")[$2000$]], [#math.equation(block: false, alt: "2500")[$2500$]], [#math.equation(block: false, alt: "3000")[$3000$]], [#math.equation(block: false, alt: "3500")[$3500$]]), [#math.equation(block: false, alt: "I")[$I$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "550")[$550$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "1350")[$1350$]], [#math.equation(block: false, alt: "1600")[$1600$]], [#math.equation(block: false, alt: "1750")[$1750$]], [#math.equation(block: false, alt: "1800")[$1800$]], [#math.equation(block: false, alt: "1750")[$1750$]], )) #figure(table( columns: 8, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "4000")[$4000$]], [#math.equation(block: false, alt: "4500")[$4500$]], [#math.equation(block: false, alt: "5000")[$5000$]], [#math.equation(block: false, alt: "5500")[$5500$]], [#math.equation(block: false, alt: "6000")[$6000$]], [#math.equation(block: false, alt: "6500")[$6500$]], [#math.equation(block: false, alt: "7000")[$7000$]]), [#math.equation(block: false, alt: "I")[$I$]], [#math.equation(block: false, alt: "1600")[$1600$]], [#math.equation(block: false, alt: "1350")[$1350$]], [#math.equation(block: false, alt: "1000")[$1000$]], [#math.equation(block: false, alt: "550")[$550$]], [#math.equation(block: false, alt: "0")[$0$]], [#math.equation(block: false, alt: "minus 650")[$− 650$]], [#math.equation(block: false, alt: "minus 1400")[$− 1400$]], )) + #math.equation(block: false, alt: "1600")[$1600$], #math.equation(block: false, alt: "1000")[$1000$], #math.equation(block: false, alt: "minus 1400")[$− 1400$] + #figure(figph[parabola], alt: "parabola", caption: none) + No increase + #math.equation(block: false, alt: "3000")[$3000$]; #math.equation(block: false, alt: "1800")[$1800$] Commercial fishermen rely on a steady supply of fish in their area. To avoid overfishing, they adjust their harvest to the size of the population. The function #math.equation(block: true, alt: "g open parenthesis x close parenthesis equals 0.4 x minus 0.0001 x squared")[$g ( x ) = 0.4 x − 0.0001 x^(2)$] gives the annual rate of growth, in tons per year, of a fish population of biomass #math.equation(block: false, alt: "x")[$x$] tons. + Make a table of values for #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 5000")[$0 ≤ x ≤ 5000$]. Use increments of #math.equation(block: false, alt: "500")[$500$] in #math.equation(block: false, alt: "x")[$x$]. + How much will a population of #math.equation(block: false, alt: "1000")[$1000$] tons increase? A population of #math.equation(block: false, alt: "3000")[$3000$] tons? A population of #math.equation(block: false, alt: "5000")[$5000$] tons? + Use your calculator to graph the annual increase, #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$], versus the size of the population, #math.equation(block: false, alt: "x")[$x$], for #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 5000")[$0 ≤ x ≤ 5000$]. + What do the #math.equation(block: false, alt: "x")[$x$]-intercepts tell us about the fish population? + Estimate the population size that results in the largest annual increase. What is that increase? For Problems 51-62, use a substitution to solve the equation. #math.equation(block: true, alt: "a to the power 4 plus a squared minus 2 equals 0")[$a^(4) + a^(2) − 2 = 0$] #math.equation(block: true, alt: "plus or minus 1")[$± 1$] #math.equation(block: true, alt: "t to the power 6 minus t cubed minus 6 equals 0")[$t^(6) − t^(3) − 6 = 0$] #math.equation(block: true, alt: "4 b to the power 6 minus 3 equals b cubed")[$4 b^(6) − 3 = b^(3)$] #math.equation(block: true, alt: "the cube root of minus 3 / 4")[$root(3, − 3 / 4)$], #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "3 x to the power 4 plus 1 equals 4 x squared")[$3 x^(4) + 1 = 4 x^(2)$] #math.equation(block: true, alt: "c to the power 2 / 3 plus 2 c to the power 1 / 3 minus 3 equals 0")[$c^(2 / 3) + 2 c^(1 / 3) − 3 = 0$] #math.equation(block: true, alt: "minus 27")[$− 27$], #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "y to the power 1 / 2 minus 3 y to the power 1 / 4 minus 4 equals 0")[$y^(1 / 2) − 3 y^(1 / 4) − 4 = 0$] #math.equation(block: true, alt: "10 to the power 2 w minus 5 times 10 to the power w plus 6 equals 0")[$10^(2 w) − 5 ⋅ 10^(w) + 6 = 0$] #math.equation(block: true, alt: "log ⁡ 2")[$log 2$], #math.equation(block: true, alt: "log ⁡ 3")[$log 3$] #math.equation(block: true, alt: "e to the power 2 x minus 5 e to the power x plus 4 equals 0")[$e^(2 x) − 5 e^(x) + 4 = 0$] #math.equation(block: true, alt: "5 to the power 2 t minus 30 times 5 to the power t plus 125 equals 0")[$5^(2 t) − 30 ⋅ 5^(t) + 125 = 0$] #math.equation(block: true, alt: "1")[$1$], #math.equation(block: true, alt: "2")[$2$] #math.equation(block: true, alt: "e to the power 4 r minus 3 e to the power 2 r plus 2 equals 0")[$e^(4 r) − 3 e^(2 r) + 2 = 0$] #math.equation(block: true, alt: "the fraction 1 over m squared plus the fraction 5 over m minus 6 equals 0")[$display(frac(1, m^(2))) + display(frac(5, m)) − 6 = 0$] #math.equation(block: true, alt: "the fraction minus 1 over 6")[$display(frac(− 1, 6))$], #math.equation(block: true, alt: "1")[$1$] #math.equation(block: true, alt: "the fraction 1 over s squared plus the fraction 4 over s minus 5 equals 0")[$display(frac(1, s^(2))) + display(frac(4, s)) − 5 = 0$] The sail in the figure is a right triangle of base and height #math.equation(block: false, alt: "x")[$x$]. It has a colored stripe along the hypotenuse and a white triangle of base and height #math.equation(block: false, alt: "y")[$y$] in the lower corner. #figure(figph[sail], alt: "sail", caption: none) + Write an expression for the area of the colored stripe. + Express the area of the stripe in factored form. + If the sail is #math.equation(block: false, alt: "7 the fraction 1 over 2")[$7 frac(1, 2)$] feet high and the white strip is #math.equation(block: false, alt: "4 the fraction 1 over 2")[$4 frac(1, 2)$] feet high, use your answer to (b) to calculate mentally the area of the stripe. + #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis x squared minus y squared close parenthesis")[$A = display(frac(1, 2)) ( x^(2) − y^(2) )$] + #math.equation(block: false, alt: "A equals the fraction 1 over 2 open parenthesis x minus y close parenthesis open parenthesis x plus y close parenthesis")[$A = display(frac(1, 2)) ( x − y ) ( x + y )$] + #math.equation(block: false, alt: "18")[$18$] sq ft An hors d'oeuvres tray has radius #math.equation(block: false, alt: "x")[$x$], and the dip container has radius #math.equation(block: false, alt: "y")[$y$], as shown in the figure. #figure(figph[tray], alt: "tray", caption: none) + Write an expression for the area for the chips (shaded region). + Express the area in factored form. + If the tray has radius #math.equation(block: false, alt: "8 the fraction 1 over 2")[$8 frac(1, 2)$] inches and the space for the dip has radius #math.equation(block: false, alt: "2 the fraction 1 over 2")[$2 frac(1, 2)$] inches, use your answer to part (b) to calculate mentally the area for chips. (Express your answer as a multiple of #math.equation(block: false, alt: "π")[$π$].)