#set document(title: "7.1 Polynomial Functions", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 7.1#h(0.6em)Polynomial Functions === Introduction We have already encountered some examples of polynomial functions. Linear functions, #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a x plus b")[$f ( x ) = a x + b$] and quadratic functions #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a x squared plus b x plus c")[$f ( x ) = a x^(2) + b x + c$] are special cases of polynomial functions. In general, we make the following definition. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Polynomial Function] A #strong[polynomial function] has the form #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals a sub n x to the power n plus a sub n minus 1 x to the power n minus 1 plus a sub n minus 2 x to the power n minus 2 plus ⋯ plus a sub 2 x squared plus a sub 1 x plus a sub 0")[$f ( x ) = a_(n) x^(n) + a_(n − 1) x^(n − 1) + a_(n − 2) x^(n − 2) + ⋯ + a_(2) x^(2) + a_(1) x + a_(0)$] where #math.equation(block: false, alt: "a sub 0")[$a_(0)$], #math.equation(block: false, alt: "a sub 1")[$a_(1)$], #math.equation(block: false, alt: "a sub 2")[$a_(2)$], #math.equation(block: false, alt: "and so on")[$…$], #math.equation(block: false, alt: "a sub n")[$a_(n)$] are constants and #math.equation(block: false, alt: "a sub n not equal to 0")[$a_(n) ≠ 0$]. The coefficient #math.equation(block: false, alt: "a sub n")[$a_(n)$] of the highest power term is called the #strong[lead coefficient]. ] Some examples of polynomials are #math.equation(block: true, alt: "f open parenthesis x close parenthesis, equals 6 x cubed minus 4 x squared plus x minus 2, g open parenthesis x close parenthesis, equals 9 x to the power 5 minus 2; p open parenthesis x close parenthesis, equals x to the power 4 plus x squared plus 1, q open parenthesis x close parenthesis, equals 2 x to the power 10 minus x to the power 7 plus 3 x to the power 6 plus 5 x cubed plus 3 x")[$f ( x ) & = 6 x^(3) − 4 x^(2) + x − 2 & g ( x ) & = 9 x^(5) − 2 \ p ( x ) & = x^(4) + x^(2) + 1 & q ( x ) & = 2 x^(10) − x^(7) + 3 x^(6) + 5 x^(3) + 3 x$] Each of the polynomials above is written in #strong[descending powers], which means that the highest-degree term comes first, and the degrees of the terms decrease from largest to smallest. Sometimes it is useful to write a polynomial in #strong[ascending powers], so that the degrees of the terms increase. For example, the polynomial #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] above would be written as #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals minus 2 plus x minus 4 x squared plus 6 x cubed")[$f ( x ) = − 2 + x − 4 x^(2) + 6 x^(3)$] in ascending powers. What is the lead coefficient of a polynomial? \_\_\_\_\_ #solutionbox[ The coefficient of the highest power ] What is the lead coefficient of a polynomial? + The first coefficient + The constant term + The largest coefficient + The coefficient of the highest power === Products of Polynomials When we multiply two or more polynomials together, we get another polynomial of higher degree. (See Algebra Skills Refresher A.7 for the definition of degree.) #examplebox("Example 1")[][ Compute the products. + #math.equation(block: false, alt: "open parenthesis x plus 2 close parenthesis open parenthesis 5 x cubed minus 3 x squared plus 4 close parenthesis")[$( x + 2 ) ( 5 x^(3) − 3 x^(2) + 4 )$] + #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 2 close parenthesis open parenthesis x minus 4 close parenthesis")[$( x − 3 ) ( x + 2 ) ( x − 4 )$] #solutionbox[ #figure(figph[diagram of polynomial multiplication], alt: "diagram of polynomial multiplication", caption: none) ] ] Multiply #math.equation(block: false, alt: "open parenthesis y plus 2 close parenthesis open parenthesis y squared minus 2 y plus 3 close parenthesis equals")[$( y + 2 ) ( y^(2) − 2 y + 3 ) =$]\_\_\_\_\_ . #solutionbox[ #math.equation(block: true, alt: "y cubed minus y plus 6")[$y^(3) − y + 6$] ] Multiply #math.equation(block: false, alt: "open parenthesis y plus 2 close parenthesis open parenthesis y squared minus 2 y plus 3 close parenthesis")[$" " ( y + 2 ) ( y^(2) − 2 y + 3 )$] #solutionbox[ #math.equation(block: true, alt: "y cubed minus y plus 6")[$y^(3) − y + 6$] ] What is the linear term of the product #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis")[$( x − 3 ) ( x + 5 )$] ? \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "2 x")[$2 x$] is the linear term of #math.equation(block: false, alt: "x squared plus 2 x minus 15")[$x^(2) + 2 x − 15$]. ] What is the linear term of the product #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis x plus 5 close parenthesis")[$( x − 3 ) ( x + 5 )$] ? + #math.equation(block: false, alt: "x squared")[$x^(2)$] + #math.equation(block: false, alt: "2 x minus 15")[$2 x − 15$] + #math.equation(block: false, alt: "minus 15")[$− 15$] + #math.equation(block: false, alt: "2 x")[$2 x$] In Examplea, we multiplied a polynomial of degree 1 by a polynomial of degree 3, and the product was a polynomial of degree 4. In Exampleb, the product of three first degree polynomials is a third-degree polynomial. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Degree of a Product] The degree of a product of nonzero polynomials is the sum of the degrees of the factors. That is, If #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] has degree #math.equation(block: false, alt: "m")[$m$] and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$] has degree #math.equation(block: false, alt: "n")[$n$], then their product #math.equation(block: false, alt: "P open parenthesis x close parenthesis Q open parenthesis x close parenthesis")[$P ( x ) Q ( x )$] has degree #math.equation(block: false, alt: "n plus m")[$n + m$]. ] #examplebox("Example 2")[][ Let #math.equation(block: false, alt: "P open parenthesis x close parenthesis equals 5 x to the power 4 minus 2 x cubed plus 6 x squared minus x plus 2")[$" " P ( x ) = 5 x^(4) − 2 x^(3) + 6 x^(2) − x + 2$], and #math.equation(block: false, alt: "Q open parenthesis x close parenthesis equals 3 x cubed minus 4 x squared plus 5 x plus 3")[$" " Q ( x ) = 3 x^(3) − 4 x^(2) + 5 x + 3$]. + What is the degree of their product? What is the coefficient of the lead term? + Find the coefficient of the #math.equation(block: false, alt: "x cubed")[$x^(3)$]-term of the product. #solutionbox[ + The degree of #math.equation(block: false, alt: "P")[$P$] is #math.equation(block: false, alt: "4")[$4$], and the degree of #math.equation(block: false, alt: "Q")[$Q$] is #math.equation(block: false, alt: "3")[$3$], so the degree of their product is #math.equation(block: false, alt: "4 plus 3 equals 7")[$4 + 3 = 7$]. The only degree #math.equation(block: false, alt: "7")[$7$] term of the product is #math.equation(block: false, alt: "open parenthesis 5 x to the power 4 close parenthesis open parenthesis 3 x cubed close parenthesis equals 15 x to the power 7")[$( 5 x^(4) ) ( 3 x^(3) ) = 15 x^(7)$], which has coefficient #math.equation(block: false, alt: "15")[$15$]. + In the product, each term of #math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$] is multiplied by each term of #math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]. We get degree #math.equation(block: false, alt: "3")[$3$] terms by multiplying together terms of degree #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "3")[$3$], or #math.equation(block: false, alt: "1")[$1$] and #math.equation(block: false, alt: "2")[$2$]. For these polynomials, the possible combinations are:#figure(table( columns: 3, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "P open parenthesis x close parenthesis")[$P ( x )$]], [#math.equation(block: false, alt: "Q open parenthesis x close parenthesis")[$Q ( x )$]], [Product]), [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3 x cubed")[$3 x^(3)$]], [#math.equation(block: false, alt: "6 x cubed")[$6 x^(3)$]], [#math.equation(block: false, alt: "minus 2 x cubed")[$− 2 x^(3)$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "minus 6 x cubed")[$− 6 x^(3)$]], [#math.equation(block: false, alt: "minus x")[$− x$]], [#math.equation(block: false, alt: "minus 4 x squared")[$− 4 x^(2)$]], [#math.equation(block: false, alt: "4 x cubed")[$4 x^(3)$]], [#math.equation(block: false, alt: "6 x squared")[$6 x^(2)$]], [#math.equation(block: false, alt: "5 x")[$5 x$]], [#math.equation(block: false, alt: "30 x cubed")[$30 x^(3)$]], )) #linebreak() The sum of the third-degree terms of the product is #math.equation(block: false, alt: "34 x cubed")[$34 x^(3)$], with coefficient #math.equation(block: false, alt: "34")[$34$]. ] ] Find the coefficient of the fourth-degree term of the product of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x to the power 6 plus 2 x to the power 4 minus x cubed plus 5 x squared plus 1")[$f ( x ) = 2 x^(6) + 2 x^(4) − x^(3) + 5 x^(2) + 1$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x to the power 5 minus 3 x to the power 4 plus 2 x cubed plus x squared minus 4 x minus 2")[$g ( x ) = x^(5) − 3 x^(4) + 2 x^(3) + x^(2) − 4 x − 2$]. Answer: \_\_\_\_\_ #solutionbox[ #math.equation(block: false, alt: "2")[$2$] is the coefficient of the #math.equation(block: false, alt: "x to the power 4")[$x^(4)$] term in the product. To compute the fourth-degree term in the product, we multiply a term of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] of degree 4, 3, 2, or 0 by a term of #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] of degree 0, 1, 2, or 4 respectively. We add those four products of terms: #math.equation(block: true, alt: "open parenthesis 2 x to the power 4 times open parenthesis minus 2 close parenthesis close parenthesis plus open parenthesis open parenthesis minus x cubed close parenthesis open parenthesis minus 4 x close parenthesis close parenthesis plus open parenthesis 5 x squared times x squared close parenthesis plus open parenthesis 1 times open parenthesis minus 3 x to the power 4 close parenthesis close parenthesis")[$( 2 x^(4) ⋅ ( − 2 ) ) + ( ( − x^(3) ) ( − 4 x ) ) + ( 5 x^(2) ⋅ x^(2) ) + ( 1 ⋅ ( − 3 x^(4) ) )$] So the fourth-degree term in the product is #math.equation(block: false, alt: "2 x to the power 4")[$2 x^(4)$], and the coefficient is #math.equation(block: false, alt: "2")[$2$]. ] Find the coefficient of the fourth-degree term of the product of #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals 2 x to the power 6 plus 2 x to the power 4 minus x cubed plus 5 x squared plus 1")[$" " f ( x ) = 2 x^(6) + 2 x^(4) − x^(3) + 5 x^(2) + 1 " "$] and #math.equation(block: false, alt: "g open parenthesis x close parenthesis equals x to the power 5 minus 3 x to the power 4 plus 2 x cubed plus x squared minus 4 x minus 2")[$" " g ( x ) = x^(5) − 3 x^(4) + 2 x^(3) + x^(2) − 4 x − 2$]. #solutionbox[ #math.equation(block: false, alt: "2")[$2$] is the coefficient of the #math.equation(block: false, alt: "x to the power 4")[$x^(4)$] term in the product. To compute the fourth-degree term in the product, we multiply a term of #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] of degree 4, 3, 2, or 0 by a term of #math.equation(block: false, alt: "g open parenthesis x close parenthesis")[$g ( x )$] of degree 0, 1, 2, or 4 respectively. We add those four products of terms: #math.equation(block: true, alt: "open parenthesis 2 x to the power 4 times open parenthesis minus 2 close parenthesis close parenthesis plus open parenthesis open parenthesis minus x cubed close parenthesis open parenthesis minus 4 x close parenthesis close parenthesis plus open parenthesis 5 x squared times x squared close parenthesis plus open parenthesis 1 times open parenthesis minus 3 x to the power 4 close parenthesis close parenthesis")[$( 2 x^(4) ⋅ ( − 2 ) ) + ( ( − x^(3) ) ( − 4 x ) ) + ( 5 x^(2) ⋅ x^(2) ) + ( 1 ⋅ ( − 3 x^(4) ) )$] So the fourth-degree term in the product is #math.equation(block: false, alt: "2 x to the power 4")[$2 x^(4)$], and the coefficient is #math.equation(block: false, alt: "2")[$2$]. ] The degree of a product of polynomials is the same as \_\_\_\_\_ #solutionbox[ the sum of the degrees of the factors. ] The degree of a product of polynomials is the same as + the product of the degrees of the factors. + the number of terms in the product. + the sum of the degrees of the factors. + the sum of the lead coefficients. === Special Products In the Algebra Skills Refresher Factoring Quadratic Trinomials, you can review the following special products involving quadratic expressions. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Special Products of Binomials] #math.equation(block: true, alt: "open parenthesis a plus b close parenthesis squared equals open parenthesis a plus b close parenthesis open parenthesis a plus b close parenthesis equals a squared plus 2 a b plus b squared; open parenthesis a minus b close parenthesis squared equals open parenthesis a minus b close parenthesis open parenthesis a minus b close parenthesis equals a squared minus 2 a b plus b squared; open parenthesis a plus b close parenthesis open parenthesis a minus b close parenthesis equals a squared minus b squared")[$& ( a + b )^(2) = ( a + b ) ( a + b ) = a^(2) + 2 a b + b^(2) \ & ( a − b )^(2) = ( a − b ) ( a − b ) = a^(2) − 2 a b + b^(2) \ & ( a + b ) ( a − b ) = a^(2) − b^(2)$] ] There are also special products resulting in cubic polynomials. In the Homework problems, you will be asked to verify the following products. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Cube of a Binomial] + #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cubed equals x cubed plus 3 x squared y plus 3 x y squared plus y cubed")[$( x + y )^(3) = x^(3) + 3 x^(2) y + 3 x y^(2) + y^(3)$] + #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis cubed equals x cubed minus 3 x squared y plus 3 x y squared minus y cubed")[$( x − y )^(3) = x^(3) − 3 x^(2) y + 3 x y^(2) − y^(3)$] ] If you become familiar with these general forms, you can use them as patterns to find specific examples of such products. #examplebox("Example 3")[][ Write #math.equation(block: false, alt: "open parenthesis 2 w minus 3 close parenthesis cubed")[$" " ( 2 w − 3 )^(3) " "$] as a polynomial. #solutionbox[ Use product above, with #math.equation(block: false, alt: "x")[$x$] replaced by #math.equation(block: false, alt: "2 w")[$2 bold(italic(w))$] and #math.equation(block: false, alt: "y")[$y$] replaced by #math.equation(block: false, alt: "3")[$3$]. #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis cubed, equals x cubed minus 3 x squared y plus 3 x y squared minus y cubed; open parenthesis 2 w minus 3 close parenthesis cubed, equals open parenthesis 2 w close parenthesis cubed minus 3 open parenthesis 2 w close parenthesis squared open parenthesis 3 close parenthesis plus 3 open parenthesis 2 w close parenthesis open parenthesis 3 close parenthesis squared minus 3 cubed, Simplify.; equals 8 w cubed minus 36 w squared plus 54 w minus 27")[$( x − y )^(3) & = x^(3) − 3 x^(2) y + 3 x y^(2) − y^(3) \ ( 2 bold(italic(w)) − 3 )^(3) & = ( 2 bold(italic(w)) )^(3) − 3 ( 2 bold(italic(w)) )^(2) ( 3 ) + 3 ( 2 bold(italic(w)) ) ( 3 )^(2) − 3^(3) & & "Simplify." \ & = 8 w^(3) − 36 w^(2) + 54 w − 27$] ] ] Of course, we can also expand the product in Example simply by polynomial multiplication and arrive at the same answer. Write #math.equation(block: false, alt: "open parenthesis 5 plus x squared close parenthesis cubed")[$attach(( 5 + x^(2) ), t: 3)$] as a polynomial. Answer: \_\_\_\_\_ #solutionbox[ #math.equation(block: true, alt: "125 plus 75 x squared plus 15 x to the power 4 plus x to the power 6")[$125 + 75 x^(2) + 15 x^(4) + x^(6)$] ] Write #math.equation(block: false, alt: "open parenthesis 5 plus x squared close parenthesis cubed")[$attach(( 5 + x^(2) ), t: 3)$] as a polynomial. #solutionbox[ #math.equation(block: true, alt: "125 plus 75 x squared plus 15 x to the power 4 plus x to the power 6")[$125 + 75 x^(2) + 15 x^(4) + x^(6)$] ] Explain the difference between the expressions #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed")[$( a + b )^(3)$] and #math.equation(block: false, alt: "a cubed plus b cubed")[$a^(3) + b^(3)$]. \_\_\_\_\_ Explain the difference between the expressions #math.equation(block: false, alt: "open parenthesis a plus b close parenthesis cubed")[$( a + b )^(3)$] and #math.equation(block: false, alt: "a cubed plus b cubed")[$a^(3) + b^(3)$]. === Factoring Cubics Another pair of products is useful for factoring cubic polynomials. In the Homework problems, you will be asked to verify the following products: #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis, equals x cubed plus y cubed; open parenthesis x minus y close parenthesis open parenthesis x squared plus x y plus y squared close parenthesis, equals x cubed minus y cubed")[$( x + y ) ( x^(2) − x y + y^(2) ) & = x^(3) + y^(3) \ ( x − y ) ( x^(2) + x y + y^(2) ) & = x^(3) − y^(3)$] Viewing these products from right to left, we have the following special factorizations for the sum and difference of two cubes. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Factoring the Sum or Difference of Two Cubes] + #math.equation(block: false, alt: "x cubed plus y cubed equals open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis")[$x^(3) + y^(3) = ( x + y ) ( x^(2) − x y + y^(2) )$] + #math.equation(block: false, alt: "x cubed minus y cubed equals open parenthesis x minus y close parenthesis open parenthesis x squared plus x y plus y squared close parenthesis")[$x^(3) − y^(3) = ( x − y ) ( x^(2) + x y + y^(2) )$] ] When we recognize a polynomial as a sum or difference of two perfect cubes, we then identify the two cubed expressions and apply the formula. #examplebox("Example 4")[][ Factor each polynomial. + #math.equation(block: false, alt: "8 a cubed plus b cubed")[$8 a^(3) + b^(3)$] + #math.equation(block: false, alt: "1 minus 27 h to the power 6")[$1 − 27 h^(6)$] #solutionbox[ + This polynomial is a sum of two cubes. The cubed expressions are #math.equation(block: false, alt: "2 a")[$2 a$], because #math.equation(block: false, alt: "open parenthesis 2 a close parenthesis cubed equals 8 a cubed")[$( 2 a )^(3) = 8 a^(3)$], and #math.equation(block: false, alt: "b")[$b$]. Use formula as a pattern, replacing #math.equation(block: false, alt: "x")[$x$] with #math.equation(block: false, alt: "2 a")[$2 bold(italic(a))$], and #math.equation(block: false, alt: "y")[$y$] with #math.equation(block: false, alt: "b")[$bold(italic(b))$]. #math.equation(block: true, alt: "x cubed plus y cubed, equals open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis; open parenthesis 2 a close parenthesis 3 plus b 3, equals open parenthesis 2 a plus b close parenthesis open parenthesis open parenthesis 2 a close parenthesis squared minus open parenthesis 2 a close parenthesis b plus b squared close parenthesis, Simplify.; equals open parenthesis 2 a plus b close parenthesis open parenthesis 4 a squared minus 2 a b plus b squared close parenthesis")[$x^(3) + y^(3) & = ( x + y ) ( x^(2) − x y + y^(2) ) \ ( 2 bold(italic(a)) ) 3 + bold(italic(b)) 3 & = ( 2 bold(italic(a)) + bold(italic(b)) ) ( ( 2 bold(italic(a)) )^(2) − ( 2 bold(italic(a)) ) bold(italic(b)) + bold(italic(b))^(2) ) & & "Simplify." \ & = ( 2 a + b ) ( 4 a^(2) − 2 a b + b^(2) )$] + This polynomial is a difference of two cubes. The cubed expressions are #math.equation(block: false, alt: "1")[$1$], because #math.equation(block: false, alt: "1 cubed equals 1")[$1^(3) = 1$], and #math.equation(block: false, alt: "3 h squared")[$3 h^(2)$], because #math.equation(block: false, alt: "open parenthesis 3 h squared close parenthesis cubed equals 27 h to the power 6")[$( 3 h^(2) )^(3) = 27 h^(6)$]. Use formula above as a pattern, replacing #math.equation(block: false, alt: "x")[$x$] by #math.equation(block: false, alt: "1")[$1$], and #math.equation(block: false, alt: "y")[$y$] by #math.equation(block: false, alt: "3 h squared")[$3 bold(italic(h))^(2)$]: #math.equation(block: true, alt: "x cubed minus y cubed, equals open parenthesis x minus y close parenthesis open parenthesis x squared plus x y plus y squared close parenthesis; 1 cubed minus open parenthesis 3 h squared close parenthesis cubed, equals open parenthesis 1 minus 3 h squared close parenthesis open parenthesis 1 squared plus 1 open parenthesis 3 h squared close parenthesis plus open parenthesis 3 h squared close parenthesis squared close parenthesis, Simplify.; equals open parenthesis 1 minus 3 h squared close parenthesis open parenthesis 1 plus 3 h squared plus 9 h to the power 4 close parenthesis")[$x^(3) − y^(3) & = ( x − y ) ( x^(2) + x y + y^(2) ) \ 1^(3) − ( 3 bold(italic(h))^(2) )^(3) & = ( 1 − 3 bold(italic(h))^(2) ) ( 1^(2) + 1 ( 3 bold(italic(h))^(2) ) + ( 3 bold(italic(h))^(2) )^(2) ) & & "Simplify." \ & = ( 1 − 3 h^(2) ) ( 1 + 3 h^(2) + 9 h^(4) )$] ] ] Which statement is true? \_\_\_\_\_ #solutionbox[ A cubic polynomial #math.equation(block: false, alt: "y equals a x cubed plus b x squared plus c x plus d")[$y = a x^(3) + b x^(2) + c x + d$] describes #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$]. ] Which statement is true? + A cubic polynomial #math.equation(block: false, alt: "y equals a x cubed plus b x squared plus c x plus d")[$" " y = a x^(3) + b x^(2) + c x + d " "$] describes #math.equation(block: false, alt: "y")[$y$] as a function of #math.equation(block: false, alt: "x")[$x$]. + We cannot factor the sum of two cubes. + #math.equation(block: false, alt: "open parenthesis 2 x plus 3 y close parenthesis cubed equals 8 x cubed plus 27 y cubed")[$( 2 x + 3 y )^(3) = 8 x^(3) + 27 y^(3)$] + A cubic polynomial must have 4 terms. Factor #math.equation(block: false, alt: "125 n cubed minus p cubed equals open parenthesis")[$125 n^(3) − p^(3) = \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis open parenthesis")[$\) \($]\_\_\_\_\_#math.equation(block: false, alt: "close parenthesis")[$\)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 n minus p close parenthesis open parenthesis 25 n squared plus 5 n p plus p squared close parenthesis")[$( 5 n − p ) ( 25 n^(2) + 5 n p + p^(2) )$] ] Factor #math.equation(block: false, alt: "125 n cubed minus p cubed")[$125 n^(3) − p^(3)$] #solutionbox[ #math.equation(block: true, alt: "open parenthesis 5 n minus p close parenthesis open parenthesis 25 n squared plus 5 n p plus p squared close parenthesis")[$( 5 n − p ) ( 25 n^(2) + 5 n p + p^(2) )$] ] Compare the formulas for factoring #math.equation(block: false, alt: "a cubed plus b cubed")[$a^(3) + b^(3)$] and #math.equation(block: false, alt: "a cubed minus b cubed")[$a^(3) − b^(3)$]. \_\_\_\_\_ Compare the formulas for factoring #math.equation(block: false, alt: "a cubed plus b cubed")[$a^(3) + b^(3)$] and #math.equation(block: false, alt: "a cubed minus b cubed")[$a^(3) − b^(3)$]. === Modeling with Polynomials Polynomials model many variable relationships, including volume and surface area. #examplebox("Example 5")[][ A closed box has a square base of length and width #math.equation(block: false, alt: "x")[$x$] inches and a height of #math.equation(block: false, alt: "8")[$8$] inches, as shown at right. #figure(figph[box with height 8 inches with both length and width of x inches], alt: "box with height 8 inches with both length and width of x inches", caption: none) + Write a polynomial function #math.equation(block: false, alt: "S open parenthesis x close parenthesis")[$S ( x )$] that gives the surface area of the box in terms of the dimensions of the base. + What is the surface area of a box of length and width #math.equation(block: false, alt: "18")[$18$] inches? #solutionbox[ + The surface area of a box is the sum of the areas of its six faces, #math.equation(block: true, alt: "S equals 2 l h plus 2 w h plus 2 l w")[$S = 2 l h + 2 w h + 2 l w$] Substituting #math.equation(block: false, alt: "x")[$x$] for #math.equation(block: false, alt: "l")[$l$] and #math.equation(block: false, alt: "w")[$w$], and #math.equation(block: false, alt: "8")[$8$] for #math.equation(block: false, alt: "h")[$h$] gives us #math.equation(block: true, alt: "S open parenthesis x close parenthesis equals 2 open parenthesis 8 close parenthesis x plus 2 open parenthesis 8 close parenthesis x plus 2 x squared equals 2 x squared plus 32 x")[$S ( x ) = 2 ( 8 ) x + 2 ( 8 ) x + 2 x^(2) = 2 x^(2) + 32 x$] + We evaluate the polynomial for #math.equation(block: false, alt: "x equals 18")[$x = 18$] to find #math.equation(block: true, alt: "S open parenthesis 18 close parenthesis equals 2 open parenthesis 18 close parenthesis squared plus 32 open parenthesis 18 close parenthesis equals 1224 square inches")[$S ( 18 ) = 2 ( 18 )^(2) + 32 ( 18 ) = 1224 " square inches"$] ] ] #figure(figph[rectangular hole 3 ft deep, x feet wide, x+8 ft long], alt: "rectangular hole 3 ft deep, x feet wide, x+8 ft long", caption: none) An empty reflecting pool is #math.equation(block: false, alt: "3")[$3$] feet deep. It is #math.equation(block: false, alt: "8")[$8$] feet longer than it is wide, as illustrated above. + Write a polynomial function #math.equation(block: false, alt: "S open parenthesis x close parenthesis")[$S ( x )$] that gives the surface area of the empty pool. #linebreak() #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals")[$S ( x ) =$]\_\_\_\_\_ + Write a polynomial function #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$] for the volume of the pool. #linebreak() #math.equation(block: false, alt: "V open parenthesis x close parenthesis equals")[$V ( x ) =$]\_\_\_\_\_ #solutionbox[ + We add the area of the bottom to the areas of each of the sides of the pool: #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals x squared plus 20 x plus 48")[$S ( x ) = x^(2) + 20 x + 48$] + We multiply depth, length, and width: #math.equation(block: false, alt: "V open parenthesis x close parenthesis equals 3 x squared plus 24 x")[$V ( x ) = 3 x^(2) + 24 x$] ] An empty reflecting pool is #math.equation(block: false, alt: "3")[$3$] feet deep. It is #math.equation(block: false, alt: "8")[$8$] feet longer than it is wide, as illustrated below. #figure(figph[rectangular hole 3 ft deep, x feet wide, x+8 ft long], alt: "rectangular hole 3 ft deep, x feet wide, x+8 ft long", caption: none) + Write a polynomial function #math.equation(block: false, alt: "S open parenthesis x close parenthesis")[$S ( x )$] that gives the surface area of the empty pool. + Write a polynomial function #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$] for the volume of the pool. #solutionbox[ + We add the area of the bottom to the areas of each of the sides of the pool: #math.equation(block: false, alt: "S open parenthesis x close parenthesis equals x squared plus 20 x plus 48")[$S ( x ) = x^(2) + 20 x + 48$] + We multiply depth, length, and width: #math.equation(block: false, alt: "V open parenthesis x close parenthesis equals 3 x squared plus 24 x")[$V ( x ) = 3 x^(2) + 24 x$] ] Explain how to find the surface area of a solid figure such as a box or a cylinder. \_\_\_\_\_ Explain how to find the surface area of a solid figure such as a box or a cylinder. Cubic polynomials are often used in economics to model cost functions. The cost of producing #math.equation(block: false, alt: "x")[$x$] items is an increasing function of #math.equation(block: false, alt: "x")[$x$], but its rate of increase is usually not constant. #examplebox("Example 6")[][ Pegasus Printing, Ltd. is launching a new magazine. The cost of printing #math.equation(block: false, alt: "x")[$x$] thousand copies is given by #math.equation(block: true, alt: "C open parenthesis x close parenthesis equals x cubed minus 24 x squared plus 195 x plus 250")[$C ( x ) = x^(3) − 24 x^(2) + 195 x + 250$] + What are the #strong[fixed costs], that is, the costs incurred before any copies are printed? + Graph the cost function in the window below and describe the graph. #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 20; Ymin, equals 0, Ymax equals 1500")[$"Xmin" & = 0 & & "Xmax" = 20 \ "Ymin" & = 0 & & "Ymax" = 1500$] + How many copies can be printed for \$1200? + What does the concavity of the graph tell you about the cost function? #solutionbox[ + Fixed costs are given by #math.equation(block: false, alt: "C open parenthesis 0 close parenthesis equals 250")[$C ( 0 ) = 250$], or \$#math.equation(block: false, alt: "250")[$250$]. The fixed costs include expenses like utility bills that must be paid even if no magazines are produced. + The graph is shown in figure (a). It is increasing from a vertical intercept of #math.equation(block: false, alt: "250")[$250$]. The graph is concave down for #math.equation(block: false, alt: "x less than 8")[$x < 8$] approximately, and concave up for #math.equation(block: false, alt: "x greater than 8")[$x > 8$].#figure(figph[GC graphs of a cubic cost function], alt: "GC graphs of a cubic cost function", caption: none) #linebreak() We must solve the equation #math.equation(block: true, alt: "x cubed minus 24 x squared plus 195 x plus 250 equals 1200")[$x^(3) − 24 x^(2) + 195 x + 250 = 1200$] We will solve the equation graphically, as shown in figure (b). Graph #math.equation(block: false, alt: "y equals 1200")[$y = 1200$] along with the cost function, and use the #emph[intersect] command to find the intersection point of the graphs, #math.equation(block: false, alt: "open parenthesis 15.319 , 1200 close parenthesis")[$( 15.319 , 1200 )$]. #math.equation(block: false, alt: "C open parenthesis x close parenthesis equals 1200")[$C ( x ) = 1200$] when #math.equation(block: false, alt: "x")[$x$] is about #math.equation(block: false, alt: "15.319")[$15.319$], so #math.equation(block: false, alt: "15 , 319")[$15 , 319$] copies can be printed for \$#math.equation(block: false, alt: "1200")[$1200$]. + Although the cost is always increasing, it increases very slowly from about #math.equation(block: false, alt: "x equals 5")[$x = 5$] to about #math.equation(block: false, alt: "x equals 11")[$x = 11$]. The flattening of the graph in this interval is a result of economy of scale: By buying supplies in bulk and using time efficiently, the cost per magazine can be minimized. However, if the production level is too large, costs begin to rise rapidly again. ] ] In Examplec, we solved a cubic equation graphically. There is a cubic formula, analogous to the quadratic formula, that allows us to solve cubic equations algebraically, but it is complicated and not often used. See the Projects for Chapter 7 if you would like to know more about the cubic formula. How do we solve cubic equations in this class? \_\_\_\_\_ #solutionbox[ By graphing. ] How do we solve cubic equations in this class? + By factoring + By taking cube roots + By using the quadratic formula + By graphing Cubic polynomials are also used to model smooth curves connecting given points. Such a curve is called a #strong[cubic spline]. Leon is flying his plane to Au Gres, Michigan. He maintains a constant altitude until he passes over a marker just outside the neighboring town of Omer, when he begins his descent for landing. During the descent, his altitude, in feet, is given by #math.equation(block: true, alt: "A open parenthesis x close parenthesis equals 128 x cubed minus 960 x squared plus 8000")[$A ( x ) = 128 x^(3) − 960 x^(2) + 8000$] where #math.equation(block: false, alt: "x")[$x$] is the number of miles Leon has traveled since passing over the marker in Omer. + What is Leon's altitude when he begins his descent? #linebreak() \_\_\_\_\_ ft + Graph #math.equation(block: false, alt: "A open parenthesis x close parenthesis")[$A ( x )$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 5; Ymin, equals 0, Ymax equals 8000")[$"Xmin" & = 0 & & "Xmax" = 5 \ "Ymin" & = 0 & & "Ymax" = 8000$] + Use the #emph[Trace] feature to discover how far from Omer Leon will travel before landing. (In other words, how far is Au Gres from Omer?) #linebreak() \_\_\_\_\_ mi + Verify your answer to part (c) algebraically. #linebreak() #math.equation(block: false, alt: "A open parenthesis 5 close parenthesis equals")[$A ( 5 ) =$]\_\_\_\_\_ #solutionbox[ + #math.equation(block: false, alt: "8000")[$8000$] ft + A graph is below. + #math.equation(block: false, alt: "5")[$5$] mi + #math.equation(block: false, alt: "A open parenthesis 5 close parenthesis equals 0")[$A ( 5 ) = 0$], that is, Leon is on the ground (at altitude 0) when he is 5 miles past Omer. ] Graph for part (b) #figure(figph[cubic], alt: "cubic", caption: none) Leon is flying his plane to Au Gres, Michigan. He maintains a constant altitude until he passes over a marker just outside the neighboring town of Omer, when he begins his descent for landing. During the descent, his altitude, in feet, is given by #math.equation(block: true, alt: "A open parenthesis x close parenthesis equals 128 x cubed minus 960 x squared plus 8000")[$A ( x ) = 128 x^(3) − 960 x^(2) + 8000$] where #math.equation(block: false, alt: "x")[$x$] is the number of miles Leon has traveled since passing over the marker in Omer. + What is Leon's altitude when he begins his descent? + Graph #math.equation(block: false, alt: "A open parenthesis x close parenthesis")[$A ( x )$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 5; Ymin, equals 0, Ymax equals 8000")[$"Xmin" & = 0 & & "Xmax" = 5 \ "Ymin" & = 0 & & "Ymax" = 8000$] + Use the #emph[Trace] feature to discover how far from Omer Leon will travel before landing. (In other words, how far is Au Gres from Omer?) + Verify your answer to part (c) algebraically. #solutionbox[ + #math.equation(block: false, alt: "8000")[$8000$] ft + #figure(figph[cubic], alt: "cubic", caption: none) + #math.equation(block: false, alt: "5")[$5$] mi + #math.equation(block: false, alt: "A open parenthesis 5 close parenthesis equals 0")[$A ( 5 ) = 0$], that is, Leon is on the ground (at altitude 0) when he is 5 miles past Omer. ] === Section Summary ==== Vocabulary Look up the definitions of new terms in the Glossary. - Polynomial function - Degree - Ascending powers - Lead coefficient - Descending powers ==== CONCEPTS + The degree of a product of nonzero polynomials is the sum of the degrees of the factors. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Cube of a Binomial] + #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cubed equals x cubed plus 3 x squared y plus 3 x y squared plus y cubed")[$( x + y )^(3) = x^(3) + 3 x^(2) y + 3 x y^(2) + y^(3)$] + #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis cubed equals x cubed minus 3 x squared y plus 3 x y squared minus y cubed")[$( x − y )^(3) = x^(3) − 3 x^(2) y + 3 x y^(2) − y^(3)$] ] + If we know the vertex of a parabola, we need only one other point to find its equation. + #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Factoring the Sum or Difference of Two Cubes] + #math.equation(block: false, alt: "x cubed plus y cubed equals open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis")[$x^(3) + y^(3) = ( x + y ) ( x^(2) − x y + y^(2) )$] + #math.equation(block: false, alt: "x cubed minus y cubed equals open parenthesis x minus y close parenthesis open parenthesis x squared plus x y plus y squared close parenthesis")[$x^(3) − y^(3) = ( x − y ) ( x^(2) + x y + y^(2) )$] ] ==== STUDY QUESTIONS + If you add two polynomials of degree #math.equation(block: false, alt: "3")[$3$], what can you say about the degree of the sum? + If you multiply a polynomial of degree #math.equation(block: false, alt: "3")[$3$] and a polynomial of degree #math.equation(block: false, alt: "4")[$4$], what can you say about the degree of the product? + If you multiply together #math.equation(block: true, alt: "open parenthesis 2 x minus 1 close parenthesis open parenthesis 2 x minus 2 close parenthesis open parenthesis 2 x minus 3 close parenthesis ⋯ open parenthesis 2 x minus 8 close parenthesis")[$( 2 x − 1 ) ( 2 x − 2 ) ( 2 x − 3 ) ⋯ ( 2 x − 8 )$] what will be the degree of the product? What will be the lead coefficient? + What are the two middle terms in the expansion of #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cubed")[$( x + y )^(3)$]? + Is it possible to factor the sum of two cubes? What about the sum of two squares? ==== SKILLS Practice each skill in the Homework problems listed. + Multiply polynomials: \#1–8 + Find specific terms of polynomial products: \#9–22 + Use the formula for the cube of a binomial: \#23–34 + Factor the sum or difference of two cubes: \#35–46 + Write and analyze polynomial models: \#47–64 === Homework 7.1 For Problems 1–8, multiply. #math.equation(block: true, alt: "open parenthesis 3 x minus 2 close parenthesis open parenthesis 4 x squared plus x minus 2 close parenthesis")[$( 3 x − 2 ) ( 4 x^(2) + x − 2 )$] #math.equation(block: true, alt: "12 x cubed minus 5 x squared minus 8 x plus 4")[$12 x^(3) − 5 x^(2) − 8 x + 4$] #math.equation(block: true, alt: "open parenthesis 2 x plus 3 close parenthesis open parenthesis 3 x squared minus 4 x plus 2 close parenthesis")[$( 2 x + 3 ) ( 3 x^(2) − 4 x + 2 )$] #math.equation(block: true, alt: "open parenthesis x minus 2 close parenthesis open parenthesis x minus 1 close parenthesis open parenthesis x minus 3 close parenthesis")[$( x − 2 ) ( x − 1 ) ( x − 3 )$] #math.equation(block: true, alt: "x cubed minus 6 x squared plus 11 x minus 6")[$x^(3) − 6 x^(2) + 11 x − 6$] #math.equation(block: true, alt: "open parenthesis z minus 5 close parenthesis open parenthesis z plus 6 close parenthesis open parenthesis z minus 1 close parenthesis")[$( z − 5 ) ( z + 6 ) ( z − 1 )$] #math.equation(block: true, alt: "open parenthesis 2 a squared minus 3 a plus 1 close parenthesis open parenthesis 3 a squared plus 2 a minus 1 close parenthesis")[$( 2 a^(2) − 3 a + 1 ) ( 3 a^(2) + 2 a − 1 )$] #math.equation(block: true, alt: "6 a to the power 4 minus 5 a cubed minus 5 a squared plus 5 a minus 1")[$6 a^(4) − 5 a^(3) − 5 a^(2) + 5 a − 1$] #math.equation(block: true, alt: "open parenthesis b squared minus 3 b plus 5 close parenthesis open parenthesis 2 b squared minus b plus 1 close parenthesis")[$( b^(2) − 3 b + 5 ) ( 2 b^(2) − b + 1 )$] #math.equation(block: true, alt: "open parenthesis y minus 2 close parenthesis open parenthesis y plus 2 close parenthesis open parenthesis y plus 4 close parenthesis open parenthesis y plus 1 close parenthesis")[$( y − 2 ) ( y + 2 ) ( y + 4 ) ( y + 1 )$] #math.equation(block: true, alt: "y to the power 4 plus 5 y cubed minus 20 y minus 16")[$y^(4) + 5 y^(3) − 20 y − 16$] #math.equation(block: true, alt: "open parenthesis z plus 3 close parenthesis open parenthesis z plus 2 close parenthesis open parenthesis z minus 1 close parenthesis open parenthesis z plus 1 close parenthesis")[$( z + 3 ) ( z + 2 ) ( z − 1 ) ( z + 1 )$] For Problems 9–12, find the first three terms of the product in ascending powers. (Do not compute the entire product!) #math.equation(block: true, alt: "open parenthesis 2 minus x plus 3 x squared close parenthesis open parenthesis 3 plus 2 x minus x squared plus 2 x to the power 4 close parenthesis")[$( 2 − x + 3 x^(2) ) ( 3 + 2 x − x^(2) + 2 x^(4) )$] #math.equation(block: true, alt: "6 plus x plus 5 x squared")[$6 + x + 5 x^(2)$] #math.equation(block: true, alt: "open parenthesis 1 plus x minus 2 x squared close parenthesis open parenthesis minus 3 plus 2 x minus 4 x cubed close parenthesis")[$( 1 + x − 2 x^(2) ) ( − 3 + 2 x − 4 x^(3) )$] #math.equation(block: true, alt: "open parenthesis 1 minus 2 x squared minus x to the power 4 close parenthesis open parenthesis 4 plus x squared minus 2 x to the power 4 close parenthesis")[$( 1 − 2 x^(2) − x^(4) ) ( 4 + x^(2) − 2 x^(4) )$] #math.equation(block: true, alt: "4 minus 7 x squared minus 8 x to the power 4")[$4 − 7 x^(2) − 8 x^(4)$] #math.equation(block: true, alt: "open parenthesis 3 plus 2 x close parenthesis open parenthesis 5 minus 2 x squared minus 3 x cubed minus x to the power 5 plus 2 x to the power 6 close parenthesis")[$( 3 + 2 x ) ( 5 − 2 x^(2) − 3 x^(3) − x^(5) + 2 x^(6) )$] For Problems 13–16, find the indicated term in each product. (Do not compute the entire product!) #math.equation(block: true, alt: "open parenthesis 4 plus 2 x minus x squared close parenthesis open parenthesis 2 minus 3 x plus 2 x squared close parenthesis")[$( 4 + 2 x − x^(2) ) ( 2 − 3 x + 2 x^(2) )$]; #math.equation(block: true, alt: "x squared")[$x^(2)$] #math.equation(block: true, alt: "0 x squared")[$0 x^(2)$] #math.equation(block: true, alt: "open parenthesis 1 minus 2 x plus 3 x squared close parenthesis open parenthesis 6 minus x minus x cubed close parenthesis")[$( 1 − 2 x + 3 x^(2) ) ( 6 − x − x^(3) )$]; #math.equation(block: true, alt: "x cubed")[$x^(3)$] #math.equation(block: true, alt: "open parenthesis 3 x plus x cubed minus 7 x to the power 5 close parenthesis open parenthesis 1 plus 4 x minus 3 x squared close parenthesis")[$( 3 x + x^(3) − 7 x^(5) ) ( 1 + 4 x − 3 x^(2) )$]; #math.equation(block: true, alt: "x cubed")[$x^(3)$] #math.equation(block: true, alt: "minus 8 x cubed")[$− 8 x^(3)$] #math.equation(block: true, alt: "open parenthesis 2 plus 3 x squared plus 2 x to the power 4 close parenthesis open parenthesis 2 minus x minus x squared minus x to the power 4 close parenthesis")[$( 2 + 3 x^(2) + 2 x^(4) ) ( 2 − x − x^(2) − x^(4) )$]; #math.equation(block: true, alt: "x to the power 4")[$x^(4)$] For Problems 17–18, without performing the multiplication, give the degree of each product. + #math.equation(block: false, alt: "open parenthesis x squared minus 4 close parenthesis open parenthesis 3 x squared minus 6 x plus 2 close parenthesis")[$( x^(2) − 4 ) ( 3 x^(2) − 6 x + 2 )$] + #math.equation(block: false, alt: "open parenthesis x minus 3 close parenthesis open parenthesis 2 x minus 5 close parenthesis open parenthesis x cubed minus x plus 2 close parenthesis")[$( x − 3 ) ( 2 x − 5 ) ( x^(3) − x + 2 )$] + #math.equation(block: false, alt: "open parenthesis 3 x squared plus 2 x close parenthesis open parenthesis x cubed plus 1 close parenthesis open parenthesis minus 2 x squared plus 8 close parenthesis")[$( 3 x^(2) + 2 x ) ( x^(3) + 1 ) ( − 2 x^(2) + 8 )$] + #math.equation(block: false, alt: "4")[$4$] + #math.equation(block: false, alt: "5")[$5$] + #math.equation(block: false, alt: "7")[$7$] + #math.equation(block: false, alt: "open parenthesis 6 x squared minus 1 close parenthesis open parenthesis 4 x squared minus 9 close parenthesis")[$( 6 x^(2) − 1 ) ( 4 x^(2) − 9 )$] + #math.equation(block: false, alt: "open parenthesis 3 x plus 4 close parenthesis open parenthesis 3 x plus 1 close parenthesis open parenthesis 2 x cubed plus x squared minus 7 close parenthesis")[$( 3 x + 4 ) ( 3 x + 1 ) ( 2 x^(3) + x^(2) − 7 )$] + #math.equation(block: false, alt: "open parenthesis x squared minus 3 close parenthesis open parenthesis 2 x cubed minus 5 x squared plus 2 close parenthesis open parenthesis minus x cubed minus 5 x close parenthesis")[$( x^(2) − 3 ) ( 2 x^(3) − 5 x^(2) + 2 ) ( − x^(3) − 5 x )$] For Problems 19–22, verify the following products discussed in the text. #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis cubed equals x cubed plus 3 x squared y plus 3 x y squared plus y cubed")[$( x + y )^(3) = x^(3) + 3 x^(2) y + 3 x y^(2) + y^(3)$] #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis cubed, equals open parenthesis x plus y close parenthesis open parenthesis x plus y close parenthesis squared; equals open parenthesis x plus y close parenthesis open parenthesis x squared plus 2 x y plus y squared close parenthesis; equals x cubed plus 2 x squared y plus x y squared plus x squared y plus 2 x y squared plus y cubed; equals x cubed plus 3 x squared y plus 3 x y squared plus y cubed")[$( x + y )^(3) & = ( x + y ) ( x + y )^(2) \ & = ( x + y ) ( x^(2) + 2 x y + y^(2) ) \ & = x^(3) + 2 x^(2) y + x y^(2) + x^(2) y + 2 x y^(2) + y^(3) \ & = x^(3) + 3 x^(2) y + 3 x y^(2) + y^(3)$] #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis cubed equals x cubed minus 3 x squared y plus 3 x y squared minus y cubed")[$( x − y )^(3) = x^(3) − 3 x^(2) y + 3 x y^(2) − y^(3)$] #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis equals x cubed plus y cubed")[$( x + y ) ( x^(2) − x y + y^(2) ) = x^(3) + y^(3)$] #math.equation(block: true, alt: "open parenthesis x plus y close parenthesis open parenthesis x squared minus x y plus y squared close parenthesis, equals x cubed minus x squared y plus x y squared plus x squared y minus x y squared plus y cubed; equals x cubed plus y cubed")[$( x + y ) ( x^(2) − x y + y^(2) ) & = x^(3) − x^(2) y + x y^(2) + x^(2) y − x y^(2) + y^(3) \ & = x^(3) + y^(3)$] #math.equation(block: true, alt: "open parenthesis x minus y close parenthesis open parenthesis x squared plus x y plus y squared close parenthesis equals x cubed minus y cubed")[$( x − y ) ( x^(2) + x y + y^(2) ) = x^(3) − y^(3)$] + As if you were addressing a classmate, explain how to remember the formula for expanding #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cubed")[$( x + y )^(3)$]. In particular, mention the exponents on each term and the numerical coefficients. + Explain how to remember the formula for expanding #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis cubed")[$( x − y )^(3)$], assuming your listener already knows the formula for #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cubed")[$( x + y )^(3)$]. + The formula begins with #math.equation(block: false, alt: "x cubed")[$x^(3)$] and ends with #math.equation(block: false, alt: "y cubed")[$y^(3)$]. As you proceed from term to term, the exponents on #math.equation(block: false, alt: "x")[$x$] decrease while the exponents on #math.equation(block: false, alt: "y")[$y$] increase, and on each term the sum of the exponents is #math.equation(block: false, alt: "3")[$3$]. The coefficients of the two middle terms are both #math.equation(block: false, alt: "3")[$3$]. + The formula is the same as for #math.equation(block: false, alt: "open parenthesis x minus y close parenthesis cubed")[$( x − y )^(3)$], except that the terms alternate in sign. + As if you were addressing a classmate, explain how to remember the formula for factoring a sum of two cubes. Pay particular attention to the placement of the variables and the signs of the terms. + Explain how to remember the formula for factoring a difference of two cubes, assuming your listener already knows how to factor a sum of two cubes. For Problems 25–28, use the formulas for the cube of a binomial to expand the products. #math.equation(block: true, alt: "open parenthesis 1 plus 2 z close parenthesis cubed")[$( 1 + 2 z )^(3)$] #math.equation(block: true, alt: "1 plus 6 z plus 12 z squared plus 8 z cubed")[$1 + 6 z + 12 z^(2) + 8 z^(3)$] #math.equation(block: true, alt: "open parenthesis 1 minus x squared close parenthesis cubed")[$( 1 − x^(2) )^(3)$] #math.equation(block: true, alt: "open parenthesis 1 minus 5 the square root of t close parenthesis cubed")[$( 1 − 5 sqrt(t) )^(3)$] #math.equation(block: true, alt: "1 minus 15 the square root of t plus 75 t minus 125 t the square root of t")[$1 − 15 sqrt(t) + 75 t − 125 t sqrt(t)$] #math.equation(block: true, alt: "open parenthesis 1 minus the fraction 3 over a close parenthesis cubed")[$attach(( 1 − display(frac(3, a)) ), t: 3)$] For Problems 29–34, write each product as a polynomial and simplify. #math.equation(block: true, alt: "open parenthesis x minus 1 close parenthesis open parenthesis x squared plus x plus 1 close parenthesis")[$( x − 1 ) ( x^(2) + x + 1 )$] #math.equation(block: true, alt: "x cubed minus 1")[$x^(3) − 1$] #math.equation(block: true, alt: "open parenthesis x plus 2 close parenthesis open parenthesis x squared minus 2 x plus 4 close parenthesis")[$( x + 2 ) ( x^(2) − 2 x + 4 )$] #math.equation(block: true, alt: "open parenthesis 2 x plus 1 close parenthesis open parenthesis 4 x squared minus 2 x plus 1 close parenthesis")[$( 2 x + 1 ) ( 4 x^(2) − 2 x + 1 )$] #math.equation(block: true, alt: "8 x cubed plus 1")[$8 x^(3) + 1$] #math.equation(block: true, alt: "open parenthesis 3 x minus 1 close parenthesis open parenthesis 9 x squared plus 3 x plus 1 close parenthesis")[$( 3 x − 1 ) ( 9 x^(2) + 3 x + 1 )$] #math.equation(block: true, alt: "open parenthesis 3 a minus 2 b close parenthesis open parenthesis 9 a squared plus 6 a b plus 4 b squared close parenthesis")[$( 3 a − 2 b ) ( 9 a^(2) + 6 a b + 4 b^(2) )$] #math.equation(block: true, alt: "27 a cubed minus 8 b cubed")[$27 a^(3) − 8 b^(3)$] #math.equation(block: true, alt: "open parenthesis 2 a plus 3 b close parenthesis open parenthesis 4 a squared minus 6 a b plus 9 b squared close parenthesis")[$( 2 a + 3 b ) ( 4 a^(2) − 6 a b + 9 b^(2) )$] For Problems 35–46, factor completely. #math.equation(block: true, alt: "x cubed plus 27")[$x^(3) + 27$] #math.equation(block: true, alt: "open parenthesis x plus 3 close parenthesis open parenthesis x squared minus 3 x plus 9 close parenthesis")[$( x + 3 ) ( x^(2) − 3 x + 9 )$] #math.equation(block: true, alt: "y cubed minus 1")[$y^(3) − 1$] #math.equation(block: true, alt: "a cubed minus 8 b cubed")[$a^(3) − 8 b^(3)$] #math.equation(block: true, alt: "open parenthesis a minus 2 b close parenthesis open parenthesis a squared plus 2 a b plus 4 b squared close parenthesis")[$( a − 2 b ) ( a^(2) + 2 a b + 4 b^(2) )$] #math.equation(block: true, alt: "27 a cubed plus b cubed")[$27 a^(3) + b^(3)$] #math.equation(block: true, alt: "x cubed y to the power 6 minus 1")[$x^(3) y^(6) − 1$] #math.equation(block: true, alt: "open parenthesis x y squared minus 1 close parenthesis open parenthesis x squared y to the power 4 plus x y squared plus 1 close parenthesis")[$( x y^(2) − 1 ) ( x^(2) y^(4) + x y^(2) + 1 )$] #math.equation(block: true, alt: "8 plus x to the power 12 y cubed")[$8 + x^(12) y^(3)$] #math.equation(block: true, alt: "27 a cubed plus 64 b cubed")[$27 a^(3) + 64 b^(3)$] #math.equation(block: true, alt: "open parenthesis 3 a plus 4 b close parenthesis open parenthesis 9 a squared minus 12 a b plus 16 b squared close parenthesis")[$( 3 a + 4 b ) ( 9 a^(2) − 12 a b + 16 b^(2) )$] #math.equation(block: true, alt: "8 a cubed minus 125 b cubed")[$8 a^(3) − 125 b^(3)$] #math.equation(block: true, alt: "125 a cubed b cubed minus 1")[$125 a^(3) b^(3) − 1$] #math.equation(block: true, alt: "open parenthesis 5 a b minus 1 close parenthesis open parenthesis 25 a squared b squared plus 5 a b plus 1 close parenthesis")[$( 5 a b − 1 ) ( 25 a^(2) b^(2) + 5 a b + 1 )$] #math.equation(block: true, alt: "64 a cubed b cubed plus 1")[$64 a^(3) b^(3) + 1$] #math.equation(block: true, alt: "64 t to the power 9 plus w to the power 6")[$64 t^(9) + w^(6)$] #math.equation(block: true, alt: "open parenthesis 4 t cubed plus w squared close parenthesis open parenthesis 16 t to the power 6 minus 4 t cubed w squared plus w to the power 4 close parenthesis")[$( 4 t^(3) + w^(2) ) ( 16 t^(6) − 4 t^(3) w^(2) + w^(4) )$] #math.equation(block: true, alt: "w to the power 15 minus 125 t to the power 9")[$w^(15) − 125 t^(9)$] + Write a polynomial function, #math.equation(block: false, alt: "A open parenthesis x close parenthesis")[$A ( x )$], that gives the area of the front face of the speaker frame (the region in color) in the figure. + If #math.equation(block: false, alt: "x equals 8")[$x = 8$] inches, find the area of the front face of the frame. #figure(figph[speaker frame], alt: "speaker frame", caption: none) + #math.equation(block: false, alt: "open parenthesis 6 minus the fraction 5 over 4 π close parenthesis x squared")[$( 6 − display(frac(5, 4)) π ) x^(2)$] + #math.equation(block: false, alt: "approximately equals 132.67")[$≈ 132.67$] square inches + A Norman window is shaped like a rectangle whose length is twice its width, with a semicircle at the top (see the figure). Write a polynomial, #math.equation(block: false, alt: "A open parenthesis x close parenthesis")[$A ( x )$], that gives its area. + If #math.equation(block: false, alt: "x equals 3")[$x = 3$] feet, find the area of the front face of the frame. #figure(figph[Norman window], alt: "Norman window", caption: none) + A grain silo is built in the shape of a cylinder with a hemisphere on top (see the figure). Write an expression for the volume of the silo in terms of the radius and height of the cylindrical portion of the silo. + If the total height of the silo is five times its radius, write a polynomial function #math.equation(block: false, alt: "V open parenthesis r close parenthesis")[$V ( r )$] in one variable for its volume. #figure(figph[grain silo], alt: "grain silo", caption: none) + #math.equation(block: false, alt: "the fraction 2 over 3 π r cubed plus π r squared h")[$display(frac(2, 3)) π r^(3) + π r^(2) h$] + #math.equation(block: false, alt: "V open parenthesis r close parenthesis equals the fraction 14 over 3 π r cubed")[$V ( r ) = display(frac(14, 3)) π r^(3)$] + A cold medication capsule is shaped like a cylinder with a hemispherical cap on each end (see the figure). Write an expression for the volume of the capsule in terms of the radius and length of the cylindrical portion. #figure(figph[capsule], alt: "capsule", caption: none) + If the radius of the capsule is one-fourth of its overall length, write a polynomial function #math.equation(block: false, alt: "V open parenthesis r close parenthesis")[$V ( r )$] in one variable for its volume. Jack invests \$#math.equation(block: false, alt: "500")[$500$] in an account bearing interest rate #math.equation(block: false, alt: "r")[$r$], compounded annually. This means that each year his account balance is increased by a factor of #math.equation(block: false, alt: "1 plus r")[$1 + r$]. + Write expressions for the amount of money in Jack's account after #math.equation(block: false, alt: "2")[$2$] years, after #math.equation(block: false, alt: "3")[$3$] years, and after #math.equation(block: false, alt: "4")[$4$] years. + Expand as polynomials the expressions you found in part (a). + How much money will be in Jack's account at the end of #math.equation(block: false, alt: "2")[$2$] years, #math.equation(block: false, alt: "3")[$3$] years, and #math.equation(block: false, alt: "4")[$4$] years if the interest rate is #math.equation(block: false, alt: "8 %")[$8 upright(%)$]? + #math.equation(block: false, alt: "500 open parenthesis 1 plus r close parenthesis squared")[$500 ( 1 + r )^(2)$]; #math.equation(block: false, alt: "500 open parenthesis 1 plus r close parenthesis cubed")[$500 ( 1 + r )^(3)$]; #math.equation(block: false, alt: "500 open parenthesis 1 plus r close parenthesis to the power 4")[$500 ( 1 + r )^(4)$] + #math.equation(block: false, alt: "500 r squared plus 1000 r plus 500")[$500 r^(2) + 1000 r + 500$]; #math.equation(block: false, alt: "500 r cubed plus 1500 r squared plus 1500 r plus 500")[$500 r^(3) + 1500 r^(2) + 1500 r + 500$]; #math.equation(block: false, alt: "500 r to the power 4 plus 2000 r cubed plus 3000 r squared plus 2000 r plus 500")[$500 r^(4) + 2000 r^(3) + 3000 r^(2) + 2000 r + 500$] + \$583.20, \$629.86, \$680.24 A small company borrows \$800 for start-up costs and agrees to repay the loan at interest rate #math.equation(block: false, alt: "r")[$r$], compounded annually. This means that each year the debt is increased by a factor of #math.equation(block: false, alt: "1 plus r")[$1 + r$]. + Write expressions for the amount of money the company will owe if it repays the loan after #math.equation(block: false, alt: "2")[$2$] years, after #math.equation(block: false, alt: "3")[$3$] years, or after #math.equation(block: false, alt: "4")[$4$] years. + Expand as polynomials the expressions you found in part (a). + How much money will the company owe after #math.equation(block: false, alt: "2")[$2$] years, after #math.equation(block: false, alt: "3")[$3$] years, or after #math.equation(block: false, alt: "4")[$4$] years at an interest rate of #math.equation(block: false, alt: "12 %")[$12 upright(%)$]? A paper company plans to make boxes without tops from sheets of cardboard #math.equation(block: false, alt: "12")[$12$] inches wide and #math.equation(block: false, alt: "16")[$16$] inches long. The company will cut out four squares of side #math.equation(block: false, alt: "x")[$x$] inches from the corners of the sheet and fold up the edges as shown in the figure. #figure(figph[box], alt: "box", caption: none) + Write expressions in terms of #math.equation(block: false, alt: "x")[$x$] for the length, width, and height of the resulting box. + Write a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of the box as a function of #math.equation(block: false, alt: "x")[$x$]. + What is the domain of the function #math.equation(block: false, alt: "V")[$V$]? (What are the largest and smallest reasonable values for #math.equation(block: false, alt: "x")[$x$]?) + Make a table of values for #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$] on its domain. + Graph your function #math.equation(block: false, alt: "V")[$V$] in a suitable window. + Use your graph to find the value of #math.equation(block: false, alt: "x")[$x$] that will yield a box with maximum possible volume. What is the maximum possible volume? + Length: #math.equation(block: false, alt: "16 minus 2 x")[$16 − 2 x$]; Width: #math.equation(block: false, alt: "12 minus 2 x")[$12 − 2 x$]; Height: #math.equation(block: false, alt: "x")[$x$] + #math.equation(block: false, alt: "V equals x open parenthesis 16 minus 2 x close parenthesis open parenthesis 12 minus 2 x close parenthesis")[$V = x ( 16 − 2 x ) ( 12 − 2 x )$] + Real numbers between #math.equation(block: false, alt: "0")[$0$] and #math.equation(block: false, alt: "6")[$6$] + #figure(table( columns: 6, align: left, inset: 6pt, table.header([#math.equation(block: false, alt: "x")[$x$]], [#math.equation(block: false, alt: "1")[$1$]], [#math.equation(block: false, alt: "2")[$2$]], [#math.equation(block: false, alt: "3")[$3$]], [#math.equation(block: false, alt: "4")[$4$]], [#math.equation(block: false, alt: "5")[$5$]]), [#math.equation(block: false, alt: "V")[$V$]], [#math.equation(block: false, alt: "140")[$140$]], [#math.equation(block: false, alt: "192")[$192$]], [#math.equation(block: false, alt: "180")[$180$]], [#math.equation(block: false, alt: "128")[$128$]], [#math.equation(block: false, alt: "60")[$60$]], )) + #figure(figph[cubic], alt: "cubic", caption: none) + #math.equation(block: false, alt: "2.26")[$2.26$] in, #math.equation(block: false, alt: "194.07")[$194.07$] cu in The paper company also plans to make boxes with tops from #math.equation(block: false, alt: "12")[$12$]-inch by #math.equation(block: false, alt: "16")[$16$]-inch sheets of cardboard by cutting out the shaded areas shown in the figure and folding along the dotted lines. #figure(figph[box], alt: "box", caption: none) + Write expressions in terms of #math.equation(block: false, alt: "x")[$x$] for the length, width, and height of the resulting box. + Write a formula for the volume, #math.equation(block: false, alt: "V")[$V$], of the box as a function of #math.equation(block: false, alt: "x")[$x$]. + What is the domain of the function #math.equation(block: false, alt: "V")[$V$]? (What are the largest and smallest reasonable values for #math.equation(block: false, alt: "x")[$x$]?) + Make a table of values for #math.equation(block: false, alt: "V open parenthesis x close parenthesis")[$V ( x )$] on its domain. + Graph your function #math.equation(block: false, alt: "V")[$V$] in a suitable window. + Use your graph to find the value of #math.equation(block: false, alt: "x")[$x$] that will yield a box with maximum possible volume. What is the maximum possible volume? Use your graphing utility to help you answer the questions in Problems 55–62. Then verify your answers algebraically. A doctor who is treating a heart patient wants to prescribe medication to lower the patient's blood pressure. The body's reaction to this medication is a function of the dose administered. If the patient takes #math.equation(block: false, alt: "x")[$x$] milliliters of the medication, his blood pressure should decrease by #math.equation(block: false, alt: "R equals f open parenthesis x close parenthesis")[$R = f ( x )$] points, where #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 3 x squared minus the fraction 1 over 3 x cubed")[$f ( x ) = 3 x^(2) − display(frac(1, 3)) x^(3)$] + For what values of #math.equation(block: false, alt: "x")[$x$] is #math.equation(block: false, alt: "R equals 0")[$R = 0$]? + Find a suitable domain for the function and explain why you chose this domain. + Graph the function #math.equation(block: false, alt: "f")[$f$] on its domain. + How much should the patient's blood pressure drop if he takes #math.equation(block: false, alt: "2")[$2$] milliliters of medication? + What is the maximum drop in blood pressure that can be achieved with this medication? + There may be risks associated with a large change in blood pressure. How many milliliters of the medication should be administered to produce half the maximum possible drop in blood pressure? + #math.equation(block: false, alt: "0 , 9")[$0 , 9$] + #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 9")[$0 ≤ x ≤ 9$]; #math.equation(block: false, alt: "R greater than or equal to 0")[$R ≥ 0$] for these values + #figure(figph[cubic], alt: "cubic", caption: none) + #math.equation(block: false, alt: "the fraction 28 over 3")[$display(frac(28, 3))$] points + #math.equation(block: false, alt: "36")[$36$] points + #math.equation(block: false, alt: "3")[$3$] ml or #math.equation(block: false, alt: "8.2")[$8.2$] ml A soup bowl has the shape of a hemisphere of radius #math.equation(block: false, alt: "6")[$6$] centimeters. The volume of the soup in the bowl, #math.equation(block: false, alt: "V equals f open parenthesis x close parenthesis")[$V = f ( x )$], is a function of the depth, #math.equation(block: false, alt: "x")[$x$], of the soup. #figure(figph[soup bowl], alt: "soup bowl", caption: none) + What is the domain of #math.equation(block: false, alt: "f")[$f$]? Why? + The function #math.equation(block: false, alt: "f")[$f$] is given by #math.equation(block: true, alt: "f open parenthesis x close parenthesis equals 6 π x squared minus the fraction π over 3 x cubed")[$f ( x ) = 6 π x^(2) − frac(π, 3) x^(3)$] Graph the function on its domain. + What is the volume of the soup if it is #math.equation(block: false, alt: "3")[$3$] centimeters deep? + What is the maximum volume of soup that the bowl can hold? + Find the depth of the soup (to within #math.equation(block: false, alt: "2")[$2$] decimal places of accuracy) when the bowl is filled to half its capacity. The population, #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$], of Cyberville has been growing according to the formula #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals t cubed minus 63 t squared plus 1403 t plus 900")[$P ( t ) = t^(3) − 63 t^(2) + 1403 t + 900$] where #math.equation(block: false, alt: "t")[$t$] is the number of years since 1970. + Graph #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 47; Ymin, equals 0, Ymax equals 20 , 000")[$"Xmin" & = 0 & & "Xmax" = 47 \ "Ymin" & = 0 & & "Ymax" = 20 , 000$] + What was the population in 1970? In 1985? In 2004? + By how much did the population grow from 1970 to 1971? From 1985 to 1986? From 2004 to 2005? + Approximately when was the population growing at the slowest rate, that is, when is the graph the least steep? + #figure(figph[cubic], alt: "cubic", caption: none) + #math.equation(block: false, alt: "900")[$900$]; #math.equation(block: false, alt: "11 , 145")[$11 , 145$]; #math.equation(block: false, alt: "15 , 078")[$15 , 078$] + #math.equation(block: false, alt: "1341")[$1341$]; #math.equation(block: false, alt: "171")[$171$]; #math.equation(block: false, alt: "627")[$627$] + Between 1990 and 1991 The annual profit, #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$], of the Enviro Company, in thousands of dollars, is given by #math.equation(block: true, alt: "P open parenthesis t close parenthesis equals 2 t cubed minus 152 t squared plus 3400 t plus 30")[$P ( t ) = 2 t^(3) − 152 t^(2) + 3400 t + 30$] where #math.equation(block: false, alt: "t")[$t$] is the number of years since 1960, the first year that the company showed a profit. + Graph #math.equation(block: false, alt: "P open parenthesis t close parenthesis")[$P ( t )$] in the window #math.equation(block: true, alt: "Xmin, equals 0, Xmax equals 94; Ymin, equals 0, Ymax equals 50 , 000")[$"Xmin" & = 0 & & "Xmax" = 94 \ "Ymin" & = 0 & & "Ymax" = 50 , 000$] + What was the profit in 1960? In 1980? In 2000? + How did the profit change from 1960 to 1961? From 1980 to 1981? From 2000 to 2001? + During which years did the profit decrease from one year to the next? The total annual cost of educating postgraduate research students at an Australian university, in thousands of dollars, is given by the function #math.equation(block: true, alt: "C open parenthesis x close parenthesis equals 0.0173 x cubed minus 0.647 x squared plus 9.587 x plus 195.366")[$C ( x ) = 0.0173 x^(3) − 0.647 x^(2) + 9.587 x + 195.366$] where #math.equation(block: false, alt: "x")[$x$] is the number of students, in hundreds. (Source: Creedy, Johnson, and Valenzuela, 2002) + Graph the function in a suitable window for up to #math.equation(block: false, alt: "3500")[$3500$] students. + Describe the concavity of the graph. For what value of #math.equation(block: false, alt: "x")[$x$] is the cost growing at the slowest rate? + Approximately how many students can be educated for \$#math.equation(block: false, alt: "350 , 000")[$350 , 000$]? + #figure(figph[cubic], alt: "cubic", caption: none) + The graph is concave down until about #math.equation(block: false, alt: "x equals 12.5")[$x = 12.5$] and is concave up afterwards. The cost is growing at the slowest rate at the inflection point at about #math.equation(block: false, alt: "x equals 12.5")[$x = 12.5$], or #math.equation(block: false, alt: "1250")[$1250$] students. + About #math.equation(block: false, alt: "2890")[$2890$] It has been proposed that certain cubic functions model the response of wheat and barley to nitrogen fertilizer. These functions exhibit a "plateau" that fits observations better than the standard quadratic model. (See Problem of Curve Fitting.) In trials in Denmark, the yield per acre was a function of the amount of nitrogen applied. A typical response function is #math.equation(block: true, alt: "Y open parenthesis x close parenthesis equals 54.45 plus 0.305 x minus 0.001655 x squared plus 2.935 times 10 to the power minus 6 x cubed")[$Y ( x ) = 54.45 + 0.305 x − 0.001655 x^(2) + 2.935 × 10^(− 6) x^(3)$] where #math.equation(block: false, alt: "x")[$x$] is the amount of fertilizer, in kilograms per acre.(Source: Beattie, Mortensen, and Knudsen, 2005) + Graph the function on the domain #math.equation(block: false, alt: "open bracket 0 , 400 close bracket")[$[ 0 , 400 ]$]. + Describe the concavity of the graph. In reality, the yield does not increase after reaching its plateau. Give a suitable domain for the model in this application. + Estimate the maximum yield attainable and the optimum application of fertilizer. During an earthquake, Nordhoff Street split in two, and one section shifted up several centimeters. Engineers created a ramp from the lower section to the upper section. In the coordinate system shown in the figure below, the ramp is part of the graph of #math.equation(block: true, alt: "y equals f open parenthesis x close parenthesis equals minus 0.00004 x cubed minus 0.006 x squared plus 20")[$y = f ( x ) = − 0.00004 x^(3) − 0.006 x^(2) + 20$] #figure(figph[ramp], alt: "ramp", caption: none) + By how much did the upper section of the street shift during the earthquake? + What is the horizontal distance from the bottom of the ramp to the raised part of the street? + #math.equation(block: false, alt: "20")[$20$] cm + #math.equation(block: false, alt: "100")[$100$] cm The off-ramp from a highway connects to a parallel one-way road. The accompanying figure shows the highway, the off-ramp, and the road. The road lies on the #math.equation(block: false, alt: "x")[$x$]-axis, and the off-ramp begins at a point on the #math.equation(block: false, alt: "y")[$y$]-axis. The offramp is part of the graph of the polynomial #math.equation(block: true, alt: "y equals f open parenthesis x close parenthesis equals 0.00006 x cubed minus 0.009 x squared plus 30")[$y = f ( x ) = 0.00006 x^(3) − 0.009 x^(2) + 30$] #figure(figph[off-ramp], alt: "off-ramp", caption: none) + How far east of the exit does the off-ramp meet the one-way road? + How far apart are the highway and the road? The number of minutes of daylight per day in Chicago is approximated by the polynomial #math.equation(block: true, alt: "H open parenthesis t close parenthesis equals 0.000000525 t to the power 4 minus 0.0213 t squared plus 864")[$H ( t ) = 0.000 #h(0.167em) 000 #h(0.167em) 525 t^(4) − 0.0213 t^(2) + 864$] where #math.equation(block: false, alt: "t")[$t$] is the number of days since the summer solstice. The approximation is valid for #math.equation(block: false, alt: "minus 74 less than t less than 74")[$− 74 < t < 74$]. (A negative value of #math.equation(block: false, alt: "t")[$t$] corresponds to a number of days before the summer solstice.) + Use a table of values with increments of #math.equation(block: false, alt: "10")[$10$] days to estimate the range of the function on its domain. + Graph the polynomial on its domain. + How many minutes of daylight are there on the summer solstice? + How much daylight is there two weeks before the solstice? + When are the days more than #math.equation(block: false, alt: "14")[$14$] hours long? + When are the days less than #math.equation(block: false, alt: "13")[$13$] hours long? + #math.equation(block: false, alt: "763.10 less than H open parenthesis t close parenthesis less than 864")[$763.10 < H ( t ) < 864$] + #figure(figph[quartic], alt: "quartic", caption: none) + #math.equation(block: false, alt: "864")[$864$] min + #math.equation(block: false, alt: "859.8")[$859.8$] min + Within #math.equation(block: false, alt: "34")[$34$] days of the summer solstice + More than #math.equation(block: false, alt: "66")[$66$] days from the summer solstice The water level (in feet) at a harbor is approximated by the polynomial #math.equation(block: true, alt: "W open parenthesis t close parenthesis equals 0.00733 t to the power 4 minus 0.332 t squared plus 9.1")[$W ( t ) = 0.00733 t^(4) − 0.332 t^(2) + 9.1$] where #math.equation(block: false, alt: "t")[$t$] is the number of hours since the high tide. The approximation is valid for #math.equation(block: false, alt: "minus 4 less than or equal to t less than or equal to 4")[$− 4 ≤ t ≤ 4$]. (A negative value of #math.equation(block: false, alt: "t")[$t$] corresponds to a number of hours before the high tide.) + Use a table of values to estimate the range of the function on its domain. + Graph the polynomial on its domain. + What is the water level at high tide? + What is the water level #math.equation(block: false, alt: "3")[$3$] hours before high tide? + When is the water level below #math.equation(block: false, alt: "8")[$8$] feet? + When is the water level above #math.equation(block: false, alt: "7")[$7$] feet?