{"id":1345,"date":"2015-11-12T18:35:30","date_gmt":"2015-11-12T18:35:30","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&#038;p=1345"},"modified":"2017-03-31T22:45:05","modified_gmt":"2017-03-31T22:45:05","slug":"solutions-42","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/chapter\/solutions-42\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"<h2>Solutions to Try Its<\/h2>\r\n1.\u00a0y-intercept [latex]\\left(0,0\\right)[\/latex]; x-intercepts [latex]\\left(0,0\\right),\\left(-5,0\\right),\\left(2,0\\right)[\/latex], and [latex]\\left(3,0\\right)[\/latex]\r\n\r\n2.\u00a0The graph has a zero of \u20135 with multiplicity 1, a zero of \u20131 with multiplicity 2, and a zero of 3 with even multiplicity.\r\n\r\n3.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201514\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" data-media-type=\"image\/jpg\"\/>\r\n\r\n4.\u00a0Because <em>f<\/em>\u00a0is a polynomial function and since [latex]f\\left(1\\right)[\/latex] is negative and [latex]f\\left(2\\right)[\/latex] is positive, there is at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].\r\n\r\n5.\u00a0[latex]f\\left(x\\right)=-\\frac{1}{8}{\\left(x - 2\\right)}^{3}{\\left(x+1\\right)}^{2}\\left(x - 4\\right)[\/latex]\r\n\r\n6.\u00a0The minimum occurs at approximately the point [latex]\\left(0,-6.5\\right)[\/latex], and the maximum occurs at approximately the point [latex]\\left(3.5,7\\right)[\/latex].\r\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\r\n1.\u00a0The <em>x-<\/em>intercept is where the graph of the function crosses the <em>x<\/em>-axis, and the zero of the function is the input value for which [latex]f\\left(x\\right)=0[\/latex].\r\n\r\n3.\u00a0If we evaluate the function at <em>a<\/em>\u00a0and at <em>b<\/em>\u00a0and the sign of the function value changes, then we know a zero exists between <em>a<\/em>\u00a0and <em>b<\/em>.\r\n\r\n5.\u00a0There will be a factor raised to an even power.\r\n\r\n7.\u00a0[latex]\\left(-2,0\\right),\\left(3,0\\right),\\left(-5,0\\right)[\/latex]\r\n\r\n9.\u00a0[latex]\\left(3,0\\right),\\left(-1,0\\right),\\left(0,0\\right)[\/latex]\r\n\r\n11.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(-5,0\\right),\\text{ }\\left(2,0\\right)[\/latex]\r\n\r\n13.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(-5,0\\right),\\text{ }\\left(4,0\\right)[\/latex]\r\n\r\n15.\u00a0[latex]\\left(2,0\\right),\\text{ }\\left(-2,0\\right),\\text{ }\\left(-1,0\\right)[\/latex]\r\n\r\n17.\u00a0[latex]\\left(-2,0\\right),\\left(2,0\\right),\\left(\\frac{1}{2},0\\right)[\/latex]\r\n\r\n19.\u00a0[latex]\\left(1,0\\right),\\text{ }\\left(-1,0\\right)[\/latex]\r\n\r\n21.\u00a0[latex]\\left(0,0\\right),\\left(\\sqrt{3},0\\right),\\left(-\\sqrt{3},0\\right)[\/latex]\r\n\r\n23.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(1,0\\right)\\text{, }\\left(-1,0\\right),\\text{ }\\left(2,0\\right),\\text{ }\\left(-2,0\\right)[\/latex]\r\n\r\n25.\u00a0[latex]f\\left(2\\right)=-10[\/latex]\u00a0and [latex]f\\left(4\\right)=28[\/latex].\u00a0Sign change confirms.\r\n\r\n27.\u00a0[latex]f\\left(1\\right)=3[\/latex]\u00a0and [latex]f\\left(3\\right)=-77[\/latex].\u00a0Sign change confirms.\r\n\r\n29.\u00a0[latex]f\\left(0.01\\right)=1.000001[\/latex]\u00a0and [latex]f\\left(0.1\\right)=-7.999[\/latex].\u00a0Sign change confirms.\r\n\r\n31.\u00a00 with multiplicity 2, [latex]-\\frac{3}{2}[\/latex]\u00a0with multiplicity 5, 4 with multiplicity 2\r\n\r\n33.\u00a00 with multiplicity 2, \u20132 with multiplicity 2\r\n\r\n35.\u00a0[latex]-\\frac{2}{3}\\text{ with multiplicity }5\\text{,}5\\text{ with multiplicity }\\text{2}[\/latex]\r\n\r\n37.\u00a0[latex]\\text{0}\\text{ with multiplicity }4\\text{,}2\\text{ with multiplicity }1\\text{,}-\\text{1}\\text{ with multiplicity }1[\/latex]\r\n\r\n39.\u00a0[latex]\\frac{3}{2}[\/latex]\u00a0with multiplicity 2, 0 with multiplicity 3\r\n\r\n41.\u00a0[latex]\\text{0}\\text{ with multiplicity }6\\text{,}\\frac{2}{3}\\text{ with multiplicity }2[\/latex]\r\n\r\n43.\u00a0<em>x<\/em>-intercepts,\u00a0[latex]\\left(1, 0\\right)[\/latex]\u00a0with multiplicity 2, [latex]\\left(-4, 0\\right)[\/latex] with multiplicity 1, <em>y-<\/em>intercept [latex]\\left(0, 4\\right)[\/latex]. As\u00a0[latex]x\\to -\\infty [\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty [\/latex] , as\u00a0[latex]x\\to \\infty [\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty [\/latex] .\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201515\/CNX_Precalc_Figure_03_04_202.jpg\" alt=\"Graph of g(x)=(x+4)(x-1)^2.\" data-media-type=\"image\/jpg\"\/>\r\n\r\n45.\u00a0<em>x<\/em>-intercepts [latex]\\left(3,0\\right)[\/latex] with multiplicity 3, [latex]\\left(2,0\\right)[\/latex] with multiplicity 2, <em>y<\/em>-intercept [latex]\\left(0,-108\\right)[\/latex] . As\u00a0[latex]x\\to -\\infty [\/latex],\u00a0[latex]f\\left(x\\right)\\to -\\infty [\/latex] , as [latex]x\\to \\infty [\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty [\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201517\/CNX_Precalc_Figure_03_04_204.jpg\" alt=\"Graph of k(x)=(x-3)^3(x-2)^2.\" data-media-type=\"image\/jpg\"\/>\r\n\r\n47.\u00a0x-intercepts [latex]\\left(0, 0\\right),\\left(-2, 0\\right),\\left(4, 0\\right)[\/latex]\u00a0with multiplicity 1, <em>y<\/em>-intercept [latex]\\left(0, 0\\right)[\/latex]. As\u00a0[latex]x\\to -\\infty [\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty [\/latex] , as\u00a0[latex]x\\to \\infty [\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty [\/latex].\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201518\/CNX_Precalc_Figure_03_04_206.jpg\" alt=\"Graph of n(x)=-3x(x+2)(x-4).\" data-media-type=\"image\/jpg\"\/>\r\n\r\n49.\u00a0[latex]f\\left(x\\right)=-\\frac{2}{9}\\left(x - 3\\right)\\left(x+1\\right)\\left(x+3\\right)[\/latex]\r\n\r\n51.\u00a0[latex]f\\left(x\\right)=\\frac{1}{4}{\\left(x+2\\right)}^{2}\\left(x - 3\\right)[\/latex]\r\n\r\n53.\u00a0\u20134, \u20132, 1, 3 with multiplicity 1\r\n\r\n55.\u00a0\u20132, 3 each with multiplicity 2\r\n\r\n57.\u00a0[latex]f\\left(x\\right)=-\\frac{2}{3}\\left(x+2\\right)\\left(x - 1\\right)\\left(x - 3\\right)[\/latex]\r\n\r\n59.\u00a0[latex]f\\left(x\\right)=\\frac{1}{3}{\\left(x - 3\\right)}^{2}{\\left(x - 1\\right)}^{2}\\left(x+3\\right)[\/latex]\r\n\r\n61.\u00a0[latex]f\\left(x\\right)=-15{\\left(x - 1\\right)}^{2}{\\left(x - 3\\right)}^{3}[\/latex]\r\n\r\n63.\u00a0[latex]f\\left(x\\right)=-2\\left(x+3\\right)\\left(x+2\\right)\\left(x - 1\\right)[\/latex]\r\n\r\n65. [latex]f\\left(x\\right)=-\\frac{3}{2}{\\left(2x - 1\\right)}^{2}\\left(x - 6\\right)\\left(x+2\\right)[\/latex]\r\n\r\n67.\u00a0local max [latex]\\left(-\\text{.58, -}.62\\right)[\/latex],\u00a0local min [latex]\\left(\\text{.58, -1}\\text{.38}\\right)[\/latex]\r\n\r\n69.\u00a0global min [latex]\\left(-\\text{.63, -}\\text{.47}\\right)[\/latex]\r\n\r\n71.\u00a0global min [latex]\\text{(}\\text{.75, }\\text{.89)}[\/latex]\r\n\r\n73.\u00a0[latex]f\\left(x\\right)={\\left(x - 500\\right)}^{2}\\left(x+200\\right)[\/latex]\r\n\r\n75.\u00a0[latex]f\\left(x\\right)=4{x}^{3}-36{x}^{2}+80x[\/latex]\r\n\r\n77.\u00a0[latex]f\\left(x\\right)=4{x}^{3}-36{x}^{2}+60x+100[\/latex]\r\n\r\n79.\u00a0[latex]f\\left(x\\right)=\\pi \\left(9{x}^{3}+45{x}^{2}+72x+36\\right)[\/latex]","rendered":"<h2>Solutions to Try Its<\/h2>\n<p>1.\u00a0y-intercept [latex]\\left(0,0\\right)[\/latex]; x-intercepts [latex]\\left(0,0\\right),\\left(-5,0\\right),\\left(2,0\\right)[\/latex], and [latex]\\left(3,0\\right)[\/latex]<\/p>\n<p>2.\u00a0The graph has a zero of \u20135 with multiplicity 1, a zero of \u20131 with multiplicity 2, and a zero of 3 with even multiplicity.<\/p>\n<p>3.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201514\/CNX_Precalc_Figure_03_04_0212.jpg\" alt=\"Graph of f(x)=(1\/4)x(x-1)^4(x+3)^3.\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>4.\u00a0Because <em>f<\/em>\u00a0is a polynomial function and since [latex]f\\left(1\\right)[\/latex] is negative and [latex]f\\left(2\\right)[\/latex] is positive, there is at least one real zero between [latex]x=1[\/latex] and [latex]x=2[\/latex].<\/p>\n<p>5.\u00a0[latex]f\\left(x\\right)=-\\frac{1}{8}{\\left(x - 2\\right)}^{3}{\\left(x+1\\right)}^{2}\\left(x - 4\\right)[\/latex]<\/p>\n<p>6.\u00a0The minimum occurs at approximately the point [latex]\\left(0,-6.5\\right)[\/latex], and the maximum occurs at approximately the point [latex]\\left(3.5,7\\right)[\/latex].<\/p>\n<h2>Solutions to Odd-Numbered Exercises<\/h2>\n<p>1.\u00a0The <em>x-<\/em>intercept is where the graph of the function crosses the <em>x<\/em>-axis, and the zero of the function is the input value for which [latex]f\\left(x\\right)=0[\/latex].<\/p>\n<p>3.\u00a0If we evaluate the function at <em>a<\/em>\u00a0and at <em>b<\/em>\u00a0and the sign of the function value changes, then we know a zero exists between <em>a<\/em>\u00a0and <em>b<\/em>.<\/p>\n<p>5.\u00a0There will be a factor raised to an even power.<\/p>\n<p>7.\u00a0[latex]\\left(-2,0\\right),\\left(3,0\\right),\\left(-5,0\\right)[\/latex]<\/p>\n<p>9.\u00a0[latex]\\left(3,0\\right),\\left(-1,0\\right),\\left(0,0\\right)[\/latex]<\/p>\n<p>11.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(-5,0\\right),\\text{ }\\left(2,0\\right)[\/latex]<\/p>\n<p>13.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(-5,0\\right),\\text{ }\\left(4,0\\right)[\/latex]<\/p>\n<p>15.\u00a0[latex]\\left(2,0\\right),\\text{ }\\left(-2,0\\right),\\text{ }\\left(-1,0\\right)[\/latex]<\/p>\n<p>17.\u00a0[latex]\\left(-2,0\\right),\\left(2,0\\right),\\left(\\frac{1}{2},0\\right)[\/latex]<\/p>\n<p>19.\u00a0[latex]\\left(1,0\\right),\\text{ }\\left(-1,0\\right)[\/latex]<\/p>\n<p>21.\u00a0[latex]\\left(0,0\\right),\\left(\\sqrt{3},0\\right),\\left(-\\sqrt{3},0\\right)[\/latex]<\/p>\n<p>23.\u00a0[latex]\\left(0,0\\right),\\text{ }\\left(1,0\\right)\\text{, }\\left(-1,0\\right),\\text{ }\\left(2,0\\right),\\text{ }\\left(-2,0\\right)[\/latex]<\/p>\n<p>25.\u00a0[latex]f\\left(2\\right)=-10[\/latex]\u00a0and [latex]f\\left(4\\right)=28[\/latex].\u00a0Sign change confirms.<\/p>\n<p>27.\u00a0[latex]f\\left(1\\right)=3[\/latex]\u00a0and [latex]f\\left(3\\right)=-77[\/latex].\u00a0Sign change confirms.<\/p>\n<p>29.\u00a0[latex]f\\left(0.01\\right)=1.000001[\/latex]\u00a0and [latex]f\\left(0.1\\right)=-7.999[\/latex].\u00a0Sign change confirms.<\/p>\n<p>31.\u00a00 with multiplicity 2, [latex]-\\frac{3}{2}[\/latex]\u00a0with multiplicity 5, 4 with multiplicity 2<\/p>\n<p>33.\u00a00 with multiplicity 2, \u20132 with multiplicity 2<\/p>\n<p>35.\u00a0[latex]-\\frac{2}{3}\\text{ with multiplicity }5\\text{,}5\\text{ with multiplicity }\\text{2}[\/latex]<\/p>\n<p>37.\u00a0[latex]\\text{0}\\text{ with multiplicity }4\\text{,}2\\text{ with multiplicity }1\\text{,}-\\text{1}\\text{ with multiplicity }1[\/latex]<\/p>\n<p>39.\u00a0[latex]\\frac{3}{2}[\/latex]\u00a0with multiplicity 2, 0 with multiplicity 3<\/p>\n<p>41.\u00a0[latex]\\text{0}\\text{ with multiplicity }6\\text{,}\\frac{2}{3}\\text{ with multiplicity }2[\/latex]<\/p>\n<p>43.\u00a0<em>x<\/em>-intercepts,\u00a0[latex]\\left(1, 0\\right)[\/latex]\u00a0with multiplicity 2, [latex]\\left(-4, 0\\right)[\/latex] with multiplicity 1, <em>y-<\/em>intercept [latex]\\left(0, 4\\right)[\/latex]. As\u00a0[latex]x\\to -\\infty[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty[\/latex] , as\u00a0[latex]x\\to \\infty[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty[\/latex] .<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201515\/CNX_Precalc_Figure_03_04_202.jpg\" alt=\"Graph of g(x)=(x+4)(x-1)^2.\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>45.\u00a0<em>x<\/em>-intercepts [latex]\\left(3,0\\right)[\/latex] with multiplicity 3, [latex]\\left(2,0\\right)[\/latex] with multiplicity 2, <em>y<\/em>-intercept [latex]\\left(0,-108\\right)[\/latex] . As\u00a0[latex]x\\to -\\infty[\/latex],\u00a0[latex]f\\left(x\\right)\\to -\\infty[\/latex] , as [latex]x\\to \\infty[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201517\/CNX_Precalc_Figure_03_04_204.jpg\" alt=\"Graph of k(x)=(x-3)^3(x-2)^2.\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>47.\u00a0x-intercepts [latex]\\left(0, 0\\right),\\left(-2, 0\\right),\\left(4, 0\\right)[\/latex]\u00a0with multiplicity 1, <em>y<\/em>-intercept [latex]\\left(0, 0\\right)[\/latex]. As\u00a0[latex]x\\to -\\infty[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to \\infty[\/latex] , as\u00a0[latex]x\\to \\infty[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty[\/latex].<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201518\/CNX_Precalc_Figure_03_04_206.jpg\" alt=\"Graph of n(x)=-3x(x+2)(x-4).\" data-media-type=\"image\/jpg\" \/><\/p>\n<p>49.\u00a0[latex]f\\left(x\\right)=-\\frac{2}{9}\\left(x - 3\\right)\\left(x+1\\right)\\left(x+3\\right)[\/latex]<\/p>\n<p>51.\u00a0[latex]f\\left(x\\right)=\\frac{1}{4}{\\left(x+2\\right)}^{2}\\left(x - 3\\right)[\/latex]<\/p>\n<p>53.\u00a0\u20134, \u20132, 1, 3 with multiplicity 1<\/p>\n<p>55.\u00a0\u20132, 3 each with multiplicity 2<\/p>\n<p>57.\u00a0[latex]f\\left(x\\right)=-\\frac{2}{3}\\left(x+2\\right)\\left(x - 1\\right)\\left(x - 3\\right)[\/latex]<\/p>\n<p>59.\u00a0[latex]f\\left(x\\right)=\\frac{1}{3}{\\left(x - 3\\right)}^{2}{\\left(x - 1\\right)}^{2}\\left(x+3\\right)[\/latex]<\/p>\n<p>61.\u00a0[latex]f\\left(x\\right)=-15{\\left(x - 1\\right)}^{2}{\\left(x - 3\\right)}^{3}[\/latex]<\/p>\n<p>63.\u00a0[latex]f\\left(x\\right)=-2\\left(x+3\\right)\\left(x+2\\right)\\left(x - 1\\right)[\/latex]<\/p>\n<p>65. [latex]f\\left(x\\right)=-\\frac{3}{2}{\\left(2x - 1\\right)}^{2}\\left(x - 6\\right)\\left(x+2\\right)[\/latex]<\/p>\n<p>67.\u00a0local max [latex]\\left(-\\text{.58, -}.62\\right)[\/latex],\u00a0local min [latex]\\left(\\text{.58, -1}\\text{.38}\\right)[\/latex]<\/p>\n<p>69.\u00a0global min [latex]\\left(-\\text{.63, -}\\text{.47}\\right)[\/latex]<\/p>\n<p>71.\u00a0global min [latex]\\text{(}\\text{.75, }\\text{.89)}[\/latex]<\/p>\n<p>73.\u00a0[latex]f\\left(x\\right)={\\left(x - 500\\right)}^{2}\\left(x+200\\right)[\/latex]<\/p>\n<p>75.\u00a0[latex]f\\left(x\\right)=4{x}^{3}-36{x}^{2}+80x[\/latex]<\/p>\n<p>77.\u00a0[latex]f\\left(x\\right)=4{x}^{3}-36{x}^{2}+60x+100[\/latex]<\/p>\n<p>79.\u00a0[latex]f\\left(x\\right)=\\pi \\left(9{x}^{3}+45{x}^{2}+72x+36\\right)[\/latex]<\/p>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1345\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":276,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1345","chapter","type-chapter","status-publish","hentry"],"part":1290,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1345","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/users\/276"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1345\/revisions"}],"predecessor-version":[{"id":2911,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1345\/revisions\/2911"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/1290"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1345\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/media?parent=1345"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1345"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1345"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/odessa-collegealgebra\/wp-json\/wp\/v2\/license?post=1345"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}