{"id":1861,"date":"2024-06-14T20:55:23","date_gmt":"2024-06-14T20:55:23","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/?post_type=chapter&#038;p=1861"},"modified":"2025-08-14T01:12:33","modified_gmt":"2025-08-14T01:12:33","slug":"analysis-of-quadratic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/chapter\/analysis-of-quadratic-functions-learn-it-2\/","title":{"raw":"Analysis of Quadratic Functions: Learn It 2","rendered":"Analysis of Quadratic Functions: Learn It 2"},"content":{"raw":"<h2>Finding the Maximum and Minimum Value of a Quadratic Function<\/h2>\r\nThere are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.\r\n\r\n<section class=\"textbox recall\">The vertex of a parabola is the highest (maximum) or lowest (minimum) point, depending on the direction the parabola opens.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170347\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/> Two parabola functions with key points and equations labeled[\/caption]\r\n\r\n<\/section><section class=\"textbox example\">A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[\/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170349\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" \/> Diagram of the garden and the backyard[\/caption]\r\n\r\nFind a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex]. Then, use the formula to answer: What dimensions should she make her garden to maximize the enclosed area?[reveal-answer q=\"958774\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"958774\"]Let\u2019s use a diagram to record the given information.Let [latex]L[\/latex] represents the length of the fence and [latex]W[\/latex] represents the width of the fence.\r\n\r\nWe know we have only [latex]80[\/latex] feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex].\r\n<p style=\"text-align: left;\">This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]W=80 - 2L[\/latex]<\/p>\r\nNow, we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width.\r\n<p style=\"text-align: center;\">[latex]A=L \\cdot W=L \\cdot (80 - 2L)=80L - 2{L}^{2}[\/latex].<\/p>\r\nThis formula represents the area of the fence in terms of the variable length [latex]L[\/latex].\r\n\r\n<strong>Thus, the function, written in general form, is[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/strong>\r\n\r\nNow, to answer the question: What dimensions should she make her garden to maximize the enclosed area?\r\n\r\nThe quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area.\r\n\r\nBased on the function above, we have [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].\r\n\r\nVertex:\r\n<p style=\"text-align: center;\">[latex]h= -\\dfrac{b}{2a} = -\\dfrac{80}{2(-2)} = -\\dfrac{80}{-4} = 20[\/latex]<\/p>\r\n<p style=\"text-align: center;\">and<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}k&amp;=A\\left(20\\right) \\\\&amp;=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\\\&amp;=800 \\end{align}[\/latex]<\/p>\r\nThus, the vertex is [latex](20, 800)[\/latex].\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"268\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170352\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"268\" height=\"262\" \/> Graph of a parabolic function with the vertex labeled[\/caption]\r\n\r\nThis problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.\r\n\r\n<strong>Interpretation of the vertex:<\/strong> The maximum value of the function is an area of [latex]800[\/latex] square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are [latex]L = 20[\/latex] feet, there is [latex]W = 80-2(20) = 40[\/latex] feet of fencing left for the longer side.\r\n\r\n<strong>So, to maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[\/latex] feet and the longer side parallel to the existing fence has length [latex]40[\/latex] feet.<\/strong>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>The problem we solved above is called a constrained <strong>optimization problem<\/strong>. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials.\r\n\r\n<section class=\"textbox tryIt\">[ohm2_question hide_question_numbers=1]24488[\/ohm2_question]<\/section>","rendered":"<h2>Finding the Maximum and Minimum Value of a Quadratic Function<\/h2>\n<p>There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.<\/p>\n<section class=\"textbox recall\">The vertex of a parabola is the highest (maximum) or lowest (minimum) point, depending on the direction the parabola opens.<\/p>\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170347\/CNX_Precalc_Figure_03_02_0092.jpg\" alt=\"Two graphs where the first graph shows the maximum value for f(x)=(x-2)^2+1 which occurs at (2, 1) and the second graph shows the minimum value for g(x)=-(x+3)^2+4 which occurs at (-3, 4).\" width=\"975\" height=\"558\" \/><figcaption class=\"wp-caption-text\">Two parabola functions with key points and equations labeled<\/figcaption><\/figure>\n<\/section>\n<section class=\"textbox example\">A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased [latex]80[\/latex] feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.<\/p>\n<figure style=\"width: 487px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170349\/CNX_Precalc_Figure_03_02_0102.jpg\" alt=\"Diagram of the garden and the backyard.\" width=\"487\" height=\"310\" \/><figcaption class=\"wp-caption-text\">Diagram of the garden and the backyard<\/figcaption><\/figure>\n<p>Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length [latex]L[\/latex]. Then, use the formula to answer: What dimensions should she make her garden to maximize the enclosed area?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q958774\">Show Answer<\/button><\/p>\n<div id=\"q958774\" class=\"hidden-answer\" style=\"display: none\">Let\u2019s use a diagram to record the given information.Let [latex]L[\/latex] represents the length of the fence and [latex]W[\/latex] represents the width of the fence.<\/p>\n<p>We know we have only [latex]80[\/latex] feet of fence available, and [latex]L+W+L=80[\/latex], or more simply, [latex]2L+W=80[\/latex].<\/p>\n<p style=\"text-align: left;\">This allows us to represent the width, [latex]W[\/latex], in terms of [latex]L[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]W=80 - 2L[\/latex]<\/p>\n<p>Now, we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width.<\/p>\n<p style=\"text-align: center;\">[latex]A=L \\cdot W=L \\cdot (80 - 2L)=80L - 2{L}^{2}[\/latex].<\/p>\n<p>This formula represents the area of the fence in terms of the variable length [latex]L[\/latex].<\/p>\n<p><strong>Thus, the function, written in general form, is[latex]A\\left(L\\right)=-2{L}^{2}+80L[\/latex].<\/strong><\/p>\n<p>Now, to answer the question: What dimensions should she make her garden to maximize the enclosed area?<\/p>\n<p>The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area.<\/p>\n<p>Based on the function above, we have [latex]a=-2,b=80[\/latex], and [latex]c=0[\/latex].<\/p>\n<p>Vertex:<\/p>\n<p style=\"text-align: center;\">[latex]h= -\\dfrac{b}{2a} = -\\dfrac{80}{2(-2)} = -\\dfrac{80}{-4} = 20[\/latex]<\/p>\n<p style=\"text-align: center;\">and<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k&=A\\left(20\\right) \\\\&=80\\left(20\\right)-2{\\left(20\\right)}^{2}\\\\&=800 \\end{align}[\/latex]<\/p>\n<p>Thus, the vertex is [latex](20, 800)[\/latex].<\/p>\n<figure style=\"width: 268px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02170352\/CNX_Precalc_Figure_03_02_0112.jpg\" alt=\"Graph of the parabolic function A(L)=-2L^2+80L, which the x-axis is labeled Length (L) and the y-axis is labeled Area (A). The vertex is at (20, 800).\" width=\"268\" height=\"262\" \/><figcaption class=\"wp-caption-text\">Graph of a parabolic function with the vertex labeled<\/figcaption><\/figure>\n<p>This problem also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function below.<\/p>\n<p><strong>Interpretation of the vertex:<\/strong> The maximum value of the function is an area of [latex]800[\/latex] square feet, which occurs when [latex]L=20[\/latex] feet. When the shorter sides are [latex]L = 20[\/latex] feet, there is [latex]W = 80-2(20) = 40[\/latex] feet of fencing left for the longer side.<\/p>\n<p><strong>So, to maximize the area, she should enclose the garden so the two shorter sides have length [latex]20[\/latex] feet and the longer side parallel to the existing fence has length [latex]40[\/latex] feet.<\/strong><\/p>\n<\/div>\n<\/div>\n<\/section>\n<p>The problem we solved above is called a constrained <strong>optimization problem<\/strong>. We can optimize our desired outcome given a constraint, which in this case was a limited amount of fencing materials.<\/p>\n<section class=\"textbox tryIt\"><iframe loading=\"lazy\" id=\"ohm24488\" class=\"resizable\" src=\"https:\/\/ohm.one.lumenlearning.com\/multiembedq.php?id=24488&theme=lumen&iframe_resize_id=ohm24488&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n","protected":false},"author":12,"menu_order":19,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":185,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1861"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/users\/12"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1861\/revisions"}],"predecessor-version":[{"id":7732,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1861\/revisions\/7732"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/parts\/185"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1861\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/media?parent=1861"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1861"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/contributor?post=1861"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/collegealgebra\/wp-json\/wp\/v2\/license?post=1861"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}