{"id":424,"date":"2025-02-13T19:44:48","date_gmt":"2025-02-13T19:44:48","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-integration-background-youll-need-2\/"},"modified":"2025-02-13T19:44:48","modified_gmt":"2025-02-13T19:44:48","slug":"applications-of-integration-background-youll-need-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/applications-of-integration-background-youll-need-2\/","title":{"raw":"Applications of Integration: Background You'll Need 2","rendered":"Applications of Integration: Background You&#8217;ll Need 2"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Find out if and where two graphs cross each other<\/li>\n<\/ul>\n<\/section>\n<h2>Determine Where Two Functions Intersect<\/h2>\n<p>Understanding where two functions intersect is a fundamental concept in algebra and calculus. This process can reveal key insights into the relationship between the functions and is crucial for graph analysis and solving system of equations.<\/p>\n<p>To find where two functions intersect, the key is to understand that the intersection points are where the outputs of both functions are equal. This involves setting the equations of the functions equal to one another and solving for the variables involved. The solutions to these equations represent the coordinates of the points where the graphs of the functions intersect.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>intersection of functions<\/h3>\n<p>The intersection of functions refers to the set of points where the graphs of two or more functions meet or cross each other. This occurs when the output values of the functions are equal at the same input value.<\/p>\n<p><br>\nTo determine where two functions intersect, set their equations equal to each other and solve for the variable. This process finds the exact points where the graphs of the functions meet.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider finding the intersection of the linear function [latex]f(x) = 2x+3[\/latex] and the quadratic function [latex]g(x)=x^2+x+1[\/latex].<\/p>\n<p>We start by setting the equations equal to one another.<\/p>\n<p style=\"text-align: center;\">[latex]2x+3=x^2+x+1[\/latex]<\/p>\n<p>Next we rearrange the equation to isolate [latex]x[\/latex] to one side and set the equation equal to zero.<\/p>\n<p style=\"text-align: center;\"><br>\n[latex]0=x^2-x-2[\/latex]<\/p>\n<p>We can solve for [latex]x[\/latex] by factoring.<\/p>\n<p style=\"text-align: center;\">[latex](x-2)(x+1)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = 2 \\text{ and } x=-1[\/latex]<\/p>\n<p>For [latex]x=2[\/latex], substituting back into [latex]f(x)[\/latex] gives [latex]y=7[\/latex]. For [latex]x=-1[\/latex], [latex]y=1[\/latex].<\/p>\n<p>Thus, the interaction points are [latex](2,7)[\/latex] and [latex](-1,1)[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><b>How to: Find Intersection Points<\/b><\/p>\n<ol>\n\t<li><strong>Set Equations Equal<\/strong>: Align the functions such that [latex]f(x)=g(x)[\/latex]. This step equates the two functions' outputs at their intersection points.<\/li>\n\t<li><strong>Rearrange the Equation<\/strong>: Simplify the equation to isolate terms involving [latex]x[\/latex] on one side. This often involves subtracting one side of the equation from the other.<\/li>\n\t<li><strong>Solve for [latex]x[\/latex]<\/strong>: Use algebraic methods such as factoring, applying the quadratic formula, or computational tools to find the value(s) of [latex]x[\/latex] where the functions intersect.<\/li>\n\t<li><strong>Find Corresponding [latex]y[\/latex] Values<\/strong>: Substitute the [latex]x[\/latex] values back into either original function to find the corresponding [latex]y[\/latex] values for each intersection point.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the points of intersection of the functions [latex]y=x+2[\/latex] and [latex]y=x^2+3x+2[\/latex].<\/p>\n<p>[reveal-answer q=\"133750\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"133750\"]<\/p>\n<p>Set the two functions equal to each other.<\/p>\n<p>So, we have:<\/p>\n<p style=\"text-align: center;\">[latex]x+2=x^2+3x+2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]0=x^2+2x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x^2+2x=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x(x+2)=0[\/latex]<\/p>\n<p>Setting [latex]x=0[\/latex] gives us an intersection point at [latex]x=0[\/latex]. To find the corresponding&nbsp;[latex]y[\/latex]-value of the point, let&nbsp;[latex]x=0[\/latex] in either function equation: [latex]y=x+2=0+2=2[\/latex].<\/p>\n<p>Setting [latex]x+2=0[\/latex] gives us an intersection point at [latex]x=-2[\/latex].&nbsp;To find the corresponding&nbsp;[latex]y[\/latex]-value of the point, let&nbsp;[latex]x=-2[\/latex] in either function equation: [latex]y=x+2=-2+2=0[\/latex].<\/p>\n<p>Notice that graphically, we can see that the line and the parabola intersect at the points [latex](0,2)[\/latex] and&nbsp;[latex](-2,0)[\/latex].<\/p>\n<p style=\"text-align: center;\"><img class=\"alignnone size-medium wp-image-469\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/20220158\/Intersect_1-300x215.jpg\" alt=\"The graph of a parabola and line intersecting at (-2,0) and (0,2)\" width=\"300\" height=\"215\"><\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the points of intersection of the functions [latex]y=x-2[\/latex] and [latex]y=2x^2-4x+1[\/latex].<\/p>\n<p>[reveal-answer q=\"133751\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"133751\"]<\/p>\n<p style=\"text-align: left;\">[latex](1,-1)[\/latex] and&nbsp;[latex](\\dfrac{3}{2}, -\\dfrac{1}{2})[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox tryIt\">\n<p>[ohm_question hide_question_numbers=1]287923[\/ohm_question]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Find out if and where two graphs cross each other<\/li>\n<\/ul>\n<\/section>\n<h2>Determine Where Two Functions Intersect<\/h2>\n<p>Understanding where two functions intersect is a fundamental concept in algebra and calculus. This process can reveal key insights into the relationship between the functions and is crucial for graph analysis and solving system of equations.<\/p>\n<p>To find where two functions intersect, the key is to understand that the intersection points are where the outputs of both functions are equal. This involves setting the equations of the functions equal to one another and solving for the variables involved. The solutions to these equations represent the coordinates of the points where the graphs of the functions intersect.<\/p>\n<section class=\"textbox keyTakeaway\">\n<h3>intersection of functions<\/h3>\n<p>The intersection of functions refers to the set of points where the graphs of two or more functions meet or cross each other. This occurs when the output values of the functions are equal at the same input value.<\/p>\n<p>\nTo determine where two functions intersect, set their equations equal to each other and solve for the variable. This process finds the exact points where the graphs of the functions meet.<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider finding the intersection of the linear function [latex]f(x) = 2x+3[\/latex] and the quadratic function [latex]g(x)=x^2+x+1[\/latex].<\/p>\n<p>We start by setting the equations equal to one another.<\/p>\n<p style=\"text-align: center;\">[latex]2x+3=x^2+x+1[\/latex]<\/p>\n<p>Next we rearrange the equation to isolate [latex]x[\/latex] to one side and set the equation equal to zero.<\/p>\n<p style=\"text-align: center;\">\n[latex]0=x^2-x-2[\/latex]<\/p>\n<p>We can solve for [latex]x[\/latex] by factoring.<\/p>\n<p style=\"text-align: center;\">[latex](x-2)(x+1)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x = 2 \\text{ and } x=-1[\/latex]<\/p>\n<p>For [latex]x=2[\/latex], substituting back into [latex]f(x)[\/latex] gives [latex]y=7[\/latex]. For [latex]x=-1[\/latex], [latex]y=1[\/latex].<\/p>\n<p>Thus, the interaction points are [latex](2,7)[\/latex] and [latex](-1,1)[\/latex].<\/p>\n<\/section>\n<section class=\"textbox questionHelp\">\n<p><b>How to: Find Intersection Points<\/b><\/p>\n<ol>\n<li><strong>Set Equations Equal<\/strong>: Align the functions such that [latex]f(x)=g(x)[\/latex]. This step equates the two functions&#8217; outputs at their intersection points.<\/li>\n<li><strong>Rearrange the Equation<\/strong>: Simplify the equation to isolate terms involving [latex]x[\/latex] on one side. This often involves subtracting one side of the equation from the other.<\/li>\n<li><strong>Solve for [latex]x[\/latex]<\/strong>: Use algebraic methods such as factoring, applying the quadratic formula, or computational tools to find the value(s) of [latex]x[\/latex] where the functions intersect.<\/li>\n<li><strong>Find Corresponding [latex]y[\/latex] Values<\/strong>: Substitute the [latex]x[\/latex] values back into either original function to find the corresponding [latex]y[\/latex] values for each intersection point.<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the points of intersection of the functions [latex]y=x+2[\/latex] and [latex]y=x^2+3x+2[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q133750\">Show Solution<\/button><\/p>\n<div id=\"q133750\" class=\"hidden-answer\" style=\"display: none\">\n<p>Set the two functions equal to each other.<\/p>\n<p>So, we have:<\/p>\n<p style=\"text-align: center;\">[latex]x+2=x^2+3x+2[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]0=x^2+2x[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x^2+2x=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x(x+2)=0[\/latex]<\/p>\n<p>Setting [latex]x=0[\/latex] gives us an intersection point at [latex]x=0[\/latex]. To find the corresponding&nbsp;[latex]y[\/latex]-value of the point, let&nbsp;[latex]x=0[\/latex] in either function equation: [latex]y=x+2=0+2=2[\/latex].<\/p>\n<p>Setting [latex]x+2=0[\/latex] gives us an intersection point at [latex]x=-2[\/latex].&nbsp;To find the corresponding&nbsp;[latex]y[\/latex]-value of the point, let&nbsp;[latex]x=-2[\/latex] in either function equation: [latex]y=x+2=-2+2=0[\/latex].<\/p>\n<p>Notice that graphically, we can see that the line and the parabola intersect at the points [latex](0,2)[\/latex] and&nbsp;[latex](-2,0)[\/latex].<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"alignnone size-medium wp-image-469\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5604\/2021\/05\/20220158\/Intersect_1-300x215.jpg\" alt=\"The graph of a parabola and line intersecting at (-2,0) and (0,2)\" width=\"300\" height=\"215\" \/><\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Find the points of intersection of the functions [latex]y=x-2[\/latex] and [latex]y=2x^2-4x+1[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q133751\">Show Solution<\/button><\/p>\n<div id=\"q133751\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: left;\">[latex](1,-1)[\/latex] and&nbsp;[latex](\\dfrac{3}{2}, -\\dfrac{1}{2})[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm287923\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=287923&theme=lumen&iframe_resize_id=ohm287923&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":3,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":421,"module-header":"","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\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/424"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/424\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/421"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/424\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=424"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=424"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=424"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=424"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}