{"id":1001,"date":"2023-07-14T15:38:40","date_gmt":"2023-07-14T15:38:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/?post_type=chapter&#038;p=1001"},"modified":"2023-08-10T22:18:31","modified_gmt":"2023-08-10T22:18:31","slug":"solving-equations-using-graphs-of-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/solving-equations-using-graphs-of-functions\/","title":{"raw":"\u25aa   Solving Equations Using Graphs of Functions","rendered":"\u25aa   Solving Equations Using Graphs of Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Solve equations with one variable using graphs of functions.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Solving Equations with One Variable using One Function<\/h2>\r\nAs we have seen, we can solve an equation [latex]f(x)=0[\/latex] by finding the [latex]x[\/latex]-intercepts of its function [latex]y=f(x)[\/latex], and the graph of [latex]y=f(x)[\/latex] makes the process much easier. For example, we can easily locate the [latex]x[\/latex]-intercept of [latex]f(x)=-\\frac{3}{4}x+3[\/latex] from its graph, and we can conclude that the solution of [latex]-\\frac{3}{4}x+3=0[\/latex] is [latex]x=4[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"306\"]<img class=\"aligncenter size-full wp-image-1020\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation.png\" alt=\"a graph of y=-3\/4x+3\" width=\"306\" height=\"300\" \/> Figure 3. Graph of [latex]f(x)=-\\frac{3}{4}x+3[\/latex][\/caption]Moreover, this method helps us to solve more complex equations as well. For example, to solve [latex]x+\\frac{3}{x}-4=0[\/latex], we should consider the graph of [latex]f(x)=x+\\frac{3}{x}-4[\/latex] and find the [latex]x[\/latex]-intercepts. As we can read on the graph, the [latex]x[\/latex]-intercepts of this function are [latex](1, 0)[\/latex] and [latex](3, 0)[\/latex]. So, the solutions of the equation are [latex]x=1, 3[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"365\"]<img class=\"aligncenter size-full wp-image-1027\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2.png\" alt=\"a graph of f(x)=x+3\/x-4\" width=\"365\" height=\"342\" \/> Figure 4. Graph of [latex]f(x)=x+\\frac{3}{x}-4[\/latex][\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving [latex]f(x)=0[\/latex] Graphically<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex]x^3-x-4x^2+4=0[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"434622\"]Show Graph[\/reveal-answer]\r\n[hidden-answer a=\"434622\"]<img class=\"aligncenter wp-image-1029\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms.png\" alt=\"graph of f(x)=x^3-x-4x^2+4\" width=\"216\" height=\"256\" \/>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"15532\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"15532\"]Let [latex]f(x)=x^3-x-4x^2+4[\/latex]. Then [latex]x=-1, 1, 4[\/latex].\u00a0<span style=\"font-size: 0.9em;\">[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex](x-1)^4-3(x-1)^2-4=0[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"371099\"]Show Graph[\/reveal-answer]\r\n[hidden-answer a=\"371099\"]<img class=\"aligncenter wp-image-1030\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form.png\" alt=\"graph of f(x)=(x-1)^4-3(x-1)^2-4\" width=\"187\" height=\"186\" \/>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"985969\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"985969\"]Let [latex]f(x)=(x-1)^4-3(x-1)^2-4[\/latex]. Then, [latex] x=-1, 3[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Solving [latex]f(x)=0[\/latex] Graphically<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex]\\sqrt[3]{(x+1)^2}-4=0[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"571288\"]Show Graph[\/reveal-answer]\r\n[hidden-answer a=\"571288\"]<img class=\"aligncenter wp-image-1028\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq.png\" alt=\"graph of f(x)=(x+1)^(2\/3)-4\" width=\"212\" height=\"126\" \/>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"541339\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"541339\"]Let [latex]f(x)=\\sqrt[3]{(x+1)^2}-4[\/latex]. Then, [latex]x=-9, 7[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex]\\frac{x-3}{x}-\\frac{1}{2}=0[\/latex]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"934285\"]Show Graph[\/reveal-answer]\r\n[hidden-answer a=\"934285\"]<img class=\"aligncenter wp-image-1031\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq.png\" alt=\"graph of f(x)=(x-3)\/x-1\/2\" width=\"215\" height=\"156\" \/>[\/hidden-answer]\r\n\r\n[reveal-answer q=\"834844\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"834844\"]Let [latex]f(x)=\\frac{x-3}{x}-\\frac{1}{2}[\/latex]. Then, [latex]x=6[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Solving Equations with One Variable using Multiple Functions<\/h2>\r\nNow we know how to solve equations graphically. But what if the given equation is not in \"[latex]f(x)=0[\/latex]\" form? How can we apply the method we used in the previous examples? There are two different ways to solve this equation:\r\n<ol>\r\n \t<li>Move all terms to the left and make it in \"[latex]f(x)=0[\/latex]\" form. Then, graph and find the [latex]x[\/latex]-intercepts.<\/li>\r\n \t<li>Let \"[latex]f(x)=[\/latex](lefthand side)\" and \"[latex]g(x)=[\/latex](righthand side).\" Then, graph and find the intersecting points of [latex]f(x)[\/latex] and [latex]g(x)[\/latex]. The [latex]x[\/latex]-coordinates are the solutions of the equation.<\/li>\r\n<\/ol>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Solving [latex]f(x)=g(x)[\/latex] Graphically<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex]2x^2-3=-2x+1[\/latex]\r\n\r\n[reveal-answer q=\"877252\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"877252\"]\r\n\r\n<span style=\"text-decoration: underline;\"><strong>Method 1<\/strong><\/span>\r\n\r\nMove all terms to the left: [latex]2x^2-3+2x-1=0[\/latex]*\r\n\r\n[latex]2x^2+2x-4=0[\/latex]\r\n\r\nLet\u00a0[latex]f(x)=2x^2+2x-4[\/latex].\r\n<p style=\"padding-left: 30px;\">* If you are using a graphing tool, you don't need to simplify the equation. Just let\u00a0[latex]f(x)=2x^2-3-(-2x+1)[\/latex] or\u00a0 [latex]f(x)=2x^2-3+2x-1[\/latex].<\/p>\r\n<img class=\"aligncenter wp-image-1041\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function.png\" alt=\"graph of 2x^2+2x-4\" width=\"178\" height=\"177\" \/>\r\n\r\nSince the [latex]x[\/latex]-intercepts are [latex](-2, 0), (1, 0)[\/latex], the solutions of\u00a0[latex]2x^2-3=-2x+1[\/latex] are [latex]x=-2, 1[\/latex].\r\n\r\n<span style=\"text-decoration: underline;\"><strong>Method 2<\/strong><\/span>\r\n\r\nLet [latex]f(x)=2x^2-3[\/latex] and [latex]g(x)=-2x+1[\/latex].\r\n\r\n<img class=\"aligncenter wp-image-1042\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions.png\" alt=\"Graphs of f(x)=2x^2-3 and g(x)=-2x+1\" width=\"181\" height=\"180\" \/>\r\n\r\nSince the intersecting points are [latex](-2, 5), (1, -1)[\/latex], the solutions of\u00a0[latex]2x^2-3=-2x+1[\/latex] are [latex]x=-2, 1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It<\/h3>\r\nSolve the equation graphically. Use a graphing tool.\r\n\r\n[latex]-2x^3+6x-3=x^2-3[\/latex]\r\n\r\n[reveal-answer q=\"106098\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"106098\"]\r\n\r\nMove all terms to the lefthand side: [latex]-2x^3+6x-3-x^2+3=0[\/latex]. Then let [latex]f(x)=-2x^3-x^2+6x[\/latex]. Since the [latex]x[\/latex]-intercepts are [latex](-2, 0), (\\frac{3}{2}, 0)[\/latex], the solutions are [latex]x=-2, \\frac{3}{2}[\/latex].\r\n<p style=\"text-align: center;\">OR<\/p>\r\nLet [latex]g(x)=-2x^3+6x-3[\/latex] and [latex]h(x)=x^2-3[\/latex]. Since the intersecting points of [latex]g(x)[\/latex] and [latex]h(x)[\/latex] are [latex](-2, 1)[\/latex] and [latex](\\frac{3}{2}, -\\frac{3}{4})[\/latex], the solutions are [latex]x=-2, \\frac{3}{2}[\/latex].\r\n<table class=\"no-lines\" style=\"border-collapse: collapse; width: 100%;\">\r\n<tbody>\r\n<tr>\r\n<td style=\"width: 33.3333%;\"><img class=\"aligncenter wp-image-1044\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_One-function.png\" alt=\"graphs of f(x)=-2x^3-x^2+6x\" width=\"180\" height=\"165\" \/><\/td>\r\n<td style=\"width: 33.3333%;\"><img class=\"aligncenter wp-image-1079\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1.png\" alt=\"graphs of g(x)=-2x^3+6x-3, h(x)=x^2-3\" width=\"174\" height=\"163\" \/><\/td>\r\n<td style=\"width: 33.3333%; vertical-align: middle;\"><img class=\"aligncenter wp-image-1077\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1.png\" alt=\"graphs of f(x)=-2x^3-x^2+6x, g(x)=-2x^3+6x-3, h(x)=x^2-3\" width=\"179\" height=\"165\" \/><\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[\/hidden-answer]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Solve equations with one variable using graphs of functions.<\/li>\n<\/ul>\n<\/div>\n<h2>Solving Equations with One Variable using One Function<\/h2>\n<p>As we have seen, we can solve an equation [latex]f(x)=0[\/latex] by finding the [latex]x[\/latex]-intercepts of its function [latex]y=f(x)[\/latex], and the graph of [latex]y=f(x)[\/latex] makes the process much easier. For example, we can easily locate the [latex]x[\/latex]-intercept of [latex]f(x)=-\\frac{3}{4}x+3[\/latex] from its graph, and we can conclude that the solution of [latex]-\\frac{3}{4}x+3=0[\/latex] is [latex]x=4[\/latex].<\/p>\n<div style=\"width: 316px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1020\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation.png\" alt=\"a graph of y=-3\/4x+3\" width=\"306\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation.png 306w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation-300x294.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation-65x64.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-linear-equation-225x221.png 225w\" sizes=\"auto, (max-width: 306px) 100vw, 306px\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Graph of [latex]f(x)=-\\frac{3}{4}x+3[\/latex]<\/p>\n<\/div>\n<p>Moreover, this method helps us to solve more complex equations as well. For example, to solve [latex]x+\\frac{3}{x}-4=0[\/latex], we should consider the graph of [latex]f(x)=x+\\frac{3}{x}-4[\/latex] and find the [latex]x[\/latex]-intercepts. As we can read on the graph, the [latex]x[\/latex]-intercepts of this function are [latex](1, 0)[\/latex] and [latex](3, 0)[\/latex]. So, the solutions of the equation are [latex]x=1, 3[\/latex].<\/p>\n<div style=\"width: 375px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-1027\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2.png\" alt=\"a graph of f(x)=x+3\/x-4\" width=\"365\" height=\"342\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2.png 365w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2-300x281.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2-65x61.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2-225x211.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-complex-equation-2-350x328.png 350w\" sizes=\"auto, (max-width: 365px) 100vw, 365px\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Graph of [latex]f(x)=x+\\frac{3}{x}-4[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Solving [latex]f(x)=0[\/latex] Graphically<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex]x^3-x-4x^2+4=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q434622\">Show Graph<\/span><\/p>\n<div id=\"q434622\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1029\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms.png\" alt=\"graph of f(x)=x^3-x-4x^2+4\" width=\"216\" height=\"256\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms.png 363w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms-253x300.png 253w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms-65x77.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms-225x267.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-2-Poly-Eq-w-4-terms-350x416.png 350w\" sizes=\"auto, (max-width: 216px) 100vw, 216px\" \/><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q15532\">Show Answer<\/span><\/p>\n<div id=\"q15532\" class=\"hidden-answer\" style=\"display: none\">Let [latex]f(x)=x^3-x-4x^2+4[\/latex]. Then [latex]x=-1, 1, 4[\/latex].\u00a0<span style=\"font-size: 0.9em;\"><\/div>\n<\/div>\n<p><\/span><\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex](x-1)^4-3(x-1)^2-4=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q371099\">Show Graph<\/span><\/p>\n<div id=\"q371099\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1030\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form.png\" alt=\"graph of f(x)=(x-1)^4-3(x-1)^2-4\" width=\"187\" height=\"186\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form.png 332w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form-150x150.png 150w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form-300x300.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form-65x65.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-3-Eq-in-Quad-Form-225x224.png 225w\" sizes=\"auto, (max-width: 187px) 100vw, 187px\" \/><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q985969\">Show Answer<\/span><\/p>\n<div id=\"q985969\" class=\"hidden-answer\" style=\"display: none\">Let [latex]f(x)=(x-1)^4-3(x-1)^2-4[\/latex]. Then, [latex]x=-1, 3[\/latex].<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Solving [latex]f(x)=0[\/latex] Graphically<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex]\\sqrt[3]{(x+1)^2}-4=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q571288\">Show Graph<\/span><\/p>\n<div id=\"q571288\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1028\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq.png\" alt=\"graph of f(x)=(x+1)^(2\/3)-4\" width=\"212\" height=\"126\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq.png 368w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq-300x179.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq-65x39.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq-225x134.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-1-Rational-Power-Eq-350x208.png 350w\" sizes=\"auto, (max-width: 212px) 100vw, 212px\" \/><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q541339\">Show Answer<\/span><\/p>\n<div id=\"q541339\" class=\"hidden-answer\" style=\"display: none\">Let [latex]f(x)=\\sqrt[3]{(x+1)^2}-4[\/latex]. Then, [latex]x=-9, 7[\/latex]<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex]\\frac{x-3}{x}-\\frac{1}{2}=0[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934285\">Show Graph<\/span><\/p>\n<div id=\"q934285\" class=\"hidden-answer\" style=\"display: none\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1031\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq.png\" alt=\"graph of f(x)=(x-3)\/x-1\/2\" width=\"215\" height=\"156\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq.png 362w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq-300x217.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq-65x47.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq-225x163.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-4-Rational-Eq-350x253.png 350w\" sizes=\"auto, (max-width: 215px) 100vw, 215px\" \/><\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834844\">Show Answer<\/span><\/p>\n<div id=\"q834844\" class=\"hidden-answer\" style=\"display: none\">Let [latex]f(x)=\\frac{x-3}{x}-\\frac{1}{2}[\/latex]. Then, [latex]x=6[\/latex]<\/div>\n<\/div>\n<\/div>\n<h2>Solving Equations with One Variable using Multiple Functions<\/h2>\n<p>Now we know how to solve equations graphically. But what if the given equation is not in &#8220;[latex]f(x)=0[\/latex]&#8221; form? How can we apply the method we used in the previous examples? There are two different ways to solve this equation:<\/p>\n<ol>\n<li>Move all terms to the left and make it in &#8220;[latex]f(x)=0[\/latex]&#8221; form. Then, graph and find the [latex]x[\/latex]-intercepts.<\/li>\n<li>Let &#8220;[latex]f(x)=[\/latex](lefthand side)&#8221; and &#8220;[latex]g(x)=[\/latex](righthand side).&#8221; Then, graph and find the intersecting points of [latex]f(x)[\/latex] and [latex]g(x)[\/latex]. The [latex]x[\/latex]-coordinates are the solutions of the equation.<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Solving [latex]f(x)=g(x)[\/latex] Graphically<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex]2x^2-3=-2x+1[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q877252\">Show Answer<\/span><\/p>\n<div id=\"q877252\" class=\"hidden-answer\" style=\"display: none\">\n<p><span style=\"text-decoration: underline;\"><strong>Method 1<\/strong><\/span><\/p>\n<p>Move all terms to the left: [latex]2x^2-3+2x-1=0[\/latex]*<\/p>\n<p>[latex]2x^2+2x-4=0[\/latex]<\/p>\n<p>Let\u00a0[latex]f(x)=2x^2+2x-4[\/latex].<\/p>\n<p style=\"padding-left: 30px;\">* If you are using a graphing tool, you don&#8217;t need to simplify the equation. Just let\u00a0[latex]f(x)=2x^2-3-(-2x+1)[\/latex] or\u00a0 [latex]f(x)=2x^2-3+2x-1[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1041\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function.png\" alt=\"graph of 2x^2+2x-4\" width=\"178\" height=\"177\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function.png 282w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function-150x150.png 150w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function-65x65.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_One-function-225x224.png 225w\" sizes=\"auto, (max-width: 178px) 100vw, 178px\" \/><\/p>\n<p>Since the [latex]x[\/latex]-intercepts are [latex](-2, 0), (1, 0)[\/latex], the solutions of\u00a0[latex]2x^2-3=-2x+1[\/latex] are [latex]x=-2, 1[\/latex].<\/p>\n<p><span style=\"text-decoration: underline;\"><strong>Method 2<\/strong><\/span><\/p>\n<p>Let [latex]f(x)=2x^2-3[\/latex] and [latex]g(x)=-2x+1[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1042\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions.png\" alt=\"Graphs of f(x)=2x^2-3 and g(x)=-2x+1\" width=\"181\" height=\"180\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions.png 283w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions-150x150.png 150w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions-65x65.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-5-quad-eq_Two-functions-225x224.png 225w\" sizes=\"auto, (max-width: 181px) 100vw, 181px\" \/><\/p>\n<p>Since the intersecting points are [latex](-2, 5), (1, -1)[\/latex], the solutions of\u00a0[latex]2x^2-3=-2x+1[\/latex] are [latex]x=-2, 1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It<\/h3>\n<p>Solve the equation graphically. Use a graphing tool.<\/p>\n<p>[latex]-2x^3+6x-3=x^2-3[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q106098\">Show Answer<\/span><\/p>\n<div id=\"q106098\" class=\"hidden-answer\" style=\"display: none\">\n<p>Move all terms to the lefthand side: [latex]-2x^3+6x-3-x^2+3=0[\/latex]. Then let [latex]f(x)=-2x^3-x^2+6x[\/latex]. Since the [latex]x[\/latex]-intercepts are [latex](-2, 0), (\\frac{3}{2}, 0)[\/latex], the solutions are [latex]x=-2, \\frac{3}{2}[\/latex].<\/p>\n<p style=\"text-align: center;\">OR<\/p>\n<p>Let [latex]g(x)=-2x^3+6x-3[\/latex] and [latex]h(x)=x^2-3[\/latex]. Since the intersecting points of [latex]g(x)[\/latex] and [latex]h(x)[\/latex] are [latex](-2, 1)[\/latex] and [latex](\\frac{3}{2}, -\\frac{3}{4})[\/latex], the solutions are [latex]x=-2, \\frac{3}{2}[\/latex].<\/p>\n<table class=\"no-lines\" style=\"border-collapse: collapse; width: 100%;\">\n<tbody>\n<tr>\n<td style=\"width: 33.3333%;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1044\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_One-function.png\" alt=\"graphs of f(x)=-2x^3-x^2+6x\" width=\"180\" height=\"165\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_One-function.png 281w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_One-function-65x60.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_One-function-225x207.png 225w\" sizes=\"auto, (max-width: 180px) 100vw, 180px\" \/><\/td>\n<td style=\"width: 33.3333%;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1079\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1.png\" alt=\"graphs of g(x)=-2x^3+6x-3, h(x)=x^2-3\" width=\"174\" height=\"163\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1.png 378w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1-300x280.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1-65x61.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1-225x210.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_Two-function-1-350x327.png 350w\" sizes=\"auto, (max-width: 174px) 100vw, 174px\" \/><\/td>\n<td style=\"width: 33.3333%; vertical-align: middle;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1077\" src=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1.png\" alt=\"graphs of f(x)=-2x^3-x^2+6x, g(x)=-2x^3+6x-3, h(x)=x^2-3\" width=\"179\" height=\"165\" srcset=\"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1.png 377w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1-300x276.png 300w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1-65x60.png 65w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1-225x207.png 225w, https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-content\/uploads\/sites\/5858\/2023\/07\/2.5-EX-6-poly-eq_all-1-350x322.png 350w\" sizes=\"auto, (max-width: 179px) 100vw, 179px\" \/><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\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-1001\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li> Solving Equations Using Graphs of Functions. <strong>Authored by<\/strong>: Michelle Eunhee Chung. <strong>Provided by<\/strong>: Georgia State University . <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/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":705214,"menu_order":31,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\" Solving Equations Using Graphs of Functions\",\"author\":\"Michelle Eunhee Chung\",\"organization\":\"Georgia State University \",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":["mchung12"],"pb_section_license":""},"chapter-type":[],"contributor":[62],"license":[],"class_list":["post-1001","chapter","type-chapter","status-publish","hentry","contributor-mchung12"],"part":91,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1001","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/users\/705214"}],"version-history":[{"count":17,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1001\/revisions"}],"predecessor-version":[{"id":1085,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1001\/revisions\/1085"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/91"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/1001\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/media?parent=1001"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=1001"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/contributor?post=1001"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/wp-json\/wp\/v2\/license?post=1001"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}