{"id":3896,"date":"2021-05-20T18:29:56","date_gmt":"2021-05-20T18:29:56","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-derivatives-and-the-shape-of-a-graph\/"},"modified":"2021-07-03T19:04:47","modified_gmt":"2021-07-03T19:04:47","slug":"review-for-derivatives-and-the-shape-of-a-graph","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-derivatives-and-the-shape-of-a-graph\/","title":{"raw":"Skills Review for Derivatives and the Shape of a Graph","rendered":"Skills Review for Derivatives and the Shape of a Graph"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify extrema and increasing and decreasing intervals<\/li>\r\n \t<li>Use a sign chart to determine where a function is positive or negative<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn the Derivatives and the Shape of a Graph section, we will analyze graphs of functions using calculus. Here we will review identifying extrema and increasing and decreasing intervals of a function along with how to create sign charts to determine where a certain equation is positive or negative.\r\n<h2>Identify Extrema and Increasing and Decreasing Intervals<\/h2>\r\nWe say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The graph below\u00a0shows examples of increasing and decreasing intervals on a function.\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\/10\/18194750\/CNX_Precalc_Figure_01_03_0042.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum.\" width=\"487\" height=\"518\" \/> The function [latex]f\\left(x\\right)={x}^{3}-12x[\/latex] is increasing on [latex]\\left(-\\infty \\text{,}-\\text{2}\\right){{\\cup }^{\\text{ }}}^{\\text{ }}\\left(2,\\infty \\right)[\/latex] and is decreasing on [latex]\\left(-2\\text{,}2\\right)[\/latex].[\/caption]This video further explains how to find where a function is increasing or decreasing.\r\n\r\nhttps:\/\/www.youtube.com\/watch?v=78b4HOMVcKM\r\n\r\nA value of the output where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a <strong>local maximum<\/strong>. If a function has more than one, we say it has local maxima. Similarly, a value of the output where a function changes from decreasing to increasing as the input variable increases is called a <strong>local minimum<\/strong>. The plural form is \"local minima.\" Together, local maxima and minima are called <strong>local extrema<\/strong>, or local extreme values, of the function. Often, the term <em>local<\/em> is replaced by the term <em>relative<\/em>.\r\n\r\nFor the function below, the local maximum is 16, and it occurs at [latex]x=-2[\/latex]. The local minimum is [latex]-16[\/latex] and it occurs at [latex]x=2[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194752\/CNX_Precalc_Figure_01_03_0142.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum. The local maximum is 16 and occurs at x = negative 2. This is the point negative 2, 16. The local minimum is negative 16 and occurs at x = 2. This is the point 2, negative 16.\" width=\"731\" height=\"467\" \/>\r\n\r\nTo locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. The graph below\u00a0illustrates these ideas for a local maximum.\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\/10\/18194754\/CNX_Precalc_Figure_01_03_0052.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum.\" width=\"487\" height=\"295\" \/> Definition of a local maximum.[\/caption]\r\n\r\nThese observations lead us to a formal definition of local extrema.\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Increasing and Decreasing Intervals on a Graph<\/h3>\r\nGiven the function [latex]p\\left(t\\right)[\/latex] in the graph below, identify the intervals on which the function appears to be increasing.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194756\/CNX_Precalc_Figure_01_03_0062.jpg\" alt=\"Graph of a polynomial. As x gets large in the negative direction, the outputs of the function get large in the positive direction. As inputs approach 1, then the function value approaches a minimum of negative one. As x approaches 3, the values increase again and between 3 and 4 decrease one last time. As x gets large in the positive direction, the function values increase without bound.\" width=\"487\" height=\"295\" \/>\r\n[reveal-answer q=\"927495\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"927495\"]\r\n\r\nWe see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from [latex]t=1[\/latex] to [latex]t=3[\/latex] and from [latex]t=4[\/latex] on.\r\n\r\nIn <strong>interval notation<\/strong>, we would say the function appears to be increasing on the interval [latex](1,3)[\/latex] and the interval [latex]\\left(4,\\infty \\right)[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nNotice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at [latex]t=1[\/latex] , [latex]t=3[\/latex] , and [latex]t=4[\/latex] . These points are the local extrema (two minima and a maximum).\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]4084[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding Local Maxima and Minima from a Graph<\/h3>\r\nFor the function [latex]f[\/latex] whose graph is shown below, find all local maxima and minima.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194804\/CNX_Precalc_Figure_01_03_0112.jpg\" alt=\"Graph of a polynomial. The line curves down to x = negative 2 and up to x = 1.\" width=\"487\" height=\"368\" \/>\r\n\r\n[reveal-answer q=\"523190\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"523190\"]\r\n\r\nObserve the graph of [latex]f[\/latex]. The graph attains a local maximum at [latex]x=1[\/latex] because it is the highest point in an open interval around [latex]x=1[\/latex]. The local maximum is the [latex]y[\/latex] -coordinate at [latex]x=1[\/latex], which is [latex]2[\/latex].\r\n\r\nThe graph attains a local minimum at [latex]\\text{ }x=-1\\text{ }[\/latex] because it is the lowest point in an open interval around [latex]x=-1[\/latex]. The local minimum is the <em>y<\/em>-coordinate at [latex]x=-1[\/latex], which is [latex]-2[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]32571[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Use a Sign Chart to Determine Where a Function is Positive or Negative<\/h2>\r\n<p id=\"fs-id1169149308017\" class=\" \">A sign chart can be created to determine where a function equation is positive or negative. Equations most often change signs only where the equation is equal to zero. Graphically, where an equation is equal to zero is an [latex]x[\/latex]-intercept. It is at these [latex]x[\/latex]-intercepts that an equation can go from being above the [latex]x[\/latex]-axis (positive) to being below the [latex]x[\/latex]-axis (negative) or vice-versa.<\/p>\r\n<p id=\"fs-id1169148867090\" class=\" \">To create a sign chart (number line), we will use the\u00a0<span id=\"term279\" class=\"no-emphasis\" data-type=\"term\">critical points (where a given equation is equal to zero or not defined)<\/span>\u00a0to divide the number line into intervals and then determine whether the function equation is positive or negative in the interval. We then determine where the function equation is positive or negative.<\/p>\r\n\r\n<div id=\"fs-id1169149024679\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example:\u00a0Determining Where a Function is Positive or Negative<\/h3>\r\nSolve [latex]x^2-x-12 \\ge{0}[\/latex] algebraically. Write the solution in interval notation.\r\n\r\n[reveal-answer q=\"3388996\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"3388996\"]\r\n\r\nWe are looking for intervals where the quadratic function is equal to 0 or positive. Since the function is quadratic, we set the inequality to 0 and find where the function equation is equal to 0. We then proceed with making the sign chart to find the desired intervals.\r\n\r\n<img id=\"21\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/fa59912abde07457c50033eebd5ed38122f3a1c1\" alt=\"This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n<img id=\"22\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/69b56648309576ead6c1e04dbf91c827df5c3d30\" alt=\"Step 2 is to determine the critical points -- the solutions to the related quadratic equation. To do this, change the inequality sign to an equal sign and then solve the equation. x squared minus x minus 12 equals 0 factors to the quantity x plus 3 times the quantity x minus 4 equals 0. Then, x plus 3 equals 0 and x minus 4 equals 0 to give x equals negative 3 and x equals 4.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n<img id=\"23\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/8a3a261a99269dcbc01f072a3e98f9e5d71f927e\" alt=\"Step 3 is to use the critical points to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n<img id=\"24\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/520308630678d93ff8a2a8e1d405f3464eecf6db\" alt=\"Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8.\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n<img id=\"25\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/6e378cd43805e2a61b896287080d1fc4f3b4fd61\" alt=\"For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3] in union with [4, infinity).\" data-media-type=\"image\/jpeg\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]x^2+2x-8 \\ge{0}[\/latex] algebraically. Write the solution in interval notation.\r\n\r\n[reveal-answer q=\"4466771\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"4466771\"]\r\n\r\n[latex]\\left(- \\infty ,-4 ] \\cup [2, \\infty \\right) [\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nSolve [latex]x^2-2x-15 \\le{0}[\/latex] algebraically. Write the solution in interval notation.\r\n\r\n[reveal-answer q=\"8877002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"8877002\"]\r\n\r\n[latex][-3,5][\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the previous example, since the expression\u00a0[latex]x^2-x-12[\/latex]\u00a0factors nicely, we can find the sign of each of the factors, and then the sign of the product. Our number line would look like this:\r\n\r\n<img id=\"26\" class=\"aligncenter\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/c848cf38f47d6069a732d9361b9e81d451093b61\" alt=\"The figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.\" data-media-type=\"image\/jpeg\" \/>\r\n<p id=\"fs-id1169146665788\" class=\" \">The result is the same as we found using the other method.<\/p>\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify extrema and increasing and decreasing intervals<\/li>\n<li>Use a sign chart to determine where a function is positive or negative<\/li>\n<\/ul>\n<\/div>\n<p>In the Derivatives and the Shape of a Graph section, we will analyze graphs of functions using calculus. Here we will review identifying extrema and increasing and decreasing intervals of a function along with how to create sign charts to determine where a certain equation is positive or negative.<\/p>\n<h2>Identify Extrema and Increasing and Decreasing Intervals<\/h2>\n<p>We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on an interval if the function values decrease as the input values increase over that interval. The graph below\u00a0shows examples of increasing and decreasing intervals on a function.<\/p>\n<div style=\"width: 497px\" 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\/10\/18194750\/CNX_Precalc_Figure_01_03_0042.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum.\" width=\"487\" height=\"518\" \/><\/p>\n<p class=\"wp-caption-text\">The function [latex]f\\left(x\\right)={x}^{3}-12x[\/latex] is increasing on [latex]\\left(-\\infty \\text{,}-\\text{2}\\right){{\\cup }^{\\text{ }}}^{\\text{ }}\\left(2,\\infty \\right)[\/latex] and is decreasing on [latex]\\left(-2\\text{,}2\\right)[\/latex].<\/p>\n<\/div>\n<p>This video further explains how to find where a function is increasing or decreasing.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Determine Where a Function is Increasing and Decreasing\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/78b4HOMVcKM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>A value of the output where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a <strong>local maximum<\/strong>. If a function has more than one, we say it has local maxima. Similarly, a value of the output where a function changes from decreasing to increasing as the input variable increases is called a <strong>local minimum<\/strong>. The plural form is &#8220;local minima.&#8221; Together, local maxima and minima are called <strong>local extrema<\/strong>, or local extreme values, of the function. Often, the term <em>local<\/em> is replaced by the term <em>relative<\/em>.<\/p>\n<p>For the function below, the local maximum is 16, and it occurs at [latex]x=-2[\/latex]. The local minimum is [latex]-16[\/latex] and it occurs at [latex]x=2[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194752\/CNX_Precalc_Figure_01_03_0142.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum and minimum. The local maximum is 16 and occurs at x = negative 2. This is the point negative 2, 16. The local minimum is negative 16 and occurs at x = 2. This is the point 2, negative 16.\" width=\"731\" height=\"467\" \/><\/p>\n<p>To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. The graph below\u00a0illustrates these ideas for a local maximum.<\/p>\n<div style=\"width: 497px\" 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\/10\/18194754\/CNX_Precalc_Figure_01_03_0052.jpg\" alt=\"Graph of a polynomial that shows the increasing and decreasing intervals and local maximum.\" width=\"487\" height=\"295\" \/><\/p>\n<p class=\"wp-caption-text\">Definition of a local maximum.<\/p>\n<\/div>\n<p>These observations lead us to a formal definition of local extrema.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Increasing and Decreasing Intervals on a Graph<\/h3>\n<p>Given the function [latex]p\\left(t\\right)[\/latex] in the graph below, identify the intervals on which the function appears to be increasing.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194756\/CNX_Precalc_Figure_01_03_0062.jpg\" alt=\"Graph of a polynomial. As x gets large in the negative direction, the outputs of the function get large in the positive direction. As inputs approach 1, then the function value approaches a minimum of negative one. As x approaches 3, the values increase again and between 3 and 4 decrease one last time. As x gets large in the positive direction, the function values increase without bound.\" width=\"487\" height=\"295\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q927495\">Show Solution<\/span><\/p>\n<div id=\"q927495\" class=\"hidden-answer\" style=\"display: none\">\n<p>We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from [latex]t=1[\/latex] to [latex]t=3[\/latex] and from [latex]t=4[\/latex] on.<\/p>\n<p>In <strong>interval notation<\/strong>, we would say the function appears to be increasing on the interval [latex](1,3)[\/latex] and the interval [latex]\\left(4,\\infty \\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at [latex]t=1[\/latex] , [latex]t=3[\/latex] , and [latex]t=4[\/latex] . These points are the local extrema (two minima and a maximum).<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm4084\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=4084&theme=oea&iframe_resize_id=ohm4084&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding Local Maxima and Minima from a Graph<\/h3>\n<p>For the function [latex]f[\/latex] whose graph is shown below, find all local maxima and minima.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18194804\/CNX_Precalc_Figure_01_03_0112.jpg\" alt=\"Graph of a polynomial. The line curves down to x = negative 2 and up to x = 1.\" width=\"487\" height=\"368\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q523190\">Show Solution<\/span><\/p>\n<div id=\"q523190\" class=\"hidden-answer\" style=\"display: none\">\n<p>Observe the graph of [latex]f[\/latex]. The graph attains a local maximum at [latex]x=1[\/latex] because it is the highest point in an open interval around [latex]x=1[\/latex]. The local maximum is the [latex]y[\/latex] -coordinate at [latex]x=1[\/latex], which is [latex]2[\/latex].<\/p>\n<p>The graph attains a local minimum at [latex]\\text{ }x=-1\\text{ }[\/latex] because it is the lowest point in an open interval around [latex]x=-1[\/latex]. The local minimum is the <em>y<\/em>-coordinate at [latex]x=-1[\/latex], which is [latex]-2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm32571\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=32571&theme=oea&iframe_resize_id=ohm32571&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Use a Sign Chart to Determine Where a Function is Positive or Negative<\/h2>\n<p id=\"fs-id1169149308017\" class=\"\">A sign chart can be created to determine where a function equation is positive or negative. Equations most often change signs only where the equation is equal to zero. Graphically, where an equation is equal to zero is an [latex]x[\/latex]-intercept. It is at these [latex]x[\/latex]-intercepts that an equation can go from being above the [latex]x[\/latex]-axis (positive) to being below the [latex]x[\/latex]-axis (negative) or vice-versa.<\/p>\n<p id=\"fs-id1169148867090\" class=\"\">To create a sign chart (number line), we will use the\u00a0<span id=\"term279\" class=\"no-emphasis\" data-type=\"term\">critical points (where a given equation is equal to zero or not defined)<\/span>\u00a0to divide the number line into intervals and then determine whether the function equation is positive or negative in the interval. We then determine where the function equation is positive or negative.<\/p>\n<div id=\"fs-id1169149024679\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example:\u00a0Determining Where a Function is Positive or Negative<\/h3>\n<p>Solve [latex]x^2-x-12 \\ge{0}[\/latex] algebraically. Write the solution in interval notation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q3388996\">Show Solution<\/span><\/p>\n<div id=\"q3388996\" class=\"hidden-answer\" style=\"display: none\">\n<p>We are looking for intervals where the quadratic function is equal to 0 or positive. Since the function is quadratic, we set the inequality to 0 and find where the function equation is equal to 0. We then proceed with making the sign chart to find the desired intervals.<\/p>\n<p><img decoding=\"async\" id=\"21\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/fa59912abde07457c50033eebd5ed38122f3a1c1\" alt=\"This figure is a table giving the instructions for solving x squared minus x minus 12 greater than or equal to 0 algebraically. It consists of 3 columns where the instructions are given in the first column, the explanation in the second, and the work in the third. Step 1 is to write the quadratic inequality in standard form. The quadratic inequality in already in standard form, so x squared minus x minus 12 greater than or equal to 0.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p><img decoding=\"async\" id=\"22\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/69b56648309576ead6c1e04dbf91c827df5c3d30\" alt=\"Step 2 is to determine the critical points -- the solutions to the related quadratic equation. To do this, change the inequality sign to an equal sign and then solve the equation. x squared minus x minus 12 equals 0 factors to the quantity x plus 3 times the quantity x minus 4 equals 0. Then, x plus 3 equals 0 and x minus 4 equals 0 to give x equals negative 3 and x equals 4.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p><img decoding=\"async\" id=\"23\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/8a3a261a99269dcbc01f072a3e98f9e5d71f927e\" alt=\"Step 3 is to use the critical points to divide the number line into intervals. Use negative 3 and 4 to divide the number line into intervals. A number line is shown that includes from left to right the values of negative 3, 0, and 4, with dotted lines on negative 3 and 4.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p><img decoding=\"async\" id=\"24\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/520308630678d93ff8a2a8e1d405f3464eecf6db\" alt=\"Step 4 says above the number line show the sign of each quadratic expression using test points from each interval substituted into the original inequality. X equals negative 5, x equals 0, and x equals 5 are chosen to test. The expression negative x squared minus x minus 12 is given with negative 5 squared minus negative 5 minus 12 underneath, which gives 18. The expression negative x squared minus x minus 12 is given with 0 squared minus 0 minus 12 underneath, which gives 12. The expression negative x squared minus x minus 12 is given with 5 squared minus 5 minus 12 underneath, which gives 8.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p><img decoding=\"async\" id=\"25\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/6e378cd43805e2a61b896287080d1fc4f3b4fd61\" alt=\"For Step 5, determine the intervals where the inequality is correct. Write the solution in interval notation. x squared minus x minus 12 greater than or equal to 0 is shown. The inequality is positive in the first and last intervals and equals 0 at the points negative 4, 3 . The solution, in interval notation, is (negative infinity, negative 3] in union with [4, infinity).\" data-media-type=\"image\/jpeg\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]x^2+2x-8 \\ge{0}[\/latex] algebraically. Write the solution in interval notation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q4466771\">Show Solution<\/span><\/p>\n<div id=\"q4466771\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(- \\infty ,-4 ] \\cup [2, \\infty \\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Solve [latex]x^2-2x-15 \\le{0}[\/latex] algebraically. Write the solution in interval notation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q8877002\">Show Solution<\/span><\/p>\n<div id=\"q8877002\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex][-3,5][\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In the previous example, since the expression\u00a0[latex]x^2-x-12[\/latex]\u00a0factors nicely, we can find the sign of each of the factors, and then the sign of the product. Our number line would look like this:<\/p>\n<p><img decoding=\"async\" id=\"26\" class=\"aligncenter\" src=\"https:\/\/openstax.org\/apps\/archive\/20210421.141058\/resources\/c848cf38f47d6069a732d9361b9e81d451093b61\" alt=\"The figure shows the expression x squared minus x minus 12 factored to the quantity of x plus 3 times the quantity of x minus 4. The image shows a number line showing dotted lines on negative 3 and 4. It shows the signs of the quantity x plus 3 to be negative, positive, positive, and the signs of the quantity x minus 4 to be negative, negative, positive. Under the number line, it shows the quantity x plus 3 times the quantity x minus 4 with the signs positive, negative, positive.\" data-media-type=\"image\/jpeg\" \/><\/p>\n<p id=\"fs-id1169146665788\" class=\"\">The result is the same as we found using the other method.<\/p>\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-3896\">\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>Modification and Revision . <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <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 class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <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":17533,"menu_order":4,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Modification and Revision \",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3896","chapter","type-chapter","status-publish","hentry"],"part":3090,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3896","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":3,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3896\/revisions"}],"predecessor-version":[{"id":4461,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3896\/revisions\/4461"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/3090"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3896\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3896"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3896"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3896"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3896"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}