{"id":2006,"date":"2016-06-30T22:19:09","date_gmt":"2016-06-30T22:19:09","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/intermediatealgebra\/?post_type=chapter&#038;p=2006"},"modified":"2018-05-17T00:45:16","modified_gmt":"2018-05-17T00:45:16","slug":"read-write-the-equation-of-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/odessa-coreq-collegealgebra\/chapter\/read-write-the-equation-of-a-linear-function\/","title":{"raw":"Equations of Linear Functions","rendered":"Equations of Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul>\r\n \t<li>Write the equation of a line\r\n<ul>\r\n \t<li>Define the equation of a line given two points<\/li>\r\n \t<li>Define the equation of a line given the slope and a point<\/li>\r\n \t<li>Write a linear function in standard form<\/li>\r\n \t<li>Identify the equations for vertical and horizontal lines<\/li>\r\n \t<li>Given a graph, write the equation of a linear function<\/li>\r\n \t<li>Given two function values and corresponding inputs, write the equation of the linear function passing through them<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t<li>Model an Application With a Linear Function\r\n<ul>\r\n \t<li>Define initial value<\/li>\r\n \t<li>Write a linear function given an initial value and a rate of change<\/li>\r\n \t<li>Write a linear function given data in a table<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<\/div>\r\nIn this section, we will learn how to write equations for linear functions given different pieces of information, including two points on the line and the graph of the line. \u00a0We will also identify the equations of horizontal and vertical lines and write linear equations from written information.\r\n<h2>Point-Slope Formula<\/h2>\r\nWe have seen that we can define the slope of a line given two points on the line, and use that information along with the y-intercept to graph the line. \u00a0If you don't know the y-intercept, or the equation for the line you can use two points to define\u00a0the equation of the line using the point-slope formula.\r\n<p style=\"text-align: center\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nThis is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.\r\n<div class=\"textbox shaded\">\r\n<h3>The Point-Slope Formula<\/h3>\r\nGiven one point and the slope, the point-slope formula will lead to the equation of a line:\r\n<p style=\"text-align: center\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n\r\n<\/div>\r\nIn our first example, we will start with the slope, then we will show how to find the equation of a line without being given the slope.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"524449\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"524449\"]\r\n\r\nUsing the point-slope formula, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/p>\r\nNote that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn our next example we will start with two points and define the equation of the line that passes through them.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"249539\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"249539\"]\r\n\r\nFirst, we calculate the slope using the slope formula and two points.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}m\\hfill=\\frac{-3 - 4}{0 - 3}\\hfill \\\\ \\hfill =\\frac{-7}{-3}\\hfill \\\\ \\hfill =\\frac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nNext, we use the point-slope formula with the slope of [latex]\\frac{7}{3}[\/latex], and either point. Let\u2019s pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y - 4=\\frac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\frac{7}{3}x - 7\\hfill&amp;\\text{Distribute the }\\frac{7}{3}.\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\r\nIn slope-intercept form, the equation is written as [latex]y=\\frac{7}{3}x - 3[\/latex].\r\n\r\nTo prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\frac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\frac{7}{3}x\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\r\nWe see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe following video examples shows how to write the equation for a line given it's slope and a point on the line.\r\n\r\nhttps:\/\/youtu.be\/vut5b2fRQQ0\r\n<h2>Standard Form of a Line<\/h2>\r\nAnother way that we can represent the equation of a line is in <strong>standard form<\/strong>. Standard form is given as\r\n<p style=\"text-align: center\">[latex]Ax+By=C[\/latex]<\/p>\r\nwhere [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x- <\/em>and <em>y-<\/em>terms are on one side of the equal sign and the constant term is on the other side.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the equation of the line with [latex]m=-6[\/latex] and passing through the point [latex]\\left(\\frac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.\r\n\r\n[reveal-answer q=\"526704\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"526704\"]\r\n\r\nWe begin using the point-slope formula.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\frac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nFrom here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\frac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/p>\r\nThis equation is now written in standard form.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Vertical and Horizontal Lines<\/h2>\r\nThe equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a <strong>vertical line<\/strong> is given as\r\n<p style=\"text-align: center\">[latex]x=c[\/latex]<\/p>\r\nwhere <em>c <\/em>is a constant. The slope of a vertical line is undefined, and regardless of the <em>y-<\/em>value of any point on the line, the <em>x-<\/em>coordinate of the point will be <em>c<\/em>.\r\n\r\nSuppose that we want to find the equation of a line containing the following points: [latex]\\left(-3,-5\\right),\\left(-3,1\\right),\\left(-3,3\\right)[\/latex], and [latex]\\left(-3,5\\right)[\/latex]. First, we will find the slope.\r\n<p style=\"text-align: center\">[latex]m=\\frac{5 - 3}{-3-\\left(-3\\right)}=\\frac{2}{0}[\/latex]<\/p>\r\nZero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the <em>x-<\/em>coordinates are the same and we find a vertical line through [latex]x=-3[\/latex].\r\n\r\nThe equation of a <strong>horizontal line<\/strong> is given as\r\n<p style=\"text-align: center\">[latex]y=c[\/latex]<\/p>\r\nwhere <em>c <\/em>is a constant. The slope of a horizontal line is zero, and for any <em>x-<\/em>value of a point on the line, the <em>y-<\/em>coordinate will be <em>c<\/em>.\r\n\r\nSuppose we want to find the equation of a line that contains the following set of points: [latex]\\left(-2,-2\\right),\\left(0,-2\\right),\\left(3,-2\\right)[\/latex], and [latex]\\left(5,-2\\right)[\/latex]. We can use the point-slope formula. First, we find the slope using any two points on the line.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}m=\\frac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ =\\frac{0}{2}\\hfill \\\\ =0\\hfill \\end{array}[\/latex]<\/p>\r\nUse any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in the formula, or use the <em>y<\/em>-intercept.\r\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/p>\r\nThe graph is a horizontal line through [latex]y=-2[\/latex]. Notice that all of the <em>y-<\/em>coordinates are the same.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/> The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.[\/caption]\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nFind the equation of the line passing through the given points: [latex]\\left(1,-3\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].\r\n\r\n[reveal-answer q=\"346281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"346281\"]\r\n\r\nThe <em>x-<\/em>coordinate of both points is 1. Therefore, we have a vertical line, [latex]x=1[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nOur last video show another example of writing the equation of a line given two points on the line.\r\n\r\nhttps:\/\/youtu.be\/ndRpJxdmZJI\r\n<h2><\/h2>\r\n<p id=\"fs-id1165137911671\">Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function <em>f<\/em>\u00a0below.<span id=\"fs-id1165135182766\">\r\n<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"347\" \/> <b>The function f passing through the points (0.7) and \u00a0(4,4) with a negative slope.<\/b>[\/caption]\r\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let\u2019s choose (0, 7)\u00a0and (4, 4). We can use these points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{4 - 7}{4 - 0}\\hfill \\\\ \\text{ }=-\\frac{3}{4} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\frac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in the slope-intercept form, we would find<\/p>\r\n\r\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}y - 4=-\\frac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\frac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\frac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/> <b>Rewrite the equation in slope intercept form.<\/b>[\/caption]\r\n<p id=\"fs-id1165137769983\">If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is 7. Therefore, <em>b<\/em> = 7.\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m<\/em>\u00a0so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into the slope-intercept form of a line.<span id=\"fs-id1165137548391\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137705273\">So the function is [latex]f\\left(x\\right)=-\\frac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\frac{3}{4}x+7[\/latex].<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1165137824881\">Given the graph of a linear function, write an equation to represent the function.<\/h3>\r\n<ol id=\"fs-id1165137803240\">\r\n \t<li>Identify two points on the line.<\/li>\r\n \t<li>Use the two points to calculate the slope.<\/li>\r\n \t<li>Determine where the line crosses the <em>y<\/em>-axis to identify the <em>y<\/em>-intercept by visual inspection.<\/li>\r\n \t<li>Substitute the slope and <em>y<\/em>-intercept into the slope-intercept form of a line equation.<\/li>\r\n<\/ol>\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWrite an equation for a linear function given a graph of <em>f<\/em>\u00a0shown below.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/>\r\n[reveal-answer q=\"728685\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"728685\"]\r\n<p id=\"fs-id1165135536538\">Identify two points on the line, such as (0, 2) and (\u20132, \u20134). Use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{-4 - 2}{-2 - 0}\\hfill \\\\ \\text{ }=\\frac{-6}{-2}\\hfill \\\\ \\text{ }=3\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n\r\n<div id=\"fs-id1165137460061\" class=\"equation unnumbered\">[latex]\\begin{array}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ \\text{ }y=3x+2\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h3>Analysis of the Solution<\/h3>\r\nThis makes sense because we can see from the graph below\u00a0that the line crosses the y-axis at the point (0, 2), which is the <em>y<\/em>-intercept, so <em>b<\/em>\u00a0= 2.\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201029\/CNX_Precalc_Figure_02_01_008b2.jpg\" alt=\"Graph of an increasing line with points at (0, 2) and (-2, -4).\" width=\"369\" height=\"378\" \/>\r\n\r\nIn the following video we show an example of how to write the equation of a line given it's graph.\r\n\r\nhttps:\/\/youtu.be\/mmWf_oLTNSQ\r\n<div id=\"fs-id1165137767515\" class=\"commentary\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nIf <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.\r\n[reveal-answer q=\"834495\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"834495\"]\r\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\r\n\r\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\r\n\r\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ \\text{ }=\\frac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\r\n\r\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\frac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n\r\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}y+2=\\frac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\frac{3}{5}x-\\frac{9}{5}\\hfill \\\\ \\text{ }y=\\frac{3}{5}x-\\frac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the video we show another example of how to write a linear function given two points written with function notation.\r\n\r\nhttps:\/\/youtu.be\/TSIcryAtCmY\r\n\r\nThe following applet graphs a line with slope [latex]m[\/latex] going through the point [latex](x_1,y_1)[\/latex], thus having a point-slope equation of [latex]y-y_1=m(x-x_1)[\/latex]. The label on the graph, displays the equation of the line in slope-intercept format.\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/cipa2nfcce\r\n\r\nIn the next applet you can see the graph of a line going through two points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex].\r\n\r\nhttps:\/\/www.desmos.com\/calculator\/j8jn4jsmg3\r\n\r\n&nbsp;\r\n<h2>Model an Application With a Linear Function<\/h2>\r\n<p id=\"fs-id1165137594074\">In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 id=\"fs-id1165137404879\">Given a linear function <em>f<\/em>\u00a0and the initial value and rate of change, evaluate <em>f<\/em>(<em>c<\/em>).<\/h3>\r\n<ol id=\"fs-id1165137660790\">\r\n \t<li>Determine the initial value and the rate of change (slope).<\/li>\r\n \t<li>Substitute the values into [latex]f\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t<li>Evaluate the function at [latex]x=c[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\nInitial value is a term that is typically used in applications of functions. \u00a0It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the y-intercept. Here are some characteristics of initial value:\r\n<ul>\r\n \t<li>The point [latex](0,y)[\/latex] is often the initial value of a linear function<\/li>\r\n \t<li>The y value of the initial value comes from b in the slope intercept form of a linear function,\u00a0[latex]f\\left(x\\right)=mx+b[\/latex]<\/li>\r\n \t<li>The initial value can be found by solving for b, or substituting 0 for x in a linear function.<\/li>\r\n<\/ul>\r\nIn our first example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection, and write an equation that represents the number of songs in the collection for any number of months, t.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nMarcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, <em>N<\/em>, in his collection as a function of time, <em>t<\/em>, the number of months. How many songs will he own in a year?\r\n[reveal-answer q=\"14550\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"14550\"]\r\n<p id=\"fs-id1165135411394\">The initial value for this function is 200 because he currently owns 200 songs, so <i>N<\/i>(0) = 200, which means that <em>b<\/em> = 200.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201031\/CNX_Precalc_Figure_02_01_0102.jpg\" alt=\"\" width=\"487\" height=\"131\" \/> <b>Figure 12<\/b>[\/caption]\r\n<p id=\"fs-id1165137738190\">The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that <em>m<\/em> = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.<span id=\"fs-id1165137417445\">\r\n<\/span><\/p>\r\n<p id=\"fs-id1165137810258\">We can write the formula [latex]N\\left(t\\right)=15t+200[\/latex].<\/p>\r\n<p id=\"fs-id1165137454711\">With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at <em>t<\/em> = 12.<\/p>\r\n\r\n<div id=\"fs-id1165137462736\" class=\"equation unnumbered\">[latex]\\begin{array}N\\left(12\\right)=15\\left(12\\right)+200\\hfill \\\\ \\text{ }=180+200\\hfill \\\\ \\text{ }=380\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137694205\">Marcus will have 380 songs in 12 months.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_02_01_10\" class=\"example\">\r\n<div id=\"fs-id1165137836747\" class=\"exercise\">\r\n<h2>Analysis of the Solution<\/h2>\r\n<p id=\"fs-id1165134065131\">Notice that <em>N<\/em> is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\r\nIn our next example, we will show that you can write the equation for a linear function given two data points. \u00a0In this case, Ilya's weekly income depends on the number of insurance policies he sells. \u00a0We are given his income for two different weeks and the number of policies sold. \u00a0We first find the rate of change and then solve for the initial value.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nWorking as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya\u2019s weekly income, <i>I<\/i>, depends on the number of new policies, <em>n<\/em>, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for <em>I<\/em>(<em>n<\/em>), and interpret the meaning of the components of the equation.\r\n[reveal-answer q=\"249315\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"249315\"]\r\n<p id=\"fs-id1165135169134\">The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding the rate of change.<\/p>\r\n\r\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}m=\\frac{920 - 760}{5 - 3}\\hfill \\\\ \\text{ }=\\frac{$160}{\\text{2 policies}}\\hfill \\\\ \\text{ }=$80\\text{ per policy}\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137855034\">Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.<\/p>\r\n<p id=\"fs-id1165137855040\">We can then solve for the initial value.<\/p>\r\n\r\n<div id=\"fs-id1165135484088\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}\\text{ }I\\left(n\\right)=80n+b\\hfill &amp; \\hfill \\\\ \\text{ }760=80\\left(3\\right)+b\\hfill &amp; \\text{when }n=3, I\\left(3\\right)=760\\hfill \\\\ 760 - 80\\left(3\\right)=b\\hfill &amp; \\hfill \\\\ \\text{ }520=b\\hfill &amp; \\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137400716\">The value of <em>b<\/em>\u00a0is the starting value for the function and represents Ilya\u2019s income when\u00a0<em>n<\/em> = 0, or when no new policies are sold. We can interpret this as Ilya\u2019s base salary for the week, which does not depend upon the number of policies sold.<\/p>\r\n<p id=\"fs-id1165135203653\">We can now write the final equation.<\/p>\r\n\r\n<div id=\"fs-id1165137506449\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\r\n<p id=\"fs-id1165137655487\">Our final interpretation is that Ilya\u2019s base salary is $520 per week and he earns an additional $80 commission for each policy sold.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Analysis of the Solution<\/h2>\r\nWe used units to help us verify that we were calculating the rate correctly. It makes sense to speak in terms of the price per policy. To calculate the initial value, we solved for b by substituting values from one of the points we were given for n and I.\r\n\r\nIn the following video example we show how to identify the initial value, slope and equation for a linear function.\r\n\r\nhttps:\/\/youtu.be\/JMQSdRFJ1S4\r\n\r\nWe will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function <em>C\u00a0<\/em>where <i>C<\/i>(<em>x<\/em>)\u00a0is the cost for <em>x<\/em>\u00a0items produced in a given month.\r\n[reveal-answer q=\"625547\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"625547\"]The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by [latex]C\\left(x\\right)=1250+37.5x[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Analysis of the Solution<\/h2>\r\n<div id=\"fs-id1165137767515\" class=\"commentary\">\r\n<p id=\"fs-id1165135511326\">It is important to note that we are writing a function based on monthly costs, so the initial cost will be $1,250 becuase Ben has to pay that amount monthly for rent. If Ben produces 100 items in a month, his monthly cost is represented by<\/p>\r\n\r\n<div id=\"fs-id1165137417815\" class=\"equation unnumbered\">[latex]\\begin{array}C\\left(100\\right)=1250+37.5\\left(100\\right)\\hfill\\text{ } \\\\ =5000\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165137862645\">So his monthly cost would be $5,000.<\/p>\r\nThe following video example show how to write a linear function that represents how many miles you can travel in a rental car for a fixed amount.\r\n\r\nhttps:\/\/youtu.be\/H8KR3w2nXqs\r\n\r\n<\/div>\r\nIn the next example we will take data that is in tabular (table) form to write an equation that describes the rate of change in the numbers of a rat population.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\n<p id=\"fs-id1165135161325\">The table below relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.<\/p>\r\n\r\n<table id=\"Table_02_01_02\" summary=\"Two rows and five columns. The first row is labeled, 'w, the numers of weeks'. The second row is labeled is labeled, 'P(w), number of rats'. Reading the remaining rows as ordered pairs (i.e., (w, P(w)), we have the following values: (0, 1000), (2, 1080), (4, 1160), and (6, 1240).\">\r\n<tbody>\r\n<tr>\r\n<td><strong><em>w<\/em>, number of weeks<\/strong><\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>P(w)<\/em>, number of rats<\/strong><\/td>\r\n<td>1000<\/td>\r\n<td>1080<\/td>\r\n<td>1160<\/td>\r\n<td>1240<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"780371\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"780371\"]\r\n<p id=\"fs-id1165137530990\">We can see from the table that the initial value for the number of rats is 1000, so <em>b<\/em> = 1000.<\/p>\r\n<p id=\"fs-id1165137935601\">Rather than solving for <em>m<\/em>, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.<\/p>\r\n\r\n<div id=\"fs-id1165137737900\" class=\"equation unnumbered\">[latex]P\\left(w\\right)=40w+1000[\/latex]<\/div>\r\n<p id=\"fs-id1165137465125\">If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240)<\/p>\r\n\r\n<div id=\"fs-id1165137627069\" class=\"equation unnumbered\">[latex]\\begin{array}m=\\frac{1240 - 1080}{6 - 2}\\hfill \\\\ \\text{ }=\\frac{160}{4}\\hfill \\\\ \\text{ }=40\\hfill \\end{array}[\/latex]<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Think About It<\/h3>\r\nIs the initial value always provided in a table of values like the table in\u00a0the previous example? Write your ideas in the textbox below before you look at the answer.\r\n\r\nIf your answer is no, give a description of how you would find the initial value.\r\n\r\n[practice-area rows=\"2\"][\/practice-area]\r\n[reveal-answer q=\"298919\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"298919\"]\r\n\r\n<em>No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into [latex]f\\left(x\\right)=mx+b[\/latex], and solve for <\/em>b<em>.<\/em>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Summary<\/h2>\r\n<h3>Slope of a line<\/h3>\r\n<\/div>\r\n<\/div>\r\n<ul>\r\n \t<li>The <strong>slope<\/strong> of a line\u00a0indicates the direction in which a line slants as well as its steepness. Slope is\u00a0defined algebraically as:[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/li>\r\n \t<li>Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<h3>Equations of Lines<\/h3>\r\n<ul>\r\n \t<li>Standard form of a line is given as\u00a0[latex]Ax+By=C[\/latex].<\/li>\r\n \t<li>The equation of a <strong>vertical line<\/strong> is given as\u00a0[latex]x=c[\/latex]<\/li>\r\n \t<li>The equation of a <strong>horizontal line<\/strong> is given as\u00a0[latex]y=c[\/latex]<\/li>\r\n \t<li>Given a graph you can write the equation of the line it represents<\/li>\r\n \t<li>This method only works for graphs that have points that are easy to verify by visual inspection<\/li>\r\n<\/ul>\r\n<h3>Modeling Applications With Linear Functions<\/h3>\r\n<div id=\"Example_02_01_10\" class=\"example\">\r\n<div id=\"fs-id1165137836747\" class=\"exercise\">\r\n<ul>\r\n \t<li>Sometimes we are given an initial value, and sometimes we have to solve for it<\/li>\r\n \t<li>Using units can help you verify that you have calculated slope correctly<\/li>\r\n \t<li>We can write the equation for a line given a slope and a data point, or from a table of data<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<h2><\/h2>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul>\n<li>Write the equation of a line\n<ul>\n<li>Define the equation of a line given two points<\/li>\n<li>Define the equation of a line given the slope and a point<\/li>\n<li>Write a linear function in standard form<\/li>\n<li>Identify the equations for vertical and horizontal lines<\/li>\n<li>Given a graph, write the equation of a linear function<\/li>\n<li>Given two function values and corresponding inputs, write the equation of the linear function passing through them<\/li>\n<\/ul>\n<\/li>\n<li>Model an Application With a Linear Function\n<ul>\n<li>Define initial value<\/li>\n<li>Write a linear function given an initial value and a rate of change<\/li>\n<li>Write a linear function given data in a table<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<p>In this section, we will learn how to write equations for linear functions given different pieces of information, including two points on the line and the graph of the line. \u00a0We will also identify the equations of horizontal and vertical lines and write linear equations from written information.<\/p>\n<h2>Point-Slope Formula<\/h2>\n<p>We have seen that we can define the slope of a line given two points on the line, and use that information along with the y-intercept to graph the line. \u00a0If you don&#8217;t know the y-intercept, or the equation for the line you can use two points to define\u00a0the equation of the line using the point-slope formula.<\/p>\n<p style=\"text-align: center\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.<\/p>\n<div class=\"textbox shaded\">\n<h3>The Point-Slope Formula<\/h3>\n<p>Given one point and the slope, the point-slope formula will lead to the equation of a line:<\/p>\n<p style=\"text-align: center\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<\/div>\n<p>In our first example, we will start with the slope, then we will show how to find the equation of a line without being given the slope.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write the equation of the line with slope [latex]m=-3[\/latex] and passing through the point [latex]\\left(4,8\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q524449\">Show Solution<\/span><\/p>\n<div id=\"q524449\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the point-slope formula, substitute [latex]-3[\/latex] for <em>m <\/em>and the point [latex]\\left(4,8\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 8=-3\\left(x - 4\\right)\\hfill \\\\ y - 8=-3x+12\\hfill \\\\ y=-3x+20\\hfill \\end{array}[\/latex]<\/p>\n<p>Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>In our next example we will start with two points and define the equation of the line that passes through them.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the equation of the line passing through the points [latex]\\left(3,4\\right)[\/latex] and [latex]\\left(0,-3\\right)[\/latex]. Write the final equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q249539\">Show Solution<\/span><\/p>\n<div id=\"q249539\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, we calculate the slope using the slope formula and two points.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}m\\hfill=\\frac{-3 - 4}{0 - 3}\\hfill \\\\ \\hfill =\\frac{-7}{-3}\\hfill \\\\ \\hfill =\\frac{7}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Next, we use the point-slope formula with the slope of [latex]\\frac{7}{3}[\/latex], and either point. Let\u2019s pick the point [latex]\\left(3,4\\right)[\/latex] for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex].<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y - 4=\\frac{7}{3}\\left(x - 3\\right)\\hfill \\\\ y - 4=\\frac{7}{3}x - 7\\hfill&\\text{Distribute the }\\frac{7}{3}.\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\n<p>In slope-intercept form, the equation is written as [latex]y=\\frac{7}{3}x - 3[\/latex].<\/p>\n<p>To prove that either point can be used, let us use the second point [latex]\\left(0,-3\\right)[\/latex] and see if we get the same equation.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-3\\right)=\\frac{7}{3}\\left(x - 0\\right)\\hfill \\\\ y+3=\\frac{7}{3}x\\hfill \\\\ y=\\frac{7}{3}x - 3\\hfill \\end{array}[\/latex]<\/p>\n<p>We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video examples shows how to write the equation for a line given it&#8217;s slope and a point on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/vut5b2fRQQ0?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2>Standard Form of a Line<\/h2>\n<p>Another way that we can represent the equation of a line is in <strong>standard form<\/strong>. Standard form is given as<\/p>\n<p style=\"text-align: center\">[latex]Ax+By=C[\/latex]<\/p>\n<p>where [latex]A[\/latex], [latex]B[\/latex], and [latex]C[\/latex] are integers. The <em>x- <\/em>and <em>y-<\/em>terms are on one side of the equal sign and the constant term is on the other side.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the equation of the line with [latex]m=-6[\/latex] and passing through the point [latex]\\left(\\frac{1}{4},-2\\right)[\/latex]. Write the equation in standard form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q526704\">Show Solution<\/span><\/p>\n<div id=\"q526704\" class=\"hidden-answer\" style=\"display: none\">\n<p>We begin using the point-slope formula.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-2\\right)=-6\\left(x-\\frac{1}{4}\\right)\\hfill \\\\ y+2=-6x+\\frac{3}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}2\\left(y+2\\right)=\\left(-6x+\\frac{3}{2}\\right)2\\hfill \\\\ 2y+4=-12x+3\\hfill \\\\ 12x+2y=-1\\hfill \\end{array}[\/latex]<\/p>\n<p>This equation is now written in standard form.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Vertical and Horizontal Lines<\/h2>\n<p>The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a <strong>vertical line<\/strong> is given as<\/p>\n<p style=\"text-align: center\">[latex]x=c[\/latex]<\/p>\n<p>where <em>c <\/em>is a constant. The slope of a vertical line is undefined, and regardless of the <em>y-<\/em>value of any point on the line, the <em>x-<\/em>coordinate of the point will be <em>c<\/em>.<\/p>\n<p>Suppose that we want to find the equation of a line containing the following points: [latex]\\left(-3,-5\\right),\\left(-3,1\\right),\\left(-3,3\\right)[\/latex], and [latex]\\left(-3,5\\right)[\/latex]. First, we will find the slope.<\/p>\n<p style=\"text-align: center\">[latex]m=\\frac{5 - 3}{-3-\\left(-3\\right)}=\\frac{2}{0}[\/latex]<\/p>\n<p>Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the <em>x-<\/em>coordinates are the same and we find a vertical line through [latex]x=-3[\/latex].<\/p>\n<p>The equation of a <strong>horizontal line<\/strong> is given as<\/p>\n<p style=\"text-align: center\">[latex]y=c[\/latex]<\/p>\n<p>where <em>c <\/em>is a constant. The slope of a horizontal line is zero, and for any <em>x-<\/em>value of a point on the line, the <em>y-<\/em>coordinate will be <em>c<\/em>.<\/p>\n<p>Suppose we want to find the equation of a line that contains the following set of points: [latex]\\left(-2,-2\\right),\\left(0,-2\\right),\\left(3,-2\\right)[\/latex], and [latex]\\left(5,-2\\right)[\/latex]. We can use the point-slope formula. First, we find the slope using any two points on the line.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}m=\\frac{-2-\\left(-2\\right)}{0-\\left(-2\\right)}\\hfill \\\\ =\\frac{0}{2}\\hfill \\\\ =0\\hfill \\end{array}[\/latex]<\/p>\n<p>Use any point for [latex]\\left({x}_{1},{y}_{1}\\right)[\/latex] in the formula, or use the <em>y<\/em>-intercept.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{array}{l}y-\\left(-2\\right)=0\\left(x - 3\\right)\\hfill \\\\ y+2=0\\hfill \\\\ y=-2\\hfill \\end{array}[\/latex]<\/p>\n<p>The graph is a horizontal line through [latex]y=-2[\/latex]. Notice that all of the <em>y-<\/em>coordinates are the same.<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/09\/25200322\/CNX_CAT_Figure_02_02_003.jpg\" alt=\"Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4. The function y = negative 2 and the line x = negative 3 are plotted.\" width=\"487\" height=\"367\" \/><\/p>\n<p class=\"wp-caption-text\">The line [latex]x=\u22123[\/latex] is a vertical line. The line [latex]y=\u22122[\/latex] is a horizontal line.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Find the equation of the line passing through the given points: [latex]\\left(1,-3\\right)[\/latex] and [latex]\\left(1,4\\right)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q346281\">Show Solution<\/span><\/p>\n<div id=\"q346281\" class=\"hidden-answer\" style=\"display: none\">\n<p>The <em>x-<\/em>coordinate of both points is 1. Therefore, we have a vertical line, [latex]x=1[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Our last video show another example of writing the equation of a line given two points on the line.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Ex:  Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ndRpJxdmZJI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<h2><\/h2>\n<p id=\"fs-id1165137911671\">Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function <em>f<\/em>\u00a0below.<span id=\"fs-id1165135182766\"><br \/>\n<\/span><\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201025\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\"><b>The function f passing through the points (0.7) and \u00a0(4,4) with a negative slope.<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137657203\">We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let\u2019s choose (0, 7)\u00a0and (4, 4). We can use these points to calculate the slope.<\/p>\n<div id=\"fs-id1165137666429\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{4 - 7}{4 - 0}\\hfill \\\\ \\text{ }=-\\frac{3}{4} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165135176567\">Now we can substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\n<div id=\"fs-id1165135450369\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 4=-\\frac{3}{4}\\left(x - 4\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137634475\">If we want to rewrite the equation in the slope-intercept form, we would find<\/p>\n<div id=\"fs-id1165137768649\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}y - 4=-\\frac{3}{4}\\left(x - 4\\right)\\hfill \\\\ y - 4=-\\frac{3}{4}x+3\\hfill \\\\ \\text{ }y=-\\frac{3}{4}x+7\\hfill \\end{array}[\/latex]<\/div>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201027\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Rewrite the equation in slope intercept form.<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137769983\">If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the <em>y<\/em>-axis when the output value is 7. Therefore, <em>b<\/em> = 7.\u00a0We now have the initial value <em>b<\/em>\u00a0and the slope <em>m<\/em>\u00a0so we can substitute <em>m<\/em>\u00a0and <em>b<\/em>\u00a0into the slope-intercept form of a line.<span id=\"fs-id1165137548391\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137705273\">So the function is [latex]f\\left(x\\right)=-\\frac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\frac{3}{4}x+7[\/latex].<\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1165137824881\">Given the graph of a linear function, write an equation to represent the function.<\/h3>\n<ol id=\"fs-id1165137803240\">\n<li>Identify two points on the line.<\/li>\n<li>Use the two points to calculate the slope.<\/li>\n<li>Determine where the line crosses the <em>y<\/em>-axis to identify the <em>y<\/em>-intercept by visual inspection.<\/li>\n<li>Substitute the slope and <em>y<\/em>-intercept into the slope-intercept form of a line equation.<\/li>\n<\/ol>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Write an equation for a linear function given a graph of <em>f<\/em>\u00a0shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201028\/CNX_Precalc_Figure_02_01_008a2.jpg\" alt=\"Graph of an increasing function with points at (-3, 0) and (0, 1).\" width=\"369\" height=\"378\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q728685\">Show Answer<\/span><\/p>\n<div id=\"q728685\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135536538\">Identify two points on the line, such as (0, 2) and (\u20132, \u20134). Use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137824250\" class=\"equation unnumbered\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{-4 - 2}{-2 - 0}\\hfill \\\\ \\text{ }=\\frac{-6}{-2}\\hfill \\\\ \\text{ }=3\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\n<div id=\"fs-id1165134305424\" class=\"equation unnumbered\">[latex]\\begin{array}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y-\\left(-4\\right)=3\\left(x-\\left(-2\\right)\\right)\\\\ y+4=3\\left(x+2\\right)\\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n<div id=\"fs-id1165137460061\" class=\"equation unnumbered\">[latex]\\begin{array}y+4=3\\left(x+2\\right)\\hfill \\\\ y+4=3x+6\\hfill \\\\ \\text{ }y=3x+2\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<h3>Analysis of the Solution<\/h3>\n<p>This makes sense because we can see from the graph below\u00a0that the line crosses the y-axis at the point (0, 2), which is the <em>y<\/em>-intercept, so <em>b<\/em>\u00a0= 2.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201029\/CNX_Precalc_Figure_02_01_008b2.jpg\" alt=\"Graph of an increasing line with points at (0, 2) and (-2, -4).\" width=\"369\" height=\"378\" \/><\/p>\n<p>In the following video we show an example of how to write the equation of a line given it&#8217;s graph.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Ex 1:  Find the Equation of a Line in Slope Intercept Form Given the Graph of a Line\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/mmWf_oLTNSQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<div id=\"fs-id1165137767515\" class=\"commentary\">\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>If <em>f<\/em>\u00a0is a linear function, with [latex]f\\left(3\\right)=-2[\/latex] , and [latex]f\\left(8\\right)=1[\/latex], find an equation for the function in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q834495\">Show Answer<\/span><\/p>\n<div id=\"q834495\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\n<div id=\"fs-id1165137639761\" class=\"equation unnumbered\">[latex]\\begin{array}f\\left(3\\right)=-2\\to \\left(3,-2\\right)\\hfill \\\\ f\\left(8\\right)=1\\to \\left(8,1\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\n<div id=\"fs-id1165137582561\" class=\"equation unnumbered\">[latex]\\begin{array} m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{ }=\\frac{1-\\left(-2\\right)}{8 - 3}\\hfill \\\\ \\text{ }=\\frac{3}{5}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\n<div id=\"fs-id1165137583929\" class=\"equation unnumbered\">[latex]\\begin{array}\\text{ }y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y-\\left(-2\\right)=\\frac{3}{5}\\left(x - 3\\right)\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n<div id=\"fs-id1165135543458\" class=\"equation unnumbered\">[latex]\\begin{array}y+2=\\frac{3}{5}\\left(x - 3\\right)\\hfill \\\\ y+2=\\frac{3}{5}x-\\frac{9}{5}\\hfill \\\\ \\text{ }y=\\frac{3}{5}x-\\frac{19}{5}\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>In the video we show another example of how to write a linear function given two points written with function notation.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Ex:  Find the Linear Function Given Two Function Values in Function Notation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/TSIcryAtCmY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>The following applet graphs a line with slope [latex]m[\/latex] going through the point [latex](x_1,y_1)[\/latex], thus having a point-slope equation of [latex]y-y_1=m(x-x_1)[\/latex]. The label on the graph, displays the equation of the line in slope-intercept format.<\/p>\n<p>https:\/\/www.desmos.com\/calculator\/cipa2nfcce<\/p>\n<p>In the next applet you can see the graph of a line going through two points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex].<\/p>\n<p>https:\/\/www.desmos.com\/calculator\/j8jn4jsmg3<\/p>\n<p>&nbsp;<\/p>\n<h2>Model an Application With a Linear Function<\/h2>\n<p id=\"fs-id1165137594074\">In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.<\/p>\n<div class=\"textbox shaded\">\n<h3 id=\"fs-id1165137404879\">Given a linear function <em>f<\/em>\u00a0and the initial value and rate of change, evaluate <em>f<\/em>(<em>c<\/em>).<\/h3>\n<ol id=\"fs-id1165137660790\">\n<li>Determine the initial value and the rate of change (slope).<\/li>\n<li>Substitute the values into [latex]f\\left(x\\right)=mx+b[\/latex].<\/li>\n<li>Evaluate the function at [latex]x=c[\/latex].<\/li>\n<\/ol>\n<\/div>\n<p>Initial value is a term that is typically used in applications of functions. \u00a0It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the y-intercept. Here are some characteristics of initial value:<\/p>\n<ul>\n<li>The point [latex](0,y)[\/latex] is often the initial value of a linear function<\/li>\n<li>The y value of the initial value comes from b in the slope intercept form of a linear function,\u00a0[latex]f\\left(x\\right)=mx+b[\/latex]<\/li>\n<li>The initial value can be found by solving for b, or substituting 0 for x in a linear function.<\/li>\n<\/ul>\n<p>In our first example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection, and write an equation that represents the number of songs in the collection for any number of months, t.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, <em>N<\/em>, in his collection as a function of time, <em>t<\/em>, the number of months. How many songs will he own in a year?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q14550\">Show Answer<\/span><\/p>\n<div id=\"q14550\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135411394\">The initial value for this function is 200 because he currently owns 200 songs, so <i>N<\/i>(0) = 200, which means that <em>b<\/em> = 200.<\/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-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201031\/CNX_Precalc_Figure_02_01_0102.jpg\" alt=\"\" width=\"487\" height=\"131\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137738190\">The number of songs increases by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that <em>m<\/em> = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line.<span id=\"fs-id1165137417445\"><br \/>\n<\/span><\/p>\n<p id=\"fs-id1165137810258\">We can write the formula [latex]N\\left(t\\right)=15t+200[\/latex].<\/p>\n<p id=\"fs-id1165137454711\">With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at <em>t<\/em> = 12.<\/p>\n<div id=\"fs-id1165137462736\" class=\"equation unnumbered\">[latex]\\begin{array}N\\left(12\\right)=15\\left(12\\right)+200\\hfill \\\\ \\text{ }=180+200\\hfill \\\\ \\text{ }=380\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137694205\">Marcus will have 380 songs in 12 months.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_01_10\" class=\"example\">\n<div id=\"fs-id1165137836747\" class=\"exercise\">\n<h2>Analysis of the Solution<\/h2>\n<p id=\"fs-id1165134065131\">Notice that <em>N<\/em> is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.<\/p>\n<p>In our next example, we will show that you can write the equation for a linear function given two data points. \u00a0In this case, Ilya&#8217;s weekly income depends on the number of insurance policies he sells. \u00a0We are given his income for two different weeks and the number of policies sold. \u00a0We first find the rate of change and then solve for the initial value.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya\u2019s weekly income, <i>I<\/i>, depends on the number of new policies, <em>n<\/em>, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for <em>I<\/em>(<em>n<\/em>), and interpret the meaning of the components of the equation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q249315\">Show Answer<\/span><\/p>\n<div id=\"q249315\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135169134\">The given information gives us two input-output pairs: (3, 760) and (5, 920). We start by finding the rate of change.<\/p>\n<div id=\"fs-id1165135195046\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}m=\\frac{920 - 760}{5 - 3}\\hfill \\\\ \\text{ }=\\frac{$160}{\\text{2 policies}}\\hfill \\\\ \\text{ }=$80\\text{ per policy}\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137855034\">Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week.<\/p>\n<p id=\"fs-id1165137855040\">We can then solve for the initial value.<\/p>\n<div id=\"fs-id1165135484088\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{array}\\text{ }I\\left(n\\right)=80n+b\\hfill & \\hfill \\\\ \\text{ }760=80\\left(3\\right)+b\\hfill & \\text{when }n=3, I\\left(3\\right)=760\\hfill \\\\ 760 - 80\\left(3\\right)=b\\hfill & \\hfill \\\\ \\text{ }520=b\\hfill & \\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137400716\">The value of <em>b<\/em>\u00a0is the starting value for the function and represents Ilya\u2019s income when\u00a0<em>n<\/em> = 0, or when no new policies are sold. We can interpret this as Ilya\u2019s base salary for the week, which does not depend upon the number of policies sold.<\/p>\n<p id=\"fs-id1165135203653\">We can now write the final equation.<\/p>\n<div id=\"fs-id1165137506449\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/div>\n<p id=\"fs-id1165137655487\">Our final interpretation is that Ilya\u2019s base salary is $520 per week and he earns an additional $80 commission for each policy sold.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Analysis of the Solution<\/h2>\n<p>We used units to help us verify that we were calculating the rate correctly. It makes sense to speak in terms of the price per policy. To calculate the initial value, we solved for b by substituting values from one of the points we were given for n and I.<\/p>\n<p>In the following video example we show how to identify the initial value, slope and equation for a linear function.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Ex: Savings Linear Function Application (Slope, Intercept Meaning)\" width=\"500\" height=\"375\" src=\"https:\/\/www.youtube.com\/embed\/JMQSdRFJ1S4?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>We will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function <em>C\u00a0<\/em>where <i>C<\/i>(<em>x<\/em>)\u00a0is the cost for <em>x<\/em>\u00a0items produced in a given month.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q625547\">Show Answer<\/span><\/p>\n<div id=\"q625547\" class=\"hidden-answer\" style=\"display: none\">The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by [latex]C\\left(x\\right)=1250+37.5x[\/latex].<\/div>\n<\/div>\n<\/div>\n<h2>Analysis of the Solution<\/h2>\n<div id=\"fs-id1165137767515\" class=\"commentary\">\n<p id=\"fs-id1165135511326\">It is important to note that we are writing a function based on monthly costs, so the initial cost will be $1,250 becuase Ben has to pay that amount monthly for rent. If Ben produces 100 items in a month, his monthly cost is represented by<\/p>\n<div id=\"fs-id1165137417815\" class=\"equation unnumbered\">[latex]\\begin{array}C\\left(100\\right)=1250+37.5\\left(100\\right)\\hfill\\text{ } \\\\ =5000\\hfill \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165137862645\">So his monthly cost would be $5,000.<\/p>\n<p>The following video example show how to write a linear function that represents how many miles you can travel in a rental car for a fixed amount.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Ex:  Linear Equation Application (Cost of a Rental Car)\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/H8KR3w2nXqs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>In the next example we will take data that is in tabular (table) form to write an equation that describes the rate of change in the numbers of a rat population.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p id=\"fs-id1165135161325\">The table below relates the number of rats in a population to time, in weeks. Use the table to write a linear equation.<\/p>\n<table id=\"Table_02_01_02\" summary=\"Two rows and five columns. The first row is labeled, 'w, the numers of weeks'. The second row is labeled is labeled, 'P(w), number of rats'. Reading the remaining rows as ordered pairs (i.e., (w, P(w)), we have the following values: (0, 1000), (2, 1080), (4, 1160), and (6, 1240).\">\n<tbody>\n<tr>\n<td><strong><em>w<\/em>, number of weeks<\/strong><\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>6<\/td>\n<\/tr>\n<tr>\n<td><strong><em>P(w)<\/em>, number of rats<\/strong><\/td>\n<td>1000<\/td>\n<td>1080<\/td>\n<td>1160<\/td>\n<td>1240<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q780371\">Show Answer<\/span><\/p>\n<div id=\"q780371\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137530990\">We can see from the table that the initial value for the number of rats is 1000, so <em>b<\/em> = 1000.<\/p>\n<p id=\"fs-id1165137935601\">Rather than solving for <em>m<\/em>, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week.<\/p>\n<div id=\"fs-id1165137737900\" class=\"equation unnumbered\">[latex]P\\left(w\\right)=40w+1000[\/latex]<\/div>\n<p id=\"fs-id1165137465125\">If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240)<\/p>\n<div id=\"fs-id1165137627069\" class=\"equation unnumbered\">[latex]\\begin{array}m=\\frac{1240 - 1080}{6 - 2}\\hfill \\\\ \\text{ }=\\frac{160}{4}\\hfill \\\\ \\text{ }=40\\hfill \\end{array}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Think About It<\/h3>\n<p>Is the initial value always provided in a table of values like the table in\u00a0the previous example? Write your ideas in the textbox below before you look at the answer.<\/p>\n<p>If your answer is no, give a description of how you would find the initial value.<\/p>\n<p><textarea aria-label=\"Your Answer\" rows=\"2\"><\/textarea><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q298919\">Show Answer<\/span><\/p>\n<div id=\"q298919\" class=\"hidden-answer\" style=\"display: none\">\n<p><em>No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into [latex]f\\left(x\\right)=mx+b[\/latex], and solve for <\/em>b<em>.<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Summary<\/h2>\n<h3>Slope of a line<\/h3>\n<\/div>\n<\/div>\n<ul>\n<li>The <strong>slope<\/strong> of a line\u00a0indicates the direction in which a line slants as well as its steepness. Slope is\u00a0defined algebraically as:[latex]m=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/li>\n<li>Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.\u00a0[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/li>\n<\/ul>\n<h3>Equations of Lines<\/h3>\n<ul>\n<li>Standard form of a line is given as\u00a0[latex]Ax+By=C[\/latex].<\/li>\n<li>The equation of a <strong>vertical line<\/strong> is given as\u00a0[latex]x=c[\/latex]<\/li>\n<li>The equation of a <strong>horizontal line<\/strong> is given as\u00a0[latex]y=c[\/latex]<\/li>\n<li>Given a graph you can write the equation of the line it represents<\/li>\n<li>This method only works for graphs that have points that are easy to verify by visual inspection<\/li>\n<\/ul>\n<h3>Modeling Applications With Linear Functions<\/h3>\n<div id=\"Example_02_01_10\" class=\"example\">\n<div id=\"fs-id1165137836747\" class=\"exercise\">\n<ul>\n<li>Sometimes we are given an initial value, and sometimes we have to solve for it<\/li>\n<li>Using units can help you verify that you have calculated slope correctly<\/li>\n<li>We can write the equation for a line given a slope and a data point, or from a table of data<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2><\/h2>\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-2006\">\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>Revision and Adaptation. <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>Ex: Savings Linear Function Application (Slope, Intercept Meaning). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/JMQSdRFJ1S4\">https:\/\/youtu.be\/JMQSdRFJ1S4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Desmos Applet Point-Slope Form. <strong>Authored by<\/strong>: Tatyana Khodorovskiy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/www.desmos.com\/calculator\/j8jn4jsmg3\">https:\/\/www.desmos.com\/calculator\/j8jn4jsmg3<\/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>Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/vut5b2fRQQ0\">https:\/\/youtu.be\/vut5b2fRQQ0<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ndRpJxdmZJI\">https:\/\/youtu.be\/ndRpJxdmZJI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>College Algebra. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free:  http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@3.278:1\/Preface<\/li><li>Ex: Linear Equation Application (Cost of a Rental Car). <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/H8KR3w2nXqs\">https:\/\/youtu.be\/H8KR3w2nXqs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex 1: Find the Equation of a Line in Slope Intercept Form Given the Graph of a Line. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/mmWf_oLTNSQ\">https:\/\/youtu.be\/mmWf_oLTNSQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Find the Linear Function Given Two Function Values in Function Notation. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/TSIcryAtCmY\">https:\/\/youtu.be\/TSIcryAtCmY<\/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":21,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Ex: Find the Equation of a Line in Point Slope and Slope Intercept Form Given the Slope and a Point\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/vut5b2fRQQ0\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Find The Equation of the Line in Point-Slope and Slope Intercept Form Given Two Points\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ndRpJxdmZJI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College 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