{"id":168,"date":"2023-06-21T13:22:40","date_gmt":"2023-06-21T13:22:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/write-a-linear-function\/"},"modified":"2023-09-21T07:34:56","modified_gmt":"2023-09-21T07:34:56","slug":"write-a-linear-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/write-a-linear-function\/","title":{"raw":"\u25aa   Writing a Linear Function","rendered":"\u25aa   Writing a Linear Function"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate slope for a linear function given two points<\/li>\r\n \t<li>Write the equation of a linear function given two points and a slope<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Calculate and Interpret Slope<\/h2>\r\nIn the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the <strong>slope<\/strong> given input and output values. Given two values for the input, [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex], and two corresponding values for the output, [latex]{y}_{1}[\/latex]\u00a0and [latex]{y}_{2}[\/latex] which can be represented by a set of points, [latex]\\left({x}_{1}\\text{, }{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2}\\text{, }{y}_{2}\\right)[\/latex], we can calculate the slope [latex]m[\/latex],\u00a0as follows:\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\nwhere [latex]\\Delta y[\/latex] is the change in output and [latex]\\Delta x[\/latex] is the change in input. In function notation, [latex]{y}_{1}=f\\left({x}_{1}\\right)[\/latex] and [latex]{y}_{2}=f\\left({x}_{2}\\right)[\/latex] so we could write:\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{f\\left({x}_{2}\\right)-f\\left({x}_{1}\\right)}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\nThe graph below indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2,}{y}_{2}\\right)[\/latex],\u00a0is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.\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\/18223100\/CNX_Precalc_Figure_02_01_005n2.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"569\" \/> The slope of a function is calculated by the change in [latex]y[\/latex] divided by the change in [latex]x[\/latex]. It does not matter which coordinate is used as the [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and which is the [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex], as long as each calculation is started with the elements from the same coordinate pair.[\/caption]\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Are the units for slope always [latex]\\frac{\\text{units for the output}}{\\text{units for the input}}[\/latex] ?<\/strong>\r\n\r\n<em>Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: CalculatING Slope<\/h3>\r\nThe slope, or rate of change, [latex]m[\/latex], of a function can be calculated according to the following:\r\n\r\n&nbsp;\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\Rightarrow \\dfrac{f(x_2)-f(x_1)}{x_2 - x_1}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nwhere [latex]{x_1}[\/latex] and [latex]x_2[\/latex] are input values and [latex]{f(x_1)}[\/latex] and [latex]f(x_2)[\/latex] are output values.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two points from a linear function, calculate and interpret the slope.<\/h3>\r\n<ol>\r\n \t<li>Determine the units for output and input values.<\/li>\r\n \t<li>Calculate the change of output values and change of input values.<\/li>\r\n \t<li>Interpret the slope as the change in output values per unit of the input value.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Slope of a Linear Function<\/h3>\r\nIf [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function and [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]\u00a0are points on the line, find the slope. Is this function increasing or decreasing?\r\n\r\n[reveal-answer q=\"899087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"899087\"]\r\n\r\nThe coordinate pairs are [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]. To find the rate of change, we divide the change in output by the change in input.\r\n\r\n[latex]m=\\frac{\\text{change in output}}{\\text{change in input}}=\\frac{1-\\left(-2\\right)}{8 - 3}=\\frac{3}{5}[\/latex]\r\n\r\nWe could also write the slope as [latex]m=0.6[\/latex]. The function is increasing because [latex]m&gt;0[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nAs noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value or <em>y<\/em>-coordinate used corresponds with the first input value or <em>x<\/em>-coordinate used.\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\nIf [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function, and [latex]\\left(2,3\\right)[\/latex]\u00a0and [latex]\\left(0,4\\right)[\/latex]\u00a0are points on the line, find the slope. Is this function increasing or decreasing?\r\n\r\n[reveal-answer q=\"538866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"538866\"]\r\n\r\n[latex]m=\\frac{4 - 3}{0 - 2}=\\frac{1}{-2}=-\\frac{1}{2}[\/latex] ; decreasing because [latex]m&lt;0[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question height=\"180\"]2100[\/ohm_question]\r\n\r\n[ohm_question height=\"180\"]2102[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>interpreting slope<\/h3>\r\nWhen interpreting slope as an average rate of change, it is customary to express the calculation as change in output\u00a0 per unit of input.\r\n\r\nEx. A vehicle traveled from mile marker 57 to mile marker 180 between 4:15 pm and 6:21 pm. Can you use the formula for slope to find the vehicle's average rate of change in distance as a function of the time it traveled?\r\n\r\n[reveal-answer q=\"437733\"]more[\/reveal-answer]\r\n[hidden-answer a=\"437733\"]\r\n\r\nTime is usually considered the input variable. We say that distance traveled depends on the amount of time spent traveling at a constant rate. But we can calculate an average rate traveled over the entire time traveled using the formula for slope. The average rate of speed will be calculated by the total distance over the total time traveled.\r\n<p style=\"text-align: center;\">[latex]\\dfrac{\\text{change in distance}}{\\text{change in time}} = \\dfrac{180\\text{ miles}-57\\text{ miles}}{6.35\\text{ hours} - 4.15\\text{ hours}} = \\dfrac{123\\text{ miles}}{2.1\\text{ hours}} = \\dfrac{123}{2.1}\\text{ miles per hour} \\approx 58.6 \\text{ mph}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Population Change from a Linear Function<\/h3>\r\nThe population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change in population per year if we assume the change was constant from 2008 to 2012.\r\n\r\n[reveal-answer q=\"146109\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"146109\"]\r\n\r\nThe rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.\r\n\r\n[latex]\\frac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\frac{\\text{people}}{\\text{year}}[\/latex]\r\n\r\nSo the population increased by 1,100 people per year.\r\n<h4>Analysis of the Solution<\/h4>\r\nBecause we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.\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\nThe population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change in population per year if we assume the change was constant from 2009 to 2012.\r\n\r\n[reveal-answer q=\"706559\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"706559\"]\r\n\r\n[latex]m=\\frac{1,868 - 1,442}{2,012 - 2,009}=\\frac{426}{3}=142\\text{ people per year}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n[ohm_question height=\"450\"]113460[\/ohm_question]\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>recall equations of lines<\/h3>\r\nThe following methods of writing the equation of a line apply to writing the equation of a function. Just remember the translation: [latex]y = f(x)[\/latex]. That is, graphs of functions behave exactly like lines. Wherever you see a y-coordinate, you can replace it with a function output.\r\n\r\n<\/div>\r\n<h2>Point-Slope Form<\/h2>\r\nUp until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Now we will learn another way to write a linear function called\u00a0<strong>point-slope form\u00a0<\/strong>which is given below:\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nwhere\u00a0[latex]m[\/latex] is the slope of the linear function and [latex]({x}_{1},{y}_{1})[\/latex] is any point which satisfies the linear function.\r\n\r\nThe point-slope form is derived from the slope formula.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill &amp; \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill &amp; \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill &amp; \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &amp;\\text{Rearrange}\\hfill \\end{array}[\/latex]<\/p>\r\nKeep in mind that slope-intercept form and point-slope form can be used to describe the same linear function. We can move from one form to another using basic algebra. For example, suppose we are given the equation [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] which is in point-slope form. We can convert it to slope-intercept form as shown below.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill &amp; \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill &amp; \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{}y=-\\frac{1}{2}x+7\\hfill &amp; \\text{Add 4 to each side}.\\hfill \\end{array}[\/latex]<\/p>\r\nTherefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].\r\n<div class=\"textbox\">\r\n<h3>A General Note: Point-Slope Form of a Linear Equation<\/h3>\r\n<strong>Point-slope form<\/strong> of a linear equation takes the form\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\nwhere [latex]m[\/latex]\u00a0is the slope and [latex]{x}_{1 }\\text{ and } {y}_{1}[\/latex]\u00a0are the [latex]x\\text{ and }y[\/latex]\u00a0coordinates of a specific point through which the line passes.\r\n\r\n<\/div>\r\nPoint-slope form is particularly useful if we know one point and the slope of a line. For example, suppose we are told that a line has a slope of 2 and passes through the point [latex]\\left(4,1\\right)[\/latex].\u00a0We know that [latex]m=2[\/latex]\u00a0and that [latex]{x}_{1}=4[\/latex]\u00a0and [latex]{y}_{1}=1[\/latex]. We can substitute these values into point-slope form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{array}[\/latex]<\/p>\r\nIf we wanted to rewrite the equation in slope-intercept form, we apply algebraic techniques.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll} y - 1=2\\left(x - 4\\right)\\hfill &amp; \\hfill \\\\ y - 1=2x - 8\\hfill &amp; \\text{Distribute the }2.\\hfill \\\\ \\text{}y=2x - 7\\hfill &amp; \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\r\nBoth equations [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex] describe the same line. The graph of the line can be seen below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201023\/CNX_Precalc_Figure_02_01_0132.jpg\" width=\"487\" height=\"386\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing Linear Equations Using a Point and the Slope<\/h3>\r\nWrite the point-slope form of an equation of a line with a slope of 3 that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite the equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"579101\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"579101\"]\r\n\r\nLet\u2019s figure out what we know from the given information. The slope is 3, so <em>m\u00a0<\/em>= 3. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into point-slope form.\r\n\r\n[latex]\\begin{array}{llll}\\text{}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill &amp; \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill &amp; \\text{Substitute known values}.\\hfill \\\\ \\text{}y+1=3\\left(x - 6\\right)\\hfill &amp; \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{array}[\/latex]\r\n\r\nWe use algebra to find slope-intercept form of the linear equation.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y+1=3\\left(x - 6\\right)\\hfill &amp; \\hfill \\\\ y+1=3x - 18\\hfill &amp; \\text{Distribute 3}.\\hfill \\\\ \\text{}y=3x - 19\\hfill &amp; \\text{Simplify to slope-intercept form}.\\hfill \\end{array}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the equation of a line in point-slope form with a slope of [latex]-2[\/latex] that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite the equation in the slope-intercept form.\r\n\r\n[reveal-answer q=\"624568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"624568\"]\r\n\r\n[latex]y - 2=-2\\left(x+2\\right)[\/latex]; [latex]y=-2x - 2[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Writing the Equation of a Line Using Two Points<\/h2>\r\nPoint-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points [latex]\\left(0,\\text{ }1\\right)[\/latex]\u00a0and [latex]\\left(3,\\text{ }2\\right)[\/latex]. We can use the coordinates of the two points to find the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ \\text{}{m}=\\frac{2 - 1}{3 - 0}\\hfill \\\\ \\text{}{m}=\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nNow we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let's use (0, 1) for our point.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=\\frac{1}{3}\\left(x - 0\\right)\\end{array}[\/latex]<\/p>\r\nAs before, we can use algebra to rewrite the equation in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y - 1=\\frac{1}{3}\\left(x - 0\\right)\\hfill &amp; \\hfill \\\\ y - 1=\\frac{1}{3}x\\hfill &amp; \\text{Distribute the }\\frac{1}{3}.\\hfill \\\\ \\text{}y=\\frac{1}{3}x+1\\hfill &amp; \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\r\nBoth equations describe the line graphed below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201024\/CNX_Precalc_Figure_02_01_0142.jpg\" width=\"487\" height=\"386\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing Linear Equations Using Two Points<\/h3>\r\nWrite the point-slope form of an equation of a line that passes through the points\u00a0[latex]\\left(5,\\text{ }1\\right)[\/latex] and [latex]\\left(8,\\text{ }7\\right)[\/latex]. Then rewrite the equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"981323\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"981323\"]\r\n\r\nLet\u2019s begin by finding the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{}{m}=\\frac{7 - 1}{8 - 5}\\hfill \\\\ \\text{}{m}=\\frac{6}{3}\\hfill \\\\ \\text{}{m}=2\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into point-slope form. We can choose either point, but we will use [latex]\\left(5,\\text{ }1\\right)[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 5\\right)\\end{array}[\/latex]<\/p>\r\nTo rewrite the equation in slope-intercept form, we use algebra.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=2\\left(x - 5\\right)\\hfill \\\\ y - 1=2x - 10\\hfill \\\\ \\text{}y=2x - 9\\hfill \\end{array}[\/latex]<\/p>\r\nThe slope-intercept form of the equation of the line is [latex]y=2x - 9[\/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\nWrite the point-slope form of an equation of a line that passes through the points [latex]\\left(-1,3\\right)[\/latex] and [latex]\\left(0,0\\right)[\/latex]. Then rewrite the equation in slope-intercept form.\r\n\r\n[reveal-answer q=\"85979\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"85979\"]\r\n\r\n[latex]y - 0=-3\\left(x - 0\\right)[\/latex] ; [latex]y=-3x[\/latex][\/hidden-answer]\r\n\r\n[ohm_question height=\"500\"]76345[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate slope for a linear function given two points<\/li>\n<li>Write the equation of a linear function given two points and a slope<\/li>\n<\/ul>\n<\/div>\n<h2>Calculate and Interpret Slope<\/h2>\n<p>In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the <strong>slope<\/strong> given input and output values. Given two values for the input, [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex], and two corresponding values for the output, [latex]{y}_{1}[\/latex]\u00a0and [latex]{y}_{2}[\/latex] which can be represented by a set of points, [latex]\\left({x}_{1}\\text{, }{y}_{1}\\right)[\/latex]\u00a0and [latex]\\left({x}_{2}\\text{, }{y}_{2}\\right)[\/latex], we can calculate the slope [latex]m[\/latex],\u00a0as follows:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<p>where [latex]\\Delta y[\/latex] is the change in output and [latex]\\Delta x[\/latex] is the change in input. In function notation, [latex]{y}_{1}=f\\left({x}_{1}\\right)[\/latex] and [latex]{y}_{2}=f\\left({x}_{2}\\right)[\/latex] so we could write:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{f\\left({x}_{2}\\right)-f\\left({x}_{1}\\right)}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<p>The graph below indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex] and [latex]\\left({x}_{2,}{y}_{2}\\right)[\/latex],\u00a0is calculated. Recall that the slope measures steepness. The greater the absolute value of the slope, the steeper the line is.<\/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\/18223100\/CNX_Precalc_Figure_02_01_005n2.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"569\" \/><\/p>\n<p class=\"wp-caption-text\">The slope of a function is calculated by the change in [latex]y[\/latex] divided by the change in [latex]x[\/latex]. It does not matter which coordinate is used as the [latex]\\left({x}_{2,\\text{ }}{y}_{2}\\right)[\/latex] and which is the [latex]\\left({x}_{1},\\text{ }{y}_{1}\\right)[\/latex], as long as each calculation is started with the elements from the same coordinate pair.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Are the units for slope always [latex]\\frac{\\text{units for the output}}{\\text{units for the input}}[\/latex] ?<\/strong><\/p>\n<p><em>Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: CalculatING Slope<\/h3>\n<p>The slope, or rate of change, [latex]m[\/latex], of a function can be calculated according to the following:<\/p>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\Rightarrow \\dfrac{f(x_2)-f(x_1)}{x_2 - x_1}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>where [latex]{x_1}[\/latex] and [latex]x_2[\/latex] are input values and [latex]{f(x_1)}[\/latex] and [latex]f(x_2)[\/latex] are output values.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two points from a linear function, calculate and interpret the slope.<\/h3>\n<ol>\n<li>Determine the units for output and input values.<\/li>\n<li>Calculate the change of output values and change of input values.<\/li>\n<li>Interpret the slope as the change in output values per unit of the input value.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Slope of a Linear Function<\/h3>\n<p>If [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function and [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]\u00a0are points on the line, find the slope. Is this function increasing or decreasing?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q899087\">Show Solution<\/span><\/p>\n<div id=\"q899087\" class=\"hidden-answer\" style=\"display: none\">\n<p>The coordinate pairs are [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]. To find the rate of change, we divide the change in output by the change in input.<\/p>\n<p>[latex]m=\\frac{\\text{change in output}}{\\text{change in input}}=\\frac{1-\\left(-2\\right)}{8 - 3}=\\frac{3}{5}[\/latex]<\/p>\n<p>We could also write the slope as [latex]m=0.6[\/latex]. The function is increasing because [latex]m>0[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>As noted earlier, the order in which we write the points does not matter when we compute the slope of the line as long as the first output value or <em>y<\/em>-coordinate used corresponds with the first input value or <em>x<\/em>-coordinate used.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>If [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function, and [latex]\\left(2,3\\right)[\/latex]\u00a0and [latex]\\left(0,4\\right)[\/latex]\u00a0are points on the line, find the slope. Is this function increasing or decreasing?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q538866\">Show Solution<\/span><\/p>\n<div id=\"q538866\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]m=\\frac{4 - 3}{0 - 2}=\\frac{1}{-2}=-\\frac{1}{2}[\/latex] ; decreasing because [latex]m<0[\/latex]\n\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm2100\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2100&theme=oea&iframe_resize_id=ohm2100&show_question_numbers\" width=\"100%\" height=\"180\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm2102\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2102&theme=oea&iframe_resize_id=ohm2102&show_question_numbers\" width=\"100%\" height=\"180\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>interpreting slope<\/h3>\n<p>When interpreting slope as an average rate of change, it is customary to express the calculation as change in output\u00a0 per unit of input.<\/p>\n<p>Ex. A vehicle traveled from mile marker 57 to mile marker 180 between 4:15 pm and 6:21 pm. Can you use the formula for slope to find the vehicle&#8217;s average rate of change in distance as a function of the time it traveled?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q437733\">more<\/span><\/p>\n<div id=\"q437733\" class=\"hidden-answer\" style=\"display: none\">\n<p>Time is usually considered the input variable. We say that distance traveled depends on the amount of time spent traveling at a constant rate. But we can calculate an average rate traveled over the entire time traveled using the formula for slope. The average rate of speed will be calculated by the total distance over the total time traveled.<\/p>\n<p style=\"text-align: center;\">[latex]\\dfrac{\\text{change in distance}}{\\text{change in time}} = \\dfrac{180\\text{ miles}-57\\text{ miles}}{6.35\\text{ hours} - 4.15\\text{ hours}} = \\dfrac{123\\text{ miles}}{2.1\\text{ hours}} = \\dfrac{123}{2.1}\\text{ miles per hour} \\approx 58.6 \\text{ mph}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Population Change from a Linear Function<\/h3>\n<p>The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change in population per year if we assume the change was constant from 2008 to 2012.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q146109\">Show Solution<\/span><\/p>\n<div id=\"q146109\" class=\"hidden-answer\" style=\"display: none\">\n<p>The rate of change relates the change in population to the change in time. The population increased by [latex]27,800-23,400=4400[\/latex] people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years.<\/p>\n<p>[latex]\\frac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\frac{\\text{people}}{\\text{year}}[\/latex]<\/p>\n<p>So the population increased by 1,100 people per year.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change in population per year if we assume the change was constant from 2009 to 2012.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q706559\">Show Solution<\/span><\/p>\n<div id=\"q706559\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]m=\\frac{1,868 - 1,442}{2,012 - 2,009}=\\frac{426}{3}=142\\text{ people per year}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm113460\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=113460&theme=oea&iframe_resize_id=ohm113460&show_question_numbers\" width=\"100%\" height=\"450\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall equations of lines<\/h3>\n<p>The following methods of writing the equation of a line apply to writing the equation of a function. Just remember the translation: [latex]y = f(x)[\/latex]. That is, graphs of functions behave exactly like lines. Wherever you see a y-coordinate, you can replace it with a function output.<\/p>\n<\/div>\n<h2>Point-Slope Form<\/h2>\n<p>Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Now we will learn another way to write a linear function called\u00a0<strong>point-slope form\u00a0<\/strong>which is given below:<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>where\u00a0[latex]m[\/latex] is the slope of the linear function and [latex]({x}_{1},{y}_{1})[\/latex] is any point which satisfies the linear function.<\/p>\n<p>The point-slope form is derived from the slope formula.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}{m}=\\frac{y-{y}_{1}}{x-{x}_{1}}\\hfill & \\text{assuming }{ x }\\ne {x}_{1}\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right)\\hfill & \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\hfill\\\\{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1}\\hfill & \\text{Simplify}\\hfill \\\\ y-{y}_{1}={ m }\\left(x-{x}_{1}\\right)\\hfill &\\text{Rearrange}\\hfill \\end{array}[\/latex]<\/p>\n<p>Keep in mind that slope-intercept form and point-slope form can be used to describe the same linear function. We can move from one form to another using basic algebra. For example, suppose we are given the equation [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] which is in point-slope form. We can convert it to slope-intercept form as shown below.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y - 4=-\\frac{1}{2}\\left(x - 6\\right)\\hfill & \\hfill \\\\ y - 4=-\\frac{1}{2}x+3\\hfill & \\text{Distribute the }-\\frac{1}{2}.\\hfill \\\\ \\text{}y=-\\frac{1}{2}x+7\\hfill & \\text{Add 4 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Point-Slope Form of a Linear Equation<\/h3>\n<p><strong>Point-slope form<\/strong> of a linear equation takes the form<\/p>\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\n<p>where [latex]m[\/latex]\u00a0is the slope and [latex]{x}_{1 }\\text{ and } {y}_{1}[\/latex]\u00a0are the [latex]x\\text{ and }y[\/latex]\u00a0coordinates of a specific point through which the line passes.<\/p>\n<\/div>\n<p>Point-slope form is particularly useful if we know one point and the slope of a line. For example, suppose we are told that a line has a slope of 2 and passes through the point [latex]\\left(4,1\\right)[\/latex].\u00a0We know that [latex]m=2[\/latex]\u00a0and that [latex]{x}_{1}=4[\/latex]\u00a0and [latex]{y}_{1}=1[\/latex]. We can substitute these values into point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{array}[\/latex]<\/p>\n<p>If we wanted to rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll} y - 1=2\\left(x - 4\\right)\\hfill & \\hfill \\\\ y - 1=2x - 8\\hfill & \\text{Distribute the }2.\\hfill \\\\ \\text{}y=2x - 7\\hfill & \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Both equations [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex] describe the same line. The graph of the line can be seen below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201023\/CNX_Precalc_Figure_02_01_0132.jpg\" width=\"487\" height=\"386\" alt=\"image\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing Linear Equations Using a Point and the Slope<\/h3>\n<p>Write the point-slope form of an equation of a line with a slope of 3 that passes through the point [latex]\\left(6,-1\\right)[\/latex]. Then rewrite the equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q579101\">Show Solution<\/span><\/p>\n<div id=\"q579101\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s figure out what we know from the given information. The slope is 3, so <em>m\u00a0<\/em>= 3. We also know one point, so we know [latex]{x}_{1}=6[\/latex] and [latex]{y}_{1}=-1[\/latex]. Now we can substitute these values into point-slope form.<\/p>\n<p>[latex]\\begin{array}{llll}\\text{}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill & \\hfill \\\\ y-\\left(-1\\right)=3\\left(x - 6\\right)\\hfill & \\text{Substitute known values}.\\hfill \\\\ \\text{}y+1=3\\left(x - 6\\right)\\hfill & \\text{Distribute }-1\\text{ to find point-slope form}.\\hfill \\end{array}[\/latex]<\/p>\n<p>We use algebra to find slope-intercept form of the linear equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{llll}y+1=3\\left(x - 6\\right)\\hfill & \\hfill \\\\ y+1=3x - 18\\hfill & \\text{Distribute 3}.\\hfill \\\\ \\text{}y=3x - 19\\hfill & \\text{Simplify to slope-intercept form}.\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the equation of a line in point-slope form with a slope of [latex]-2[\/latex] that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite the equation in the slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q624568\">Show Solution<\/span><\/p>\n<div id=\"q624568\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y - 2=-2\\left(x+2\\right)[\/latex]; [latex]y=-2x - 2[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Writing the Equation of a Line Using Two Points<\/h2>\n<p>Point-slope form of an equation is also useful if we know any two points through which a line passes. Suppose, for example, we know that a line passes through the points [latex]\\left(0,\\text{ }1\\right)[\/latex]\u00a0and [latex]\\left(3,\\text{ }2\\right)[\/latex]. We can use the coordinates of the two points to find the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ \\text{}{m}=\\frac{2 - 1}{3 - 0}\\hfill \\\\ \\text{}{m}=\\frac{1}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let&#8217;s use (0, 1) for our point.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=\\frac{1}{3}\\left(x - 0\\right)\\end{array}[\/latex]<\/p>\n<p>As before, we can use algebra to rewrite the equation in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{lll}y - 1=\\frac{1}{3}\\left(x - 0\\right)\\hfill & \\hfill \\\\ y - 1=\\frac{1}{3}x\\hfill & \\text{Distribute the }\\frac{1}{3}.\\hfill \\\\ \\text{}y=\\frac{1}{3}x+1\\hfill & \\text{Add 1 to each side}.\\hfill \\end{array}[\/latex]<\/p>\n<p>Both equations describe the line graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201024\/CNX_Precalc_Figure_02_01_0142.jpg\" width=\"487\" height=\"386\" alt=\"image\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing Linear Equations Using Two Points<\/h3>\n<p>Write the point-slope form of an equation of a line that passes through the points\u00a0[latex]\\left(5,\\text{ }1\\right)[\/latex] and [latex]\\left(8,\\text{ }7\\right)[\/latex]. Then rewrite the equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q981323\">Show Solution<\/span><\/p>\n<div id=\"q981323\" class=\"hidden-answer\" style=\"display: none\">\n<p>Let\u2019s begin by finding the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\hfill \\\\ \\text{}{m}=\\frac{7 - 1}{8 - 5}\\hfill \\\\ \\text{}{m}=\\frac{6}{3}\\hfill \\\\ \\text{}{m}=2\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into point-slope form. We can choose either point, but we will use [latex]\\left(5,\\text{ }1\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 5\\right)\\end{array}[\/latex]<\/p>\n<p>To rewrite the equation in slope-intercept form, we use algebra.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y - 1=2\\left(x - 5\\right)\\hfill \\\\ y - 1=2x - 10\\hfill \\\\ \\text{}y=2x - 9\\hfill \\end{array}[\/latex]<\/p>\n<p>The slope-intercept form of the equation of the line is [latex]y=2x - 9[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the point-slope form of an equation of a line that passes through the points [latex]\\left(-1,3\\right)[\/latex] and [latex]\\left(0,0\\right)[\/latex]. Then rewrite the equation in slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q85979\">Show Solution<\/span><\/p>\n<div id=\"q85979\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y - 0=-3\\left(x - 0\\right)[\/latex] ; [latex]y=-3x[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"ohm76345\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=76345&theme=oea&iframe_resize_id=ohm76345&show_question_numbers\" width=\"100%\" height=\"500\"><\/iframe><\/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-168\">\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>Question ID 113460, 76345. <strong>Authored 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>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/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 at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 2100, 2102. <strong>Authored by<\/strong>: Morales,Lawrence. <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>: IMathAS Community License CC-BY + GPL<\/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":395986,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen 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