{"id":13798,"date":"2018-08-24T19:30:58","date_gmt":"2018-08-24T19:30:58","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13798"},"modified":"2025-02-05T05:18:33","modified_gmt":"2025-02-05T05:18:33","slug":"linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/linear-functions\/","title":{"raw":"Linear Functions","rendered":"Linear Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Identify and Interpret the Slope and Vertical Intercept of a Line<\/li>\r\n \t<li>Calculate and interpret the slope of a line.<\/li>\r\n \t<li>Determine the equation of a linear function.<\/li>\r\n<\/ul>\r\n<\/div>\r\n\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010637\/CNX_Precalc_02_00_012.jpg\" alt=\"An upward view of bamboo trees.\" \/><\/span>\r\n<p id=\"fs-id1165137705130\">Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour. [footnote]http:\/\/www.guinnessworldrecords.com\/records-3000\/fastest-growing-plant\/[\/footnote] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function.<\/p>\r\n<p id=\"fs-id1165137605883\">Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data.<\/p>\r\n\r\n<figure id=\"CNX_Precalc_Figure_02_01_001\">\r\n\r\n[caption id=\"\" align=\"alignright\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0112.jpg\" alt=\"Front view of a subway train, the maglev train.\" width=\"325\" height=\"432\" \/> Shanghai MagLev Train (credit: \"kanegen\"\/Flickr)[\/caption]<\/figure>\r\n<p id=\"fs-id1165137697072\">Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train. It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[footnote]<a href=\"http:\/\/www.chinahighlights.com\/shanghai\/transportation\/maglev-train.htm\" target=\"_blank\" rel=\"noopener\">http:\/\/www.chinahighlights.com\/shanghai\/transportation\/maglev-train.htm<\/a>[\/footnote]<\/p>\r\n<p id=\"fs-id1165134340085\">Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train\u2019s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train\u2019s distance from the station at a given point in time.<\/p>\r\n\r\n<h2 style=\"text-align: left;\">Representing Linear Functions<\/h2>\r\n<p id=\"fs-id1165137573850\">The function describing the train\u2019s motion is a <strong>linear function<\/strong>, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train\u2019s motion as a function using each method.<\/p>\r\n\r\n<section id=\"fs-id1165137759903\">\r\n<h3 style=\"text-align: center;\">Representing a Linear Function in Word Form<\/h3>\r\n<p id=\"fs-id1165137588695\">Let\u2019s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.<\/p>\r\n\r\n<ul id=\"fs-id1165135526954\">\r\n \t<li><em>The train\u2019s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.<\/em><\/li>\r\n<\/ul>\r\n<p id=\"fs-id1165135188466\">The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.<\/p>\r\n\r\n<\/section><section id=\"fs-id1165135639903\">\r\n<h3 style=\"text-align: center;\">Representing a Linear Function in Function Notation<\/h3>\r\n<p id=\"fs-id1165137833100\">Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the <strong>slope-intercept form<\/strong> of a line, where [latex]x[\/latex] is the input value, [latex]m[\/latex] is the rate of change, and [latex]b[\/latex] is the initial value of the dependent variable.<\/p>\r\n\r\n<div id=\"Equation_02_01_01\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{align}&amp;\\text{Equation form} &amp;&amp; y=mx+b \\\\ &amp;\\text{Function notation} &amp;&amp; f(x)=mx+b\\\\&amp; \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137411219\">In the example of the train, we might use the notation [latex]D\\left(t\\right)[\/latex]\u00a0in which the total distance [latex]D[\/latex]\r\nis a function of the time [latex]t[\/latex].\u00a0The rate, [latex]m[\/latex],\u00a0is 83 meters per second. The initial value of the dependent variable [latex]b[\/latex]\u00a0is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.<\/p>\r\n\r\n<div id=\"fs-id1165137559254\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]D(t)=83t+250[\/latex]<\/div>\r\n<\/section><section id=\"fs-id1165135415800\">\r\n<h3 style=\"text-align: center;\">Representing a Linear Function in Tabular Form<\/h3>\r\n<p id=\"fs-id1165137438406\">A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in the table in Figure 1. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.<\/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\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0152.jpg\" alt=\"Table with the first row, labeled t, containing the seconds from 0 to 3, and with the second row, labeled D(t), containing the meters 250 to 499. The first row goes up by 1 second, and the second row goes up by 83 meters.\" width=\"487\" height=\"161\" \/> <b>Figure 1.<\/b> Tabular representation of the function D showing selected input and output values[\/caption]\r\n\r\n<div id=\"fs-id1165137482942\" class=\"note precalculus qa textbox\">\r\n<h3 id=\"fs-id1165137828205\"><strong>Q &amp; A<\/strong><\/h3>\r\n<strong>Can the input in the previous example be any real number?<\/strong>\r\n<p id=\"fs-id1165135209002\"><em>No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.<\/em><\/p>\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137619188\">\r\n<h3 style=\"text-align: center;\">Representing a Linear Function in Graphical Form<\/h3>\r\n<p id=\"fs-id1165137827353\">Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, [latex]D(t)=83t+250[\/latex], to draw a graph, represented in the graph in Figure 2. Notice the graph is a line. When we plot a linear function, the graph is always a line.<\/p>\r\n<p id=\"fs-id1165137451297\">The rate of change, which is constant, determines the slant, or <strong>slope<\/strong> of the line. The point at which the input value is zero is the vertical intercept, or <strong><em>y<\/em>-intercept<\/strong>, of the line. We can see from the graph that the <em>y<\/em>-intercept in the train example we just saw is [latex]\\left(0,250\\right)[\/latex]\u00a0and represents the distance of the train from the station when it began moving at a constant speed.<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0122.jpg\" alt=\"A graph of an increasing function with points at (-2, -4) and (0, 2).\" width=\"487\" height=\"289\" \/>\r\n<p style=\"text-align: center;\"><strong>Figure 2.<\/strong> The graph of [latex]D(t)=83t+250[\/latex]. Graphs of linear functions are lines because the rate of change is constant.<\/p>\r\n<p id=\"fs-id1165137715509\">Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line [latex]f(x)=2{x}_{}+1[\/latex].\u00a0Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.<\/p>\r\n\r\n<div id=\"fs-id1165137726089\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Linear Function<\/h3>\r\n<p id=\"fs-id1165137454496\">A <strong>linear function<\/strong> is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line<\/p>\r\n\r\n<div id=\"Equation_02_01_02\" class=\"equation\" style=\"text-align: center;\">[latex]f(x)=mx+b[\/latex]<\/div>\r\n<p id=\"fs-id1165137784222\">where [latex]b[\/latex]\u00a0is the initial or starting value of the function (when input, [latex]x=0[\/latex]), and [latex]m[\/latex]\u00a0is the constant rate of change, or <strong>slope<\/strong> of the function. The <strong><em>y<\/em>-intercept<\/strong> is at [latex]\\left(0,b\\right)[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_02_01_01\" class=\"example\">\r\n<div id=\"fs-id1165137583894\" class=\"exercise\">\r\n<div id=\"fs-id1165135209144\" class=\"problem textbox shaded\">\r\n<h3>Example 1: Using a Linear Function to Find the Pressure on a Diver<\/h3>\r\nThe pressure, [latex]P[\/latex],\u00a0in pounds per square inch (PSI) on the diver in Figure 3\u00a0depends upon her depth below the water surface, [latex]d[\/latex], in feet. This relationship may be modeled by the equation, [latex]P(d)=0.434d+14.696[\/latex]. Restate this function in words.\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\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0032.jpg\" alt=\"Scuba diver.\" width=\"487\" height=\"366\" \/> <b>Figure 3.<\/b> (credit: Ilse Reijs and Jan-Noud Hutten)[\/caption]\r\n\r\n[reveal-answer q=\"129605\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"129605\"]\r\n\r\nTo restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137740917\">The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137738187\" class=\"commentary\"><\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<h2 style=\"text-align: center;\">Determining Whether a Linear Function is Increasing, Decreasing, or Constant<\/h2>\r\n<section id=\"fs-id1165137749252\">\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_004abc2.jpg\" alt=\"Three graphs depicting an increasing function, a decreasing function, and a constant function.\" width=\"975\" height=\"375\" \/> <b>Figure 4<\/b>[\/caption]\r\n<p id=\"fs-id1165135482019\">The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant.<\/p>\r\nFor an <strong>increasing function<\/strong>, as with the train example,\r\n<p style=\"text-align: center;\"><strong><em>the output values increase as the input values increase. <\/em><\/strong><\/p>\r\nThe graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in <strong>(a)<\/strong>.\r\n\r\nFor a <strong>decreasing function<\/strong>, the slope is negative.\r\n<p style=\"text-align: center;\"><strong><em>The output values decrease as the input values increase. <\/em><\/strong><\/p>\r\nA line with a negative slope slants downward from left to right as in <strong>(b)<\/strong>. If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in <strong>(c)<\/strong>.<span id=\"fs-id1165137453957\">\r\n<\/span>\r\n<div id=\"fs-id1165137446154\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Increasing and Decreasing Functions<\/h3>\r\n<p id=\"fs-id1165134085973\">The slope determines if the function is an <strong>increasing linear function<\/strong>, a <strong>decreasing linear function<\/strong>, or a constant function.<\/p>\r\n\r\n<ul id=\"eip-643\">\r\n \t<li>[latex]f(x)=mx+b\\text{ is an increasing function if }m&gt;0[\/latex].<\/li>\r\n \t<li>[latex]f(x)=mx+b\\text{ is an decreasing function if }m&lt;0[\/latex].<\/li>\r\n \t<li>[latex]f(x)=mx+b\\text{ is a constant function if }m=0[\/latex].<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"Example_02_01_02\" class=\"example\">\r\n<div id=\"fs-id1165137405281\" class=\"exercise\">\r\n<div id=\"fs-id1165135571684\" class=\"problem textbox shaded\">\r\n<h3>Example 2: Deciding whether a Function Is Increasing, Decreasing, or Constant<\/h3>\r\n<p id=\"fs-id1165137400625\">Some recent studies suggest that a teenager sends an average of 60 texts per day.[footnote]<a href=\"http:\/\/www.cbsnews.com\/8301-501465_162-57400228-501465\/teens-are-sending-60-texts-a-day-study-says\/\" target=\"_blank\" rel=\"noopener\">http:\/\/www.cbsnews.com\/8301-501465_162-57400228-501465\/teens-are-sending-60-texts-a-day-study-says\/<\/a>[\/footnote]\u00a0For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.<\/p>\r\n\r\n<ol id=\"fs-id1165137807449\">\r\n \t<li>The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.<\/li>\r\n \t<li>A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.<\/li>\r\n \t<li>A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"539033\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"539033\"]\r\n<p id=\"fs-id1165137508045\">Analyze each function.<\/p>\r\n\r\n<ol id=\"fs-id1165137778959\">\r\n \t<li>The function can be represented as [latex]f(x)=60x[\/latex] where [latex]x[\/latex] is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.<\/li>\r\n \t<li>The function can be represented as [latex]f(x)=500 - 60x[\/latex] where [latex]x[\/latex] is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after [latex]x[\/latex] days.<\/li>\r\n \t<li>The cost function can be represented as [latex]f(x)=50[\/latex] because the number of days does not affect the total cost. The slope is 0 so the function is constant.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137837055\">\r\n<h2 id=\"fs-id1165137410528\" class=\"note precalculus try\" style=\"text-align: center;\">Calculating and Interpreting Slope<\/h2>\r\n<p id=\"fs-id1165137573268\">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] \u2014which 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]\u2014we can calculate the slope [latex]m[\/latex],\u00a0as follows<\/p>\r\n\r\n<div id=\"fs-id1165137757690\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137767606\">where [latex]\\Delta y[\/latex] is the vertical displacement and [latex]\\Delta x[\/latex] is the horizontal displacement. Note in function notation two corresponding values for the output [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] for the function [latex]f[\/latex], [latex]{y}_{1}=f\\left({x}_{1}\\right)[\/latex] and [latex]{y}_{2}=f\\left({x}_{2}\\right)[\/latex], so we could equivalently write<\/p>\r\n\r\n<div id=\"fs-id1165137438737\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\frac{f\\left({x}_{2}\\right)-f\\left({x}_{1}\\right)}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\nThe graph in Figure 5\u00a0indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex]\r\nand [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-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_005n2.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"569\" \/> <b>Figure 5<\/b>[\/caption]\r\n\r\n<em>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.<\/em>\r\n<div id=\"fs-id1165137831952\" class=\"note precalculus qa textbox\">\r\n<h3 id=\"fs-id1165137501372\">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<p id=\"fs-id1165137462769\"><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>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135161219\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Calculate Slope<\/h3>\r\n<p id=\"fs-id1165134069119\">The slope, or rate of change, of a function [latex]m[\/latex] can be calculated according to the following:<\/p>\r\n<p style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex][\/latex]<\/p>\r\n<p id=\"fs-id1165135414432\">where [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex] are input values, [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] are output values.<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137539010\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137675584\">How To: Given two points from a linear function, calculate and interpret the slope.<\/h3>\r\n<ol id=\"fs-id1165137820100\">\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 id=\"Example_02_01_03\" class=\"example\">\r\n<div id=\"fs-id1165137641892\" class=\"exercise\">\r\n<div id=\"fs-id1165137811167\" class=\"problem textbox shaded\">\r\n<h3>Example 3: Finding the Slope of a Linear Function<\/h3>\r\n<p id=\"fs-id1165137724908\">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>\r\n[reveal-answer q=\"603714\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"603714\"]\r\n<p id=\"fs-id1165137714054\">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>\r\n<p style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output}}{\\text{change in input}}=\\frac{1-\\left(-2\\right)}{8 - 3}=\\frac{3}{5}[\/latex]<\/p>\r\n<p id=\"fs-id1165137469338\">We could also write the slope as [latex]m=0.6[\/latex]. The function is increasing because [latex]m&gt;0[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137842413\">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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137451238\">If [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function, and [latex]\\left(2,\\text{ }3\\right)[\/latex]\u00a0and [latex]\\left(0,\\text{ }4\\right)[\/latex]\u00a0are points on the line, find the slope. Is this function increasing or decreasing?<\/p>\r\n[reveal-answer q=\"786191\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"786191\"]\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<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]169760[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"Example_02_01_04\" class=\"example\">\r\n<div id=\"fs-id1165135307935\" class=\"exercise\">\r\n<div id=\"fs-id1165135307937\" class=\"problem textbox shaded\">\r\n<h3>Example 4: Finding the Population Change from a Linear Function<\/h3>\r\n<p id=\"fs-id1165135386495\">The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012.<\/p>\r\n[reveal-answer q=\"870587\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"870587\"]\r\n<p id=\"fs-id1165137415969\">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>\r\n<p style=\"text-align: center;\">[latex]\\frac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\frac{\\text{people}}{\\text{year}}[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex][\/latex]<\/p>\r\n<p id=\"fs-id1165137705484\">So the population increased by 1,100 people per year.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137451439\">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>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3 id=\"fs-id1165137425642\">Try It<\/h3>\r\nThe population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012.\r\n\r\n[reveal-answer q=\"468046\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"468046\"]\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<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]57147[\/ohm_question]\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2 style=\"text-align: center;\">Writing the Point-Slope Form of a Linear Equation<\/h2>\r\n<p id=\"fs-id1165137639245\">Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the <strong>point-slope form<\/strong>.<\/p>\r\n\r\n<div id=\"fs-id1165137452508\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137551238\">The point-slope form is derived from the slope formula.<\/p>\r\n\r\n<div id=\"eip-301\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;{m}=\\frac{y-{y}_{1}}{x-{x}_{1}} &amp;&amp; \\text{assuming }{ x }\\ne {x}_{1} \\\\ &amp;{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right) &amp;&amp; \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\\\ &amp;{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1} &amp;&amp; \\text{Simplify} \\\\ &amp;y-{y}_{1}={ m }\\left(x-{x}_{1}\\right) &amp;&amp;\\text{Rearrange} \\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137844021\">Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] . We can convert it to the slope-intercept form as shown.<\/p>\r\n\r\n<div id=\"eip-424\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;y - 4=-\\frac{1}{2}\\left(x - 6\\right) \\\\ &amp;y - 4=-\\frac{1}{2}x+3 &amp;&amp; \\text{Distribute the }-\\frac{1}{2}. \\\\ &amp;y=-\\frac{1}{2}x+7 &amp;&amp; \\text{Add 4 to each side}. \\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137409234\">Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137461941\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Point-Slope Form of a Linear Equation<\/h3>\r\n<p id=\"fs-id1165137647838\">The <strong>point-slope form<\/strong> of a linear equation takes the form<\/p>\r\n<p style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/p>\r\n<p id=\"fs-id1165137663648\">where [latex]m[\/latex]\u00a0is the slope, [latex]{x}_{1 }[\/latex] and [latex] {y}_{1}[\/latex]\u00a0are the [latex]x[\/latex] and [latex]y[\/latex]\u00a0coordinates of a specific point through which the line passes.<\/p>\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137745298\">\r\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Using a Point and the Slope<\/h2>\r\n<p id=\"fs-id1165137444576\">The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, 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 the general point-slope equation.<\/p>\r\n\r\n<div id=\"fs-id1165134380388\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{gathered}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137639550\">If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\r\n\r\n<div id=\"eip-420\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;y - 1=2\\left(x - 4\\right) \\\\ &amp;y - 1=2x - 8 &amp;&amp; \\text{Distribute the }2 \\\\ &amp;y=2x - 7 &amp;&amp; \\text{Add 1 to each side} \\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165135333698\">Both equations, [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex], describe the same line. See Figure 6.<span id=\"fs-id1165137925529\">\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\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_0132.jpg\" alt=\"\" width=\"487\" height=\"386\" \/> <b>Figure 6<\/b>[\/caption]\r\n\r\n<div id=\"Example_02_01_05\" class=\"example\">\r\n<div id=\"fs-id1165137423582\" class=\"exercise\">\r\n<div id=\"fs-id1165137423584\" class=\"problem textbox shaded\">\r\n<h3>Example 5: Writing Linear Equations Using a Point and the Slope<\/h3>\r\n<p id=\"fs-id1165137736482\">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 it in the slope-intercept form.<\/p>\r\n[reveal-answer q=\"500465\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"500465\"]\r\n<p id=\"fs-id1165137530790\">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 the general point-slope equation.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ &amp;y-\\left(-1\\right)=3\\left(x - 6\\right) &amp;&amp; \\text{Substitute known values} \\\\ &amp;y+1=3\\left(x - 6\\right) &amp;&amp; \\text{Distribute }-1\\text{ to find point-slope form} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137594974\">Then we use algebra to find the slope-intercept form.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;y+1=3\\left(x - 6\\right) \\\\ &amp;y+1=3x - 18 &amp;&amp; \\text{Distribute 3} \\\\ &amp;y=3x - 19 &amp;&amp; \\text{Simplify to slope-intercept form} \\end{align}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137731407\">Write the point-slope form of an equation of a line with a slope of \u20132 that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\r\n[reveal-answer q=\"484667\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"484667\"]\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<\/section><section id=\"fs-id1165137804818\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]103509[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Using Two Points<\/h2>\r\n<p id=\"fs-id1165137413780\">The 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>\r\n\r\n<div id=\"fs-id1165137566741\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{m}&amp;=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ &amp;=\\frac{2 - 1}{3 - 0} \\\\ &amp;=\\frac{1}{3} \\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165135431090\">Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.<\/p>\r\n\r\n<div id=\"fs-id1165137468831\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\ y - 1&amp;=\\frac{1}{3}\\left(x - 0\\right)\\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165137638900\">As before, we can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n\r\n<div id=\"eip-403\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;y - 1=\\frac{1}{3}\\left(x - 0\\right) \\\\ &amp;y - 1=\\frac{1}{3}x &amp;&amp; \\text{Distribute the }\\frac{1}{3} \\\\ &amp;y=\\frac{1}{3}x+1 &amp;&amp; \\text{Add 1 to each side} \\end{align}[\/latex]<\/div>\r\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\r\n<p id=\"fs-id1165135456723\">Both equations describe the line shown in Figure 7.<\/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\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_0142.jpg\" alt=\"\" width=\"487\" height=\"386\" \/> <b>Figure 7<\/b>[\/caption]\r\n\r\n<div id=\"Example_02_01_06\" class=\"example\">\r\n<div id=\"fs-id1165137736140\" class=\"exercise\">\r\n<div id=\"fs-id1165137736142\" class=\"problem textbox shaded\">\r\n<h3>Example 6: Writing Linear Equations Using Two Points<\/h3>\r\n<p id=\"fs-id1165137634384\">Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.<\/p>\r\n[reveal-answer q=\"515596\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"515596\"]\r\n<p id=\"fs-id1165137749763\">Let\u2019s begin by finding the slope.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}&amp;=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &amp;=\\frac{7 - 1}{8 - 5} \\\\ &amp;=\\frac{6}{3} \\\\ &amp;=2 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135530557\">So [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use [latex]\\left(5,1\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}y-{y}_{1}&amp;=m\\left(x-{x}_{1}\\right)\\\\ y - 1&amp;=2\\left(x - 5\\right)\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135431079\">The point-slope equation of the line is [latex]{y}_{2}-1=2\\left({x}_{2}-5\\right)[\/latex]. To rewrite the equation in slope-intercept form, we use algebra.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 1=2\\left(x - 5\\right) \\\\ y - 1=2x - 10 \\\\ y=2x - 9 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165135445782\">The slope-intercept equation of the line is [latex]y=2x - 9[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137863827\">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 it in the slope-intercept form.<\/p>\r\n[reveal-answer q=\"451031\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"451031\"]\r\n\r\n[latex]y - 0=-3\\left(x - 0\\right)[\/latex] ; [latex]y=-3x[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 style=\"text-align: center;\">Writing and Interpreting an Equation for a Linear Function<\/h2>\r\n<section id=\"fs-id1165137770240\">\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>\u00a0in Figure 8.<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\/1227\/2015\/04\/03010640\/CNX_Precalc_Figure_02_01_0062.jpg\" alt=\"Graph depicting how to calculate the slope of a line\" width=\"487\" height=\"347\" \/> <b>Figure 8<\/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{align} m&amp;=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &amp;=\\frac{4 - 7}{4 - 0} \\\\ &amp;=-\\frac{3}{4} \\end{align}[\/latex]\r\n[latex][\/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{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y - 4=-\\frac{3}{4}\\left(x - 4\\right) \\end{gathered}[\/latex]\r\n[latex][\/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{gathered}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{gathered}[\/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\/1227\/2015\/04\/03010640\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/> <b>Figure 9<\/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(xt)=-\\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 id=\"fs-id1165137824876\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137824881\">How To: 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<div id=\"Example_02_01_07\" class=\"example\">\r\n<div id=\"fs-id1165137602269\" class=\"exercise\">\r\n<div id=\"fs-id1165137602271\" class=\"problem textbox shaded\">\r\n<h3>Example 7: Writing an Equation for a Linear Function<\/h3>\r\nWrite an equation for a linear function given a graph of <em>f<\/em>\u00a0shown in Figure 10.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"369\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010640\/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\" \/> <b>Figure 10<\/b>[\/caption]\r\n\r\n<div class=\"mceTemp\"><\/div>\r\n[reveal-answer q=\"721194\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"721194\"]\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<p style=\"text-align: center;\">[latex]\\begin{align} m&amp;=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &amp;=\\frac{-4 - 2}{-2 - 0} \\\\ &amp;=\\frac{-6}{-2} \\\\ &amp;=3 \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{gathered}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{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y+4=3\\left(x+2\\right) \\\\ y+4=3x+6 \\\\ y=3x+2 \\end{gathered}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis makes sense because we can see from Figure 11\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[caption id=\"\" align=\"aligncenter\" width=\"369\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010640\/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\" \/> <b>Figure 11<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_02_01_08\" class=\"example\">\r\n<div id=\"fs-id1165137426936\" class=\"exercise\">\r\n<div id=\"fs-id1165137757806\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Writing an Equation for a Linear Cost Function<\/h3>\r\n<p id=\"fs-id1165137638162\">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>\r\n[reveal-answer q=\"116259\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"116259\"]\r\n\r\nThe 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].\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135511326\">If Ben produces 100 items in a month, his monthly cost is represented by<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(100\\right)&amp;=1250+37.5\\left(100\\right) \\\\ &amp;=5000 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137862645\">So his monthly cost would be $5,000.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_02_01_09\" class=\"example\">\r\n<div id=\"fs-id1165137459731\" class=\"exercise\">\r\n<div id=\"fs-id1165135259656\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Writing an Equation for a Linear Function Given Two Points<\/h3>\r\n<p id=\"fs-id1165135259662\">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>\r\n[reveal-answer q=\"666963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"666963\"]\r\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(3\\right)=-2\\to \\left(3,-2\\right) \\\\ &amp;f\\left(8\\right)=1\\to \\left(8,1\\right) \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align} m=&amp;\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &amp;=\\frac{1-\\left(-2\\right)}{8 - 3} \\\\ &amp;=\\frac{3}{5} \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y-\\left(-2\\right)=\\frac{3}{5}\\left(x - 3\\right) \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y+2=\\frac{3}{5}\\left(x - 3\\right) \\\\ y+2=\\frac{3}{5}x-\\frac{9}{5} \\\\ y=\\frac{3}{5}x-\\frac{19}{5} \\end{gathered}[\/latex]<\/p>\r\nAnd since the function is\u00a0<em>f<\/em> we write it with function notation:\r\n<p style=\"text-align: center;\">[latex]f(x)=\\frac{3}{5}x-\\frac{19}{5}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135456746\">If [latex]f\\left(x\\right)[\/latex] is a linear function, with [latex]f\\left(2\\right)=-11[\/latex], and [latex]f\\left(4\\right)=-25[\/latex], find an equation for the function in slope-intercept form.<\/p>\r\n[reveal-answer q=\"177290\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"177290\"]\r\n\r\n[latex]f(x)=-7x+3[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137894282\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try it 9<\/h3>\r\n[ohm_question hide_question_numbers=1]169761[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 style=\"text-align: center;\">Modeling Real-World Problems with Linear Functions<\/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 id=\"fs-id1165137900351\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137404879\">How To: 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\n<div id=\"Example_02_01_10\" class=\"example\">\r\n<div id=\"fs-id1165137836747\" class=\"exercise\">\r\n<div id=\"fs-id1165137836750\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Using a Linear Function to Determine the Number of Songs in a Music Collection<\/h3>\r\n<p id=\"fs-id1165137740859\">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>\r\n[reveal-answer q=\"296335\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"296335\"]\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\/1227\/2015\/04\/03010641\/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<p style=\"text-align: center;\">[latex]\\begin{align}N\\left(12\\right)&amp;=15\\left(12\\right)+200 \\\\ &amp;=180+200 \\\\ &amp;=380 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137694205\">Marcus will have 380 songs in 12 months.<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\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\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_02_01_11\" class=\"example\">\r\n<div id=\"fs-id1165137803740\" class=\"exercise\">\r\n<div id=\"fs-id1165137766973\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Using a Linear Function to Calculate Salary Plus Commission<\/h3>\r\n<p id=\"fs-id1165137742312\">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>\r\n[reveal-answer q=\"91726\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"91726\"]\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<p style=\"text-align: center;\">[latex]\\begin{align}m&amp;=\\frac{920 - 760}{5 - 3} \\\\ &amp;=\\frac{$160}{\\text{2 policies}} \\\\ &amp;=$80\\text{ per policy} \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{align}I\\left(n\\right)=&amp;80n+b \\\\ 760=&amp;80\\left(3\\right)+b &amp;&amp; \\text{when }n=3, I\\left(3\\right)=760 \\\\ 760 - &amp;80\\left(3\\right)=b \\\\ 520=&amp;b \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/p>\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<\/div>\r\n<\/div>\r\n<div id=\"Example_02_01_12\" class=\"example\">\r\n<div id=\"fs-id1165137705959\" class=\"exercise\">\r\n<div id=\"fs-id1165137705961\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Using Tabular Form to Write an Equation for a Linear Function<\/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=\"428388\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"428388\"]\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\" style=\"text-align: center;\">[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<p style=\"text-align: center;\">[latex]\\begin{align}m&amp;=\\frac{1240 - 1080}{6 - 2} \\\\ &amp;=\\frac{160}{4} \\\\ &amp;=40 \\end{align}[\/latex]<\/p>\r\n\r\n<div><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137725510\" class=\"note precalculus qa textbox\">\r\n<h3 id=\"fs-id1165137920800\">Q &amp; A<\/h3>\r\n<strong>Is the initial value always provided in a table of values like the table in Example 12?<\/strong>\r\n<p id=\"fs-id1165137762000\"><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>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137645254\">A new plant food was introduced to a young tree to test its effect on the height of the tree. The table below\u00a0shows the height of the tree, in feet, <em>x<\/em>\u00a0months since the measurements began. Write a linear function, <em>H<\/em>(<em>x<\/em>), where <em>x<\/em>\u00a0is the number of months since the start of the experiment.<\/p>\r\n\r\n<table id=\"Table_02_01_03\" summary=\"Two rows and six columns. The first row is labeled, 'x'. The second row is labeled is labeled, 'H(x)'. Reading the remaining rows as ordered pairs (i.e., (x, H(x)), we have the following values: (0, 12.5), (2, 13.5), (4, 14.5), (8, 16.5), and (12, 18.5).\">\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>0<\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>8<\/td>\r\n<td>12<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><em><strong>H<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\r\n<td>12.5<\/td>\r\n<td>13.5<\/td>\r\n<td>14.5<\/td>\r\n<td>16.5<\/td>\r\n<td>18.5<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"389435\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"389435\"]\r\n\r\n[latex]H\\left(x\\right)=0.5x+12.5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137784950\" class=\"key-equations\">\r\n<h2>Key Equations<\/h2>\r\n<table id=\"fs-id1165137784956\" summary=\"...\">\r\n<tbody>\r\n<tr>\r\n<td>slope-intercept form of a line<\/td>\r\n<td>[latex]f\\left(x\\right)=mx+b[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>slope<\/td>\r\n<td>[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>point-slope form of a line<\/td>\r\n<td>[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<\/section><section id=\"fs-id1165135696154\" class=\"key-concepts\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165137736447\">\r\n \t<li>The ordered pairs given by a linear function represent points on a line.<\/li>\r\n \t<li>Linear functions can be represented in words, function notation, tabular form, and graphical form.<\/li>\r\n \t<li>The rate of change of a linear function is also known as the slope.<\/li>\r\n \t<li>An equation in the slope-intercept form of a line includes the slope and the initial value of the function.<\/li>\r\n \t<li>The initial value, or <em>y<\/em>-intercept, is the output value when the input of a linear function is zero. It is the <em>y<\/em>-value of the point at which the line crosses the <em>y<\/em>-axis.<\/li>\r\n \t<li>An increasing linear function results in a graph that slants upward from left to right and has a positive slope.<\/li>\r\n \t<li>A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.<\/li>\r\n \t<li>A constant linear function results in a graph that is a horizontal line.<\/li>\r\n \t<li>Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant.<\/li>\r\n \t<li>The slope of a linear function can be calculated by dividing the difference between <em>y<\/em>-values by the difference in corresponding <em>x<\/em>-values of any two points on the line.<\/li>\r\n \t<li>The slope and initial value can be determined given a graph or any two points on the line.<\/li>\r\n \t<li>One type of function notation is the slope-intercept form of an equation.<\/li>\r\n \t<li>The point-slope form is useful for finding a linear equation when given the slope of a line and one point.<\/li>\r\n \t<li>The point-slope form is also convenient for finding a linear equation when given two points through which a line passes.<\/li>\r\n \t<li>The equation for a linear function can be written if the slope <em>m<\/em>\u00a0and initial value <em>b\u00a0<\/em>are known.<\/li>\r\n \t<li>A linear function can be used to solve real-world problems.<\/li>\r\n \t<li>A linear function can be written from tabular form.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165137405111\" class=\"definition\">\r\n \t<dt>decreasing linear function<\/dt>\r\n \t<dd id=\"fs-id1165137405116\">a function with a negative slope: If [latex]f\\left(x\\right)=mx+b, \\text{then} m&lt;0[\/latex].<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137863356\" class=\"definition\">\r\n \t<dt>increasing linear function<\/dt>\r\n \t<dd id=\"fs-id1165135188274\">a function with a positive slope: If [latex]f\\left(x\\right)=mx+b, \\text{then} m&gt;0[\/latex].<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt>linear function<\/dt>\r\n \t<dd id=\"fs-id1165135429394\">a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165134389091\" class=\"definition\">\r\n \t<dt>point-slope form<\/dt>\r\n \t<dd id=\"fs-id1165134389097\">the equation for a line that represents a linear function of the form [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137635132\" class=\"definition\">\r\n \t<dt>slope<\/dt>\r\n \t<dd id=\"fs-id1165137635137\">the ratio of the change in output values to the change in input values; a measure of the steepness of a line<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137817449\" class=\"definition\">\r\n \t<dt>slope-intercept form<\/dt>\r\n \t<dd id=\"fs-id1165137817454\">the equation for a line that represents a linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135195656\" class=\"definition\">\r\n \t<dt><em>y<\/em>-intercept<\/dt>\r\n \t<dd id=\"fs-id1165137635107\">the value of a function when the input value is zero; also known as initial value<\/dd>\r\n<\/dl>\r\n<\/section><\/section><\/section><\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Identify and Interpret the Slope and Vertical Intercept of a Line<\/li>\n<li>Calculate and interpret the slope of a line.<\/li>\n<li>Determine the equation of a linear function.<\/li>\n<\/ul>\n<\/div>\n<p><img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010637\/CNX_Precalc_02_00_012.jpg\" alt=\"An upward view of bamboo trees.\" \/><\/p>\n<p id=\"fs-id1165137705130\">Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour. <a class=\"footnote\" title=\"http:\/\/www.guinnessworldrecords.com\/records-3000\/fastest-growing-plant\/\" id=\"return-footnote-13798-1\" href=\"#footnote-13798-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function.<\/p>\n<p id=\"fs-id1165137605883\">Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data.<\/p>\n<figure id=\"CNX_Precalc_Figure_02_01_001\">\n<div style=\"width: 335px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0112.jpg\" alt=\"Front view of a subway train, the maglev train.\" width=\"325\" height=\"432\" \/><\/p>\n<p class=\"wp-caption-text\">Shanghai MagLev Train (credit: &#8220;kanegen&#8221;\/Flickr)<\/p>\n<\/div>\n<\/figure>\n<p id=\"fs-id1165137697072\">Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train. It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.<a class=\"footnote\" title=\"http:\/\/www.chinahighlights.com\/shanghai\/transportation\/maglev-train.htm\" id=\"return-footnote-13798-2\" href=\"#footnote-13798-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a><\/p>\n<p id=\"fs-id1165134340085\">Suppose a maglev train were to travel a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train\u2019s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train\u2019s distance from the station at a given point in time.<\/p>\n<h2 style=\"text-align: left;\">Representing Linear Functions<\/h2>\n<p id=\"fs-id1165137573850\">The function describing the train\u2019s motion is a <strong>linear function<\/strong>, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train\u2019s motion as a function using each method.<\/p>\n<section id=\"fs-id1165137759903\">\n<h3 style=\"text-align: center;\">Representing a Linear Function in Word Form<\/h3>\n<p id=\"fs-id1165137588695\">Let\u2019s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.<\/p>\n<ul id=\"fs-id1165135526954\">\n<li><em>The train\u2019s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.<\/em><\/li>\n<\/ul>\n<p id=\"fs-id1165135188466\">The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.<\/p>\n<\/section>\n<section id=\"fs-id1165135639903\">\n<h3 style=\"text-align: center;\">Representing a Linear Function in Function Notation<\/h3>\n<p id=\"fs-id1165137833100\">Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the form known as the <strong>slope-intercept form<\/strong> of a line, where [latex]x[\/latex] is the input value, [latex]m[\/latex] is the rate of change, and [latex]b[\/latex] is the initial value of the dependent variable.<\/p>\n<div id=\"Equation_02_01_01\" class=\"equation\" style=\"text-align: center;\">[latex]\\begin{align}&\\text{Equation form} && y=mx+b \\\\ &\\text{Function notation} && f(x)=mx+b\\\\& \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137411219\">In the example of the train, we might use the notation [latex]D\\left(t\\right)[\/latex]\u00a0in which the total distance [latex]D[\/latex]<br \/>\nis a function of the time [latex]t[\/latex].\u00a0The rate, [latex]m[\/latex],\u00a0is 83 meters per second. The initial value of the dependent variable [latex]b[\/latex]\u00a0is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.<\/p>\n<div id=\"fs-id1165137559254\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]D(t)=83t+250[\/latex]<\/div>\n<\/section>\n<section id=\"fs-id1165135415800\">\n<h3 style=\"text-align: center;\">Representing a Linear Function in Tabular Form<\/h3>\n<p id=\"fs-id1165137438406\">A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in the table in Figure 1. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time.<\/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\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0152.jpg\" alt=\"Table with the first row, labeled t, containing the seconds from 0 to 3, and with the second row, labeled D(t), containing the meters 250 to 499. The first row goes up by 1 second, and the second row goes up by 83 meters.\" width=\"487\" height=\"161\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 1.<\/b> Tabular representation of the function D showing selected input and output values<\/p>\n<\/div>\n<div id=\"fs-id1165137482942\" class=\"note precalculus qa textbox\">\n<h3 id=\"fs-id1165137828205\"><strong>Q &amp; A<\/strong><\/h3>\n<p><strong>Can the input in the previous example be any real number?<\/strong><\/p>\n<p id=\"fs-id1165135209002\"><em>No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers.<\/em><\/p>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137619188\">\n<h3 style=\"text-align: center;\">Representing a Linear Function in Graphical Form<\/h3>\n<p id=\"fs-id1165137827353\">Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, [latex]D(t)=83t+250[\/latex], to draw a graph, represented in the graph in Figure 2. Notice the graph is a line. When we plot a linear function, the graph is always a line.<\/p>\n<p id=\"fs-id1165137451297\">The rate of change, which is constant, determines the slant, or <strong>slope<\/strong> of the line. The point at which the input value is zero is the vertical intercept, or <strong><em>y<\/em>-intercept<\/strong>, of the line. We can see from the graph that the <em>y<\/em>-intercept in the train example we just saw is [latex]\\left(0,250\\right)[\/latex]\u00a0and represents the distance of the train from the station when it began moving at a constant speed.<\/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\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0122.jpg\" alt=\"A graph of an increasing function with points at (-2, -4) and (0, 2).\" width=\"487\" height=\"289\" \/><\/p>\n<p style=\"text-align: center;\"><strong>Figure 2.<\/strong> The graph of [latex]D(t)=83t+250[\/latex]. Graphs of linear functions are lines because the rate of change is constant.<\/p>\n<p id=\"fs-id1165137715509\">Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line [latex]f(x)=2{x}_{}+1[\/latex].\u00a0Ask yourself what numbers can be input to the function, that is, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product.<\/p>\n<div id=\"fs-id1165137726089\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Linear Function<\/h3>\n<p id=\"fs-id1165137454496\">A <strong>linear function<\/strong> is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line<\/p>\n<div id=\"Equation_02_01_02\" class=\"equation\" style=\"text-align: center;\">[latex]f(x)=mx+b[\/latex]<\/div>\n<p id=\"fs-id1165137784222\">where [latex]b[\/latex]\u00a0is the initial or starting value of the function (when input, [latex]x=0[\/latex]), and [latex]m[\/latex]\u00a0is the constant rate of change, or <strong>slope<\/strong> of the function. The <strong><em>y<\/em>-intercept<\/strong> is at [latex]\\left(0,b\\right)[\/latex].<\/p>\n<\/div>\n<div id=\"Example_02_01_01\" class=\"example\">\n<div id=\"fs-id1165137583894\" class=\"exercise\">\n<div id=\"fs-id1165135209144\" class=\"problem textbox shaded\">\n<h3>Example 1: Using a Linear Function to Find the Pressure on a Diver<\/h3>\n<p>The pressure, [latex]P[\/latex],\u00a0in pounds per square inch (PSI) on the diver in Figure 3\u00a0depends upon her depth below the water surface, [latex]d[\/latex], in feet. This relationship may be modeled by the equation, [latex]P(d)=0.434d+14.696[\/latex]. Restate this function in words.<\/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\/1227\/2015\/04\/03010638\/CNX_Precalc_Figure_02_01_0032.jpg\" alt=\"Scuba diver.\" width=\"487\" height=\"366\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 3.<\/b> (credit: Ilse Reijs and Jan-Noud Hutten)<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q129605\">Show Solution<\/span><\/p>\n<div id=\"q129605\" class=\"hidden-answer\" style=\"display: none\">\n<p>To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137740917\">The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot. This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137738187\" class=\"commentary\"><\/div>\n<\/div>\n<\/div>\n<\/section>\n<h2 style=\"text-align: center;\">Determining Whether a Linear Function is Increasing, Decreasing, or Constant<\/h2>\n<section id=\"fs-id1165137749252\">\n<div style=\"width: 985px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_004abc2.jpg\" alt=\"Three graphs depicting an increasing function, a decreasing function, and a constant function.\" width=\"975\" height=\"375\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 4<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165135482019\">The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant.<\/p>\n<p>For an <strong>increasing function<\/strong>, as with the train example,<\/p>\n<p style=\"text-align: center;\"><strong><em>the output values increase as the input values increase. <\/em><\/strong><\/p>\n<p>The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in <strong>(a)<\/strong>.<\/p>\n<p>For a <strong>decreasing function<\/strong>, the slope is negative.<\/p>\n<p style=\"text-align: center;\"><strong><em>The output values decrease as the input values increase. <\/em><\/strong><\/p>\n<p>A line with a negative slope slants downward from left to right as in <strong>(b)<\/strong>. If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in <strong>(c)<\/strong>.<span id=\"fs-id1165137453957\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1165137446154\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Increasing and Decreasing Functions<\/h3>\n<p id=\"fs-id1165134085973\">The slope determines if the function is an <strong>increasing linear function<\/strong>, a <strong>decreasing linear function<\/strong>, or a constant function.<\/p>\n<ul id=\"eip-643\">\n<li>[latex]f(x)=mx+b\\text{ is an increasing function if }m>0[\/latex].<\/li>\n<li>[latex]f(x)=mx+b\\text{ is an decreasing function if }m<0[\/latex].<\/li>\n<li>[latex]f(x)=mx+b\\text{ is a constant function if }m=0[\/latex].<\/li>\n<\/ul>\n<\/div>\n<div id=\"Example_02_01_02\" class=\"example\">\n<div id=\"fs-id1165137405281\" class=\"exercise\">\n<div id=\"fs-id1165135571684\" class=\"problem textbox shaded\">\n<h3>Example 2: Deciding whether a Function Is Increasing, Decreasing, or Constant<\/h3>\n<p id=\"fs-id1165137400625\">Some recent studies suggest that a teenager sends an average of 60 texts per day.<a class=\"footnote\" title=\"http:\/\/www.cbsnews.com\/8301-501465_162-57400228-501465\/teens-are-sending-60-texts-a-day-study-says\/\" id=\"return-footnote-13798-3\" href=\"#footnote-13798-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>\u00a0For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant.<\/p>\n<ol id=\"fs-id1165137807449\">\n<li>The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent.<\/li>\n<li>A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month.<\/li>\n<li>A teen has an unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q539033\">Show Solution<\/span><\/p>\n<div id=\"q539033\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137508045\">Analyze each function.<\/p>\n<ol id=\"fs-id1165137778959\">\n<li>The function can be represented as [latex]f(x)=60x[\/latex] where [latex]x[\/latex] is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day.<\/li>\n<li>The function can be represented as [latex]f(x)=500 - 60x[\/latex] where [latex]x[\/latex] is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after [latex]x[\/latex] days.<\/li>\n<li>The cost function can be represented as [latex]f(x)=50[\/latex] because the number of days does not affect the total cost. The slope is 0 so the function is constant.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137837055\">\n<h2 id=\"fs-id1165137410528\" class=\"note precalculus try\" style=\"text-align: center;\">Calculating and Interpreting Slope<\/h2>\n<p id=\"fs-id1165137573268\">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] \u2014which 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]\u2014we can calculate the slope [latex]m[\/latex],\u00a0as follows<\/p>\n<div id=\"fs-id1165137757690\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137767606\">where [latex]\\Delta y[\/latex] is the vertical displacement and [latex]\\Delta x[\/latex] is the horizontal displacement. Note in function notation two corresponding values for the output [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] for the function [latex]f[\/latex], [latex]{y}_{1}=f\\left({x}_{1}\\right)[\/latex] and [latex]{y}_{2}=f\\left({x}_{2}\\right)[\/latex], so we could equivalently write<\/p>\n<div id=\"fs-id1165137438737\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\frac{f\\left({x}_{2}\\right)-f\\left({x}_{1}\\right)}{{x}_{2}-{x}_{1}}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p>The graph in Figure 5\u00a0indicates how the slope of the line between the points, [latex]\\left({x}_{1,}{y}_{1}\\right)[\/latex]<br \/>\nand [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-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010639\/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\"><b>Figure 5<\/b><\/p>\n<\/div>\n<p><em>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.<\/em><\/p>\n<div id=\"fs-id1165137831952\" class=\"note precalculus qa textbox\">\n<h3 id=\"fs-id1165137501372\">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 id=\"fs-id1165137462769\"><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 id=\"fs-id1165135161219\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Calculate Slope<\/h3>\n<p id=\"fs-id1165134069119\">The slope, or rate of change, of a function [latex]m[\/latex] can be calculated according to the following:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex][\/latex]<\/p>\n<p id=\"fs-id1165135414432\">where [latex]{x}_{1}[\/latex] and [latex]{x}_{2}[\/latex] are input values, [latex]{y}_{1}[\/latex] and [latex]{y}_{2}[\/latex] are output values.<\/p>\n<\/div>\n<div id=\"fs-id1165137539010\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137675584\">How To: Given two points from a linear function, calculate and interpret the slope.<\/h3>\n<ol id=\"fs-id1165137820100\">\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 id=\"Example_02_01_03\" class=\"example\">\n<div id=\"fs-id1165137641892\" class=\"exercise\">\n<div id=\"fs-id1165137811167\" class=\"problem textbox shaded\">\n<h3>Example 3: Finding the Slope of a Linear Function<\/h3>\n<p id=\"fs-id1165137724908\">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=\"q603714\">Show Solution<\/span><\/p>\n<div id=\"q603714\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137714054\">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 style=\"text-align: center;\">[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 id=\"fs-id1165137469338\">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 id=\"fs-id1165137842413\">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>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137451238\">If [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function, and [latex]\\left(2,\\text{ }3\\right)[\/latex]\u00a0and [latex]\\left(0,\\text{ }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=\"q786191\">Show Solution<\/span><\/p>\n<div id=\"q786191\" 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<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169760\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169760&theme=oea&iframe_resize_id=ohm169760\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"Example_02_01_04\" class=\"example\">\n<div id=\"fs-id1165135307935\" class=\"exercise\">\n<div id=\"fs-id1165135307937\" class=\"problem textbox shaded\">\n<h3>Example 4: Finding the Population Change from a Linear Function<\/h3>\n<p id=\"fs-id1165135386495\">The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of 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=\"q870587\">Show Solution<\/span><\/p>\n<div id=\"q870587\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137415969\">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 style=\"text-align: center;\">[latex]\\frac{4,400\\text{ people}}{4\\text{ years}}=1,100\\text{ }\\frac{\\text{people}}{\\text{year}}[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex][\/latex]<\/p>\n<p id=\"fs-id1165137705484\">So the population increased by 1,100 people per year.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137451439\">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>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3 id=\"fs-id1165137425642\">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 of 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=\"q468046\">Show Solution<\/span><\/p>\n<div id=\"q468046\" 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<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm57147\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=57147&theme=oea&iframe_resize_id=ohm57147&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2 style=\"text-align: center;\">Writing the Point-Slope Form of a Linear Equation<\/h2>\n<p id=\"fs-id1165137639245\">Up until now, we have been using the slope-intercept form of a linear equation to describe linear functions. Here, we will learn another way to write a linear function, the <strong>point-slope form<\/strong>.<\/p>\n<div id=\"fs-id1165137452508\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137551238\">The point-slope form is derived from the slope formula.<\/p>\n<div id=\"eip-301\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&{m}=\\frac{y-{y}_{1}}{x-{x}_{1}} && \\text{assuming }{ x }\\ne {x}_{1} \\\\ &{ m }\\left(x-{x}_{1}\\right)=\\frac{y-{y}_{1}}{x-{x}_{1}}\\left(x-{x}_{1}\\right) && \\text{Multiply both sides by }\\left(x-{x}_{1}\\right)\\\\ &{ m }\\left(x-{x}_{1}\\right)=y-{y}_{1} && \\text{Simplify} \\\\ &y-{y}_{1}={ m }\\left(x-{x}_{1}\\right) &&\\text{Rearrange} \\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137844021\">Keep in mind that the slope-intercept form and the point-slope form can be used to describe the same function. We can move from one form to another using basic algebra. For example, suppose we are given an equation in point-slope form, [latex]y - 4=-\\frac{1}{2}\\left(x - 6\\right)[\/latex] . We can convert it to the slope-intercept form as shown.<\/p>\n<div id=\"eip-424\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&y - 4=-\\frac{1}{2}\\left(x - 6\\right) \\\\ &y - 4=-\\frac{1}{2}x+3 && \\text{Distribute the }-\\frac{1}{2}. \\\\ &y=-\\frac{1}{2}x+7 && \\text{Add 4 to each side}. \\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137409234\">Therefore, the same line can be described in slope-intercept form as [latex]y=-\\frac{1}{2}x+7[\/latex].<\/p>\n<div id=\"fs-id1165137461941\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Point-Slope Form of a Linear Equation<\/h3>\n<p id=\"fs-id1165137647838\">The <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 id=\"fs-id1165137663648\">where [latex]m[\/latex]\u00a0is the slope, [latex]{x}_{1 }[\/latex] and [latex]{y}_{1}[\/latex]\u00a0are the [latex]x[\/latex] and [latex]y[\/latex]\u00a0coordinates of a specific point through which the line passes.<\/p>\n<\/div>\n<section id=\"fs-id1165137745298\">\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Using a Point and the Slope<\/h2>\n<p id=\"fs-id1165137444576\">The point-slope form is particularly useful if we know one point and the slope of a line. Suppose, for example, 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 the general point-slope equation.<\/p>\n<div id=\"fs-id1165134380388\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\\\ y - 1=2\\left(x - 4\\right)\\end{gathered}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137639550\">If we wanted to then rewrite the equation in slope-intercept form, we apply algebraic techniques.<\/p>\n<div id=\"eip-420\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&y - 1=2\\left(x - 4\\right) \\\\ &y - 1=2x - 8 && \\text{Distribute the }2 \\\\ &y=2x - 7 && \\text{Add 1 to each side} \\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165135333698\">Both equations, [latex]y - 1=2\\left(x - 4\\right)[\/latex]\u00a0and [latex]y=2x - 7[\/latex], describe the same line. See Figure 6.<span id=\"fs-id1165137925529\"><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\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_0132.jpg\" alt=\"\" width=\"487\" height=\"386\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 6<\/b><\/p>\n<\/div>\n<div id=\"Example_02_01_05\" class=\"example\">\n<div id=\"fs-id1165137423582\" class=\"exercise\">\n<div id=\"fs-id1165137423584\" class=\"problem textbox shaded\">\n<h3>Example 5: Writing Linear Equations Using a Point and the Slope<\/h3>\n<p id=\"fs-id1165137736482\">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 it in the slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q500465\">Show Solution<\/span><\/p>\n<div id=\"q500465\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137530790\">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 the general point-slope equation.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ &y-\\left(-1\\right)=3\\left(x - 6\\right) && \\text{Substitute known values} \\\\ &y+1=3\\left(x - 6\\right) && \\text{Distribute }-1\\text{ to find point-slope form} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137594974\">Then we use algebra to find the slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&y+1=3\\left(x - 6\\right) \\\\ &y+1=3x - 18 && \\text{Distribute 3} \\\\ &y=3x - 19 && \\text{Simplify to slope-intercept form} \\end{align}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137731407\">Write the point-slope form of an equation of a line with a slope of \u20132 that passes through the point [latex]\\left(-2,\\text{ }2\\right)[\/latex]. Then rewrite it in the slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q484667\">Show Solution<\/span><\/p>\n<div id=\"q484667\" 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<\/section>\n<section id=\"fs-id1165137804818\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm103509\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=103509&theme=oea&iframe_resize_id=ohm103509\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Using Two Points<\/h2>\n<p id=\"fs-id1165137413780\">The 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<div id=\"fs-id1165137566741\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}{m}&=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\\ &=\\frac{2 - 1}{3 - 0} \\\\ &=\\frac{1}{3} \\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165135431090\">Now we can use the slope we found and the coordinates of one of the points to find the equation for the line. Let use (0, 1) for our point.<\/p>\n<div id=\"fs-id1165137468831\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\ y - 1&=\\frac{1}{3}\\left(x - 0\\right)\\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165137638900\">As before, we can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n<div id=\"eip-403\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&y - 1=\\frac{1}{3}\\left(x - 0\\right) \\\\ &y - 1=\\frac{1}{3}x && \\text{Distribute the }\\frac{1}{3} \\\\ &y=\\frac{1}{3}x+1 && \\text{Add 1 to each side} \\end{align}[\/latex]<\/div>\n<div style=\"text-align: center;\">[latex][\/latex]<\/div>\n<p id=\"fs-id1165135456723\">Both equations describe the line shown in Figure 7.<\/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\/1227\/2015\/04\/03010639\/CNX_Precalc_Figure_02_01_0142.jpg\" alt=\"\" width=\"487\" height=\"386\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 7<\/b><\/p>\n<\/div>\n<div id=\"Example_02_01_06\" class=\"example\">\n<div id=\"fs-id1165137736140\" class=\"exercise\">\n<div id=\"fs-id1165137736142\" class=\"problem textbox shaded\">\n<h3>Example 6: Writing Linear Equations Using Two Points<\/h3>\n<p id=\"fs-id1165137634384\">Write the point-slope form of an equation of a line that passes through the points (5, 1) and (8, 7). Then rewrite it in the slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q515596\">Show Solution<\/span><\/p>\n<div id=\"q515596\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137749763\">Let\u2019s begin by finding the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}&=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &=\\frac{7 - 1}{8 - 5} \\\\ &=\\frac{6}{3} \\\\ &=2 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135530557\">So [latex]m=2[\/latex]. Next, we substitute the slope and the coordinates for one of the points into the general point-slope equation. We can choose either point, but we will use [latex]\\left(5,1\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}y-{y}_{1}&=m\\left(x-{x}_{1}\\right)\\\\ y - 1&=2\\left(x - 5\\right)\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135431079\">The point-slope equation of the line is [latex]{y}_{2}-1=2\\left({x}_{2}-5\\right)[\/latex]. To rewrite the equation in slope-intercept form, we use algebra.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y - 1=2\\left(x - 5\\right) \\\\ y - 1=2x - 10 \\\\ y=2x - 9 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165135445782\">The slope-intercept equation of the line is [latex]y=2x - 9[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137863827\">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 it in the slope-intercept form.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q451031\">Show Solution<\/span><\/p>\n<div id=\"q451031\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y - 0=-3\\left(x - 0\\right)[\/latex] ; [latex]y=-3x[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 style=\"text-align: center;\">Writing and Interpreting an Equation for a Linear Function<\/h2>\n<section id=\"fs-id1165137770240\">\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>\u00a0in Figure 8.<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\/1227\/2015\/04\/03010640\/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>Figure 8<\/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{align} m&=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &=\\frac{4 - 7}{4 - 0} \\\\ &=-\\frac{3}{4} \\end{align}[\/latex]<br \/>\n[latex][\/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{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y - 4=-\\frac{3}{4}\\left(x - 4\\right) \\end{gathered}[\/latex]<br \/>\n[latex][\/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{gathered}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{gathered}[\/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\/1227\/2015\/04\/03010640\/CNX_Precalc_Figure_02_01_0072.jpg\" alt=\"\" width=\"487\" height=\"155\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 9<\/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(xt)=-\\frac{3}{4}x+7[\/latex],\u00a0and the linear equation would be [latex]y=-\\frac{3}{4}x+7[\/latex].<\/p>\n<div id=\"fs-id1165137824876\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137824881\">How To: 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<div id=\"Example_02_01_07\" class=\"example\">\n<div id=\"fs-id1165137602269\" class=\"exercise\">\n<div id=\"fs-id1165137602271\" class=\"problem textbox shaded\">\n<h3>Example 7: Writing an Equation for a Linear Function<\/h3>\n<p>Write an equation for a linear function given a graph of <em>f<\/em>\u00a0shown in Figure 10.<\/p>\n<div style=\"width: 379px\" 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\/1227\/2015\/04\/03010640\/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<p class=\"wp-caption-text\"><b>Figure 10<\/b><\/p>\n<\/div>\n<div class=\"mceTemp\"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q721194\">Show Solution<\/span><\/p>\n<div id=\"q721194\" 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<p style=\"text-align: center;\">[latex]\\begin{align} m&=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &=\\frac{-4 - 2}{-2 - 0} \\\\ &=\\frac{-6}{-2} \\\\ &=3 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137732234\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}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{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137731540\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y+4=3\\left(x+2\\right) \\\\ y+4=3x+6 \\\\ y=3x+2 \\end{gathered}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This makes sense because we can see from Figure 11\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<div style=\"width: 379px\" 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\/1227\/2015\/04\/03010640\/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 class=\"wp-caption-text\"><b>Figure 11<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_01_08\" class=\"example\">\n<div id=\"fs-id1165137426936\" class=\"exercise\">\n<div id=\"fs-id1165137757806\" class=\"problem textbox shaded\">\n<h3>Example 8: Writing an Equation for a Linear Cost Function<\/h3>\n<p id=\"fs-id1165137638162\">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=\"q116259\">Show Solution<\/span><\/p>\n<div id=\"q116259\" class=\"hidden-answer\" style=\"display: none\">\n<p>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].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135511326\">If Ben produces 100 items in a month, his monthly cost is represented by<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(100\\right)&=1250+37.5\\left(100\\right) \\\\ &=5000 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137862645\">So his monthly cost would be $5,000.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_01_09\" class=\"example\">\n<div id=\"fs-id1165137459731\" class=\"exercise\">\n<div id=\"fs-id1165135259656\" class=\"problem textbox shaded\">\n<h3>Example 9: Writing an Equation for a Linear Function Given Two Points<\/h3>\n<p id=\"fs-id1165135259662\">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=\"q666963\">Show Solution<\/span><\/p>\n<div id=\"q666963\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137730075\">We can write the given points using coordinates.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&f\\left(3\\right)=-2\\to \\left(3,-2\\right) \\\\ &f\\left(8\\right)=1\\to \\left(8,1\\right) \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137603531\">We can then use the points to calculate the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align} m=&\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}} \\\\ &=\\frac{1-\\left(-2\\right)}{8 - 3} \\\\ &=\\frac{3}{5} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137851970\">Substitute the slope and the coordinates of one of the points into the point-slope form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y-\\left(-2\\right)=\\frac{3}{5}\\left(x - 3\\right) \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137629473\">We can use algebra to rewrite the equation in the slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}y+2=\\frac{3}{5}\\left(x - 3\\right) \\\\ y+2=\\frac{3}{5}x-\\frac{9}{5} \\\\ y=\\frac{3}{5}x-\\frac{19}{5} \\end{gathered}[\/latex]<\/p>\n<p>And since the function is\u00a0<em>f<\/em> we write it with function notation:<\/p>\n<p style=\"text-align: center;\">[latex]f(x)=\\frac{3}{5}x-\\frac{19}{5}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135456746\">If [latex]f\\left(x\\right)[\/latex] is a linear function, with [latex]f\\left(2\\right)=-11[\/latex], and [latex]f\\left(4\\right)=-25[\/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=\"q177290\">Show Solution<\/span><\/p>\n<div id=\"q177290\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)=-7x+3[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137894282\">\n<div class=\"textbox key-takeaways\">\n<h3>Try it 9<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169761\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169761&theme=oea&iframe_resize_id=ohm169761\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 style=\"text-align: center;\">Modeling Real-World Problems with Linear Functions<\/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 id=\"fs-id1165137900351\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137404879\">How To: 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<div id=\"Example_02_01_10\" class=\"example\">\n<div id=\"fs-id1165137836747\" class=\"exercise\">\n<div id=\"fs-id1165137836750\" class=\"problem textbox shaded\">\n<h3>Example 10: Using a Linear Function to Determine the Number of Songs in a Music Collection<\/h3>\n<p id=\"fs-id1165137740859\">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=\"q296335\">Show Solution<\/span><\/p>\n<div id=\"q296335\" 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\/1227\/2015\/04\/03010641\/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<p style=\"text-align: center;\">[latex]\\begin{align}N\\left(12\\right)&=15\\left(12\\right)+200 \\\\ &=180+200 \\\\ &=380 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137694205\">Marcus will have 380 songs in 12 months.<\/p>\n<h4>Analysis of the Solution<\/h4>\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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_01_11\" class=\"example\">\n<div id=\"fs-id1165137803740\" class=\"exercise\">\n<div id=\"fs-id1165137766973\" class=\"problem textbox shaded\">\n<h3>Example 11: Using a Linear Function to Calculate Salary Plus Commission<\/h3>\n<p id=\"fs-id1165137742312\">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=\"q91726\">Show Solution<\/span><\/p>\n<div id=\"q91726\" 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<p style=\"text-align: center;\">[latex]\\begin{align}m&=\\frac{920 - 760}{5 - 3} \\\\ &=\\frac{$160}{\\text{2 policies}} \\\\ &=$80\\text{ per policy} \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]\\begin{align}I\\left(n\\right)=&80n+b \\\\ 760=&80\\left(3\\right)+b && \\text{when }n=3, I\\left(3\\right)=760 \\\\ 760 - &80\\left(3\\right)=b \\\\ 520=&b \\end{align}[\/latex]<\/p>\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<p style=\"text-align: center;\">[latex]I\\left(n\\right)=80n+520[\/latex]<\/p>\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<\/div>\n<\/div>\n<div id=\"Example_02_01_12\" class=\"example\">\n<div id=\"fs-id1165137705959\" class=\"exercise\">\n<div id=\"fs-id1165137705961\" class=\"problem textbox shaded\">\n<h3>Example 12: Using Tabular Form to Write an Equation for a Linear Function<\/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=\"q428388\">Show Solution<\/span><\/p>\n<div id=\"q428388\" 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\" style=\"text-align: center;\">[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<p style=\"text-align: center;\">[latex]\\begin{align}m&=\\frac{1240 - 1080}{6 - 2} \\\\ &=\\frac{160}{4} \\\\ &=40 \\end{align}[\/latex]<\/p>\n<div><\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137725510\" class=\"note precalculus qa textbox\">\n<h3 id=\"fs-id1165137920800\">Q &amp; A<\/h3>\n<p><strong>Is the initial value always provided in a table of values like the table in Example 12?<\/strong><\/p>\n<p id=\"fs-id1165137762000\"><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 class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137645254\">A new plant food was introduced to a young tree to test its effect on the height of the tree. The table below\u00a0shows the height of the tree, in feet, <em>x<\/em>\u00a0months since the measurements began. Write a linear function, <em>H<\/em>(<em>x<\/em>), where <em>x<\/em>\u00a0is the number of months since the start of the experiment.<\/p>\n<table id=\"Table_02_01_03\" summary=\"Two rows and six columns. The first row is labeled, 'x'. The second row is labeled is labeled, 'H(x)'. Reading the remaining rows as ordered pairs (i.e., (x, H(x)), we have the following values: (0, 12.5), (2, 13.5), (4, 14.5), (8, 16.5), and (12, 18.5).\">\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>8<\/td>\n<td>12<\/td>\n<\/tr>\n<tr>\n<td><em><strong>H<\/strong><\/em><strong>(<em>x<\/em>)<\/strong><\/td>\n<td>12.5<\/td>\n<td>13.5<\/td>\n<td>14.5<\/td>\n<td>16.5<\/td>\n<td>18.5<\/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=\"q389435\">Show Solution<\/span><\/p>\n<div id=\"q389435\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]H\\left(x\\right)=0.5x+12.5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137784950\" class=\"key-equations\">\n<h2>Key Equations<\/h2>\n<table id=\"fs-id1165137784956\" summary=\"...\">\n<tbody>\n<tr>\n<td>slope-intercept form of a line<\/td>\n<td>[latex]f\\left(x\\right)=mx+b[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>slope<\/td>\n<td>[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/td>\n<\/tr>\n<tr>\n<td>point-slope form of a line<\/td>\n<td>[latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/section>\n<section id=\"fs-id1165135696154\" class=\"key-concepts\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165137736447\">\n<li>The ordered pairs given by a linear function represent points on a line.<\/li>\n<li>Linear functions can be represented in words, function notation, tabular form, and graphical form.<\/li>\n<li>The rate of change of a linear function is also known as the slope.<\/li>\n<li>An equation in the slope-intercept form of a line includes the slope and the initial value of the function.<\/li>\n<li>The initial value, or <em>y<\/em>-intercept, is the output value when the input of a linear function is zero. It is the <em>y<\/em>-value of the point at which the line crosses the <em>y<\/em>-axis.<\/li>\n<li>An increasing linear function results in a graph that slants upward from left to right and has a positive slope.<\/li>\n<li>A decreasing linear function results in a graph that slants downward from left to right and has a negative slope.<\/li>\n<li>A constant linear function results in a graph that is a horizontal line.<\/li>\n<li>Analyzing the slope within the context of a problem indicates whether a linear function is increasing, decreasing, or constant.<\/li>\n<li>The slope of a linear function can be calculated by dividing the difference between <em>y<\/em>-values by the difference in corresponding <em>x<\/em>-values of any two points on the line.<\/li>\n<li>The slope and initial value can be determined given a graph or any two points on the line.<\/li>\n<li>One type of function notation is the slope-intercept form of an equation.<\/li>\n<li>The point-slope form is useful for finding a linear equation when given the slope of a line and one point.<\/li>\n<li>The point-slope form is also convenient for finding a linear equation when given two points through which a line passes.<\/li>\n<li>The equation for a linear function can be written if the slope <em>m<\/em>\u00a0and initial value <em>b\u00a0<\/em>are known.<\/li>\n<li>A linear function can be used to solve real-world problems.<\/li>\n<li>A linear function can be written from tabular form.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165137405111\" class=\"definition\">\n<dt>decreasing linear function<\/dt>\n<dd id=\"fs-id1165137405116\">a function with a negative slope: If [latex]f\\left(x\\right)=mx+b, \\text{then} m<0[\/latex].<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137863356\" class=\"definition\">\n<dt>increasing linear function<\/dt>\n<dd id=\"fs-id1165135188274\">a function with a positive slope: If [latex]f\\left(x\\right)=mx+b, \\text{then} m>0[\/latex].<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt>linear function<\/dt>\n<dd id=\"fs-id1165135429394\">a function with a constant rate of change that is a polynomial of degree 1, and whose graph is a straight line<\/dd>\n<\/dl>\n<dl id=\"fs-id1165134389091\" class=\"definition\">\n<dt>point-slope form<\/dt>\n<dd id=\"fs-id1165134389097\">the equation for a line that represents a linear function of the form [latex]y-{y}_{1}=m\\left(x-{x}_{1}\\right)[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137635132\" class=\"definition\">\n<dt>slope<\/dt>\n<dd id=\"fs-id1165137635137\">the ratio of the change in output values to the change in input values; a measure of the steepness of a line<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137817449\" class=\"definition\">\n<dt>slope-intercept form<\/dt>\n<dd id=\"fs-id1165137817454\">the equation for a line that represents a linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135195656\" class=\"definition\">\n<dt><em>y<\/em>-intercept<\/dt>\n<dd id=\"fs-id1165137635107\">the value of a function when the input value is zero; also known as initial value<\/dd>\n<\/dl>\n<\/section>\n<\/section>\n<\/section>\n<\/section>\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-13798\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Specific attribution<\/div><ul class=\"citation-list\"><li>Precalculus. <strong>Authored by<\/strong>: OpenStax College. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-13798-1\">http:\/\/www.guinnessworldrecords.com\/records-3000\/fastest-growing-plant\/ <a href=\"#return-footnote-13798-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-13798-2\"><a href=\"http:\/\/www.chinahighlights.com\/shanghai\/transportation\/maglev-train.htm\" target=\"_blank\" rel=\"noopener\">http:\/\/www.chinahighlights.com\/shanghai\/transportation\/maglev-train.htm<\/a> <a href=\"#return-footnote-13798-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-13798-3\"><a href=\"http:\/\/www.cbsnews.com\/8301-501465_162-57400228-501465\/teens-are-sending-60-texts-a-day-study-says\/\" target=\"_blank\" rel=\"noopener\">http:\/\/www.cbsnews.com\/8301-501465_162-57400228-501465\/teens-are-sending-60-texts-a-day-study-says\/<\/a> <a href=\"#return-footnote-13798-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":23588,"menu_order":1,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc-attribution\",\"description\":\"Precalculus\",\"author\":\"OpenStax College\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1\/Preface\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-13798","chapter","type-chapter","status-publish","hentry"],"part":10717,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/23588"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/revisions"}],"predecessor-version":[{"id":16049,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/revisions\/16049"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/10717"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=13798"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13798"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=13798"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=13798"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}