{"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":"2020-05-21T06:44:43","modified_gmt":"2020-05-21T06:44:43","slug":"linear-functions-and-their-graphs","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/chapter\/linear-functions-and-their-graphs\/","title":{"raw":"Section 1.4: Linear Functions and Their Graphs","rendered":"Section 1.4: Linear Functions and Their Graphs"},"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 \t<li>Graph linear functions.<\/li>\r\n \t<li>Write the equation for a linear function from the graph of a line.<\/li>\r\n<\/ul>\r\n<\/div>\r\nIntroduction to Linear Functions\r\n\r\nA bamboo forest in China (credit: \"JFXie\"\/Flickr)\r\n\r\n<span id=\"fs-id2381652\">\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_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 8: 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<\/h3>\r\n[ohm_question hide_question_numbers=1]169761[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Graphing Linear Functions<\/h2>\r\n<div>\r\n<p id=\"fs-id1165137806314\">As we have seen, the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics.<\/p>\r\n<p id=\"fs-id1165135310597\">There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the <em>y-<\/em>intercept and slope. And the third is by using transformations of the identity function [latex]f(x)=x[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div><section id=\"fs-id1165134224961\">\r\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Graphing a Function by Plotting Points<\/span><\/h2>\r\n<p id=\"fs-id1165137640062\">To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, [latex]f(x)=2x[\/latex], we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.<\/p>\r\n\r\n<\/section><\/div>\r\n<div id=\"fs-id1165134235818\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165132976455\">How To: Given a linear function, graph by plotting points.<\/h3>\r\n<ol id=\"fs-id1165137863963\">\r\n \t<li>Choose a minimum of two input values.<\/li>\r\n \t<li>Evaluate the function at each input value.<\/li>\r\n \t<li>Use the resulting output values to identify coordinate pairs.<\/li>\r\n \t<li>Plot the coordinate pairs on a grid.<\/li>\r\n \t<li>Draw a line through the points.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_01\" class=\"example\">\r\n<div id=\"fs-id1165137784347\" class=\"exercise\">\r\n<div id=\"fs-id1165137456612\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Graphing by Plotting Points<\/h3>\r\n<p id=\"fs-id1165137559100\">Graph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] by plotting points.<\/p>\r\n[reveal-answer q=\"472717\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"472717\"]\r\n<p id=\"fs-id1165137574896\">Begin by choosing input values. This function includes a fraction with a denominator of 3, so let\u2019s choose multiples of 3 as input values. We will choose 0, 3, and 6.<\/p>\r\n<p id=\"fs-id1165135514710\">Evaluate the function at each input value, and use the output value to identify coordinate pairs.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}&amp;x=0 &amp;&amp; f\\left(0\\right)=-\\frac{2}{3}\\left(0\\right)+5=5\\Rightarrow \\left(0,5\\right)\\\\[1mm] &amp;x=3 &amp;&amp; f\\left(3\\right)=-\\frac{2}{3}\\left(3\\right)+5=3\\Rightarrow \\left(3,3\\right) \\\\[1mm] &amp;x=6 &amp;&amp; f\\left(6\\right)=-\\frac{2}{3}\\left(6\\right)+5=1\\Rightarrow \\left(6,1\\right)\\end{align}[\/latex]<\/p>\r\nPlot the coordinate pairs and draw a line through the points. Figure 12 shows\u00a0the graph of the function [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010644\/CNX_Precalc_Figure_02_02_0012.jpg\" alt=\"The graph of the linear function [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex].\" width=\"400\" height=\"347\" \/> <b>Figure 12<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135508515\">The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative constant rate of change in the equation for the function.<\/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-id1165137410246\">Graph [latex]f\\left(x\\right)=-\\frac{3}{4}x+6[\/latex] by plotting points.<\/p>\r\n[reveal-answer q=\"254452\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"254452\"]\r\n\r\n<img class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010644\/CNX_Precalc_Figure_02_02_0022.jpg\" alt=\"Graph of the line y = (3\/4)x + 6, with the points (0,6), (4,3) and (8,0) labeled.\" width=\"487\" height=\"316\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<section id=\"fs-id1165137470730\">\r\n<h2 style=\"text-align: center\">Graphing a Linear Function Using <em>y-<\/em>intercept and Slope<\/h2>\r\n<p id=\"fs-id1165137566712\">Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its <em>y-<\/em>intercept, which is the point at which the input value is zero. To find the <strong><em>y-<\/em>intercept<\/strong>, we can set <em>x<\/em> = 0 in the equation.<\/p>\r\n<p id=\"fs-id1165135242882\">The other characteristic of the linear function is its slope <em>m<\/em>, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the <em>y-<\/em>intercept and the slope in Linear Functions.<\/p>\r\n<p id=\"fs-id1165137472540\">Let\u2019s consider the following function.<\/p>\r\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=\\frac{1}{2}x+1[\/latex]<\/p>\r\n\r\n<div class=\"equation unnumbered\">The slope is [latex]\\frac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The <em>y-<\/em>intercept is the point on the graph when <em>x\u00a0<\/em>= 0. The graph crosses the <em>y<\/em>-axis at (0, 1). Now we know the slope and the <em>y<\/em>-intercept. We can begin graphing by plotting the point (0,1) We know that the slope is rise over run, [latex]m=\\frac{\\text{rise}}{\\text{run}}[\/latex]. From our example, we have [latex]m=\\frac{1}{2}[\/latex], which means that the rise is 1 and the run is 2. So starting from our <em>y<\/em>-intercept (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 14.<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"617\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010644\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/> <b>Figure 14<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165135570225\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Graphical Interpretation of a Linear Function<\/h3>\r\n<p id=\"fs-id1165137732688\">In the equation [latex]f\\left(x\\right)=mx+b[\/latex]<\/p>\r\n\r\n<ul id=\"fs-id1165137422713\">\r\n \t<li><em>b<\/em>\u00a0is the <em>y<\/em>-intercept of the graph and indicates the point (0, <em>b<\/em>) at which the graph crosses the <em>y<\/em>-axis.<\/li>\r\n \t<li><em>m<\/em>\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\r\n<\/ul>\r\n<div id=\"eip-988\" 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>\r\n<div id=\"fs-id1165137427698\" class=\"note precalculus qa textbox\">\r\n<h3 id=\"fs-id1165137538874\">Q &amp; A<\/h3>\r\n<strong>Do all linear functions have <em>y<\/em>-intercepts?<\/strong>\r\n<p id=\"fs-id1165135168195\"><em>Yes. All linear functions cross the y-axis and therefore have y-intercepts.<\/em> (Note: <em>A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.<\/em>)<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137761726\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137675970\">How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope.<\/h3>\r\n<ol id=\"fs-id1165137605269\">\r\n \t<li>Evaluate the function at an input value of zero to find the <em>y-<\/em>intercept.<\/li>\r\n \t<li>Identify the slope as the rate of change of the input value.<\/li>\r\n \t<li>Plot the point represented by the <em>y-<\/em>intercept.<\/li>\r\n \t<li>Use [latex]\\frac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\r\n \t<li>Sketch the line that passes through the points.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_02\" class=\"example\">\r\n<div id=\"fs-id1165135180117\" class=\"exercise\">\r\n<div id=\"fs-id1165137705133\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Graphing by Using the <em>y-<\/em>intercept and Slope<\/h3>\r\n<p id=\"fs-id1165135545818\">Graph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] using the <em>y-<\/em>intercept and slope.<\/p>\r\n[reveal-answer q=\"908667\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"908667\"]\r\n<p id=\"fs-id1165137842403\">Evaluate the function at <em>x\u00a0<\/em>= 0 to find the <em>y-<\/em>intercept. The output value when <em>x\u00a0<\/em>= 0 is 5, so the graph will cross the <em>y<\/em>-axis at (0,5).<\/p>\r\nAccording to the equation for the function, the slope of the line is [latex]-\\frac{2}{3}[\/latex]. This tells us that for each vertical decrease in the \"rise\" of \u20132 units, the \"run\" increases by 3 units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in Figure 3. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points.\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\/03010645\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/> <b>Figure 15<\/b>[\/caption]\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137387381\">The graph slants downward from left to right, which means it has a negative slope as expected.<\/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-id1165135322023\">Find a point on the graph we drew in Example 10 that has a negative <em>x<\/em>-value.<\/p>\r\n[reveal-answer q=\"858176\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"858176\"]\r\n\r\nPossible answers include [latex]\\left(-3,7\\right)[\/latex], [latex]\\left(-6,9\\right)[\/latex], or [latex]\\left(-9,11\\right)[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137543411\">\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question hide_question_numbers=1]169763[\/ohm_question]\r\n\r\n<\/div>\r\n<h2 style=\"text-align: center\">Graphing a Linear Function Using Transformations<\/h2>\r\n<p id=\"fs-id1165137695235\">Another option for graphing is to use <strong>transformations<\/strong> of the identity function [latex]f(x)=x[\/latex] . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.<\/p>\r\n\r\n<section id=\"fs-id1165137662254\">\r\n<h3>Vertical Stretch or Compression<\/h3>\r\n<p id=\"fs-id1165137444518\">In the equation [latex]f(x)=mx[\/latex], the <em>m<\/em>\u00a0is acting as the <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the identity function. When <em>m<\/em>\u00a0is negative, there is also a vertical reflection of the graph. Notice in Figure 16 that multiplying the equation of [latex]f(x)=x[\/latex] by <em>m<\/em>\u00a0stretches the graph of <i>f<\/i>\u00a0by a factor of <em>m<\/em>\u00a0units if <em>m\u00a0<\/em>&gt; 1 and compresses the graph of <em>f<\/em>\u00a0by a factor of <em>m<\/em>\u00a0units if 0 &lt; <em>m\u00a0<\/em>&lt; 1. This means the larger the absolute value of <em>m<\/em>, the steeper the slope.<\/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\/03010645\/CNX_Precalc_Figure_02_02_0052.jpg\" alt=\"Graph with several linear functions including y = 3x, y = 2x, y = x, y = (1\/2)x, y = (1\/3)x, y = (-1\/2)x, y = -x, and y = -2x\" width=\"900\" height=\"759\" \/>\r\n\r\n<\/section><\/section>\r\n<p style=\"text-align: center\"><strong>Figure 16.<\/strong> Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].<\/p>\r\n\r\n<section id=\"fs-id1165137543411\"><section id=\"fs-id1165135667863\">\r\n<h3>Vertical Shift<\/h3>\r\n<p id=\"fs-id1165137600044\">In [latex]f\\left(x\\right)=mx+b[\/latex], the <em>b<\/em>\u00a0acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice in Figure 17 that adding a value of <em>b<\/em>\u00a0to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of\u00a0<em>f<\/em>\u00a0a total of <em>b<\/em>\u00a0units up if <em>b<\/em>\u00a0is positive and\u00a0|<em>b<\/em>| units down if <em>b<\/em>\u00a0is negative.<\/p>\r\n<span id=\"fs-id1165137634286\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010645\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"graph showing y = x , y = x+2, y = x+4, y = x-2, y = x-4\" width=\"900\" height=\"759\" \/><\/span>\r\n\r\n<\/section><\/section>\r\n<p style=\"text-align: center\"><strong>Figure 17.<\/strong> This graph illustrates vertical shifts of the function [latex]f(x)=x[\/latex].<\/p>\r\n\r\n<section id=\"fs-id1165137543411\"><section id=\"fs-id1165135667863\">\r\n<p id=\"fs-id1165137564772\">Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.<\/p>\r\n\r\n<div id=\"fs-id1165137641217\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137680349\">How To: Given the equation of a linear function, use transformations to graph the linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex].<\/h3>\r\n<ol id=\"fs-id1165135449594\">\r\n \t<li>Graph [latex]f\\left(x\\right)=x[\/latex].<\/li>\r\n \t<li>Vertically stretch or compress the graph by a factor <em>m<\/em>.<\/li>\r\n \t<li>Shift the graph up or down <em>b<\/em>\u00a0units.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_03\" class=\"example\">\r\n<div id=\"fs-id1165137456438\" class=\"exercise\">\r\n<div id=\"fs-id1165137434794\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Graphing by Using Transformations<\/h3>\r\n<p id=\"fs-id1165135570273\">Graph [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex] using transformations.<\/p>\r\n[reveal-answer q=\"653846\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"653846\"]\r\n<p id=\"fs-id1165135192082\">The equation for the function shows that [latex]m=\\frac{1}{2}[\/latex] so the identity function is vertically compressed by [latex]\\frac{1}{2}[\/latex]. The equation for the function also shows that <em>b\u00a0<\/em>= \u20133 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression.<\/p>\r\n<span id=\"fs-id1165135245753\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010645\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\" \/><\/span>\r\n<p style=\"text-align: center\"><strong>Figure 18.\u00a0<\/strong>The function, <em>y\u00a0<\/em>= <em>x<\/em>, compressed by a factor of [latex]\\frac{1}{2}[\/latex].<\/p>\r\nThen show the vertical shift.\r\n<p style=\"text-align: center\"><span id=\"fs-id1165137610735\"> <img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010646\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Graph showing the lines y = (1\/2)x, and y = (1\/2) + 3\" width=\"487\" height=\"377\" \/><\/span>\r\n<strong>Figure 19.<\/strong> The function [latex]y=\\frac{1}{2}x[\/latex], shifted down 3 units.<\/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-id1165137823624\">Graph [latex]f\\left(x\\right)=4+2x[\/latex], using transformations.<\/p>\r\n[reveal-answer q=\"713974\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"713974\"]\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\/03010646\/CNX_Precalc_Figure_02_02_0092.jpg\" alt=\"\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135176280\" class=\"note precalculus qa textbox\">\r\n<h3 id=\"fs-id1165137603576\">Q &amp; A<\/h3>\r\n<strong>In Example 3, could we have sketched the graph by reversing the order of the transformations?<\/strong>\r\n<p id=\"fs-id1165137730398\"><em>No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.<\/em><\/p>\r\n\r\n<div id=\"fs-id1165137619677\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}f(2)&amp;=\\frac{1}{2}(2)-3 \\\\ &amp;=1-3 \\\\ &amp;=-2 \\end{align}[\/latex]<\/div>\r\n<\/div>\r\n<h2>Writing the Equation for a Function from the Graph of a Line<\/h2>\r\n<section id=\"fs-id1165137531122\">\r\n<p id=\"fs-id1165135408570\">Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 21. We can see right away that the graph crosses the <em>y<\/em>-axis at the point (0, 4) so this is the <em>y<\/em>-intercept.<span id=\"fs-id1165137629251\">\r\n<\/span><\/p>\r\n\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\/03010646\/CNX_Precalc_Figure_02_02_0102.jpg\" alt=\"\" width=\"369\" height=\"378\" \/> <b>Figure 21<\/b>[\/caption]\r\n<p id=\"fs-id1165135501156\">Then we can calculate the slope by finding the rise and run. We can choose any two points, but let\u2019s look at the point (\u20132, 0). To get from this point to the <em>y-<\/em>intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be<\/p>\r\n\r\n<div id=\"fs-id1165137526424\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]m=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{2}=2[\/latex]<\/div>\r\n<p id=\"fs-id1165135684358\">Substituting the slope and <em>y-<\/em>intercept into the slope-intercept form of a line gives<\/p>\r\n\r\n<div id=\"fs-id1165135316180\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y=2x+4[\/latex]<\/div>\r\n<div id=\"fs-id1165137836529\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137760034\">How To: Given a graph of linear function, find the equation to describe the function.<\/h3>\r\n<ol id=\"fs-id1165137769882\">\r\n \t<li>Identify the <em>y-<\/em>intercept of an equation.<\/li>\r\n \t<li>Choose two points to determine the slope.<\/li>\r\n \t<li>Substitute the <em>y-<\/em>intercept and slope into the slope-intercept form of a line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_04\" class=\"example\">\r\n<div id=\"fs-id1165134377971\" class=\"exercise\">\r\n<div id=\"fs-id1165134377973\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Matching Linear Functions to Their Graphs<\/h3>\r\n<p id=\"fs-id1165135397960\">Match each equation of the linear functions with one of the lines in Figure 22.<\/p>\r\n\r\n<ol id=\"fs-id1165134104054\">\r\n \t<li>[latex]f\\left(x\\right)=2x+3[\/latex]<\/li>\r\n \t<li>[latex]g\\left(x\\right)=2x - 3[\/latex]<\/li>\r\n \t<li>[latex]h\\left(x\\right)=-2x+3[\/latex]<\/li>\r\n \t<li>[latex]j\\left(x\\right)=\\frac{1}{2}x+3[\/latex]<\/li>\r\n<\/ol>\r\n&nbsp;\r\n<figure id=\"CNX_Precalc_Figure_02_02_011\" class=\"small\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"393\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010647\/CNX_Precalc_Figure_02_02_0112.jpg\" alt=\"Graph of three lines, line 1) passes through (0,3) and (-2, -1), line 2) passes through (0,3) and (-6,0), line 3) passes through (0,-3) and (2,1)\" width=\"393\" height=\"305\" \/> <b>Figure 22<\/b>[\/caption]<\/figure>\r\n[reveal-answer q=\"35922\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"35922\"]\r\n<p id=\"fs-id1165135309831\">Analyze the information for each function.<\/p>\r\n\r\n<ol id=\"fs-id1165135161122\">\r\n \t<li>This function has a slope of 2 and a <em>y<\/em>-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function <em>g<\/em>\u00a0has the same slope, but a different <em>y-<\/em>intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so <em>f<\/em>\u00a0must be represented by line I.<\/li>\r\n \t<li>This function also has a slope of 2, but a <em>y<\/em>-intercept of \u20133. It must pass through the point (0, \u20133) and slant upward from left to right. It must be represented by line III.<\/li>\r\n \t<li>This function has a slope of \u20132 and a <em>y-<\/em>intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.<\/li>\r\n \t<li>This function has a slope of [latex]\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of <em>j<\/em>\u00a0is less than the slope of <em>f<\/em>\u00a0so the line for <em>j<\/em>\u00a0must be flatter. This function is represented by Line II.<\/li>\r\n<\/ol>\r\nNow we can re-label the lines as in Figure 23.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"489\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010647\/CNX_Precalc_Figure_02_02_0122.jpg\" alt=\"\" width=\"489\" height=\"374\" \/> <b>Figure 23<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section>\r\n<figure class=\"small\"><\/figure>\r\n<section id=\"fs-id1165137767695\">\r\n<h2 style=\"text-align: center\">Finding the <em>x<\/em>-intercept of a Line<\/h2>\r\n<p id=\"fs-id1165137665075\">So far, we have been finding the <em>y-<\/em>intercepts of a function: the point at which the graph of the function crosses the <em>y<\/em>-axis. A function may also have an <strong><em>x<\/em><\/strong><strong>-intercept,<\/strong> which is the <em>x<\/em>-coordinate of the point where the graph of the function crosses the <em>x<\/em>-axis. In other words, it is the input value when the output value is zero.<\/p>\r\n<p id=\"fs-id1165135528375\">To find the <em>x<\/em>-intercept, set a function <em>f<\/em>(<em>x<\/em>) equal to zero and solve for the value of <em>x<\/em>. For example, consider the function shown.<\/p>\r\n\r\n<div id=\"eip-901\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=3x - 6[\/latex]<\/div>\r\n<p id=\"fs-id1165137549960\">Set the function equal to 0 and solve for <em>x<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1165137595415\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&amp;0=3x - 6 \\\\ &amp;6=3x \\\\ &amp;2=x \\\\ &amp;x=2 \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165135149818\">The graph of the function crosses the <em>x<\/em>-axis at the point (2, 0).<\/p>\r\n\r\n<div id=\"fs-id1165137705101\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165137705106\"><strong>Do all linear functions have <em>x<\/em>-intercepts?<\/strong><\/p>\r\n<p id=\"fs-id1165137827599\"><em>No. However, linear functions of the form <\/em>y\u00a0<em>= <\/em>c<em>, where <\/em>c<em> is a nonzero real number are the only examples of linear functions with no <\/em>x<em>-intercept. For example, <\/em>y\u00a0<em>= 5 is a horizontal line 5 units above the <\/em>x<em>-axis. This function has no <\/em>x<em>-intercepts<\/em>.<\/p>\r\n\r\n<figure id=\"CNX_Precalc_Figure_02_02_026\" class=\"medium\"><\/figure>\r\n<\/div>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"421\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010647\/CNX_Precalc_Figure_02_02_0262.jpg\" alt=\"Graph of y = 5.\" width=\"421\" height=\"231\" \/> <b>Figure 24<\/b>[\/caption]\r\n\r\n<div id=\"fs-id1165137653298\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: <em>x<\/em>-intercept<\/h3>\r\n<p id=\"fs-id1165137663549\">The <strong><em>x<\/em>-intercept<\/strong> of the function is the point where the graph crosses the\u00a0<em>x<\/em>-axis. Points on the <em>x<\/em>-axis have the form (<em>x<\/em>,0) so we can find <em>x<\/em>-intercepts by setting\u00a0<em>f<\/em>(<em>x<\/em>) = 0. For a linear function, we solve the equation <em>mx\u00a0<\/em>+ <em>b\u00a0<\/em>= 0<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_02_02_05\" class=\"example\">\r\n<div id=\"fs-id1165137805711\" class=\"exercise\">\r\n<div id=\"fs-id1165137805713\" class=\"problem textbox shaded\">\r\n<h3>Example 13: Finding an <em>x<\/em>-intercept<\/h3>\r\n<p id=\"fs-id1165137663560\">Find the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].<\/p>\r\n[reveal-answer q=\"383732\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"383732\"]\r\n<p id=\"fs-id1165137424379\">Set the function equal to zero to solve for <em>x<\/em>.<\/p>\r\n<p style=\"text-align: center\">[latex]\\begin{align}0&amp;=\\frac{1}{2}x - 3\\\\[1mm] 3&amp;=\\frac{1}{2}x\\\\[1mm] 6&amp;=x\\\\[1mm] x&amp;=6\\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137415633\">The graph crosses the <em>x<\/em>-axis at the point (6,0).<\/p>\r\n\r\n<div id=\"Example_02_02_05\" class=\"example\">\r\n<div id=\"fs-id1165137805711\" class=\"exercise\">\r\n<div id=\"fs-id1165135450383\" class=\"commentary\">\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135450388\">A graph of the function is shown in Figure 25. We can see that the <em>x<\/em>-intercept is (6, 0) as we expected.<\/p>\r\n<span id=\"fs-id1165137424274\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0132.jpg\" alt=\"\" \/><\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137727385\" class=\"note precalculus try\">\r\n<div id=\"ti_02_02_04\" class=\"exercise\">\r\n<div id=\"fs-id1165134389962\" class=\"problem\">\r\n<p style=\"text-align: center\"><strong>Figure 25.\u00a0<\/strong>The graph of the linear function [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"fs-id1165137727385\" class=\"note precalculus try\">\r\n<div id=\"ti_02_02_04\" class=\"exercise\">\r\n<div id=\"fs-id1165134389962\" class=\"problem\">\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165134389964\">Find the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{4}x - 4[\/latex].<\/p>\r\n[reveal-answer q=\"946091\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"946091\"]\r\n\r\n[latex]\\left(16,\\text{ 0}\\right)[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165137761836\">\r\n<h2 style=\"text-align: center\">Describing Horizontal and Vertical Lines<\/h2>\r\nThere are two special cases of lines on a graph\u2014horizontal and vertical lines. A <strong>horizontal line<\/strong> indicates a constant output, or <em>y<\/em>-value. In Figure 26, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use <em>m\u00a0<\/em>= 0 in the equation [latex]f(x)=mx+b[\/latex], the equation simplifies to [latex]f(x)=b[\/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f(x)=2[\/latex].\r\n\r\n<span id=\"fs-id1165137914048\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0142.jpg\" alt=\"\" \/><\/span>\r\n\r\n<\/section>\r\n<p style=\"text-align: center\"><strong>Figure 26.\u00a0<\/strong>A horizontal line representing the function [latex]f(x)=2[\/latex].<\/p>\r\n\r\n<section id=\"fs-id1165137761836\">\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\/03010648\/CNX_Precalc_Figure_02_02_0152.jpg\" alt=\"M equals change of output divided by change of input. The numerator is a non-zero real number, and the change of input is zero.\" width=\"487\" height=\"99\" \/> <b>Figure 27<\/b>[\/caption]\r\n<p id=\"fs-id1165137891303\">A <strong>vertical line<\/strong> indicates a constant input, or <em>x<\/em>-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.<span id=\"fs-id1165135547417\">\r\n<\/span><\/p>\r\nNotice that a vertical line, such as the one in Figure 28<strong>,<\/strong> has an <em>x<\/em>-intercept, but no <em>y-<\/em>intercept unless it\u2019s the line <em>x<\/em> = 0. This graph represents the line <em>x<\/em> = 2.\r\n\r\n<span id=\"fs-id1165137771701\"> <img class=\" aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0162.jpg\" alt=\"\" \/><\/span>\r\n\r\n<\/section>\r\n<p style=\"text-align: center\"><strong>Figure 28.<\/strong>\u00a0The vertical line, <em>x\u00a0<\/em>= 2, which does not represent a function.<\/p>\r\n\r\n<section id=\"fs-id1165137761836\">\r\n<div id=\"fs-id1165137432282\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Horizontal and Vertical Lines<\/h3>\r\n<p id=\"fs-id1165137698131\">Lines can be horizontal or vertical.<\/p>\r\n<p id=\"fs-id1165137698134\">A <strong>horizontal line<\/strong> is a line defined by an equation in the form [latex]f(x)=b[\/latex].<\/p>\r\n<p id=\"fs-id1165137602054\">A <strong>vertical line<\/strong> is a line defined by an equation in the form [latex]x=a[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_02_02_06\" class=\"example\">\r\n<div id=\"fs-id1165137697917\" class=\"exercise\">\r\n<div id=\"fs-id1165137697920\" class=\"problem textbox shaded\">\r\n<h3>Example 14: Writing the Equation of a Horizontal Line<\/h3>\r\nWrite the equation of the line graphed in Figure 29.\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\/03010649\/CNX_Precalc_Figure_02_02_0172.jpg\" alt=\"Graph of x = 7.\" width=\"369\" height=\"378\" \/> <b>Figure 29<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"483529\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"483529\"]\r\n\r\nFor any <em>x<\/em>-value, the <em>y<\/em>-value is \u20134, so the equation is <em>y\u00a0<\/em>= \u20134.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<div id=\"Example_02_02_07\" class=\"example\">\r\n<div id=\"fs-id1165137611023\" class=\"exercise\">\r\n<div id=\"fs-id1165137611025\" class=\"problem textbox shaded\">\r\n<h3>Example 15: Writing the Equation of a Vertical Line<\/h3>\r\n<p id=\"fs-id1165137871492\">Write the equation of the line graphed in Figure 30.<span id=\"fs-id1165137645052\">\r\n<\/span><\/p>\r\n\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\/03010649\/CNX_Precalc_Figure_02_02_0182.jpg\" alt=\"Graph of two functions where the baby blue line is y = -2\/3x + 7, and the blue line is y = -x + 1.\" width=\"369\" height=\"378\" \/> <b>Figure 30<\/b>[\/caption]\r\n\r\n[reveal-answer q=\"729018\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"729018\"]\r\n\r\nThe constant <em>x<\/em>-value is 7, so the equation is <em>x\u00a0<\/em>= 7.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/div>\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 \t<li>Linear functions may be graphed by plotting points or by using the <em>y<\/em>-intercept and slope.<\/li>\r\n \t<li>Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections.<\/li>\r\n \t<li>The <em>y<\/em>-intercept and slope of a line may be used to write the equation of a line.<\/li>\r\n \t<li>The <em>x<\/em>-intercept is the point at which the graph of a linear function crosses the <em>x<\/em>-axis.<\/li>\r\n \t<li>Horizontal lines are written in the form, <em>f<\/em>(<em>x<\/em>) = <em>b<\/em>.<\/li>\r\n \t<li>Vertical lines are written in the form, <em>x\u00a0<\/em>= <em>b<\/em>.<\/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-id1165137572723\" class=\"definition\">\r\n \t<dt><strong>horizontal line<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137572728\">a line defined by [latex]f(x)=b[\/latex], where <em>b<\/em>\u00a0is a real number. The slope of a horizontal line is 0.<\/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-id1165135186597\" class=\"definition\">\r\n \t<dt><strong>vertical line<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137757647\">a line defined by <em>x<\/em> = <em>a<\/em>, where <em>a<\/em>\u00a0is a real number. The slope of a vertical line is undefined.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165137757668\" class=\"definition\">\r\n \t<dt><strong><em>x<\/em>-intercept<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165137782278\">the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the horizontal axis<\/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>&nbsp;\r\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Section 1.4 Homework Exercises<\/span><\/h2>\r\n1. Terry is skiing down a steep hill. Terry's elevation, <em>E<\/em>(<em>t<\/em>), in feet after <em>t<\/em>\u00a0seconds is given by [latex]E\\left(t\\right)=3000-70t[\/latex]. Write a complete sentence describing Terry\u2019s starting elevation and how it is changing over time.\r\n\r\n2.\u00a0Maria is climbing a mountain. Maria's elevation,\u00a0<em>E<\/em>(<em>t<\/em>), in feet after <em>t<\/em>\u00a0minutes is given by [latex]E\\left(t\\right)=1200+40t[\/latex]. Write a complete sentence describing Maria\u2019s starting elevation and how it is changing over time.\r\n\r\n3. Jessica is walking home from a friend\u2019s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?\r\n\r\n4.\u00a0Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now.\r\n\r\n5. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.\r\n\r\n6.\u00a0Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding [latex]n[\/latex] rides?\r\n\r\nFor the following exercises, determine whether the equation of the curve can be written as a linear function.\r\n\r\n7. [latex]y=\\frac{1}{4}x+6[\/latex]\r\n\r\n8.\u00a0[latex]y=3x - 5[\/latex]\r\n\r\n9. [latex]y=3{x}^{2}-2[\/latex]\r\n\r\n10.\u00a0[latex]3x+5y=15[\/latex]\r\n\r\n11. [latex]3{x}^{2}+5y=15[\/latex]\r\n\r\n12.\u00a0[latex]3x+5{y}^{2}=15[\/latex]\r\n\r\n13. [latex]-2{x}^{2}+3{y}^{2}=6[\/latex]\r\n\r\n14.\u00a0[latex]-\\frac{x - 3}{5}=2y[\/latex]\r\n\r\nFor the following exercises, determine whether each function is increasing or decreasing.\r\n\r\n15. [latex]f\\left(x\\right)=4x+3[\/latex]\r\n\r\n16.\u00a0[latex]g\\left(x\\right)=5x+6[\/latex]\r\n\r\n17. [latex]a\\left(x\\right)=5 - 2x[\/latex]\r\n\r\n18.\u00a0[latex]b\\left(x\\right)=8 - 3x[\/latex]\r\n\r\n19. [latex]h\\left(x\\right)=-2x+4[\/latex]\r\n\r\n20.\u00a0[latex]k\\left(x\\right)=-4x+1[\/latex]\r\n\r\n21. [latex]j\\left(x\\right)=\\frac{1}{2}x - 3[\/latex]\r\n\r\n22.\u00a0[latex]p\\left(x\\right)=\\frac{1}{4}x - 5[\/latex]\r\n\r\n23. [latex]n\\left(x\\right)=-\\frac{1}{3}x - 2[\/latex]\r\n\r\n24.\u00a0[latex]m\\left(x\\right)=-\\frac{3}{8}x+3[\/latex]\r\n\r\nFor the following exercises, find the slope of the line that passes through the two given points.\r\n\r\n25. [latex]\\left(2,\\text{ }4\\right)[\/latex] and [latex]\\left(4,\\text{ 10}\\right)[\/latex]\r\n\r\n26.\u00a0[latex]\\left(1,\\text{ 5}\\right)[\/latex] and [latex]\\left(4,\\text{ 11}\\right)[\/latex]\r\n\r\n27. [latex]\\left(-1,\\text{4}\\right)[\/latex] and [latex]\\left(5,\\text{2}\\right)[\/latex]\r\n\r\n28. [latex]\\left(8,-2\\right)[\/latex] and [latex]\\left(4,6\\right)[\/latex]\r\n\r\n29. [latex]\\left(6,\\text{ }11\\right)[\/latex] and [latex]\\left(-4, 3\\right)[\/latex]\r\n\r\nFor the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.\r\n\r\n30. [latex]f\\left(-5\\right)=-4[\/latex],\u00a0and [latex]f\\left(5\\right)=2[\/latex]\r\n\r\n31. [latex]f\\left(-1\\right)=4[\/latex] and [latex]f\\left(5\\right)=1[\/latex]\r\n\r\n32.\u00a0[latex]\\left(2,4\\right)[\/latex] and [latex]\\left(4,10\\right)[\/latex]\r\n\r\n33. Passes through [latex]\\left(1,5\\right)[\/latex] and [latex]\\left(4,11\\right)[\/latex]\r\n\r\n34.\u00a0Passes through [latex]\\left(-1,\\text{ 4}\\right)[\/latex] and [latex]\\left(5,\\text{ 2}\\right)[\/latex]\r\n\r\n35. Passes through [latex]\\left(-2,\\text{ 8}\\right)[\/latex] and [latex]\\left(4,\\text{ 6}\\right)[\/latex]\r\n\r\n36.\u00a0<em>x<\/em> intercept at [latex]\\left(-2,\\text{ 0}\\right)[\/latex] and <em>y<\/em> intercept at [latex]\\left(0,-3\\right)[\/latex]\r\n\r\n37.<em> x<\/em> intercept at [latex]\\left(-5,\\text{ 0}\\right)[\/latex] and <em>y<\/em> intercept at [latex]\\left(0,\\text{ 4}\\right)[\/latex]\r\n\r\nFor the following exercises, find the slope of the lines graphed.\r\n\r\n38.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005131\/CNX_Precalc_Figure_02_01_201.jpg\" alt=\"\" \/>\r\n\r\n39.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_202.jpg\" alt=\"\" \/>\r\n\r\n40.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_203.jpg\" alt=\"\" \/>\r\n\r\nFor the following exercises, write an equation for the lines graphed.\r\n\r\n41.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_205.jpg\" alt=\"\" \/>\r\n\r\n42.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_206.jpg\" alt=\"\" \/>\r\n\r\n43.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_207.jpg\" alt=\"\" \/>\r\n\r\n44.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_208.jpg\" alt=\"\" \/>\r\n\r\n45.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_209.jpg\" alt=\"\" \/>\r\n\r\n46.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005134\/CNX_Precalc_Figure_02_01_210.jpg\" alt=\"\" \/>\r\n\r\nFor the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.\r\n\r\n47.\r\n<table id=\"Table_02_01_04\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 5), (5, -10), (10, -25), and (15, -40).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong><em>x<\/em><\/strong><\/td>\r\n<td>0<\/td>\r\n<td>5<\/td>\r\n<td>10<\/td>\r\n<td>15<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>g<\/em>(<em>x<\/em>)<\/strong><\/td>\r\n<td>5<\/td>\r\n<td>\u201310<\/td>\r\n<td>\u201325<\/td>\r\n<td>\u201340<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n48.\r\n<table id=\"Table_02_01_05\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (0, 5), (5, 30), (10, 105), and (15, 230).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong><em>x<\/em><\/strong><\/td>\r\n<td>0<\/td>\r\n<td>5<\/td>\r\n<td>10<\/td>\r\n<td>15<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>h<\/em>(<em>x<\/em>)<\/strong><\/td>\r\n<td>5<\/td>\r\n<td>30<\/td>\r\n<td>105<\/td>\r\n<td>230<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n49.\r\n<table id=\"Table_02_01_06\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (0,- 5), (5, 20), (10, 45), and (15, 70).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong><em>x<\/em><\/strong><\/td>\r\n<td>0<\/td>\r\n<td>5<\/td>\r\n<td>10<\/td>\r\n<td>15<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\r\n<td>\u20135<\/td>\r\n<td>20<\/td>\r\n<td>45<\/td>\r\n<td>70<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n50.\r\n<table id=\"Table_02_01_07\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (5, 13), (10, 28), (20, 58), and (25, 73).\"><colgroup> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong><em>x<\/em><\/strong><\/td>\r\n<td>5<\/td>\r\n<td>10<\/td>\r\n<td>20<\/td>\r\n<td>25<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>k<\/em>(<em>x<\/em>)<\/strong><\/td>\r\n<td>28<\/td>\r\n<td>13<\/td>\r\n<td>58<\/td>\r\n<td>73<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n51.\r\n<table id=\"Table_02_01_08\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 6), (5, -10), (10, -25), and (15, -40).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\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>6<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>g<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\r\n<td>6<\/td>\r\n<td>\u201319<\/td>\r\n<td>\u201344<\/td>\r\n<td>\u201369<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n52.\r\n<table id=\"Table_02_01_09\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (2, 13), (4, 23), (8, 43), and (10, 53).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><em><strong>x<\/strong><\/em><\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>6<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\r\n<td>\u20134<\/td>\r\n<td>16<\/td>\r\n<td>36<\/td>\r\n<td>56<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n53.\r\n<table id=\"Table_02_01_10\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (2, -4), (4, 16), (6, 36), and (8, 56).\"><colgroup> <col \/> <col \/> <col \/> <col \/> <col \/><\/colgroup>\r\n<tbody>\r\n<tr>\r\n<td><strong><em>x<\/em><\/strong><\/td>\r\n<td>2<\/td>\r\n<td>4<\/td>\r\n<td>6<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\r\n<td>\u20134<\/td>\r\n<td>16<\/td>\r\n<td>36<\/td>\r\n<td>56<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n54.\r\n<table id=\"Table_02_01_11\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (0, 6), (2, 31), (6, 106), and (8, 231).\"><colgroup> <col \/> <col \/><\/colgroup>\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>6<\/td>\r\n<td>8<\/td>\r\n<\/tr>\r\n<tr>\r\n<td><strong><em>k<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\r\n<td>6<\/td>\r\n<td>31<\/td>\r\n<td>106<\/td>\r\n<td>231<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<div id=\"fs-id1165137502934\" class=\"exercise\">\r\n<div id=\"fs-id1165137678561\" class=\"solution\"><\/div>\r\n<\/div>\r\n<div id=\"fs-id1165135319509\" class=\"exercise\">\r\n<div id=\"fs-id1165137935676\" class=\"problem\">\r\n<div style=\"padding-left: 60px\"><\/div>\r\n55.\u00a0Find the value of <em>x<\/em>\u00a0if a linear function goes through the following points and has the following slope: [latex]\\left(x,2\\right),\\left(-4,6\\right),m=3[\/latex]\r\n\r\n56.\u00a0Find the value of <em>y<\/em> if a linear function goes through the following points and has the following slope: [latex]\\left(10,y\\right),\\left(25,100\\right),m=-5[\/latex]\r\n\r\n57. Find the equation of the line that passes through the following points: [latex]\\left(a,\\text{ }b\\right)[\/latex] and [latex]\\left(a,\\text{ }b+1\\right)[\/latex]\r\n\r\n58.\u00a0Find the equation of the line that passes through the following points: [latex]\\left(2a,\\text{ }b\\right)[\/latex] and [latex]\\left(a,\\text{ }b+1\\right)[\/latex]\r\n\r\n59. Find the equation of the line that passes through the following points: [latex]\\left(a,\\text{ }0\\right)[\/latex] and [latex]\\left(c,\\text{ }d\\right)[\/latex]\r\n\r\nFor the following exercises, match the given linear equation with its graph.\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\/03005147\/CNX_Precalc_Figure_02_02_201.jpg\" alt=\"\" \/>\r\n\r\n60. [latex]f\\left(x\\right)=-x - 1[\/latex]\r\n\r\n61. [latex]f\\left(x\\right)=-2x - 1[\/latex]\r\n\r\n62.\u00a0[latex]f\\left(x\\right)=-\\frac{1}{2}x - 1[\/latex]\r\n\r\n63. [latex]f\\left(x\\right)=2[\/latex]\r\n\r\n64.\u00a0[latex]f\\left(x\\right)=2+x[\/latex]\r\n\r\n65. [latex]f\\left(x\\right)=3x+2[\/latex]\r\n\r\nFor the following exercises, sketch a line with the given features.\r\n\r\n66. An x-intercept of [latex]\\left(-\\text{4},\\text{ 0}\\right)[\/latex] and y-intercept of [latex]\\left(0,\\text{ -2}\\right)[\/latex]\r\n\r\n67. An x-intercept of [latex]\\left(-\\text{2},\\text{ 0}\\right)[\/latex] and y-intercept of [latex]\\left(0,\\text{ 4}\\right)[\/latex]\r\n\r\n68.\u00a0A y-intercept of [latex]\\left(0,\\text{ 7}\\right)[\/latex] and slope [latex]-\\frac{3}{2}[\/latex]\r\n\r\n69. A y-intercept of [latex]\\left(0,\\text{ 3}\\right)[\/latex] and slope [latex]\\frac{2}{5}[\/latex]\r\n\r\n70.\u00a0Passing through the points [latex]\\left(-\\text{6},\\text{ -2}\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ -6}\\right)[\/latex]\r\n\r\n71. Passing through the points [latex]\\left(-\\text{3},\\text{ -4}\\right)[\/latex] and [latex]\\left(\\text{3},\\text{ 0}\\right)[\/latex]\r\n\r\nFor the following exercises, sketch the graph of each equation.\r\n\r\n72. [latex]f\\left(x\\right)=-2x - 1[\/latex]\r\n\r\n73. [latex]g\\left(x\\right)=-3x+2[\/latex]\r\n\r\n74. [latex]h\\left(x\\right)=\\frac{1}{3}x+2[\/latex]\r\n\r\n75. [latex]k\\left(x\\right)=\\frac{2}{3}x - 3[\/latex]\r\n\r\n76. [latex]f\\left(t\\right)=3+2t[\/latex]\r\n\r\n77. [latex]p\\left(t\\right)=-2+3t[\/latex]\r\n\r\n78.\u00a0[latex]x=3[\/latex]\r\n\r\n79. [latex]x=-2[\/latex]\r\n\r\n80. [latex]r\\left(x\\right)=4[\/latex]\r\n\r\n81. [latex]q\\left(x\\right)=3[\/latex]\r\n\r\n82. [latex]4x=-9y+36[\/latex]\r\n\r\n83. [latex]\\frac{x}{3}-\\frac{y}{4}=1[\/latex]\r\n\r\n84. [latex]3x - 5y=15[\/latex]\r\n\r\n85. [latex]3x=15[\/latex]\r\n\r\n86. [latex]3y=12[\/latex]\r\n\r\n87. If [latex]g\\left(x\\right)[\/latex] is the transformation of [latex]f\\left(x\\right)=x[\/latex] after a vertical compression by [latex]\\frac{3}{4}[\/latex], a shift right by 2, and a shift down by 4\r\n<p style=\"padding-left: 60px\">a. Write an equation for [latex]g\\left(x\\right)[\/latex].<\/p>\r\n<p style=\"padding-left: 60px\">b. What is the slope of this line?<\/p>\r\n<p style=\"padding-left: 60px\">c. Find the y-intercept of this line.<\/p>\r\n88.\u00a0If [latex]g\\left(x\\right)[\/latex] is the transformation of [latex]f\\left(x\\right)=x[\/latex] after a vertical compression by [latex]\\frac{1}{3}[\/latex], a shift left by 1, and a shift up by 3\r\n<p style=\"padding-left: 60px\">a. Write an equation for [latex]g\\left(x\\right)[\/latex].<\/p>\r\n<p style=\"padding-left: 60px\">b. What is the slope of this line?<\/p>\r\n<p style=\"padding-left: 60px\">c. Find the y-intercept of this line.<\/p>\r\nFor the following exercises, write the equation of the line shown in the graph.\r\n\r\n89.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005149\/CNX_Precalc_Figure_02_02_222.jpg\" alt=\"\" \/>\r\n\r\n90.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_223.jpg\" alt=\"\" \/>\r\n\r\n91.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_224.jpg\" alt=\"\" \/>\r\n\r\n92.\r\n<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_225.jpg\" alt=\"\" \/>\r\n\r\n<\/div>\r\n<\/div>\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n<\/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<li>Graph linear functions.<\/li>\n<li>Write the equation for a linear function from the graph of a line.<\/li>\n<\/ul>\n<\/div>\n<p>Introduction to Linear Functions<\/p>\n<p>A bamboo forest in China (credit: &#8220;JFXie&#8221;\/Flickr)<\/p>\n<p><span id=\"fs-id2381652\"><br \/>\n<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.\" \/><\/span><\/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_09\" class=\"example\">\n<div id=\"fs-id1165137459731\" class=\"exercise\">\n<div id=\"fs-id1165135259656\" class=\"problem textbox shaded\">\n<h3>Example 8: 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<\/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>Graphing Linear Functions<\/h2>\n<div>\n<p id=\"fs-id1165137806314\">As we have seen, the graph of a linear function is a straight line. We were also able to see the points of the function as well as the initial value from a graph. By graphing two functions, then, we can more easily compare their characteristics.<\/p>\n<p id=\"fs-id1165135310597\">There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the <em>y-<\/em>intercept and slope. And the third is by using transformations of the identity function [latex]f(x)=x[\/latex].<\/p>\n<\/div>\n<div>\n<section id=\"fs-id1165134224961\">\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Graphing a Function by Plotting Points<\/span><\/h2>\n<p id=\"fs-id1165137640062\">To find points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs. We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, [latex]f(x)=2x[\/latex], we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error.<\/p>\n<\/section>\n<\/div>\n<div id=\"fs-id1165134235818\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165132976455\">How To: Given a linear function, graph by plotting points.<\/h3>\n<ol id=\"fs-id1165137863963\">\n<li>Choose a minimum of two input values.<\/li>\n<li>Evaluate the function at each input value.<\/li>\n<li>Use the resulting output values to identify coordinate pairs.<\/li>\n<li>Plot the coordinate pairs on a grid.<\/li>\n<li>Draw a line through the points.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_01\" class=\"example\">\n<div id=\"fs-id1165137784347\" class=\"exercise\">\n<div id=\"fs-id1165137456612\" class=\"problem textbox shaded\">\n<h3>Example 9: Graphing by Plotting Points<\/h3>\n<p id=\"fs-id1165137559100\">Graph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] by plotting points.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q472717\">Show Solution<\/span><\/p>\n<div id=\"q472717\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137574896\">Begin by choosing input values. This function includes a fraction with a denominator of 3, so let\u2019s choose multiples of 3 as input values. We will choose 0, 3, and 6.<\/p>\n<p id=\"fs-id1165135514710\">Evaluate the function at each input value, and use the output value to identify coordinate pairs.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}&x=0 && f\\left(0\\right)=-\\frac{2}{3}\\left(0\\right)+5=5\\Rightarrow \\left(0,5\\right)\\\\[1mm] &x=3 && f\\left(3\\right)=-\\frac{2}{3}\\left(3\\right)+5=3\\Rightarrow \\left(3,3\\right) \\\\[1mm] &x=6 && f\\left(6\\right)=-\\frac{2}{3}\\left(6\\right)+5=1\\Rightarrow \\left(6,1\\right)\\end{align}[\/latex]<\/p>\n<p>Plot the coordinate pairs and draw a line through the points. Figure 12 shows\u00a0the graph of the function [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex].<\/p>\n<div style=\"width: 410px\" 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\/03010644\/CNX_Precalc_Figure_02_02_0012.jpg\" alt=\"The graph of the linear function &#091;latex&#093;f\\left(x\\right)=-\\frac{2}{3}x+5&#091;\/latex&#093;.\" width=\"400\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 12<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135508515\">The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative constant rate of change in the equation for the function.<\/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-id1165137410246\">Graph [latex]f\\left(x\\right)=-\\frac{3}{4}x+6[\/latex] by plotting points.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q254452\">Show Solution<\/span><\/p>\n<div id=\"q254452\" class=\"hidden-answer\" style=\"display: none\">\n<p><img loading=\"lazy\" decoding=\"async\" class=\"alignnone\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010644\/CNX_Precalc_Figure_02_02_0022.jpg\" alt=\"Graph of the line y = (3\/4)x + 6, with the points (0,6), (4,3) and (8,0) labeled.\" width=\"487\" height=\"316\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<section id=\"fs-id1165137470730\">\n<h2 style=\"text-align: center\">Graphing a Linear Function Using <em>y-<\/em>intercept and Slope<\/h2>\n<p id=\"fs-id1165137566712\">Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its <em>y-<\/em>intercept, which is the point at which the input value is zero. To find the <strong><em>y-<\/em>intercept<\/strong>, we can set <em>x<\/em> = 0 in the equation.<\/p>\n<p id=\"fs-id1165135242882\">The other characteristic of the linear function is its slope <em>m<\/em>, which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a function is equal to the ratio of the change in outputs to the change in inputs. Another way to think about the slope is by dividing the vertical difference, or rise, by the horizontal difference, or run. We encountered both the <em>y-<\/em>intercept and the slope in Linear Functions.<\/p>\n<p id=\"fs-id1165137472540\">Let\u2019s consider the following function.<\/p>\n<p style=\"text-align: center\">[latex]f\\left(x\\right)=\\frac{1}{2}x+1[\/latex]<\/p>\n<div class=\"equation unnumbered\">The slope is [latex]\\frac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right. The <em>y-<\/em>intercept is the point on the graph when <em>x\u00a0<\/em>= 0. The graph crosses the <em>y<\/em>-axis at (0, 1). Now we know the slope and the <em>y<\/em>-intercept. We can begin graphing by plotting the point (0,1) We know that the slope is rise over run, [latex]m=\\frac{\\text{rise}}{\\text{run}}[\/latex]. From our example, we have [latex]m=\\frac{1}{2}[\/latex], which means that the rise is 1 and the run is 2. So starting from our <em>y<\/em>-intercept (0, 1), we can rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 14.<\/div>\n<div style=\"width: 627px\" 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\/03010644\/CNX_Precalc_Figure_02_02_0032.jpg\" alt=\"graph of the line y = (1\/2)x +1 showing the &quot;rise&quot;, or change in the y direction as 1 and the &quot;run&quot;, or change in x direction as 2, and the y-intercept at (0,1)\" width=\"617\" height=\"340\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 14<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165135570225\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Graphical Interpretation of a Linear Function<\/h3>\n<p id=\"fs-id1165137732688\">In the equation [latex]f\\left(x\\right)=mx+b[\/latex]<\/p>\n<ul id=\"fs-id1165137422713\">\n<li><em>b<\/em>\u00a0is the <em>y<\/em>-intercept of the graph and indicates the point (0, <em>b<\/em>) at which the graph crosses the <em>y<\/em>-axis.<\/li>\n<li><em>m<\/em>\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\n<\/ul>\n<div id=\"eip-988\" 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>\n<div id=\"fs-id1165137427698\" class=\"note precalculus qa textbox\">\n<h3 id=\"fs-id1165137538874\">Q &amp; A<\/h3>\n<p><strong>Do all linear functions have <em>y<\/em>-intercepts?<\/strong><\/p>\n<p id=\"fs-id1165135168195\"><em>Yes. All linear functions cross the y-axis and therefore have y-intercepts.<\/em> (Note: <em>A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.<\/em>)<\/p>\n<\/div>\n<div id=\"fs-id1165137761726\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137675970\">How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope.<\/h3>\n<ol id=\"fs-id1165137605269\">\n<li>Evaluate the function at an input value of zero to find the <em>y-<\/em>intercept.<\/li>\n<li>Identify the slope as the rate of change of the input value.<\/li>\n<li>Plot the point represented by the <em>y-<\/em>intercept.<\/li>\n<li>Use [latex]\\frac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\n<li>Sketch the line that passes through the points.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_02\" class=\"example\">\n<div id=\"fs-id1165135180117\" class=\"exercise\">\n<div id=\"fs-id1165137705133\" class=\"problem textbox shaded\">\n<h3>Example 10: Graphing by Using the <em>y-<\/em>intercept and Slope<\/h3>\n<p id=\"fs-id1165135545818\">Graph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] using the <em>y-<\/em>intercept and slope.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q908667\">Show Solution<\/span><\/p>\n<div id=\"q908667\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137842403\">Evaluate the function at <em>x\u00a0<\/em>= 0 to find the <em>y-<\/em>intercept. The output value when <em>x\u00a0<\/em>= 0 is 5, so the graph will cross the <em>y<\/em>-axis at (0,5).<\/p>\n<p>According to the equation for the function, the slope of the line is [latex]-\\frac{2}{3}[\/latex]. This tells us that for each vertical decrease in the &#8220;rise&#8221; of \u20132 units, the &#8220;run&#8221; increases by 3 units in the horizontal direction. We can now graph the function by first plotting the <em>y<\/em>-intercept in Figure 3. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to the left and right by repeating, and then draw a line through the points.<\/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\/03010645\/CNX_Precalc_Figure_02_02_0042.jpg\" alt=\"graph of the line y = (-2\/3)x + 5 showing the change of -2 in y and change of 3 in x.\" width=\"487\" height=\"318\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 15<\/b><\/p>\n<\/div>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137387381\">The graph slants downward from left to right, which means it has a negative slope as expected.<\/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-id1165135322023\">Find a point on the graph we drew in Example 10 that has a negative <em>x<\/em>-value.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q858176\">Show Solution<\/span><\/p>\n<div id=\"q858176\" class=\"hidden-answer\" style=\"display: none\">\n<p>Possible answers include [latex]\\left(-3,7\\right)[\/latex], [latex]\\left(-6,9\\right)[\/latex], or [latex]\\left(-9,11\\right)[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137543411\">\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm169763\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169763&theme=oea&iframe_resize_id=ohm169763\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2 style=\"text-align: center\">Graphing a Linear Function Using Transformations<\/h2>\n<p id=\"fs-id1165137695235\">Another option for graphing is to use <strong>transformations<\/strong> of the identity function [latex]f(x)=x[\/latex] . A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression.<\/p>\n<section id=\"fs-id1165137662254\">\n<h3>Vertical Stretch or Compression<\/h3>\n<p id=\"fs-id1165137444518\">In the equation [latex]f(x)=mx[\/latex], the <em>m<\/em>\u00a0is acting as the <strong>vertical stretch<\/strong> or <strong>compression<\/strong> of the identity function. When <em>m<\/em>\u00a0is negative, there is also a vertical reflection of the graph. Notice in Figure 16 that multiplying the equation of [latex]f(x)=x[\/latex] by <em>m<\/em>\u00a0stretches the graph of <i>f<\/i>\u00a0by a factor of <em>m<\/em>\u00a0units if <em>m\u00a0<\/em>&gt; 1 and compresses the graph of <em>f<\/em>\u00a0by a factor of <em>m<\/em>\u00a0units if 0 &lt; <em>m\u00a0<\/em>&lt; 1. This means the larger the absolute value of <em>m<\/em>, the steeper the slope.<\/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\/03010645\/CNX_Precalc_Figure_02_02_0052.jpg\" alt=\"Graph with several linear functions including y = 3x, y = 2x, y = x, y = (1\/2)x, y = (1\/3)x, y = (-1\/2)x, y = -x, and y = -2x\" width=\"900\" height=\"759\" \/><\/p>\n<\/section>\n<\/section>\n<p style=\"text-align: center\"><strong>Figure 16.<\/strong> Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<section id=\"fs-id1165137543411\">\n<section id=\"fs-id1165135667863\">\n<h3>Vertical Shift<\/h3>\n<p id=\"fs-id1165137600044\">In [latex]f\\left(x\\right)=mx+b[\/latex], the <em>b<\/em>\u00a0acts as the <strong>vertical shift<\/strong>, moving the graph up and down without affecting the slope of the line. Notice in Figure 17 that adding a value of <em>b<\/em>\u00a0to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of\u00a0<em>f<\/em>\u00a0a total of <em>b<\/em>\u00a0units up if <em>b<\/em>\u00a0is positive and\u00a0|<em>b<\/em>| units down if <em>b<\/em>\u00a0is negative.<\/p>\n<p><span id=\"fs-id1165137634286\"> <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\/03010645\/CNX_Precalc_Figure_02_02_0062.jpg\" alt=\"graph showing y = x , y = x+2, y = x+4, y = x-2, y = x-4\" width=\"900\" height=\"759\" \/><\/span><\/p>\n<\/section>\n<\/section>\n<p style=\"text-align: center\"><strong>Figure 17.<\/strong> This graph illustrates vertical shifts of the function [latex]f(x)=x[\/latex].<\/p>\n<section id=\"fs-id1165137543411\">\n<section id=\"fs-id1165135667863\">\n<p id=\"fs-id1165137564772\">Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method.<\/p>\n<div id=\"fs-id1165137641217\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137680349\">How To: Given the equation of a linear function, use transformations to graph the linear function in the form [latex]f\\left(x\\right)=mx+b[\/latex].<\/h3>\n<ol id=\"fs-id1165135449594\">\n<li>Graph [latex]f\\left(x\\right)=x[\/latex].<\/li>\n<li>Vertically stretch or compress the graph by a factor <em>m<\/em>.<\/li>\n<li>Shift the graph up or down <em>b<\/em>\u00a0units.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_03\" class=\"example\">\n<div id=\"fs-id1165137456438\" class=\"exercise\">\n<div id=\"fs-id1165137434794\" class=\"problem textbox shaded\">\n<h3>Example 11: Graphing by Using Transformations<\/h3>\n<p id=\"fs-id1165135570273\">Graph [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex] using transformations.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q653846\">Show Solution<\/span><\/p>\n<div id=\"q653846\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135192082\">The equation for the function shows that [latex]m=\\frac{1}{2}[\/latex] so the identity function is vertically compressed by [latex]\\frac{1}{2}[\/latex]. The equation for the function also shows that <em>b\u00a0<\/em>= \u20133 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression.<\/p>\n<p><span id=\"fs-id1165135245753\"> <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\/03010645\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\" \/><\/span><\/p>\n<p style=\"text-align: center\"><strong>Figure 18.\u00a0<\/strong>The function, <em>y\u00a0<\/em>= <em>x<\/em>, compressed by a factor of [latex]\\frac{1}{2}[\/latex].<\/p>\n<p>Then show the vertical shift.<\/p>\n<p style=\"text-align: center\"><span id=\"fs-id1165137610735\"> <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\/03010646\/CNX_Precalc_Figure_02_02_0082.jpg\" alt=\"Graph showing the lines y = (1\/2)x, and y = (1\/2) + 3\" width=\"487\" height=\"377\" \/><\/span><br \/>\n<strong>Figure 19.<\/strong> The function [latex]y=\\frac{1}{2}x[\/latex], shifted down 3 units.<\/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-id1165137823624\">Graph [latex]f\\left(x\\right)=4+2x[\/latex], using transformations.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q713974\">Show Solution<\/span><\/p>\n<div id=\"q713974\" class=\"hidden-answer\" style=\"display: none\">\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\/03010646\/CNX_Precalc_Figure_02_02_0092.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135176280\" class=\"note precalculus qa textbox\">\n<h3 id=\"fs-id1165137603576\">Q &amp; A<\/h3>\n<p><strong>In Example 3, could we have sketched the graph by reversing the order of the transformations?<\/strong><\/p>\n<p id=\"fs-id1165137730398\"><em>No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2.<\/em><\/p>\n<div id=\"fs-id1165137619677\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}f(2)&=\\frac{1}{2}(2)-3 \\\\ &=1-3 \\\\ &=-2 \\end{align}[\/latex]<\/div>\n<\/div>\n<h2>Writing the Equation for a Function from the Graph of a Line<\/h2>\n<section id=\"fs-id1165137531122\">\n<p id=\"fs-id1165135408570\">Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 21. We can see right away that the graph crosses the <em>y<\/em>-axis at the point (0, 4) so this is the <em>y<\/em>-intercept.<span id=\"fs-id1165137629251\"><br \/>\n<\/span><\/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\/03010646\/CNX_Precalc_Figure_02_02_0102.jpg\" alt=\"\" width=\"369\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 21<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165135501156\">Then we can calculate the slope by finding the rise and run. We can choose any two points, but let\u2019s look at the point (\u20132, 0). To get from this point to the <em>y-<\/em>intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be<\/p>\n<div id=\"fs-id1165137526424\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]m=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{2}=2[\/latex]<\/div>\n<p id=\"fs-id1165135684358\">Substituting the slope and <em>y-<\/em>intercept into the slope-intercept form of a line gives<\/p>\n<div id=\"fs-id1165135316180\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]y=2x+4[\/latex]<\/div>\n<div id=\"fs-id1165137836529\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137760034\">How To: Given a graph of linear function, find the equation to describe the function.<\/h3>\n<ol id=\"fs-id1165137769882\">\n<li>Identify the <em>y-<\/em>intercept of an equation.<\/li>\n<li>Choose two points to determine the slope.<\/li>\n<li>Substitute the <em>y-<\/em>intercept and slope into the slope-intercept form of a line.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_04\" class=\"example\">\n<div id=\"fs-id1165134377971\" class=\"exercise\">\n<div id=\"fs-id1165134377973\" class=\"problem textbox shaded\">\n<h3>Example 12: Matching Linear Functions to Their Graphs<\/h3>\n<p id=\"fs-id1165135397960\">Match each equation of the linear functions with one of the lines in Figure 22.<\/p>\n<ol id=\"fs-id1165134104054\">\n<li>[latex]f\\left(x\\right)=2x+3[\/latex]<\/li>\n<li>[latex]g\\left(x\\right)=2x - 3[\/latex]<\/li>\n<li>[latex]h\\left(x\\right)=-2x+3[\/latex]<\/li>\n<li>[latex]j\\left(x\\right)=\\frac{1}{2}x+3[\/latex]<\/li>\n<\/ol>\n<p>&nbsp;<\/p>\n<figure id=\"CNX_Precalc_Figure_02_02_011\" class=\"small\">\n<div style=\"width: 403px\" 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\/03010647\/CNX_Precalc_Figure_02_02_0112.jpg\" alt=\"Graph of three lines, line 1) passes through (0,3) and (-2, -1), line 2) passes through (0,3) and (-6,0), line 3) passes through (0,-3) and (2,1)\" width=\"393\" height=\"305\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 22<\/b><\/p>\n<\/div>\n<\/figure>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q35922\">Show Solution<\/span><\/p>\n<div id=\"q35922\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135309831\">Analyze the information for each function.<\/p>\n<ol id=\"fs-id1165135161122\">\n<li>This function has a slope of 2 and a <em>y<\/em>-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function <em>g<\/em>\u00a0has the same slope, but a different <em>y-<\/em>intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so <em>f<\/em>\u00a0must be represented by line I.<\/li>\n<li>This function also has a slope of 2, but a <em>y<\/em>-intercept of \u20133. It must pass through the point (0, \u20133) and slant upward from left to right. It must be represented by line III.<\/li>\n<li>This function has a slope of \u20132 and a <em>y-<\/em>intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.<\/li>\n<li>This function has a slope of [latex]\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of <em>j<\/em>\u00a0is less than the slope of <em>f<\/em>\u00a0so the line for <em>j<\/em>\u00a0must be flatter. This function is represented by Line II.<\/li>\n<\/ol>\n<p>Now we can re-label the lines as in Figure 23.<\/p>\n<div style=\"width: 499px\" 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\/03010647\/CNX_Precalc_Figure_02_02_0122.jpg\" alt=\"\" width=\"489\" height=\"374\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 23<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<figure class=\"small\"><\/figure>\n<section id=\"fs-id1165137767695\">\n<h2 style=\"text-align: center\">Finding the <em>x<\/em>-intercept of a Line<\/h2>\n<p id=\"fs-id1165137665075\">So far, we have been finding the <em>y-<\/em>intercepts of a function: the point at which the graph of the function crosses the <em>y<\/em>-axis. A function may also have an <strong><em>x<\/em><\/strong><strong>-intercept,<\/strong> which is the <em>x<\/em>-coordinate of the point where the graph of the function crosses the <em>x<\/em>-axis. In other words, it is the input value when the output value is zero.<\/p>\n<p id=\"fs-id1165135528375\">To find the <em>x<\/em>-intercept, set a function <em>f<\/em>(<em>x<\/em>) equal to zero and solve for the value of <em>x<\/em>. For example, consider the function shown.<\/p>\n<div id=\"eip-901\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]f\\left(x\\right)=3x - 6[\/latex]<\/div>\n<p id=\"fs-id1165137549960\">Set the function equal to 0 and solve for <em>x<\/em>.<\/p>\n<div id=\"fs-id1165137595415\" class=\"equation unnumbered\" style=\"text-align: center\">[latex]\\begin{align}&0=3x - 6 \\\\ &6=3x \\\\ &2=x \\\\ &x=2 \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165135149818\">The graph of the function crosses the <em>x<\/em>-axis at the point (2, 0).<\/p>\n<div id=\"fs-id1165137705101\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165137705106\"><strong>Do all linear functions have <em>x<\/em>-intercepts?<\/strong><\/p>\n<p id=\"fs-id1165137827599\"><em>No. However, linear functions of the form <\/em>y\u00a0<em>= <\/em>c<em>, where <\/em>c<em> is a nonzero real number are the only examples of linear functions with no <\/em>x<em>-intercept. For example, <\/em>y\u00a0<em>= 5 is a horizontal line 5 units above the <\/em>x<em>-axis. This function has no <\/em>x<em>-intercepts<\/em>.<\/p>\n<figure id=\"CNX_Precalc_Figure_02_02_026\" class=\"medium\"><\/figure>\n<\/div>\n<div style=\"width: 431px\" 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\/03010647\/CNX_Precalc_Figure_02_02_0262.jpg\" alt=\"Graph of y = 5.\" width=\"421\" height=\"231\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 24<\/b><\/p>\n<\/div>\n<div id=\"fs-id1165137653298\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: <em>x<\/em>-intercept<\/h3>\n<p id=\"fs-id1165137663549\">The <strong><em>x<\/em>-intercept<\/strong> of the function is the point where the graph crosses the\u00a0<em>x<\/em>-axis. Points on the <em>x<\/em>-axis have the form (<em>x<\/em>,0) so we can find <em>x<\/em>-intercepts by setting\u00a0<em>f<\/em>(<em>x<\/em>) = 0. For a linear function, we solve the equation <em>mx\u00a0<\/em>+ <em>b\u00a0<\/em>= 0<\/p>\n<\/div>\n<div id=\"Example_02_02_05\" class=\"example\">\n<div id=\"fs-id1165137805711\" class=\"exercise\">\n<div id=\"fs-id1165137805713\" class=\"problem textbox shaded\">\n<h3>Example 13: Finding an <em>x<\/em>-intercept<\/h3>\n<p id=\"fs-id1165137663560\">Find the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q383732\">Show Solution<\/span><\/p>\n<div id=\"q383732\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137424379\">Set the function equal to zero to solve for <em>x<\/em>.<\/p>\n<p style=\"text-align: center\">[latex]\\begin{align}0&=\\frac{1}{2}x - 3\\\\[1mm] 3&=\\frac{1}{2}x\\\\[1mm] 6&=x\\\\[1mm] x&=6\\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137415633\">The graph crosses the <em>x<\/em>-axis at the point (6,0).<\/p>\n<div id=\"Example_02_02_05\" class=\"example\">\n<div id=\"fs-id1165137805711\" class=\"exercise\">\n<div id=\"fs-id1165135450383\" class=\"commentary\">\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135450388\">A graph of the function is shown in Figure 25. We can see that the <em>x<\/em>-intercept is (6, 0) as we expected.<\/p>\n<p><span id=\"fs-id1165137424274\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0132.jpg\" alt=\"\" \/><\/span><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137727385\" class=\"note precalculus try\">\n<div id=\"ti_02_02_04\" class=\"exercise\">\n<div id=\"fs-id1165134389962\" class=\"problem\">\n<p style=\"text-align: center\"><strong>Figure 25.\u00a0<\/strong>The graph of the linear function [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137727385\" class=\"note precalculus try\">\n<div id=\"ti_02_02_04\" class=\"exercise\">\n<div id=\"fs-id1165134389962\" class=\"problem\">\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165134389964\">Find the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{4}x - 4[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q946091\">Show Solution<\/span><\/p>\n<div id=\"q946091\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(16,\\text{ 0}\\right)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165137761836\">\n<h2 style=\"text-align: center\">Describing Horizontal and Vertical Lines<\/h2>\n<p>There are two special cases of lines on a graph\u2014horizontal and vertical lines. A <strong>horizontal line<\/strong> indicates a constant output, or <em>y<\/em>-value. In Figure 26, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use <em>m\u00a0<\/em>= 0 in the equation [latex]f(x)=mx+b[\/latex], the equation simplifies to [latex]f(x)=b[\/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f(x)=2[\/latex].<\/p>\n<p><span id=\"fs-id1165137914048\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0142.jpg\" alt=\"\" \/><\/span><\/p>\n<\/section>\n<p style=\"text-align: center\"><strong>Figure 26.\u00a0<\/strong>A horizontal line representing the function [latex]f(x)=2[\/latex].<\/p>\n<section id=\"fs-id1165137761836\">\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\/03010648\/CNX_Precalc_Figure_02_02_0152.jpg\" alt=\"M equals change of output divided by change of input. The numerator is a non-zero real number, and the change of input is zero.\" width=\"487\" height=\"99\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 27<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165137891303\">A <strong>vertical line<\/strong> indicates a constant input, or <em>x<\/em>-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined.<span id=\"fs-id1165135547417\"><br \/>\n<\/span><\/p>\n<p>Notice that a vertical line, such as the one in Figure 28<strong>,<\/strong> has an <em>x<\/em>-intercept, but no <em>y-<\/em>intercept unless it\u2019s the line <em>x<\/em> = 0. This graph represents the line <em>x<\/em> = 2.<\/p>\n<p><span id=\"fs-id1165137771701\"> <img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010648\/CNX_Precalc_Figure_02_02_0162.jpg\" alt=\"\" \/><\/span><\/p>\n<\/section>\n<p style=\"text-align: center\"><strong>Figure 28.<\/strong>\u00a0The vertical line, <em>x\u00a0<\/em>= 2, which does not represent a function.<\/p>\n<section id=\"fs-id1165137761836\">\n<div id=\"fs-id1165137432282\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Horizontal and Vertical Lines<\/h3>\n<p id=\"fs-id1165137698131\">Lines can be horizontal or vertical.<\/p>\n<p id=\"fs-id1165137698134\">A <strong>horizontal line<\/strong> is a line defined by an equation in the form [latex]f(x)=b[\/latex].<\/p>\n<p id=\"fs-id1165137602054\">A <strong>vertical line<\/strong> is a line defined by an equation in the form [latex]x=a[\/latex].<\/p>\n<\/div>\n<div id=\"Example_02_02_06\" class=\"example\">\n<div id=\"fs-id1165137697917\" class=\"exercise\">\n<div id=\"fs-id1165137697920\" class=\"problem textbox shaded\">\n<h3>Example 14: Writing the Equation of a Horizontal Line<\/h3>\n<p>Write the equation of the line graphed in Figure 29.<\/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\/03010649\/CNX_Precalc_Figure_02_02_0172.jpg\" alt=\"Graph of x = 7.\" width=\"369\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 29<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q483529\">Show Solution<\/span><\/p>\n<div id=\"q483529\" class=\"hidden-answer\" style=\"display: none\">\n<p>For any <em>x<\/em>-value, the <em>y<\/em>-value is \u20134, so the equation is <em>y\u00a0<\/em>= \u20134.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_02_07\" class=\"example\">\n<div id=\"fs-id1165137611023\" class=\"exercise\">\n<div id=\"fs-id1165137611025\" class=\"problem textbox shaded\">\n<h3>Example 15: Writing the Equation of a Vertical Line<\/h3>\n<p id=\"fs-id1165137871492\">Write the equation of the line graphed in Figure 30.<span id=\"fs-id1165137645052\"><br \/>\n<\/span><\/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\/03010649\/CNX_Precalc_Figure_02_02_0182.jpg\" alt=\"Graph of two functions where the baby blue line is y = -2\/3x + 7, and the blue line is y = -x + 1.\" width=\"369\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 30<\/b><\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q729018\">Show Solution<\/span><\/p>\n<div id=\"q729018\" class=\"hidden-answer\" style=\"display: none\">\n<p>The constant <em>x<\/em>-value is 7, so the equation is <em>x\u00a0<\/em>= 7.<\/p>\n<\/div>\n<\/div>\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<li>Linear functions may be graphed by plotting points or by using the <em>y<\/em>-intercept and slope.<\/li>\n<li>Graphs of linear functions may be transformed by using shifts up, down, left, or right, as well as through stretches, compressions, and reflections.<\/li>\n<li>The <em>y<\/em>-intercept and slope of a line may be used to write the equation of a line.<\/li>\n<li>The <em>x<\/em>-intercept is the point at which the graph of a linear function crosses the <em>x<\/em>-axis.<\/li>\n<li>Horizontal lines are written in the form, <em>f<\/em>(<em>x<\/em>) = <em>b<\/em>.<\/li>\n<li>Vertical lines are written in the form, <em>x\u00a0<\/em>= <em>b<\/em>.<\/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-id1165137572723\" class=\"definition\">\n<dt><strong>horizontal line<\/strong><\/dt>\n<dd id=\"fs-id1165137572728\">a line defined by [latex]f(x)=b[\/latex], where <em>b<\/em>\u00a0is a real number. The slope of a horizontal line is 0.<\/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-id1165135186597\" class=\"definition\">\n<dt><strong>vertical line<\/strong><\/dt>\n<dd id=\"fs-id1165137757647\">a line defined by <em>x<\/em> = <em>a<\/em>, where <em>a<\/em>\u00a0is a real number. The slope of a vertical line is undefined.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165137757668\" class=\"definition\">\n<dt><strong><em>x<\/em>-intercept<\/strong><\/dt>\n<dd id=\"fs-id1165137782278\">the point on the graph of a linear function when the output value is 0; the point at which the graph crosses the horizontal axis<\/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<p>&nbsp;<\/p>\n<h2 style=\"text-align: center\"><span style=\"text-decoration: underline\">Section 1.4 Homework Exercises<\/span><\/h2>\n<p>1. Terry is skiing down a steep hill. Terry&#8217;s elevation, <em>E<\/em>(<em>t<\/em>), in feet after <em>t<\/em>\u00a0seconds is given by [latex]E\\left(t\\right)=3000-70t[\/latex]. Write a complete sentence describing Terry\u2019s starting elevation and how it is changing over time.<\/p>\n<p>2.\u00a0Maria is climbing a mountain. Maria&#8217;s elevation,\u00a0<em>E<\/em>(<em>t<\/em>), in feet after <em>t<\/em>\u00a0minutes is given by [latex]E\\left(t\\right)=1200+40t[\/latex]. Write a complete sentence describing Maria\u2019s starting elevation and how it is changing over time.<\/p>\n<p>3. Jessica is walking home from a friend\u2019s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour?<\/p>\n<p>4.\u00a0Sonya is currently 10 miles from home and is walking farther away at 2 miles per hour. Write an equation for her distance from home t hours from now.<\/p>\n<p>5. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours.<\/p>\n<p>6.\u00a0Timmy goes to the fair with $40. Each ride costs $2. How much money will he have left after riding [latex]n[\/latex] rides?<\/p>\n<p>For the following exercises, determine whether the equation of the curve can be written as a linear function.<\/p>\n<p>7. [latex]y=\\frac{1}{4}x+6[\/latex]<\/p>\n<p>8.\u00a0[latex]y=3x - 5[\/latex]<\/p>\n<p>9. [latex]y=3{x}^{2}-2[\/latex]<\/p>\n<p>10.\u00a0[latex]3x+5y=15[\/latex]<\/p>\n<p>11. [latex]3{x}^{2}+5y=15[\/latex]<\/p>\n<p>12.\u00a0[latex]3x+5{y}^{2}=15[\/latex]<\/p>\n<p>13. [latex]-2{x}^{2}+3{y}^{2}=6[\/latex]<\/p>\n<p>14.\u00a0[latex]-\\frac{x - 3}{5}=2y[\/latex]<\/p>\n<p>For the following exercises, determine whether each function is increasing or decreasing.<\/p>\n<p>15. [latex]f\\left(x\\right)=4x+3[\/latex]<\/p>\n<p>16.\u00a0[latex]g\\left(x\\right)=5x+6[\/latex]<\/p>\n<p>17. [latex]a\\left(x\\right)=5 - 2x[\/latex]<\/p>\n<p>18.\u00a0[latex]b\\left(x\\right)=8 - 3x[\/latex]<\/p>\n<p>19. [latex]h\\left(x\\right)=-2x+4[\/latex]<\/p>\n<p>20.\u00a0[latex]k\\left(x\\right)=-4x+1[\/latex]<\/p>\n<p>21. [latex]j\\left(x\\right)=\\frac{1}{2}x - 3[\/latex]<\/p>\n<p>22.\u00a0[latex]p\\left(x\\right)=\\frac{1}{4}x - 5[\/latex]<\/p>\n<p>23. [latex]n\\left(x\\right)=-\\frac{1}{3}x - 2[\/latex]<\/p>\n<p>24.\u00a0[latex]m\\left(x\\right)=-\\frac{3}{8}x+3[\/latex]<\/p>\n<p>For the following exercises, find the slope of the line that passes through the two given points.<\/p>\n<p>25. [latex]\\left(2,\\text{ }4\\right)[\/latex] and [latex]\\left(4,\\text{ 10}\\right)[\/latex]<\/p>\n<p>26.\u00a0[latex]\\left(1,\\text{ 5}\\right)[\/latex] and [latex]\\left(4,\\text{ 11}\\right)[\/latex]<\/p>\n<p>27. [latex]\\left(-1,\\text{4}\\right)[\/latex] and [latex]\\left(5,\\text{2}\\right)[\/latex]<\/p>\n<p>28. [latex]\\left(8,-2\\right)[\/latex] and [latex]\\left(4,6\\right)[\/latex]<\/p>\n<p>29. [latex]\\left(6,\\text{ }11\\right)[\/latex] and [latex]\\left(-4, 3\\right)[\/latex]<\/p>\n<p>For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible.<\/p>\n<p>30. [latex]f\\left(-5\\right)=-4[\/latex],\u00a0and [latex]f\\left(5\\right)=2[\/latex]<\/p>\n<p>31. [latex]f\\left(-1\\right)=4[\/latex] and [latex]f\\left(5\\right)=1[\/latex]<\/p>\n<p>32.\u00a0[latex]\\left(2,4\\right)[\/latex] and [latex]\\left(4,10\\right)[\/latex]<\/p>\n<p>33. Passes through [latex]\\left(1,5\\right)[\/latex] and [latex]\\left(4,11\\right)[\/latex]<\/p>\n<p>34.\u00a0Passes through [latex]\\left(-1,\\text{ 4}\\right)[\/latex] and [latex]\\left(5,\\text{ 2}\\right)[\/latex]<\/p>\n<p>35. Passes through [latex]\\left(-2,\\text{ 8}\\right)[\/latex] and [latex]\\left(4,\\text{ 6}\\right)[\/latex]<\/p>\n<p>36.\u00a0<em>x<\/em> intercept at [latex]\\left(-2,\\text{ 0}\\right)[\/latex] and <em>y<\/em> intercept at [latex]\\left(0,-3\\right)[\/latex]<\/p>\n<p>37.<em> x<\/em> intercept at [latex]\\left(-5,\\text{ 0}\\right)[\/latex] and <em>y<\/em> intercept at [latex]\\left(0,\\text{ 4}\\right)[\/latex]<\/p>\n<p>For the following exercises, find the slope of the lines graphed.<\/p>\n<p>38.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005131\/CNX_Precalc_Figure_02_01_201.jpg\" alt=\"\" \/><\/p>\n<p>39.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_202.jpg\" alt=\"\" \/><\/p>\n<p>40.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_203.jpg\" alt=\"\" \/><\/p>\n<p>For the following exercises, write an equation for the lines graphed.<\/p>\n<p>41.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005132\/CNX_Precalc_Figure_02_01_205.jpg\" alt=\"\" \/><\/p>\n<p>42.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_206.jpg\" alt=\"\" \/><\/p>\n<p>43.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_207.jpg\" alt=\"\" \/><\/p>\n<p>44.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_208.jpg\" alt=\"\" \/><\/p>\n<p>45.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005133\/CNX_Precalc_Figure_02_01_209.jpg\" alt=\"\" \/><\/p>\n<p>46.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005134\/CNX_Precalc_Figure_02_01_210.jpg\" alt=\"\" \/><\/p>\n<p>For the following exercises, which of the tables could represent a linear function? For each that could be linear, find a linear equation that models the data.<\/p>\n<p>47.<\/p>\n<table id=\"Table_02_01_04\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 5), (5, -10), (10, -25), and (15, -40).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong><em>x<\/em><\/strong><\/td>\n<td>0<\/td>\n<td>5<\/td>\n<td>10<\/td>\n<td>15<\/td>\n<\/tr>\n<tr>\n<td><strong><em>g<\/em>(<em>x<\/em>)<\/strong><\/td>\n<td>5<\/td>\n<td>\u201310<\/td>\n<td>\u201325<\/td>\n<td>\u201340<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>48.<\/p>\n<table id=\"Table_02_01_05\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (0, 5), (5, 30), (10, 105), and (15, 230).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong><em>x<\/em><\/strong><\/td>\n<td>0<\/td>\n<td>5<\/td>\n<td>10<\/td>\n<td>15<\/td>\n<\/tr>\n<tr>\n<td><strong><em>h<\/em>(<em>x<\/em>)<\/strong><\/td>\n<td>5<\/td>\n<td>30<\/td>\n<td>105<\/td>\n<td>230<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>49.<\/p>\n<table id=\"Table_02_01_06\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (0,- 5), (5, 20), (10, 45), and (15, 70).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong><em>x<\/em><\/strong><\/td>\n<td>0<\/td>\n<td>5<\/td>\n<td>10<\/td>\n<td>15<\/td>\n<\/tr>\n<tr>\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\n<td>\u20135<\/td>\n<td>20<\/td>\n<td>45<\/td>\n<td>70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>50.<\/p>\n<table id=\"Table_02_01_07\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (5, 13), (10, 28), (20, 58), and (25, 73).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong><em>x<\/em><\/strong><\/td>\n<td>5<\/td>\n<td>10<\/td>\n<td>20<\/td>\n<td>25<\/td>\n<\/tr>\n<tr>\n<td><strong><em>k<\/em>(<em>x<\/em>)<\/strong><\/td>\n<td>28<\/td>\n<td>13<\/td>\n<td>58<\/td>\n<td>73<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>51.<\/p>\n<table id=\"Table_02_01_08\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'g(x)'. Reading the remaining rows as ordered pairs (i.e., (x, g(x)), we have the following values: (0, 6), (5, -10), (10, -25), and (15, -40).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\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>6<\/td>\n<\/tr>\n<tr>\n<td><strong><em>g<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\n<td>6<\/td>\n<td>\u201319<\/td>\n<td>\u201344<\/td>\n<td>\u201369<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>52.<\/p>\n<table id=\"Table_02_01_09\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'h(x)'. Reading the remaining rows as ordered pairs (i.e., (x, h(x)), we have the following values: (2, 13), (4, 23), (8, 43), and (10, 53).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>6<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\n<td>\u20134<\/td>\n<td>16<\/td>\n<td>36<\/td>\n<td>56<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>53.<\/p>\n<table id=\"Table_02_01_10\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'f(x)'. Reading the remaining rows as ordered pairs (i.e., (x, f(x)), we have the following values: (2, -4), (4, 16), (6, 36), and (8, 56).\">\n<colgroup>\n<col \/>\n<col \/>\n<col \/>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><strong><em>x<\/em><\/strong><\/td>\n<td>2<\/td>\n<td>4<\/td>\n<td>6<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong><em>f<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\n<td>\u20134<\/td>\n<td>16<\/td>\n<td>36<\/td>\n<td>56<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>54.<\/p>\n<table id=\"Table_02_01_11\" class=\"unnumbered\" summary=\"Two columns and five rows. The first column is labeled, 'x'. The second column is labeled, 'k(x)'. Reading the remaining rows as ordered pairs (i.e., (x, k(x)), we have the following values: (0, 6), (2, 31), (6, 106), and (8, 231).\">\n<colgroup>\n<col \/>\n<col \/><\/colgroup>\n<tbody>\n<tr>\n<td><em><strong>x<\/strong><\/em><\/td>\n<td>0<\/td>\n<td>2<\/td>\n<td>6<\/td>\n<td>8<\/td>\n<\/tr>\n<tr>\n<td><strong><em>k<\/em>(<em>x<\/em>)\u00a0<\/strong><\/td>\n<td>6<\/td>\n<td>31<\/td>\n<td>106<\/td>\n<td>231<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div id=\"fs-id1165137502934\" class=\"exercise\">\n<div id=\"fs-id1165137678561\" class=\"solution\"><\/div>\n<\/div>\n<div id=\"fs-id1165135319509\" class=\"exercise\">\n<div id=\"fs-id1165137935676\" class=\"problem\">\n<div style=\"padding-left: 60px\"><\/div>\n<p>55.\u00a0Find the value of <em>x<\/em>\u00a0if a linear function goes through the following points and has the following slope: [latex]\\left(x,2\\right),\\left(-4,6\\right),m=3[\/latex]<\/p>\n<p>56.\u00a0Find the value of <em>y<\/em> if a linear function goes through the following points and has the following slope: [latex]\\left(10,y\\right),\\left(25,100\\right),m=-5[\/latex]<\/p>\n<p>57. Find the equation of the line that passes through the following points: [latex]\\left(a,\\text{ }b\\right)[\/latex] and [latex]\\left(a,\\text{ }b+1\\right)[\/latex]<\/p>\n<p>58.\u00a0Find the equation of the line that passes through the following points: [latex]\\left(2a,\\text{ }b\\right)[\/latex] and [latex]\\left(a,\\text{ }b+1\\right)[\/latex]<\/p>\n<p>59. Find the equation of the line that passes through the following points: [latex]\\left(a,\\text{ }0\\right)[\/latex] and [latex]\\left(c,\\text{ }d\\right)[\/latex]<\/p>\n<p>For the following exercises, match the given linear equation with its graph.<br \/>\n<img decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005147\/CNX_Precalc_Figure_02_02_201.jpg\" alt=\"\" \/><\/p>\n<p>60. [latex]f\\left(x\\right)=-x - 1[\/latex]<\/p>\n<p>61. [latex]f\\left(x\\right)=-2x - 1[\/latex]<\/p>\n<p>62.\u00a0[latex]f\\left(x\\right)=-\\frac{1}{2}x - 1[\/latex]<\/p>\n<p>63. [latex]f\\left(x\\right)=2[\/latex]<\/p>\n<p>64.\u00a0[latex]f\\left(x\\right)=2+x[\/latex]<\/p>\n<p>65. [latex]f\\left(x\\right)=3x+2[\/latex]<\/p>\n<p>For the following exercises, sketch a line with the given features.<\/p>\n<p>66. An x-intercept of [latex]\\left(-\\text{4},\\text{ 0}\\right)[\/latex] and y-intercept of [latex]\\left(0,\\text{ -2}\\right)[\/latex]<\/p>\n<p>67. An x-intercept of [latex]\\left(-\\text{2},\\text{ 0}\\right)[\/latex] and y-intercept of [latex]\\left(0,\\text{ 4}\\right)[\/latex]<\/p>\n<p>68.\u00a0A y-intercept of [latex]\\left(0,\\text{ 7}\\right)[\/latex] and slope [latex]-\\frac{3}{2}[\/latex]<\/p>\n<p>69. A y-intercept of [latex]\\left(0,\\text{ 3}\\right)[\/latex] and slope [latex]\\frac{2}{5}[\/latex]<\/p>\n<p>70.\u00a0Passing through the points [latex]\\left(-\\text{6},\\text{ -2}\\right)[\/latex] and [latex]\\left(\\text{6},\\text{ -6}\\right)[\/latex]<\/p>\n<p>71. Passing through the points [latex]\\left(-\\text{3},\\text{ -4}\\right)[\/latex] and [latex]\\left(\\text{3},\\text{ 0}\\right)[\/latex]<\/p>\n<p>For the following exercises, sketch the graph of each equation.<\/p>\n<p>72. [latex]f\\left(x\\right)=-2x - 1[\/latex]<\/p>\n<p>73. [latex]g\\left(x\\right)=-3x+2[\/latex]<\/p>\n<p>74. [latex]h\\left(x\\right)=\\frac{1}{3}x+2[\/latex]<\/p>\n<p>75. [latex]k\\left(x\\right)=\\frac{2}{3}x - 3[\/latex]<\/p>\n<p>76. [latex]f\\left(t\\right)=3+2t[\/latex]<\/p>\n<p>77. [latex]p\\left(t\\right)=-2+3t[\/latex]<\/p>\n<p>78.\u00a0[latex]x=3[\/latex]<\/p>\n<p>79. [latex]x=-2[\/latex]<\/p>\n<p>80. [latex]r\\left(x\\right)=4[\/latex]<\/p>\n<p>81. [latex]q\\left(x\\right)=3[\/latex]<\/p>\n<p>82. [latex]4x=-9y+36[\/latex]<\/p>\n<p>83. [latex]\\frac{x}{3}-\\frac{y}{4}=1[\/latex]<\/p>\n<p>84. [latex]3x - 5y=15[\/latex]<\/p>\n<p>85. [latex]3x=15[\/latex]<\/p>\n<p>86. [latex]3y=12[\/latex]<\/p>\n<p>87. If [latex]g\\left(x\\right)[\/latex] is the transformation of [latex]f\\left(x\\right)=x[\/latex] after a vertical compression by [latex]\\frac{3}{4}[\/latex], a shift right by 2, and a shift down by 4<\/p>\n<p style=\"padding-left: 60px\">a. Write an equation for [latex]g\\left(x\\right)[\/latex].<\/p>\n<p style=\"padding-left: 60px\">b. What is the slope of this line?<\/p>\n<p style=\"padding-left: 60px\">c. Find the y-intercept of this line.<\/p>\n<p>88.\u00a0If [latex]g\\left(x\\right)[\/latex] is the transformation of [latex]f\\left(x\\right)=x[\/latex] after a vertical compression by [latex]\\frac{1}{3}[\/latex], a shift left by 1, and a shift up by 3<\/p>\n<p style=\"padding-left: 60px\">a. Write an equation for [latex]g\\left(x\\right)[\/latex].<\/p>\n<p style=\"padding-left: 60px\">b. What is the slope of this line?<\/p>\n<p style=\"padding-left: 60px\">c. Find the y-intercept of this line.<\/p>\n<p>For the following exercises, write the equation of the line shown in the graph.<\/p>\n<p>89.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005149\/CNX_Precalc_Figure_02_02_222.jpg\" alt=\"\" \/><\/p>\n<p>90.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_223.jpg\" alt=\"\" \/><\/p>\n<p>91.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_224.jpg\" alt=\"\" \/><\/p>\n<p>92.<br \/>\n<img decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005150\/CNX_Precalc_Figure_02_02_225.jpg\" alt=\"\" \/><\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\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":4,"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":10705,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/users\/23588"}],"version-history":[{"count":26,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/revisions"}],"predecessor-version":[{"id":17561,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/revisions\/17561"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/parts\/10705"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapters\/13798\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/media?parent=13798"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13798"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/contributor?post=13798"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/csn-precalculus\/wp-json\/wp\/v2\/license?post=13798"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}