{"id":13807,"date":"2018-08-24T19:49:05","date_gmt":"2018-08-24T19:49:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/precalcone\/?post_type=chapter&#038;p=13807"},"modified":"2025-02-05T05:18:37","modified_gmt":"2025-02-05T05:18:37","slug":"graphs-of-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/precalculus\/chapter\/graphs-of-linear-functions\/","title":{"raw":"Graphs of Linear Functions","rendered":"Graphs of Linear Functions"},"content":{"raw":"<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\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 \t<li>Write the equation of a line parallel or perpendicular to a given line.<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1165135609321\">Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of comparing functions using graphs.<\/p>\r\n\r\n<h2>Graphing Linear Functions<\/h2>\r\n<div>\r\n<p id=\"fs-id1165137806314\">In Linear Functions, we saw that that 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 1: 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 1 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 1<\/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 2.<\/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 2<\/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 2: 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 3<\/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 2\u00a0that 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 4\u00a0that 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 4.<\/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 5\u00a0that 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 5.<\/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 3: 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 6.\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 7.<\/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\">Recall that in Linear Functions, 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 8. 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 8<\/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 4: 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 9.<\/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 9<\/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 10.\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 10<\/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 11<\/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 5: 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 12. 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 12.\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 13, 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 13.\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 14<\/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 15<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 15.<\/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 6: Writing the Equation of a Horizontal Line<\/h3>\r\nWrite the equation of the line graphed in Figure 16.\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 16<\/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 7: Writing the Equation of a Vertical Line<\/h3>\r\n<p id=\"fs-id1165137871492\">Write the equation of the line graphed in Figure 17.<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 17<\/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<h2 style=\"text-align: center;\">Determining Whether Lines are Parallel or Perpendicular<\/h2>\r\nThe two lines in Figure 18\u00a0are <strong>parallel<\/strong> <strong>lines<\/strong>: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the <em>y<\/em>-intercept. If we shifted one line vertically toward the <em>y<\/em>-intercept of the other, they would become the same line.\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\/03010649\/CNX_Precalc_Figure_02_02_019n2.jpg\" alt=\"Graph of two functions where the blue line is y = -2\/3x + 1, and the baby blue line is y = -2\/3x +7. Notice that they are parallel lines.\" width=\"487\" height=\"410\" \/> <b>Figure 18\u00a0<\/b>Parallel lines.[\/caption]\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"437\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010649\/CNX_Precalc_EQ_02_02_001n2.jpg\" alt=\"The functions 2x plus 6 and negative 2x minus 4 are parallel. The functions 3x plus 2 and 2x plus 2 are not parallel.\" width=\"437\" height=\"58\" \/> <b>Figure 19<\/b>[\/caption]\r\n<p id=\"fs-id1165135499959\">We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the <em>y<\/em>-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.<span id=\"eip-id1165134117274\">\r\n<\/span><\/p>\r\nUnlike parallel lines,<strong> perpendicular lines<\/strong> do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 20\u00a0are perpendicular.\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\/03010650\/CNX_Precalc_Figure_02_02_020n2.jpg\" alt=\"Graph of two functions where the blue line is perpendicular to the orange line.\" width=\"487\" height=\"441\" \/> <b>Figure 20.<\/b> Perpendicular lines.[\/caption]\r\n<p id=\"fs-id1165137731752\">Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if [latex]{m}_{1}\\text{ and }{m}_{2}[\/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[\/latex].<\/p>\r\n\r\n<div id=\"fs-id1165137786218\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{1}{m}_{2}=-1[\/latex]<\/div>\r\n<p id=\"fs-id1165137892275\">To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\\frac{1}{8}[\/latex], and the reciprocal of [latex]\\frac{1}{8}[\/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.<\/p>\r\n<p id=\"fs-id1165137611863\">As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.<\/p>\r\n\r\n<div id=\"fs-id1165137605494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(x\\right)=\\frac{1}{4}x+2 &amp;&amp; \\text{negative reciprocal of}\\frac{1}{4}\\text{ is }-4 \\\\ &amp;f\\left(x\\right)=-4x+3 &amp;&amp; \\text{negative reciprocal of}-4\\text{ is }\\frac{1}{4} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137419406\">The product of the slopes is \u20131.<\/p>\r\n\r\n<div id=\"fs-id1165135570237\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-4\\left(\\frac{1}{4}\\right)=-1[\/latex]<\/div>\r\n<div id=\"fs-id1165137722848\" class=\"note textbox\">\r\n<h3 class=\"title\">A General Note: Parallel and Perpendicular Lines<\/h3>\r\n<p id=\"fs-id1165137722856\">Two lines are <strong>parallel lines<\/strong> if they do not intersect. The slopes of the lines are the same.<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are parallel if }{m}_{1}={m}_{2}[\/latex].<\/p>\r\n<p id=\"fs-id1165135541604\">If and only if [latex]{b}_{1}={b}_{2}[\/latex] and [latex]{m}_{1}={m}_{2}[\/latex], we say the lines coincide. Coincident lines are the same line.<\/p>\r\n<p id=\"fs-id1165137782453\">Two lines are <strong>perpendicular lines<\/strong> if they intersect at right angles.<\/p>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are perpendicular if }{m}_{1}{m}_{2}=-1,\\text{ and so }{m}_{2}=-\\frac{1}{{m}_{1}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"Example_02_02_08\" class=\"example\">\r\n<div id=\"fs-id1165137596422\" class=\"exercise\">\r\n<div id=\"fs-id1165137596424\" class=\"problem textbox shaded\">\r\n<h3>Example 8: Identifying Parallel and Perpendicular Lines<\/h3>\r\n<p id=\"fs-id1165137470055\">Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}&amp;f\\left(x\\right)=2x+3 &amp;&amp;h\\left(x\\right)=-2x+2 \\\\ &amp;g\\left(x\\right)=\\frac{1}{2}x - 4 &amp;&amp; j\\left(x\\right)=2x - 6 \\end{align}[\/latex]<\/p>\r\n[reveal-answer q=\"825485\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"825485\"]\r\n\r\nParallel lines have the same slope. Because the functions [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because \u22122 and [latex]\\frac{1}{2}[\/latex] are negative reciprocals, the equations, [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] represent perpendicular lines.\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the lines is shown in Figure 21.\r\n\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\/03010650\/CNX_Precalc_Figure_02_02_0212.jpg\" alt=\"Graph of four functions where the blue line is h(x) = -2x + 2, the orange line is f(x) = 2x + 3, the green line is j(x) = 2x - 6, and the red line is g(x) = 1\/2x - 4.\" width=\"487\" height=\"428\" \/>\r\n<p id=\"fs-id1165137407484\" style=\"text-align: center;\"><strong>Figure 21.<\/strong> The graph shows that the lines [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] are parallel, and the lines [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] are perpendicular.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135187508\" class=\"commentary\">\r\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Parallel or Perpendicular to a Given Line<\/h2>\r\n<p id=\"fs-id1165137812926\">If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\r\n\r\n<section id=\"fs-id1165137812931\">\r\n<h3 style=\"text-align: center;\">Writing Equations of Parallel Lines<\/h3>\r\n<p id=\"fs-id1165135503943\">Suppose for example, we are given the following equation.<\/p>\r\n\r\n<div id=\"fs-id1165137678988\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\/latex]<\/div>\r\n<p id=\"fs-id1165137714860\">We know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).<\/p>\r\n\r\n<div id=\"fs-id1165137871544\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;g\\left(x\\right)=3x+6 \\\\ &amp;h\\left(x\\right)=3x+1 \\\\ &amp;p\\left(x\\right)=3x+\\frac{2}{3} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137465914\">Suppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.<\/p>\r\n\r\n<div id=\"fs-id1165137611864\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y - 7=3\\left(x - 1\\right) \\\\ y - 7=3x - 3 \\\\ y=3x+4 \\end{gathered}[\/latex]<\/div>\r\n<p id=\"fs-id1165137760890\">So [latex]g(x)=3x+4[\/latex] is parallel to [latex]f(x)=3x+1[\/latex] and passes through the point (1,7).<\/p>\r\n\r\n<div id=\"fs-id1165135531520\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135531526\">How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\r\n<ol id=\"fs-id1165137602390\">\r\n \t<li>Find the slope of the function.<\/li>\r\n \t<li>Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_09\" class=\"example\">\r\n<div id=\"fs-id1165137432599\" class=\"exercise\">\r\n<div id=\"fs-id1165137432601\" class=\"problem textbox shaded\">\r\n<h3>Example 9: Finding a Line Parallel to a Given Line<\/h3>\r\n<p id=\"fs-id1165137770237\">Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3,0).<\/p>\r\n[reveal-answer q=\"471532\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471532\"]\r\n<p id=\"fs-id1165135190486\">The slope of the given line is 3. If we choose the slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and <em>f<\/em>(<em>x<\/em>) = 0 into the slope-intercept form to find the <em>y-<\/em>intercept.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=3x+b \\\\ 0=3\\left(3\\right)+b \\\\ b=-9 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137643164\">The line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165137433991\">We can confirm that the two lines are parallel by graphing them. Figure 22\u00a0shows that the two lines will never intersect.<span id=\"fs-id1165135168455\">\r\n<\/span><\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010650\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/> <b>Figure 22<\/b>[\/caption]\r\n\r\n[\/hidden-answer]<b><\/b>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><section id=\"fs-id1165134093077\">\r\n<h3 style=\"text-align: center;\">Writing Equations of Perpendicular Lines<\/h3>\r\n<p id=\"fs-id1165134093082\">We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\r\n\r\n<div id=\"fs-id1165137443640\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/div>\r\n<p id=\"fs-id1165135696186\">The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).<\/p>\r\n\r\n<div id=\"fs-id1165135394319\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&amp;g\\left(x\\right)=-\\frac{1}{2}x+4 \\\\ &amp;h\\left(x\\right)=-\\frac{1}{2}x+2 \\\\ &amp;p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2} \\end{align}[\/latex]<\/div>\r\n<p id=\"fs-id1165137453371\">As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.<\/p>\r\n\r\n<div id=\"fs-id1165137645414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=mx+b \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b \\\\ 0=-2+b \\\\ 2=b \\\\ b=2 \\end{gathered}[\/latex]<\/div>\r\n<p id=\"fs-id1165135422935\">The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is<\/p>\r\n\r\n<div id=\"fs-id1165137760043\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/div>\r\n<p id=\"fs-id1165137725186\">So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4,0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\r\n\r\n<div id=\"fs-id1165137601744\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165137737863\"><strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\r\n<p id=\"fs-id1165137737871\"><em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1165137715408\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165137715414\">How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\r\n<ol id=\"fs-id1165137871694\">\r\n \t<li>Find the slope of the function.<\/li>\r\n \t<li>Determine the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t<li>Solve for <em>b<\/em>.<\/li>\r\n \t<li>Write the equation for the line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_10\" class=\"example\">\r\n<div id=\"fs-id1165135512529\" class=\"exercise\">\r\n<div id=\"fs-id1165135512531\" class=\"problem textbox shaded\">\r\n<h3>Example 10: Finding the Equation of a Perpendicular Line<\/h3>\r\n<p id=\"fs-id1165135512536\">Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3[\/latex] that passes through the point (3,0).<\/p>\r\n[reveal-answer q=\"657332\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"657332\"]\r\n<p id=\"fs-id1165137757791\">The original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=-\\frac{1}{3}x+b \\\\ 0=-\\frac{1}{3}\\left(3\\right)+b \\\\ 1=b \\\\ b=1 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165137415244\">The line perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\n<p id=\"fs-id1165135175331\">A graph of the two lines is shown in Figure 23.<span id=\"fs-id1165137936715\">\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\/03010650\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/> <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<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165135454012\">Given the function [latex]h\\left(x\\right)=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\r\n<p style=\"padding-left: 60px;\">a. parallel to <em>h<\/em>(<em>x<\/em>)\r\nb. perpendicular to <em>h<\/em>(<em>x<\/em>)<\/p>\r\n[reveal-answer q=\"398341\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"398341\"]\r\n\r\na.\u00a0[latex]f\\left(x\\right)=2x[\/latex]\r\nb. [latex]g\\left(x\\right)=-\\frac{1}{2}x[\/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 hide_question_numbers=1]169764[\/ohm_question]\r\n\r\n<\/div>\r\n<div id=\"fs-id1165135509152\" class=\"note precalculus howto textbox\">\r\n<h3 id=\"fs-id1165135509158\">How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\r\n<ol id=\"fs-id1165137676542\">\r\n \t<li>Determine the slope of the line passing through the points.<\/li>\r\n \t<li>Find the negative reciprocal of the slope.<\/li>\r\n \t<li>Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div id=\"Example_02_02_11\" class=\"example\">\r\n<div id=\"fs-id1165134267817\" class=\"exercise\">\r\n<div id=\"fs-id1165135332504\" class=\"problem textbox shaded\">\r\n<h3>Example 11: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\r\n<p id=\"fs-id1165135332509\">A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\r\n[reveal-answer q=\"196054\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"196054\"]\r\n\r\nFrom the two points of the given line, we can calculate the slope of that line.\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}_{1}&amp;=\\frac{5 - 6}{4-\\left(-2\\right)} \\\\ &amp;=\\frac{-1}{6} \\\\ &amp;=-\\frac{1}{6} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135532423\">Find the negative reciprocal of the slope.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}_{2}&amp;=\\frac{-1}{-\\frac{1}{6}} \\\\ &amp;=-1\\left(-\\frac{6}{1}\\right) \\\\ &amp;=6 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137768497\">We can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=6x+b \\\\ 5=6\\left(4\\right)+b \\\\ 5=24+b \\\\ -19=b \\\\ b=-19 \\end{gathered}[\/latex]<\/p>\r\n<p id=\"fs-id1165135159921\">The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is<\/p>\r\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/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-id1165137437225\">A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\r\n[reveal-answer q=\"137755\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"137755\"]\r\n\r\n[latex]y=-\\frac{1}{3}x+6[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 style=\"text-align: center;\">Solving a System of Linear Equations Using a Graph<\/h2>\r\n<p id=\"fs-id1165137627910\">A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system.<\/p>\r\n<p id=\"fs-id1165137812669\">To find this point when the equations are given as functions, we can solve for an input value so that [latex]f\\left(x\\right)=g\\left(x\\right)[\/latex]. In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.<\/p>\r\n\r\n<div id=\"Example_02_02_12\" class=\"example\">\r\n<div id=\"fs-id1165137896187\" class=\"exercise\">\r\n<div id=\"fs-id1165137896189\" class=\"problem textbox shaded\">\r\n<h3>Example 12: Finding a Point of Intersection Algebraically<\/h3>\r\n<p id=\"fs-id1165135693776\">Find the point of intersection of the lines [latex]h\\left(t\\right)=3t - 4[\/latex] and [latex]j\\left(t\\right)=5-t[\/latex].<\/p>\r\n[reveal-answer q=\"703225\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"703225\"]\r\n<p id=\"fs-id1165137838172\">Set [latex]h\\left(t\\right)=j\\left(t\\right)[\/latex].<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}3t - 4&amp;=5-t \\\\ 4t&amp;=9 \\\\ t&amp;=\\frac{9}{4} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137644219\">This tells us the lines intersect when the input is [latex]\\frac{9}{4}[\/latex].<\/p>\r\n<p id=\"fs-id1165137812337\">We can then find the output value of the intersection point by evaluating either function at this input.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}j\\left(\\frac{9}{4}\\right)&amp;=5-\\frac{9}{4} \\\\ &amp;=\\frac{11}{4} \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165137745176\">These lines intersect at the point [latex]\\left(\\frac{9}{4},\\frac{11}{4}\\right)[\/latex].<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nLooking at Figure 24, this result seems reasonable.\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\/03010651\/CNX_Precalc_Figure_02_02_0242.jpg\" alt=\"Graph of two functions h(t) = 3t - 4 and j(t) = t +5 and their intersection at (9\/4, 11\/4).\" width=\"487\" height=\"441\" \/> <b>Figure 24<\/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=\"fs-id1165137603219\" class=\"note precalculus qa textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<p id=\"fs-id1165137447025\"><strong>If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions?<\/strong><\/p>\r\n<p id=\"fs-id1165137447030\"><em>Yes. After setting the two equations equal to one another, the result would be the contradiction \"0 = non-zero real number\".<\/em><\/p>\r\n\r\n<\/div>\r\n<div class=\"bcc-box bcc-success\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1165137832295\">Using the graph in the Analysis of the Solution for Example 12,\u00a0identify the following for the function [latex]j\\left(t\\right)[\/latex]:<\/p>\r\n<p style=\"padding-left: 60px;\">a. y-intercept<\/p>\r\n<p style=\"padding-left: 60px;\">b. x-intercept(s)<\/p>\r\n<p style=\"padding-left: 60px;\">c. slope<\/p>\r\n<p style=\"padding-left: 60px;\">d. Is [latex]j\\left(t\\right)[\/latex] parallel or perpendicular to [latex]h\\left(t\\right)[\/latex] (or neither)?<\/p>\r\n<p style=\"padding-left: 60px;\">e. Is [latex]j\\left(t\\right)[\/latex] an increasing or decreasing function (or neither)?<\/p>\r\n<p style=\"padding-left: 60px;\">f. Write a transformation description for [latex]j\\left(t\\right)[\/latex] from the identity toolkit function [latex]f\\left(x\\right)=x[\/latex].<\/p>\r\n[reveal-answer q=\"382866\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"382866\"]\r\n<p style=\"padding-left: 60px;\">a.\u00a0[latex]\\left(0,5\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">b. [latex]\\left(5,\\text{ 0}\\right)[\/latex]<\/p>\r\n<p style=\"padding-left: 60px;\">c. Slope -1<\/p>\r\n<p style=\"padding-left: 60px;\">d. Neither parallel nor perpendicular<\/p>\r\n<p style=\"padding-left: 60px;\">e. Decreasing function<\/p>\r\n<p style=\"padding-left: 60px;\">f. Given the identity function, perform a vertical flip (over the <em>t<\/em>-axis) and shift up 5 units.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"Example_02_02_13\" class=\"example\">\r\n<div id=\"fs-id1165137761773\" class=\"exercise\">\r\n<div id=\"fs-id1165137761775\" class=\"problem textbox shaded\">\r\n<h3>Example 13: Finding a Break-Even Point<\/h3>\r\n<p id=\"fs-id1165137761781\">A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.<\/p>\r\n\r\n<ol id=\"fs-id1165137870987\">\r\n \t<li>Find the cost function, <em>C<\/em>, to produce <em>x<\/em>\u00a0helmets, in dollars.<\/li>\r\n \t<li>Find the revenue function, <em>R<\/em>, from the sales of <em>x<\/em>\u00a0helmets, in dollars.<\/li>\r\n \t<li>Find the break-even point, the point of intersection of the two graphs <em>C\u00a0<\/em>and <em>R<\/em>.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"818513\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"818513\"]\r\n<ol id=\"fs-id1165137653646\">\r\n \t<li>he cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.\r\n<div id=\"eip-id1885657\" class=\"equation unnumbered\">[latex]C\\left(x\\right)=120x+250,000[\/latex]<\/div><\/li>\r\n \t<li>The revenue function is the total revenue from the sale of [latex]x[\/latex] helmets, [latex]R\\left(x\\right)=140x[\/latex].<\/li>\r\n \t<li>The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the <em>x<\/em>-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for <em>x<\/em>.<\/li>\r\n<\/ol>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(x\\right)&amp;=R\\left(x\\right) \\\\ 250,000+120x&amp;=140x \\\\ 250,000&amp;=20x \\\\ 12,500&amp;=x \\\\ x&amp;=12,500 \\end{align}[\/latex]<\/p>\r\n<p id=\"eip-id1165134183817\">To find [latex]y[\/latex], evaluate either the revenue or the cost function at 12,500.<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{align}R\\left(20\\right)&amp;=140\\left(12,500\\right) \\\\ &amp;=$1,750,000 \\end{align}[\/latex]<\/p>\r\n<p id=\"fs-id1165135496435\">The break-even point is (12,500, 1,750,000).<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nThis means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03005146\/CNX_Precalc_Figure_02_02_025.jpg\" alt=\"Graph of the two functions, C(x) and R(x) where it shows that below (12500, 1750000) the company loses money and above that point the company makes a profit.\" width=\"731\" height=\"695\" \/> <b>Figure 25<\/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<div class=\"note precalculus media\">\r\n<h2>Key Concepts<\/h2>\r\n<ul id=\"fs-id1165134190780\">\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 \t<li>Parallel lines have the same slope.<\/li>\r\n \t<li>Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.<\/li>\r\n \t<li>A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the <em>x<\/em>- and <em>y<\/em>-values of the given point into the equation, [latex]f(x)=mx+b[\/latex], and using the <em>b<\/em>\u00a0that results. Similarly, the point-slope form of an equation can also be used.<\/li>\r\n \t<li>A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.<\/li>\r\n \t<li>A system of linear equations may be solved setting the two equations equal to one another and solving for <em>x<\/em>. The <em>y<\/em>-value may be found by evaluating either one of the original equations using this <em>x<\/em>-value.<\/li>\r\n \t<li>A system of linear equations may also be solved by finding the point of intersection on a graph.<\/li>\r\n<\/ul>\r\n<div>\r\n<h2>Glossary<\/h2>\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-id1165135330621\" class=\"definition\">\r\n \t<dt><strong>parallel lines<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135186582\">two or more lines with the same slope<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135186586\" class=\"definition\">\r\n \t<dt><strong>perpendicular lines<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135186592\">two lines that intersect at right angles and have slopes that are negative reciprocals of each other<\/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<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/section><\/section><\/section>","rendered":"<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph linear functions.<\/li>\n<li>Write the equation for a linear function from the graph of a line.<\/li>\n<li>Write the equation of a line parallel or perpendicular to a given line.<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1165135609321\">Two competing telephone companies offer different payment plans. The two plans charge the same rate per long distance minute, but charge a different monthly flat fee. A consumer wants to determine whether the two plans will ever cost the same amount for a given number of long distance minutes used. The total cost of each payment plan can be represented by a linear function. To solve the problem, we will need to compare the functions. In this section, we will consider methods of comparing functions using graphs.<\/p>\n<h2>Graphing Linear Functions<\/h2>\n<div>\n<p id=\"fs-id1165137806314\">In Linear Functions, we saw that that 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 1: 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 1 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 1<\/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 2.<\/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 2<\/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 2: 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 3<\/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 2\u00a0that 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 4\u00a0that 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 4.<\/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 5\u00a0that 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 5.<\/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 3: 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 6.\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 7.<\/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\">Recall that in Linear Functions, 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 8. 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 8<\/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 4: 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 9.<\/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 9<\/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 10.<\/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 10<\/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 11<\/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 5: 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 12. 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 12.\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 13, 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 13.\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 14<\/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 15<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 15.<\/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 6: Writing the Equation of a Horizontal Line<\/h3>\n<p>Write the equation of the line graphed in Figure 16.<\/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 16<\/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 7: Writing the Equation of a Vertical Line<\/h3>\n<p id=\"fs-id1165137871492\">Write the equation of the line graphed in Figure 17.<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 17<\/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<h2 style=\"text-align: center;\">Determining Whether Lines are Parallel or Perpendicular<\/h2>\n<p>The two lines in Figure 18\u00a0are <strong>parallel<\/strong> <strong>lines<\/strong>: they will never intersect. Notice that they have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the <em>y<\/em>-intercept. If we shifted one line vertically toward the <em>y<\/em>-intercept of the other, they would become the same line.<\/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\/03010649\/CNX_Precalc_Figure_02_02_019n2.jpg\" alt=\"Graph of two functions where the blue line is y = -2\/3x + 1, and the baby blue line is y = -2\/3x +7. Notice that they are parallel lines.\" width=\"487\" height=\"410\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 18\u00a0<\/b>Parallel lines.<\/p>\n<\/div>\n<div style=\"width: 447px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/1227\/2015\/04\/03010649\/CNX_Precalc_EQ_02_02_001n2.jpg\" alt=\"The functions 2x plus 6 and negative 2x minus 4 are parallel. The functions 3x plus 2 and 2x plus 2 are not parallel.\" width=\"437\" height=\"58\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 19<\/b><\/p>\n<\/div>\n<p id=\"fs-id1165135499959\">We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the <em>y<\/em>-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.<span id=\"eip-id1165134117274\"><br \/>\n<\/span><\/p>\n<p>Unlike parallel lines,<strong> perpendicular lines<\/strong> do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 20\u00a0are perpendicular.<\/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\/03010650\/CNX_Precalc_Figure_02_02_020n2.jpg\" alt=\"Graph of two functions where the blue line is perpendicular to the orange line.\" width=\"487\" height=\"441\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 20.<\/b> Perpendicular lines.<\/p>\n<\/div>\n<p id=\"fs-id1165137731752\">Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if [latex]{m}_{1}\\text{ and }{m}_{2}[\/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[\/latex].<\/p>\n<div id=\"fs-id1165137786218\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]{m}_{1}{m}_{2}=-1[\/latex]<\/div>\n<p id=\"fs-id1165137892275\">To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\\frac{1}{8}[\/latex], and the reciprocal of [latex]\\frac{1}{8}[\/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.<\/p>\n<p id=\"fs-id1165137611863\">As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor perpendicular. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.<\/p>\n<div id=\"fs-id1165137605494\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&f\\left(x\\right)=\\frac{1}{4}x+2 && \\text{negative reciprocal of}\\frac{1}{4}\\text{ is }-4 \\\\ &f\\left(x\\right)=-4x+3 && \\text{negative reciprocal of}-4\\text{ is }\\frac{1}{4} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137419406\">The product of the slopes is \u20131.<\/p>\n<div id=\"fs-id1165135570237\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]-4\\left(\\frac{1}{4}\\right)=-1[\/latex]<\/div>\n<div id=\"fs-id1165137722848\" class=\"note textbox\">\n<h3 class=\"title\">A General Note: Parallel and Perpendicular Lines<\/h3>\n<p id=\"fs-id1165137722856\">Two lines are <strong>parallel lines<\/strong> if they do not intersect. The slopes of the lines are the same.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are parallel if }{m}_{1}={m}_{2}[\/latex].<\/p>\n<p id=\"fs-id1165135541604\">If and only if [latex]{b}_{1}={b}_{2}[\/latex] and [latex]{m}_{1}={m}_{2}[\/latex], we say the lines coincide. Coincident lines are the same line.<\/p>\n<p id=\"fs-id1165137782453\">Two lines are <strong>perpendicular lines<\/strong> if they intersect at right angles.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are perpendicular if }{m}_{1}{m}_{2}=-1,\\text{ and so }{m}_{2}=-\\frac{1}{{m}_{1}}[\/latex].<\/p>\n<\/div>\n<div id=\"Example_02_02_08\" class=\"example\">\n<div id=\"fs-id1165137596422\" class=\"exercise\">\n<div id=\"fs-id1165137596424\" class=\"problem textbox shaded\">\n<h3>Example 8: Identifying Parallel and Perpendicular Lines<\/h3>\n<p id=\"fs-id1165137470055\">Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}&f\\left(x\\right)=2x+3 &&h\\left(x\\right)=-2x+2 \\\\ &g\\left(x\\right)=\\frac{1}{2}x - 4 && j\\left(x\\right)=2x - 6 \\end{align}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q825485\">Show Solution<\/span><\/p>\n<div id=\"q825485\" class=\"hidden-answer\" style=\"display: none\">\n<p>Parallel lines have the same slope. Because the functions [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because \u22122 and [latex]\\frac{1}{2}[\/latex] are negative reciprocals, the equations, [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] represent perpendicular lines.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the lines is shown in Figure 21.<\/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\/03010650\/CNX_Precalc_Figure_02_02_0212.jpg\" alt=\"Graph of four functions where the blue line is h(x) = -2x + 2, the orange line is f(x) = 2x + 3, the green line is j(x) = 2x - 6, and the red line is g(x) = 1\/2x - 4.\" width=\"487\" height=\"428\" \/><\/p>\n<p id=\"fs-id1165137407484\" style=\"text-align: center;\"><strong>Figure 21.<\/strong> The graph shows that the lines [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] are parallel, and the lines [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] are perpendicular.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165135187508\" class=\"commentary\">\n<h2 style=\"text-align: center;\">Writing the Equation of a Line Parallel or Perpendicular to a Given Line<\/h2>\n<p id=\"fs-id1165137812926\">If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n<section id=\"fs-id1165137812931\">\n<h3 style=\"text-align: center;\">Writing Equations of Parallel Lines<\/h3>\n<p id=\"fs-id1165135503943\">Suppose for example, we are given the following equation.<\/p>\n<div id=\"fs-id1165137678988\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\/latex]<\/div>\n<p id=\"fs-id1165137714860\">We know that the slope of the line formed by the function is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).<\/p>\n<div id=\"fs-id1165137871544\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&g\\left(x\\right)=3x+6 \\\\ &h\\left(x\\right)=3x+1 \\\\ &p\\left(x\\right)=3x+\\frac{2}{3} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137465914\">Suppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form.<\/p>\n<div id=\"fs-id1165137611864\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}y-{y}_{1}=m\\left(x-{x}_{1}\\right) \\\\ y - 7=3\\left(x - 1\\right) \\\\ y - 7=3x - 3 \\\\ y=3x+4 \\end{gathered}[\/latex]<\/div>\n<p id=\"fs-id1165137760890\">So [latex]g(x)=3x+4[\/latex] is parallel to [latex]f(x)=3x+1[\/latex] and passes through the point (1,7).<\/p>\n<div id=\"fs-id1165135531520\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135531526\">How To: Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point.<\/h3>\n<ol id=\"fs-id1165137602390\">\n<li>Find the slope of the function.<\/li>\n<li>Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_09\" class=\"example\">\n<div id=\"fs-id1165137432599\" class=\"exercise\">\n<div id=\"fs-id1165137432601\" class=\"problem textbox shaded\">\n<h3>Example 9: Finding a Line Parallel to a Given Line<\/h3>\n<p id=\"fs-id1165137770237\">Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3,0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q471532\">Show Solution<\/span><\/p>\n<div id=\"q471532\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165135190486\">The slope of the given line is 3. If we choose the slope-intercept form, we can substitute <em>m\u00a0<\/em>= 3, <em>x\u00a0<\/em>= 3, and <em>f<\/em>(<em>x<\/em>) = 0 into the slope-intercept form to find the <em>y-<\/em>intercept.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=3x+b \\\\ 0=3\\left(3\\right)+b \\\\ b=-9 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137643164\">The line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165137433991\">We can confirm that the two lines are parallel by graphing them. Figure 22\u00a0shows that the two lines will never intersect.<span id=\"fs-id1165135168455\"><br \/>\n<\/span><\/p>\n<div style=\"width: 741px\" 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\/03010650\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 22<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-id1165134093077\">\n<h3 style=\"text-align: center;\">Writing Equations of Perpendicular Lines<\/h3>\n<p id=\"fs-id1165134093082\">We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\n<div id=\"fs-id1165137443640\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/div>\n<p id=\"fs-id1165135696186\">The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). So the lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).<\/p>\n<div id=\"fs-id1165135394319\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{align}&g\\left(x\\right)=-\\frac{1}{2}x+4 \\\\ &h\\left(x\\right)=-\\frac{1}{2}x+2 \\\\ &p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2} \\end{align}[\/latex]<\/div>\n<p id=\"fs-id1165137453371\">As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.<\/p>\n<div id=\"fs-id1165137645414\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=mx+b \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b \\\\ 0=-2+b \\\\ 2=b \\\\ b=2 \\end{gathered}[\/latex]<\/div>\n<p id=\"fs-id1165135422935\">The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is<\/p>\n<div id=\"fs-id1165137760043\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/div>\n<p id=\"fs-id1165137725186\">So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4,0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n<div id=\"fs-id1165137601744\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165137737863\"><strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n<p id=\"fs-id1165137737871\"><em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n<\/div>\n<div id=\"fs-id1165137715408\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165137715414\">How To: Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line.<\/h3>\n<ol id=\"fs-id1165137871694\">\n<li>Find the slope of the function.<\/li>\n<li>Determine the negative reciprocal of the slope.<\/li>\n<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from the coordinate pair provided into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\n<li>Solve for <em>b<\/em>.<\/li>\n<li>Write the equation for the line.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_10\" class=\"example\">\n<div id=\"fs-id1165135512529\" class=\"exercise\">\n<div id=\"fs-id1165135512531\" class=\"problem textbox shaded\">\n<h3>Example 10: Finding the Equation of a Perpendicular Line<\/h3>\n<p id=\"fs-id1165135512536\">Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3[\/latex] that passes through the point (3,0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q657332\">Show Solution<\/span><\/p>\n<div id=\"q657332\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137757791\">The original line has slope <em>m\u00a0<\/em>= 3, so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=-\\frac{1}{3}x+b \\\\ 0=-\\frac{1}{3}\\left(3\\right)+b \\\\ 1=b \\\\ b=1 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165137415244\">The line perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p id=\"fs-id1165135175331\">A graph of the two lines is shown in Figure 23.<span id=\"fs-id1165137936715\"><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\/03010650\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/><\/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<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165135454012\">Given the function [latex]h\\left(x\\right)=2x - 4[\/latex], write an equation for the line passing through (0, 0) that is<\/p>\n<p style=\"padding-left: 60px;\">a. parallel to <em>h<\/em>(<em>x<\/em>)<br \/>\nb. perpendicular to <em>h<\/em>(<em>x<\/em>)<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q398341\">Show Solution<\/span><\/p>\n<div id=\"q398341\" class=\"hidden-answer\" style=\"display: none\">\n<p>a.\u00a0[latex]f\\left(x\\right)=2x[\/latex]<br \/>\nb. [latex]g\\left(x\\right)=-\\frac{1}{2}x[\/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=\"ohm169764\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=169764&theme=oea&iframe_resize_id=ohm169764\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div id=\"fs-id1165135509152\" class=\"note precalculus howto textbox\">\n<h3 id=\"fs-id1165135509158\">How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\n<ol id=\"fs-id1165137676542\">\n<li>Determine the slope of the line passing through the points.<\/li>\n<li>Find the negative reciprocal of the slope.<\/li>\n<li>Use the slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div id=\"Example_02_02_11\" class=\"example\">\n<div id=\"fs-id1165134267817\" class=\"exercise\">\n<div id=\"fs-id1165135332504\" class=\"problem textbox shaded\">\n<h3>Example 11: Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point<\/h3>\n<p id=\"fs-id1165135332509\">A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q196054\">Show Solution<\/span><\/p>\n<div id=\"q196054\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the two points of the given line, we can calculate the slope of that line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}_{1}&=\\frac{5 - 6}{4-\\left(-2\\right)} \\\\ &=\\frac{-1}{6} \\\\ &=-\\frac{1}{6} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135532423\">Find the negative reciprocal of the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}{m}_{2}&=\\frac{-1}{-\\frac{1}{6}} \\\\ &=-1\\left(-\\frac{6}{1}\\right) \\\\ &=6 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137768497\">We can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{gathered}g\\left(x\\right)=6x+b \\\\ 5=6\\left(4\\right)+b \\\\ 5=24+b \\\\ -19=b \\\\ b=-19 \\end{gathered}[\/latex]<\/p>\n<p id=\"fs-id1165135159921\">The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is<\/p>\n<p style=\"text-align: center;\">[latex]y=6x - 19[\/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-id1165137437225\">A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q137755\">Show Solution<\/span><\/p>\n<div id=\"q137755\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=-\\frac{1}{3}x+6[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 style=\"text-align: center;\">Solving a System of Linear Equations Using a Graph<\/h2>\n<p id=\"fs-id1165137627910\">A system of linear equations includes two or more linear equations. The graphs of two lines will intersect at a single point if they are not parallel. Two parallel lines can also intersect if they are coincident, which means they are the same line and they intersect at every point. For two lines that are not parallel, the single point of intersection will satisfy both equations and therefore represent the solution to the system.<\/p>\n<p id=\"fs-id1165137812669\">To find this point when the equations are given as functions, we can solve for an input value so that [latex]f\\left(x\\right)=g\\left(x\\right)[\/latex]. In other words, we can set the formulas for the lines equal to one another, and solve for the input that satisfies the equation.<\/p>\n<div id=\"Example_02_02_12\" class=\"example\">\n<div id=\"fs-id1165137896187\" class=\"exercise\">\n<div id=\"fs-id1165137896189\" class=\"problem textbox shaded\">\n<h3>Example 12: Finding a Point of Intersection Algebraically<\/h3>\n<p id=\"fs-id1165135693776\">Find the point of intersection of the lines [latex]h\\left(t\\right)=3t - 4[\/latex] and [latex]j\\left(t\\right)=5-t[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q703225\">Show Solution<\/span><\/p>\n<div id=\"q703225\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165137838172\">Set [latex]h\\left(t\\right)=j\\left(t\\right)[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}3t - 4&=5-t \\\\ 4t&=9 \\\\ t&=\\frac{9}{4} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137644219\">This tells us the lines intersect when the input is [latex]\\frac{9}{4}[\/latex].<\/p>\n<p id=\"fs-id1165137812337\">We can then find the output value of the intersection point by evaluating either function at this input.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}j\\left(\\frac{9}{4}\\right)&=5-\\frac{9}{4} \\\\ &=\\frac{11}{4} \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165137745176\">These lines intersect at the point [latex]\\left(\\frac{9}{4},\\frac{11}{4}\\right)[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Looking at Figure 24, this result seems reasonable.<\/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\/03010651\/CNX_Precalc_Figure_02_02_0242.jpg\" alt=\"Graph of two functions h(t) = 3t - 4 and j(t) = t +5 and their intersection at (9\/4, 11\/4).\" width=\"487\" height=\"441\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 24<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1165137603219\" class=\"note precalculus qa textbox\">\n<h3>Q &amp; A<\/h3>\n<p id=\"fs-id1165137447025\"><strong>If we were asked to find the point of intersection of two distinct parallel lines, should something in the solution process alert us to the fact that there are no solutions?<\/strong><\/p>\n<p id=\"fs-id1165137447030\"><em>Yes. After setting the two equations equal to one another, the result would be the contradiction &#8220;0 = non-zero real number&#8221;.<\/em><\/p>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1165137832295\">Using the graph in the Analysis of the Solution for Example 12,\u00a0identify the following for the function [latex]j\\left(t\\right)[\/latex]:<\/p>\n<p style=\"padding-left: 60px;\">a. y-intercept<\/p>\n<p style=\"padding-left: 60px;\">b. x-intercept(s)<\/p>\n<p style=\"padding-left: 60px;\">c. slope<\/p>\n<p style=\"padding-left: 60px;\">d. Is [latex]j\\left(t\\right)[\/latex] parallel or perpendicular to [latex]h\\left(t\\right)[\/latex] (or neither)?<\/p>\n<p style=\"padding-left: 60px;\">e. Is [latex]j\\left(t\\right)[\/latex] an increasing or decreasing function (or neither)?<\/p>\n<p style=\"padding-left: 60px;\">f. Write a transformation description for [latex]j\\left(t\\right)[\/latex] from the identity toolkit function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q382866\">Show Solution<\/span><\/p>\n<div id=\"q382866\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"padding-left: 60px;\">a.\u00a0[latex]\\left(0,5\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">b. [latex]\\left(5,\\text{ 0}\\right)[\/latex]<\/p>\n<p style=\"padding-left: 60px;\">c. Slope -1<\/p>\n<p style=\"padding-left: 60px;\">d. Neither parallel nor perpendicular<\/p>\n<p style=\"padding-left: 60px;\">e. Decreasing function<\/p>\n<p style=\"padding-left: 60px;\">f. Given the identity function, perform a vertical flip (over the <em>t<\/em>-axis) and shift up 5 units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"Example_02_02_13\" class=\"example\">\n<div id=\"fs-id1165137761773\" class=\"exercise\">\n<div id=\"fs-id1165137761775\" class=\"problem textbox shaded\">\n<h3>Example 13: Finding a Break-Even Point<\/h3>\n<p id=\"fs-id1165137761781\">A company sells sports helmets. The company incurs a one-time fixed cost for $250,000. Each helmet costs $120 to produce, and sells for $140.<\/p>\n<ol id=\"fs-id1165137870987\">\n<li>Find the cost function, <em>C<\/em>, to produce <em>x<\/em>\u00a0helmets, in dollars.<\/li>\n<li>Find the revenue function, <em>R<\/em>, from the sales of <em>x<\/em>\u00a0helmets, in dollars.<\/li>\n<li>Find the break-even point, the point of intersection of the two graphs <em>C\u00a0<\/em>and <em>R<\/em>.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q818513\">Show Solution<\/span><\/p>\n<div id=\"q818513\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1165137653646\">\n<li>he cost function in the sum of the fixed cost, $125,000, and the variable cost, $120 per helmet.\n<div id=\"eip-id1885657\" class=\"equation unnumbered\">[latex]C\\left(x\\right)=120x+250,000[\/latex]<\/div>\n<\/li>\n<li>The revenue function is the total revenue from the sale of [latex]x[\/latex] helmets, [latex]R\\left(x\\right)=140x[\/latex].<\/li>\n<li>The break-even point is the point of intersection of the graph of the cost and revenue functions. To find the <em>x<\/em>-coordinate of the coordinate pair of the point of intersection, set the two equations equal, and solve for <em>x<\/em>.<\/li>\n<\/ol>\n<p style=\"text-align: center;\">[latex]\\begin{align}C\\left(x\\right)&=R\\left(x\\right) \\\\ 250,000+120x&=140x \\\\ 250,000&=20x \\\\ 12,500&=x \\\\ x&=12,500 \\end{align}[\/latex]<\/p>\n<p id=\"eip-id1165134183817\">To find [latex]y[\/latex], evaluate either the revenue or the cost function at 12,500.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}R\\left(20\\right)&=140\\left(12,500\\right) \\\\ &=$1,750,000 \\end{align}[\/latex]<\/p>\n<p id=\"fs-id1165135496435\">The break-even point is (12,500, 1,750,000).<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>This means if the company sells 12,500 helmets, they break even; both the sales and cost incurred equaled 1.75 million dollars.<\/p>\n<div style=\"width: 741px\" 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\/03005146\/CNX_Precalc_Figure_02_02_025.jpg\" alt=\"Graph of the two functions, C(x) and R(x) where it shows that below (12500, 1750000) the company loses money and above that point the company makes a profit.\" width=\"731\" height=\"695\" \/><\/p>\n<p class=\"wp-caption-text\"><b>Figure 25<\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><b><\/b><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/section>\n<div class=\"note precalculus media\">\n<h2>Key Concepts<\/h2>\n<ul id=\"fs-id1165134190780\">\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<li>Parallel lines have the same slope.<\/li>\n<li>Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.<\/li>\n<li>A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the <em>x<\/em>&#8211; and <em>y<\/em>-values of the given point into the equation, [latex]f(x)=mx+b[\/latex], and using the <em>b<\/em>\u00a0that results. Similarly, the point-slope form of an equation can also be used.<\/li>\n<li>A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.<\/li>\n<li>A system of linear equations may be solved setting the two equations equal to one another and solving for <em>x<\/em>. The <em>y<\/em>-value may be found by evaluating either one of the original equations using this <em>x<\/em>-value.<\/li>\n<li>A system of linear equations may also be solved by finding the point of intersection on a graph.<\/li>\n<\/ul>\n<div>\n<h2>Glossary<\/h2>\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-id1165135330621\" class=\"definition\">\n<dt><strong>parallel lines<\/strong><\/dt>\n<dd id=\"fs-id1165135186582\">two or more lines with the same slope<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135186586\" class=\"definition\">\n<dt><strong>perpendicular lines<\/strong><\/dt>\n<dd id=\"fs-id1165135186592\">two lines that intersect at right angles and have slopes that are negative reciprocals of each other<\/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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\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-13807\">\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>","protected":false},"author":23588,"menu_order":2,"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-13807","chapter","type-chapter","status-publish","hentry"],"part":10717,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13807","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/users\/23588"}],"version-history":[{"count":8,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13807\/revisions"}],"predecessor-version":[{"id":15832,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13807\/revisions\/15832"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/parts\/10717"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapters\/13807\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/media?parent=13807"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/pressbooks\/v2\/chapter-type?post=13807"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/contributor?post=13807"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/precalculus\/wp-json\/wp\/v2\/license?post=13807"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}