{"id":5280,"date":"2021-09-30T18:01:04","date_gmt":"2021-09-30T18:01:04","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-graphs-of-linear-functions\/"},"modified":"2022-11-08T18:27:29","modified_gmt":"2022-11-08T18:27:29","slug":"introduction-graphs-of-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/chapter\/introduction-graphs-of-linear-functions\/","title":{"raw":"Graphs of Linear Functions","rendered":"Graphs of Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph linear functions by plotting points, using the slope and y-intercept, and using transformations.(CLO #1, #3, #6)<\/li>\r\n \t<li>Write the equation of a linear function given its graph.\u00a0(CLO #1, #3, #6)<\/li>\r\n \t<li>Match linear functions with their graphs.\u00a0(CLO #1, #3, #6)<\/li>\r\n \t<li>Find the x-intercept of a function given its equation.\u00a0(CLO #1, #3, #6)<\/li>\r\n \t<li>Find the equations of vertical and horizontal lines.\u00a0(CLO #1, #3, #6)<\/li>\r\n<\/ul>\r\n<\/div>\r\nWe\u00a0can now describe a variety of characteristics that explain the behavior of\u00a0linear functions. We will use this information to\u00a0analyze a graphed line and write an equation based on its observable properties. From evaluating the graph, what can you determine about this linear function?\r\n\r\n<img class=\"wp-image-4213 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/14214712\/Screen-Shot-2017-04-14-at-2.46.28-PM-300x298.png\" alt=\"Graph of the function f(x)= frac {2}{3} x plus 1\" width=\"287\" height=\"285\" \/>\r\n<ul>\r\n \t<li>initial value (y-intercept)?<\/li>\r\n \t<li>one or two points?<\/li>\r\n \t<li>slope?<\/li>\r\n \t<li>increasing or decreasing?<\/li>\r\n \t<li>vertical or horizontal?<\/li>\r\n<\/ul>\r\nIn this section, you will practice writing linear function equations using the\u00a0information you've gathered. We will\u00a0also practice graphing linear functions using different methods and predict how the graphs of linear functions will change when parts of the equation are altered.\r\n<h2><\/h2>\r\n<h2>[latex]\\\\[\/latex]<\/h2>\r\n<h2>Graphing Linear Functions<\/h2>\r\nWe previously 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.\r\n\r\nThere 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. The third is applying transformations to the identity function [latex]f\\left(x\\right)=x[\/latex].\r\n<h3>Graphing a Function by Plotting Points<\/h3>\r\nTo 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 of the function. For example, given the function [latex]f\\left(x\\right)=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.\r\n<div class=\"textbox\">\r\n<h3>How To: Given a linear function, graph by plotting points.<\/h3>\r\n<ol>\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 class=\"textbox exercises\">\r\n<h3>Example: Graphing by Plotting Points<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] by plotting points.\r\n\r\n[reveal-answer q=\"589508\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"589508\"]\r\n\r\nBegin 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.\r\n\r\nEvaluate the function at each input value and use the output value to identify coordinate pairs.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{llllll}x=0&amp; &amp; f\\left(0\\right)=-\\frac{2}{3}\\left(0\\right)+5=5\\Rightarrow \\left(0,5\\right)\\\\ x=3&amp; &amp; f\\left(3\\right)=-\\frac{2}{3}\\left(3\\right)+5=3\\Rightarrow \\left(3,3\\right)\\\\ x=6&amp; &amp; f\\left(6\\right)=-\\frac{2}{3}\\left(6\\right)+5=1\\Rightarrow \\left(6,1\\right)\\end{array}[\/latex]<\/p>\r\nPlot the coordinate pairs and draw a line through the points. The graph below is of\u00a0the function [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex].\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184320\/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\" \/>\r\n<h4>Analysis of the Solution<\/h4>\r\nThe 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.\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\nGraph [latex]f\\left(x\\right)=-\\frac{3}{4}x+6[\/latex] by plotting points.\r\n\r\n[reveal-answer q=\"156351\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"156351\"]\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15184550\/CNX_Precalc_Figure_02_02_0022.jpg\"><img class=\"aligncenter size-full wp-image-2803\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15184550\/CNX_Precalc_Figure_02_02_0022.jpg\" alt=\"cnx_precalc_figure_02_02_0022\" width=\"487\" height=\"316\" \/><\/a>\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69981&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"500\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Graphing a Linear Function Using y-intercept and Slope<\/h3>\r\nAnother 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 [latex]x=0[\/latex] in the equation.\r\n\r\nThe other characteristic of the linear function is its slope,\u00a0<em>m<\/em>,\u00a0which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a linear 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, between any two points by the horizontal difference, or run. The slope of a linear function will be the same between any two points. We encountered both the <em>y-<\/em>intercept and the slope in Linear Functions.\r\n\r\nLet\u2019s consider the following function.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=\\frac{1}{2}x+1[\/latex]<\/p>\r\nThe 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. 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 multiple points, and then we draw a line through the points as shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184323\/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\" \/>\r\n<div class=\"textbox\">\r\n<h3>A General Note: Graphical Interpretation of a Linear Function<\/h3>\r\nIn the equation [latex]f\\left(x\\right)=mx+b[\/latex]\r\n<ul>\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<p style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Do all linear functions have <em>y<\/em>-intercepts?<\/strong>\r\n\r\n<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. Keep in mind that a vertical line is the only line that is not a function.)<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope.<\/h3>\r\n<ol>\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.<\/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>Draw a line which passes through the points.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Graphing by Using the <em>y-<\/em>intercept and Slope<\/h3>\r\nGraph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] using the <em>y-<\/em>intercept and slope.\r\n\r\n[reveal-answer q=\"507667\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"507667\"]\r\n\r\nEvaluate 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).\r\n\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 [latex]\u20132[\/latex] 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. 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<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184325\/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\" \/>\r\n<h4>Analysis of the Solution<\/h4>\r\nThe graph slants downward from left to right which means it has a negative slope as expected.\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\nFind a point on the graph we drew in the previous example: Graphing by Using the <em>y<\/em>-intercept and Slope\u00a0that has a negative <em>x<\/em>-value.\r\n\r\n[reveal-answer q=\"572211\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"572211\"]\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<iframe id=\"mom100\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=88183&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Graphing a Linear Function Using Transformations<\/h3>\r\nAnother option for graphing is to use <strong>transformations<\/strong> on the identity function [latex]f\\left(x\\right)=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.\r\n<h3>Vertical Stretch or Compression<\/h3>\r\nIn the equation [latex]f\\left(x\\right)=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 that multiplying the equation [latex]f\\left(x\\right)=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.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184329\/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\" \/> Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].[\/caption]\r\n<h3>Vertical Shift<\/h3>\r\nIn [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 that adding a value of <em>b<\/em>\u00a0to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of\u00a0<em>f<\/em>\u00a0a total of <em>b<\/em>\u00a0units up if <em>b<\/em>\u00a0is positive and\u00a0|<em>b<\/em>| units down if <em>b<\/em>\u00a0is negative.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184332\/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\" \/> This graph illustrates vertical shifts of the function [latex]f\\left(x\\right)=x[\/latex].[\/caption]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.\r\n\r\nHere is a video showing how to graph a linear function using transformations on the graph of the identity function f(x) = x.\u00a0\u00a0<a href=\"https:\/\/youtu.be\/h9zn_ODlgbM\" target=\"_blank\" rel=\"noopener\">Graphing a Linear Function Using Transformations on the graph of f(x) = x<\/a>\r\n<div class=\"textbox\">\r\n<h3>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>\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 class=\"textbox exercises\">\r\n<h3>Example: Graphing by Using Transformations<\/h3>\r\nGraph [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex] using transformations.\r\n\r\n[reveal-answer q=\"750947\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"750947\"]\r\n\r\nThe 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 [latex]b=-3[\/latex], so the identity function is vertically shifted down 3 units.\r\n\r\nFirst, graph the identity function, and show the vertical compression.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184335\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\" \/> The function [latex]y=x[\/latex] compressed by a factor of [latex]\\frac{1}{2}[\/latex].[\/caption]Then, show the vertical shift.[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184338\/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\" \/> The function [latex]y=\\frac{1}{2}x[\/latex] shifted down 3 units.[\/caption][\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nGraph [latex]f\\left(x\\right)=4+2x[\/latex], using transformations.\r\n\r\n[reveal-answer q=\"350962\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"350962\"]\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15185737\/CNX_Precalc_Figure_02_02_0092.jpg\"><img class=\"aligncenter size-full wp-image-2804\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15185737\/CNX_Precalc_Figure_02_02_0092.jpg\" alt=\"cnx_precalc_figure_02_02_0092\" width=\"510\" height=\"520\" \/><\/a>[\/hidden-answer]\r\n<iframe id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114584&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"500\"><\/iframe>\r\n<iframe id=\"mom700\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114587&amp;theme=oea&amp;iframe_resize_id=mom700\" width=\"100%\" height=\"500\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations?<\/strong>\r\n\r\n<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 order of operations, let the input be 2.<\/em>\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\text{(2)}=\\frac{\\text{1}}{\\text{2}}\\text{(2)}-\\text{3}\\hfill \\\\ =\\text{1}-\\text{3}\\hfill \\\\ =-\\text{2}\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<h2>Writing Equations of Linear Functions<\/h2>\r\nWe previously wrote\u00a0the 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 the graph below. 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.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184341\/CNX_Precalc_Figure_02_02_0102.jpg\" width=\"369\" height=\"378\" \/>\r\n\r\nThen 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:\r\n<p style=\"text-align: center;\">[latex]m=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{2}=2[\/latex]<\/p>\r\nSubstituting the slope and <em>y-<\/em>intercept into slope-intercept form of a line gives:\r\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\r\nHere is a video showing how to find the equation of a line using the graph of the line.\u00a0\u00a0<a href=\"https:\/\/youtu.be\/mmWf_oLTNSQ\" target=\"_blank\" rel=\"noopener\">Find the equation of the line given the graph<\/a>\r\n<div class=\"textbox\">\r\n<h3>How To: Given THE graph of A linear function, find the equation to describe the function.<\/h3>\r\n<ol>\r\n \t<li>Identify the <em>y-<\/em>intercept from the graph.<\/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 slope-intercept form of a line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Matching Linear Functions to Their Graphs<\/h3>\r\nMatch each equation of a linear function with one of the lines in the graph below.\r\n<ol>\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<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184343\/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\" \/>\r\n[reveal-answer q=\"659573\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"659573\"]\r\n\r\nAnalyze the information for each function.\r\n<ol>\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.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184346\/CNX_Precalc_Figure_02_02_0122.jpg\" width=\"489\" height=\"374\" \/>\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<iframe id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1440&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"350\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Finding the <em>x<\/em>-intercept of a Line<\/h3>\r\nSo far we have been finding the <em>y-<\/em>intercepts of functions: the point at which the graph of a 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 a function crosses the <em>x<\/em>-axis. In other words, it is the input value when the output value is zero.\r\n\r\nTo find the <em>x<\/em>-intercept, set the 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:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x - 6[\/latex]<\/p>\r\nSet the function equal to 0 and solve for <em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=3x - 6\\hfill \\\\ 6=3x\\hfill \\\\ 2=x\\hfill \\\\ x=2\\hfill \\end{array}[\/latex]<\/p>\r\nThe graph of the function crosses the <em>x<\/em>-axis at the point (2, 0).\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>Do all linear functions have <em>x<\/em>-intercepts?<\/strong>\r\n\r\n<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>.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184348\/CNX_Precalc_Figure_02_02_0262.jpg\" alt=\"Graph of y = 5.\" width=\"421\" height=\"231\" \/>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>A General Note: <em>x<\/em>-intercept<\/h3>\r\nThe <strong><em>x<\/em>-intercept<\/strong> of a function is the value of <em>x<\/em>\u00a0where\u00a0<em>f<\/em>(<em>x<\/em>) = 0. It can be found by solving the equation 0 = <em>mx\u00a0<\/em>+ <em>b<\/em>.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding an <em>x<\/em>-intercept<\/h3>\r\nFind the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].\r\n\r\n[reveal-answer q=\"400055\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"400055\"]\r\n\r\nSet the function equal to zero to solve for <em>x<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=\\frac{1}{2}x - 3\\\\ 3=\\frac{1}{2}x\\\\ 6=x\\\\ x=6\\end{array}[\/latex]<\/p>\r\nThe graph crosses the <em>x<\/em>-axis at the point (6, 0).\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the function is shown below. We can see that the <em>x<\/em>-intercept is (6, 0) as expected.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"369\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184351\/CNX_Precalc_Figure_02_02_0132.jpg\" width=\"369\" height=\"378\" \/> The graph of the linear function [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].[\/caption]<strong>\u00a0<\/strong>[\/hidden-answer]<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nFind the <em>x<\/em>-intercept of [latex]f\\left(x\\right)=\\frac{1}{4}x - 4[\/latex].\r\n\r\n[reveal-answer q=\"406982\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"406982\"]\r\n\r\n[latex]\\left(16,\\text{ 0}\\right)[\/latex][\/hidden-answer]\r\n<iframe id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79757&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"200\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>Describing Horizontal and Vertical Lines<\/h3>\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 the graph below, we see that the output has a value of 2 for every input value. The change in outputs between any two points 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\\left(x\\right)=mx+b[\/latex], the equation simplifies to [latex]f\\left(x\\right)=b[\/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f\\left(x\\right)=2[\/latex].\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184353\/CNX_Precalc_Figure_02_02_0142.jpg\" width=\"487\" height=\"473\" \/> A horizontal line representing the function [latex]f\\left(x\\right)=2[\/latex].[\/caption]&nbsp;\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184355\/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\" \/>\r\n\r\nA <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.\r\n\r\nNotice that a vertical line 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[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184358\/CNX_Precalc_Figure_02_02_0162.jpg\" width=\"487\" height=\"473\" \/> The vertical line [latex]x=2[\/latex] which does not represent a function.[\/caption]\r\n<div class=\"textbox\">\r\n<h3>A General Note: Horizontal and Vertical Lines<\/h3>\r\nLines can be horizontal or vertical.\r\n\r\nA <strong>horizontal line<\/strong> is a line defined by an equation of the form [latex]f\\left(x\\right)=b[\/latex] where\u00a0[latex]b[\/latex] is a constant.\r\n\r\nA <strong>vertical line<\/strong> is a line defined by an equation of the form [latex]x=a[\/latex]\u00a0where\u00a0[latex]a[\/latex] is a constant.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Equation of a Horizontal Line<\/h3>\r\nWrite the equation of the line graphed below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184401\/CNX_Precalc_Figure_02_02_0172.jpg\" alt=\"Graph of x = 7.\" width=\"369\" height=\"378\" \/>\r\n[reveal-answer q=\"891444\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"891444\"]\r\n\r\nFor any <em>x<\/em>-value, the <em>y<\/em>-value is [latex]\u20134[\/latex], so the equation is [latex]y=\u20134[\/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<iframe id=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15599&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"200\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing the Equation of a Vertical Line<\/h3>\r\nWrite the equation of the line graphed below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184404\/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\" \/>\r\n\r\n[reveal-answer q=\"178822\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178822\"]\r\n\r\nThe constant <em>x<\/em>-value is 7, so the equation is [latex]x=7[\/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<iframe id=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114592&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"500\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<ul>\r\n \t<li>Write the equation of the function passing through the points [latex](2,6)[\/latex] and [latex](4,4)[\/latex] in slope-intercept form.<\/li>\r\n \t<li>Write the equation of a function whose slope is 2 and passes through the point [latex](-1,0)[\/latex]<\/li>\r\n \t<li>Write the equation of a function whose slope is undefined.<\/li>\r\n<\/ul>\r\nSolutions:\u00a0 y = -x + 8,\u00a0 y = 2x + 2,\u00a0 \u00a0x = 4 (this could be any number - vertical lines have undefined slopes)\r\n\r\n<\/div>\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 shifting the graph up, down, left, or right as well as using 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, [latex]f(x)=b[\/latex].<\/li>\r\n \t<li>Vertical lines are written in the form, [latex]x=b[\/latex].<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt><strong>horizontal line<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">a line defined by [latex]f\\left(x\\right)=b[\/latex] where <em>b<\/em> is a real number. The slope of a horizontal line is 0.<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt><strong>vertical line<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">a line defined by [latex]x=a[\/latex] 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-id1165135429388\" class=\"definition\">\r\n \t<dt><strong><em>x<\/em>-intercept<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph linear functions by plotting points, using the slope and y-intercept, and using transformations.(CLO #1, #3, #6)<\/li>\n<li>Write the equation of a linear function given its graph.\u00a0(CLO #1, #3, #6)<\/li>\n<li>Match linear functions with their graphs.\u00a0(CLO #1, #3, #6)<\/li>\n<li>Find the x-intercept of a function given its equation.\u00a0(CLO #1, #3, #6)<\/li>\n<li>Find the equations of vertical and horizontal lines.\u00a0(CLO #1, #3, #6)<\/li>\n<\/ul>\n<\/div>\n<p>We\u00a0can now describe a variety of characteristics that explain the behavior of\u00a0linear functions. We will use this information to\u00a0analyze a graphed line and write an equation based on its observable properties. From evaluating the graph, what can you determine about this linear function?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-4213 alignleft\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/14214712\/Screen-Shot-2017-04-14-at-2.46.28-PM-300x298.png\" alt=\"Graph of the function f(x)= frac {2}{3} x plus 1\" width=\"287\" height=\"285\" \/><\/p>\n<ul>\n<li>initial value (y-intercept)?<\/li>\n<li>one or two points?<\/li>\n<li>slope?<\/li>\n<li>increasing or decreasing?<\/li>\n<li>vertical or horizontal?<\/li>\n<\/ul>\n<p>In this section, you will practice writing linear function equations using the\u00a0information you&#8217;ve gathered. We will\u00a0also practice graphing linear functions using different methods and predict how the graphs of linear functions will change when parts of the equation are altered.<\/p>\n<h2><\/h2>\n<h2>[latex]\\\\[\/latex]<\/h2>\n<h2>Graphing Linear Functions<\/h2>\n<p>We previously 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.<\/p>\n<p>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. The third is applying transformations to the identity function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<h3>Graphing a Function by Plotting Points<\/h3>\n<p>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 of the function. For example, given the function [latex]f\\left(x\\right)=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<div class=\"textbox\">\n<h3>How To: Given a linear function, graph by plotting points.<\/h3>\n<ol>\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 class=\"textbox exercises\">\n<h3>Example: Graphing by Plotting Points<\/h3>\n<p>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=\"q589508\">Show Solution<\/span><\/p>\n<div id=\"q589508\" class=\"hidden-answer\" style=\"display: none\">\n<p>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>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{array}{llllll}x=0& & f\\left(0\\right)=-\\frac{2}{3}\\left(0\\right)+5=5\\Rightarrow \\left(0,5\\right)\\\\ x=3& & f\\left(3\\right)=-\\frac{2}{3}\\left(3\\right)+5=3\\Rightarrow \\left(3,3\\right)\\\\ x=6& & f\\left(6\\right)=-\\frac{2}{3}\\left(6\\right)+5=1\\Rightarrow \\left(6,1\\right)\\end{array}[\/latex]<\/p>\n<p>Plot the coordinate pairs and draw a line through the points. The graph below is of\u00a0the function [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184320\/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<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q156351\">Show Solution<\/span><\/p>\n<div id=\"q156351\" class=\"hidden-answer\" style=\"display: none\">\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15184550\/CNX_Precalc_Figure_02_02_0022.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2803\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15184550\/CNX_Precalc_Figure_02_02_0022.jpg\" alt=\"cnx_precalc_figure_02_02_0022\" width=\"487\" height=\"316\" \/><\/a><\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69981&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"500\"><\/iframe><\/p>\n<\/div>\n<h3>Graphing a Linear Function Using y-intercept and Slope<\/h3>\n<p>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 [latex]x=0[\/latex] in the equation.<\/p>\n<p>The other characteristic of the linear function is its slope,\u00a0<em>m<\/em>,\u00a0which is a measure of its steepness. Recall that the slope is the rate of change of the function. The slope of a linear 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, between any two points by the horizontal difference, or run. The slope of a linear function will be the same between any two points. We encountered both the <em>y-<\/em>intercept and the slope in Linear Functions.<\/p>\n<p>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<p>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. 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 multiple points, and then we draw a line through the points as shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184323\/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<div class=\"textbox\">\n<h3>A General Note: Graphical Interpretation of a Linear Function<\/h3>\n<p>In the equation [latex]f\\left(x\\right)=mx+b[\/latex]<\/p>\n<ul>\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<p style=\"text-align: center;\">[latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Do all linear functions have <em>y<\/em>-intercepts?<\/strong><\/p>\n<p><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. Keep in mind that a vertical line is the only line that is not a function.)<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the equation for a linear function, graph the function using the <em>y<\/em>-intercept and slope.<\/h3>\n<ol>\n<li>Evaluate the function at an input value of zero to find the <em>y-<\/em>intercept.<\/li>\n<li>Identify the slope.<\/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>Draw a line which passes through the points.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Graphing by Using the <em>y-<\/em>intercept and Slope<\/h3>\n<p>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=\"q507667\">Show Solution<\/span><\/p>\n<div id=\"q507667\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 [latex]\u20132[\/latex] 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. 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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184325\/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<h4>Analysis of the Solution<\/h4>\n<p>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 class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Find a point on the graph we drew in the previous example: Graphing by Using the <em>y<\/em>-intercept and Slope\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=\"q572211\">Show Solution<\/span><\/p>\n<div id=\"q572211\" 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<p><iframe loading=\"lazy\" id=\"mom100\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=88183&amp;theme=oea&amp;iframe_resize_id=mom100\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<h3>Graphing a Linear Function Using Transformations<\/h3>\n<p>Another option for graphing is to use <strong>transformations<\/strong> on the identity function [latex]f\\left(x\\right)=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<h3>Vertical Stretch or Compression<\/h3>\n<p>In the equation [latex]f\\left(x\\right)=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 that multiplying the equation [latex]f\\left(x\\right)=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<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184329\/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<p class=\"wp-caption-text\">Vertical stretches and compressions and reflections on the function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<\/div>\n<h3>Vertical Shift<\/h3>\n<p>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 that adding a value of <em>b<\/em>\u00a0to the equation of [latex]f\\left(x\\right)=x[\/latex] shifts the graph of\u00a0<em>f<\/em>\u00a0a total of <em>b<\/em>\u00a0units up if <em>b<\/em>\u00a0is positive and\u00a0|<em>b<\/em>| units down if <em>b<\/em>\u00a0is negative.<\/p>\n<div style=\"width: 910px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184332\/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\" \/><\/p>\n<p class=\"wp-caption-text\">This graph illustrates vertical shifts of the function [latex]f\\left(x\\right)=x[\/latex].<\/p>\n<\/div>\n<p>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<p>Here is a video showing how to graph a linear function using transformations on the graph of the identity function f(x) = x.\u00a0\u00a0<a href=\"https:\/\/youtu.be\/h9zn_ODlgbM\" target=\"_blank\" rel=\"noopener\">Graphing a Linear Function Using Transformations on the graph of f(x) = x<\/a><\/p>\n<div class=\"textbox\">\n<h3>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>\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 class=\"textbox exercises\">\n<h3>Example: Graphing by Using Transformations<\/h3>\n<p>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=\"q750947\">Show Solution<\/span><\/p>\n<div id=\"q750947\" class=\"hidden-answer\" style=\"display: none\">\n<p>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 [latex]b=-3[\/latex], so the identity function is vertically shifted down 3 units.<\/p>\n<p>First, graph the identity function, and show the vertical compression.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184335\/CNX_Precalc_Figure_02_02_0072.jpg\" alt=\"graph showing the lines y = x and y = (1\/2)x\" width=\"487\" height=\"378\" \/><\/p>\n<p class=\"wp-caption-text\">The function [latex]y=x[\/latex] compressed by a factor of [latex]\\frac{1}{2}[\/latex].<\/p>\n<\/div>\n<p>Then, show the vertical shift.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184338\/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\" \/><\/p>\n<p class=\"wp-caption-text\">The function [latex]y=\\frac{1}{2}x[\/latex] shifted down 3 units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q350962\">Show Solution<\/span><\/p>\n<div id=\"q350962\" class=\"hidden-answer\" style=\"display: none\">\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15185737\/CNX_Precalc_Figure_02_02_0092.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-2804\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/15185737\/CNX_Precalc_Figure_02_02_0092.jpg\" alt=\"cnx_precalc_figure_02_02_0092\" width=\"510\" height=\"520\" \/><\/a><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114584&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"500\"><\/iframe><br \/>\n<iframe loading=\"lazy\" id=\"mom700\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114587&amp;theme=oea&amp;iframe_resize_id=mom700\" width=\"100%\" height=\"500\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>In Example: Graphing by Using Transformations, could we have sketched the graph by reversing the order of the transformations?<\/strong><\/p>\n<p><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 order of operations, let the input be 2.<\/em><\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\text{(2)}=\\frac{\\text{1}}{\\text{2}}\\text{(2)}-\\text{3}\\hfill \\\\ =\\text{1}-\\text{3}\\hfill \\\\ =-\\text{2}\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<h2>Writing Equations of Linear Functions<\/h2>\n<p>We previously wrote\u00a0the 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 the graph below. 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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184341\/CNX_Precalc_Figure_02_02_0102.jpg\" width=\"369\" height=\"378\" alt=\"image\" \/><\/p>\n<p>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<p style=\"text-align: center;\">[latex]m=\\frac{\\text{rise}}{\\text{run}}=\\frac{4}{2}=2[\/latex]<\/p>\n<p>Substituting the slope and <em>y-<\/em>intercept into slope-intercept form of a line gives:<\/p>\n<p style=\"text-align: center;\">[latex]y=2x+4[\/latex]<\/p>\n<p>Here is a video showing how to find the equation of a line using the graph of the line.\u00a0\u00a0<a href=\"https:\/\/youtu.be\/mmWf_oLTNSQ\" target=\"_blank\" rel=\"noopener\">Find the equation of the line given the graph<\/a><\/p>\n<div class=\"textbox\">\n<h3>How To: Given THE graph of A linear function, find the equation to describe the function.<\/h3>\n<ol>\n<li>Identify the <em>y-<\/em>intercept from the graph.<\/li>\n<li>Choose two points to determine the slope.<\/li>\n<li>Substitute the <em>y-<\/em>intercept and slope into slope-intercept form of a line.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Matching Linear Functions to Their Graphs<\/h3>\n<p>Match each equation of a linear function with one of the lines in the graph below.<\/p>\n<ol>\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><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184343\/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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q659573\">Show Solution<\/span><\/p>\n<div id=\"q659573\" class=\"hidden-answer\" style=\"display: none\">\n<p>Analyze the information for each function.<\/p>\n<ol>\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.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184346\/CNX_Precalc_Figure_02_02_0122.jpg\" width=\"489\" height=\"374\" alt=\"image\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=1440&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"350\"><\/iframe><\/p>\n<\/div>\n<h3>Finding the <em>x<\/em>-intercept of a Line<\/h3>\n<p>So far we have been finding the <em>y-<\/em>intercepts of functions: the point at which the graph of a 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 a function crosses the <em>x<\/em>-axis. In other words, it is the input value when the output value is zero.<\/p>\n<p>To find the <em>x<\/em>-intercept, set the 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<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x - 6[\/latex]<\/p>\n<p>Set the function equal to 0 and solve for <em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=3x - 6\\hfill \\\\ 6=3x\\hfill \\\\ 2=x\\hfill \\\\ x=2\\hfill \\end{array}[\/latex]<\/p>\n<p>The graph of the function crosses the <em>x<\/em>-axis at the point (2, 0).<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>Do all linear functions have <em>x<\/em>-intercepts?<\/strong><\/p>\n<p><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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184348\/CNX_Precalc_Figure_02_02_0262.jpg\" alt=\"Graph of y = 5.\" width=\"421\" height=\"231\" \/><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: <em>x<\/em>-intercept<\/h3>\n<p>The <strong><em>x<\/em>-intercept<\/strong> of a function is the value of <em>x<\/em>\u00a0where\u00a0<em>f<\/em>(<em>x<\/em>) = 0. It can be found by solving the equation 0 = <em>mx\u00a0<\/em>+ <em>b<\/em>.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding an <em>x<\/em>-intercept<\/h3>\n<p>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=\"q400055\">Show Solution<\/span><\/p>\n<div id=\"q400055\" class=\"hidden-answer\" style=\"display: none\">\n<p>Set the function equal to zero to solve for <em>x<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=\\frac{1}{2}x - 3\\\\ 3=\\frac{1}{2}x\\\\ 6=x\\\\ x=6\\end{array}[\/latex]<\/p>\n<p>The graph crosses the <em>x<\/em>-axis at the point (6, 0).<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the function is shown below. We can see that the <em>x<\/em>-intercept is (6, 0) as expected.<\/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\/wp-content\/uploads\/sites\/896\/2016\/10\/21184351\/CNX_Precalc_Figure_02_02_0132.jpg\" width=\"369\" height=\"378\" alt=\"image\" \/><\/p>\n<p class=\"wp-caption-text\">The graph of the linear function [latex]f\\left(x\\right)=\\frac{1}{2}x - 3[\/latex].<\/p>\n<\/div>\n<p><strong>\u00a0<\/strong><\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>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=\"q406982\">Show Solution<\/span><\/p>\n<div id=\"q406982\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\left(16,\\text{ 0}\\right)[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=79757&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"200\"><\/iframe><\/p>\n<\/div>\n<h3>Describing Horizontal and Vertical Lines<\/h3>\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 the graph below, we see that the output has a value of 2 for every input value. The change in outputs between any two points 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\\left(x\\right)=mx+b[\/latex], the equation simplifies to [latex]f\\left(x\\right)=b[\/latex]. In other words, the value of the function is a constant. This graph represents the function [latex]f\\left(x\\right)=2[\/latex].<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184353\/CNX_Precalc_Figure_02_02_0142.jpg\" width=\"487\" height=\"473\" alt=\"image\" \/><\/p>\n<p class=\"wp-caption-text\">A horizontal line representing the function [latex]f\\left(x\\right)=2[\/latex].<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184355\/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>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.<\/p>\n<p>Notice that a vertical line 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<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184358\/CNX_Precalc_Figure_02_02_0162.jpg\" width=\"487\" height=\"473\" alt=\"image\" \/><\/p>\n<p class=\"wp-caption-text\">The vertical line [latex]x=2[\/latex] which does not represent a function.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>A General Note: Horizontal and Vertical Lines<\/h3>\n<p>Lines can be horizontal or vertical.<\/p>\n<p>A <strong>horizontal line<\/strong> is a line defined by an equation of the form [latex]f\\left(x\\right)=b[\/latex] where\u00a0[latex]b[\/latex] is a constant.<\/p>\n<p>A <strong>vertical line<\/strong> is a line defined by an equation of the form [latex]x=a[\/latex]\u00a0where\u00a0[latex]a[\/latex] is a constant.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Equation of a Horizontal Line<\/h3>\n<p>Write the equation of the line graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184401\/CNX_Precalc_Figure_02_02_0172.jpg\" alt=\"Graph of x = 7.\" width=\"369\" height=\"378\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q891444\">Show Solution<\/span><\/p>\n<div id=\"q891444\" class=\"hidden-answer\" style=\"display: none\">\n<p>For any <em>x<\/em>-value, the <em>y<\/em>-value is [latex]\u20134[\/latex], so the equation is [latex]y=\u20134[\/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=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=15599&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"200\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing the Equation of a Vertical Line<\/h3>\n<p>Write the equation of the line graphed below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/21184404\/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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q178822\">Show Solution<\/span><\/p>\n<div id=\"q178822\" class=\"hidden-answer\" style=\"display: none\">\n<p>The constant <em>x<\/em>-value is 7, so the equation is [latex]x=7[\/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=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114592&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"500\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<ul>\n<li>Write the equation of the function passing through the points [latex](2,6)[\/latex] and [latex](4,4)[\/latex] in slope-intercept form.<\/li>\n<li>Write the equation of a function whose slope is 2 and passes through the point [latex](-1,0)[\/latex]<\/li>\n<li>Write the equation of a function whose slope is undefined.<\/li>\n<\/ul>\n<p>Solutions:\u00a0 y = -x + 8,\u00a0 y = 2x + 2,\u00a0 \u00a0x = 4 (this could be any number &#8211; vertical lines have undefined slopes)<\/p>\n<\/div>\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 shifting the graph up, down, left, or right as well as using 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, [latex]f(x)=b[\/latex].<\/li>\n<li>Vertical lines are written in the form, [latex]x=b[\/latex].<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong>horizontal line<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">a line defined by [latex]f\\left(x\\right)=b[\/latex] where <em>b<\/em> is a real number. The slope of a horizontal line is 0.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong>vertical line<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">a line defined by [latex]x=a[\/latex] where <em>a<\/em>\u00a0is a real number. The slope of a vertical line is undefined.<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong><em>x<\/em>-intercept<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">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\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-5280\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 114584, 114587, 79757, 114592. <strong>Authored by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra. <strong>Authored by<\/strong>: Abramson, Jay et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\">http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2<\/li><li>Question ID 69981. <strong>Authored by<\/strong>: Majerus, Ryan. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 88183. <strong>Authored by<\/strong>: Shahbazian, Roy. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Precalculus. <strong>Authored by<\/strong>: Jay Abramson, et al.. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\">http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.<\/li><li>Question ID 1440. <strong>Authored by<\/strong>: unknown, mb Lippman,David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 15599. <strong>Authored by<\/strong>: Johns, Bryan . <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 469. <strong>Authored by<\/strong>: WebWork-Rochester, mb Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 60791. <strong>Authored by<\/strong>: Day, Alyson. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>Question ID 40657. <strong>Authored by<\/strong>: Michael Jenck. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":167848,"menu_order":11,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"College Algebra\",\"author\":\"Abramson, Jay et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download for free at http:\/\/cnx.org\/contents\/9b08c294-057f-4201-9f48-5d6ad992740d@5.2\"},{\"type\":\"cc\",\"description\":\"Question ID 69981\",\"author\":\"Majerus, Ryan\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 88183\",\"author\":\"Shahbazian, Roy\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"original\",\"description\":\"Question ID 114584, 114587, 79757, 114592\",\"author\":\"Lumen Learning\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"Jay Abramson, et al.\",\"organization\":\"OpenStax\",\"url\":\"http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"Download For Free at : http:\/\/cnx.org\/contents\/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175.\"},{\"type\":\"cc\",\"description\":\"Question ID 1440\",\"author\":\"unknown, mb Lippman,David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 15599\",\"author\":\"Johns, Bryan \",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 469\",\"author\":\"WebWork-Rochester, mb Lippman, David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 60791\",\"author\":\"Day, Alyson\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"},{\"type\":\"cc\",\"description\":\"Question ID 40657\",\"author\":\"Michael Jenck\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-5280","chapter","type-chapter","status-publish","hentry"],"part":5266,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5280","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/users\/167848"}],"version-history":[{"count":9,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5280\/revisions"}],"predecessor-version":[{"id":5788,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5280\/revisions\/5788"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/parts\/5266"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapters\/5280\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/media?parent=5280"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/pressbooks\/v2\/chapter-type?post=5280"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/contributor?post=5280"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/lcudd-tulsacc-collegealgebra\/wp-json\/wp\/v2\/license?post=5280"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}