{"id":171,"date":"2023-06-21T13:22:40","date_gmt":"2023-06-21T13:22:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/graph-linear-functions\/"},"modified":"2023-09-21T07:52:10","modified_gmt":"2023-09-21T07:52:10","slug":"graph-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/gsu-collegealgebra\/chapter\/graph-linear-functions\/","title":{"raw":"\u25aa   Graphing Linear Functions","rendered":"\u25aa   Graphing Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Graph a linear function by plotting points<\/li>\r\n \t<li>Graph a linear function using the slope and y-intercept<\/li>\r\n \t<li>Graph a linear function using transformations<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>recall graphing functions<\/h3>\r\nRecall that you can graph a linear function by evaluating it at chosen inputs to obtain a table of points that exist on the line, then plotting those points and drawing a line between them. Since it only takes two points to describe a line, this method works very well for graphing linear functions that yield integer output for integer input.\r\n\r\nThis section provides this method along with two other good ways to draw a quick, accurate sketch of a linear function.\r\n\r\n<\/div>\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<h2>Graphing a Function by Plotting Points<\/h2>\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\r\n[ohm_question height=\"460\"]69981[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Graphing a Linear Function Using y-intercept and Slope<\/h2>\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<div class=\"textbox examples\">\r\n<h3>tip for success<\/h3>\r\nKeep in mind that if a function has a y-intercept, we can always find it by setting [latex]x=0[\/latex] and then solving for [latex]y[\/latex].\r\n\r\n<\/div>\r\nThe other characteristic of the linear function is its slope [latex]m[\/latex], which 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 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\r\n[ohm_question height=\"620\"]88183[\/ohm_question]\r\n\r\n<\/div>\r\n<h2>Graphing a Linear Function Using Transformations<\/h2>\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<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\r\n[ohm_question height=\"440\"]114584[\/ohm_question]\r\n\r\n[ohm_question height=\"465\"]114587[\/ohm_question]\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>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Graph a linear function by plotting points<\/li>\n<li>Graph a linear function using the slope and y-intercept<\/li>\n<li>Graph a linear function using transformations<\/li>\n<\/ul>\n<\/div>\n<div class=\"textbox examples\">\n<h3>recall graphing functions<\/h3>\n<p>Recall that you can graph a linear function by evaluating it at chosen inputs to obtain a table of points that exist on the line, then plotting those points and drawing a line between them. Since it only takes two points to describe a line, this method works very well for graphing linear functions that yield integer output for integer input.<\/p>\n<p>This section provides this method along with two other good ways to draw a quick, accurate sketch of a linear function.<\/p>\n<\/div>\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<h2>Graphing a Function by Plotting Points<\/h2>\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=\"ohm69981\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=69981&theme=oea&iframe_resize_id=ohm69981&show_question_numbers\" width=\"100%\" height=\"460\"><\/iframe><\/p>\n<\/div>\n<h2>Graphing a Linear Function Using y-intercept and Slope<\/h2>\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<div class=\"textbox examples\">\n<h3>tip for success<\/h3>\n<p>Keep in mind that if a function has a y-intercept, we can always find it by setting [latex]x=0[\/latex] and then solving for [latex]y[\/latex].<\/p>\n<\/div>\n<p>The other characteristic of the linear function is its slope [latex]m[\/latex], which 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 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=\"ohm88183\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=88183&theme=oea&iframe_resize_id=ohm88183&show_question_numbers\" width=\"100%\" height=\"620\"><\/iframe><\/p>\n<\/div>\n<h2>Graphing a Linear Function Using Transformations<\/h2>\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<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=\"ohm114584\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114584&theme=oea&iframe_resize_id=ohm114584&show_question_numbers\" width=\"100%\" height=\"440\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"ohm114587\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=114587&theme=oea&iframe_resize_id=ohm114587&show_question_numbers\" width=\"100%\" height=\"465\"><\/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\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-171\">\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. <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><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":395986,"menu_order":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 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