{"id":1076,"date":"2016-10-21T01:21:05","date_gmt":"2016-10-21T01:21:05","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakercollegealgebra\/?post_type=chapter&#038;p=1076"},"modified":"2023-04-04T18:59:07","modified_gmt":"2023-04-04T18:59:07","slug":"introduction-graphs-of-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-collegealgebra\/chapter\/introduction-graphs-of-linear-functions\/","title":{"raw":"Graphing and Writing Equations of Linear Functions","rendered":"Graphing and Writing Equations 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.<\/li>\r\n \t<li>Write the equation of a linear function given its graph.<\/li>\r\n \t<li>Match linear functions with their graphs.<\/li>\r\n \t<li>Find the x-intercept of a function given its equation.<\/li>\r\n \t<li>Find the equations of vertical and horizontal lines.<\/li>\r\n \t<li>Determine whether lines are parallel or perpendicular given their equations.<\/li>\r\n \t<li>Find equations of lines that are parallel or perpendicular to a given line.<\/li>\r\n \t<li>Graph an absolute value function.<\/li>\r\n \t<li>Find the intercepts of an absolute value function.<\/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\/multiembedq2.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\/multiembedq2.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<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\/multiembedq2.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\/multiembedq2.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\n\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\/multiembedq2.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\/multiembedq2.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\/multiembedq2.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\/multiembedq2.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\n<\/div>\r\n<h2>Parallel and Perpendicular Lines<\/h2>\r\nThe two lines in the graph below\u00a0are <strong>parallel lines<\/strong>: they will never intersect. Notice that they have exactly the same steepness which means their slopes are identical. The only difference between the two lines is the <em>y<\/em>-intercept. If we shifted one line vertically toward the <em>y<\/em>-intercept of the other, they would become the same line.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201110\/CNX_Precalc_Figure_02_02_019n2.jpg\" alt=\"Graph of two functions where the blue line is y = -2\/3x + 1, and the baby blue line is y = -2\/3x +7. Notice that they are parallel lines.\" width=\"487\" height=\"410\" \/> Parallel lines.[\/caption]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201111\/CNX_Precalc_EQ_02_02_001n2.jpg\" alt=\"The functions 2x plus 6 and negative 2x minus 4 are parallel. The functions 3x plus 2 and 2x plus 2 are not parallel.\" width=\"535\" height=\"71\" \/>\r\n\r\nWe can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the <em>y<\/em>-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.\r\n\r\nUnlike parallel lines,<strong> perpendicular lines<\/strong> do intersect. Their intersection forms a right or 90-degree angle. The two lines below are perpendicular.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201113\/CNX_Precalc_Figure_02_02_020n2.jpg\" alt=\"Graph of two functions where the blue line is perpendicular to the orange line.\" width=\"487\" height=\"441\" \/> Perpendicular lines.[\/caption]\r\n\r\nPerpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. If [latex]{m}_{1}\\text{ and }{m}_{2}[\/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[\/latex].\r\n<p style=\"text-align: center;\">[latex]{m}_{1}*{m}_{2}=-1[\/latex]<\/p>\r\nTo find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\\frac{1}{8}[\/latex], and the reciprocal of [latex]\\frac{1}{8}[\/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.\r\n\r\nAs with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}f\\left(x\\right)=\\frac{1}{4}x+2\\hfill &amp; \\text{negative reciprocal of }\\frac{1}{4}\\text{ is }-4\\hfill \\\\ f\\left(x\\right)=-4x+3\\hfill &amp; \\text{negative reciprocal of }-4\\text{ is }\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/p>\r\nThe product of the slopes is \u20131.\r\n<p style=\"text-align: center;\">[latex]-4\\left(\\frac{1}{4}\\right)=-1[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n<h3>A General Note: Parallel and Perpendicular Lines<\/h3>\r\nTwo lines are <strong>parallel lines<\/strong> if they do not intersect. The slopes of the lines are the same.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are parallel if }{m}_{1}={m}_{2}[\/latex].<\/p>\r\nIf and only if [latex]{b}_{1}={b}_{2}[\/latex] and [latex]{m}_{1}={m}_{2}[\/latex], we say the lines coincide. Coincident lines are the same line.\r\n\r\nTwo lines are <strong>perpendicular lines<\/strong> if they intersect at right angles.\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are perpendicular if }{m}_{1}*{m}_{2}=-1,\\text{ and }{m}_{2}=-\\frac{1}{{m}_{1}}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Identifying Parallel and Perpendicular Lines<\/h3>\r\nGiven the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(x\\right)=2x+3\\hfill &amp; \\hfill &amp; h\\left(x\\right)=-2x+2\\hfill \\\\ g\\left(x\\right)=\\frac{1}{2}x - 4\\hfill &amp; \\hfill &amp; j\\left(x\\right)=2x - 6\\hfill \\end{array}[\/latex]<\/p>\r\n[reveal-answer q=\"391905\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"391905\"]\r\n\r\nParallel lines have the same slope. Because the functions [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because \u22122 and [latex]\\frac{1}{2}[\/latex] are negative reciprocals, the equations, [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] represent perpendicular lines.\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the lines is shown below.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201114\/CNX_Precalc_Figure_02_02_0212.jpg\" alt=\"Graph of four functions where the blue line is h(x) = -2x + 2, the orange line is f(x) = 2x + 3, the green line is j(x) = 2x - 6, and the red line is g(x) = 1\/2x - 4.\" width=\"487\" height=\"428\" \/> The graph shows that the lines [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] are parallel, and the lines [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] are perpendicular.[\/caption][\/hidden-answer]<\/div>\r\n<h3>Writing Equations of Parallel Lines<\/h3>\r\nIf we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.\r\n\r\nSuppose we are given the following function:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\/latex]<\/p>\r\nWe know that the slope of the line is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). The lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\r\nSuppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin by using point-slope form of an equation for a line. We can then rewrite it in slope-intercept form.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\r\nSo [latex]g\\left(x\\right)=3x+4[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1[\/latex] and passes through the point (1, 7).\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is parallel to the given line.<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the function.<\/li>\r\n \t<li>Substitute the slope and given point into point-slope or slope-intercept form.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\r\nFind a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"961163\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"961163\"]\r\n\r\nThe slope of the given line is 3. If we choose slope-intercept form, we can substitute [latex]m=3[\/latex],\u00a0[latex]x=3[\/latex], and [latex]f(x)=0[\/latex] into slope-intercept form to find the <em>y-<\/em>intercept.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{}0=3\\left(3\\right)+b\\hfill \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\r\nThe line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nWe can confirm that the two lines are parallel by graphing them. The figure below shows that the two lines will never intersect.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201115\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Writing Equations of Perpendicular Lines<\/h3>\r\nWe can use a very similar process to write the equation of a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/p>\r\nThe slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). The lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\r\nAs before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose that we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is\r\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/p>\r\nSo [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.\r\n<div class=\"textbox\">\r\n<h3>Q &amp; A<\/h3>\r\n<strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong>\r\n\r\n<em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is Perpendicular to the given line.<\/h3>\r\n<ol>\r\n \t<li>Find the slope of the given function.<\/li>\r\n \t<li>Determine the negative reciprocal of the slope.<\/li>\r\n \t<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from given point into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\r\n \t<li>Solve for <em>b<\/em>.<\/li>\r\n \t<li>Write the equation of the line.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\r\nFind the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3[\/latex] that passes through the point (3, 0).\r\n\r\n[reveal-answer q=\"668867\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"668867\"]\r\n\r\nThe original line has slope [latex]m=3[\/latex], so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=-\\frac{1}{3}x+b\\hfill \\\\ 0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ b=1\\hfill \\end{array}[\/latex]<\/p>\r\nThe line perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nA graph of the two lines is shown below.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201117\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/>\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<ol>\r\n \t<li>For what y-intercept will the graph of[latex]f(x)[\/latex] pass through the point [latex](-2,5)[\/latex]?<\/li>\r\n \t<li>Add a new function that uses the slope <em>m<\/em> that will create a line that is perpendicular to the function [latex]f(x)=mx-2[\/latex].<\/li>\r\n \t<li>For what y-intercept will the new function pass through the point [latex](4,1)[\/latex], and still be perpendicular to [latex]f(x)[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"201131\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"201131\"]\r\n<ol>\r\n \t<li>When the y-intercept is [latex](0,3)[\/latex] the function will be [latex]f(x) = mx+3[\/latex] and the function will pass through the point [latex](-2,5)[\/latex].<\/li>\r\n \t<li>[latex]f(x)=\\frac{-1}{m}x-2[\/latex] for example. Any value for the y-intercept will work.<\/li>\r\n \t<li>The y-intercept [latex](0,-3)[\/latex] will give a line perpendicular to [latex]f(x)[\/latex] that passes through the point [latex](4,1)[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\r\n<ol>\r\n \t<li>Determine the slope of the line passing through the points.<\/li>\r\n \t<li>Find the negative reciprocal of the slope.<\/li>\r\n \t<li>Use slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\r\n \t<li>Simplify.<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Equation of a Line going through a point and Perpendicular to a Given Line<\/h3>\r\nA line passes through the points (\u20132, 6) and (4, 5). Find the equation of a line that is perpendicular and passes through the point (4, 5).\r\n\r\n[reveal-answer q=\"289053\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"289053\"]\r\n\r\nFrom the two points of the given line, we can calculate the slope of that line.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ {m}_{1}=\\frac{-1}{6}\\hfill \\\\ {m}_{1}=-\\frac{1}{6}\\hfill \\end{array}[\/latex]<\/p>\r\nFind the negative reciprocal of the slope.\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ {m}_{2}=-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ {m}_{2}=6\\hfill \\end{array}[\/latex]<\/p>\r\nWe can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{array}[\/latex]<\/p>\r\nThe equation of the line that passes through the point (4, 5) and is perpendicular to the line passing through the two given points is [latex]y=6x - 19[\/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\nA line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).\r\n\r\n[reveal-answer q=\"678591\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"678591\"]\r\n\r\n[latex]y=-\\frac{1}{3}x+6[\/latex][\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Absolute Value Functions<\/h2>\r\n&nbsp;\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015138\/CNX_Precalc_Figure_01_06_001n2.jpg\" alt=\"The Milky Way.\" width=\"488\" height=\"338\" \/> Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: \"s58y\"\/Flickr)[\/caption]\r\n\r\nUntil the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate <strong>absolute value functions<\/strong>.\r\n<h3>Understanding Absolute Value<\/h3>\r\nRecall that in its basic form [latex]\\displaystyle{f}\\left({x}\\right)={|x|}[\/latex], the absolute value function, is one of our toolkit functions. The <strong>absolute value<\/strong> function is commonly thought of as providing the distance a number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.\r\n<div class=\"textbox\">\r\n<h3>A General Note: Absolute Value Function<\/h3>\r\nThe absolute value function can be defined as a piecewise function\r\n<p style=\"text-align: center;\">$latex f(x) =\\begin{cases}x ,\\ x \\geq 0 \\\\ -x , x &lt; 0\\\\ \\end{cases} $<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Determine a Number within a Prescribed Distance<\/h3>\r\nDescribe all values [latex]x[\/latex] within or including a distance of 4 from the number 5.\r\n\r\n[reveal-answer q=\"68130\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"68130\"]\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015140\/CNX_Precalc_Figure_01_06_0022.jpg\" alt=\"Number line describing the difference of the distance of 4 away from 5.\" width=\"487\" height=\"81\" \/>\r\n\r\nWe want the distance between [latex]x[\/latex] and 5 to be less than or equal to 4. We can draw a number line\u00a0to represent the condition to be satisfied.\r\n\r\nThe distance from [latex]x[\/latex] to 5 can be represented using [latex]|x - 5|[\/latex]. We want the values of [latex]x[\/latex] that satisfy the condition [latex]|x - 5|\\le 4[\/latex].\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that\r\n\r\n[latex]\\displaystyle{-4}\\le{x - 5}[\/latex]\r\n[latex]\\displaystyle{1}\\le{x}[\/latex]\r\nAnd:\r\n[latex]\\displaystyle{x-5}\\le{4}[\/latex]\r\n[latex]\\displaystyle{x}\\le{9}[\/latex]\r\n\r\nSo [latex]|x - 5|\\le 4[\/latex] is equal to [latex]1\\le x\\le 9[\/latex].\r\n\r\nHowever, mathematicians generally prefer absolute value notation.\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\nDescribe all values [latex]x[\/latex] within a distance of 3 from the number 2.\r\n\r\n[reveal-answer q=\"98953\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"98953\"]\r\n\r\n[latex]|x - 2|\\le 3[\/latex]\r\n\r\n[\/hidden-answer]\r\n<iframe id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=469&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"400\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Resistance of a Resistor<\/h3>\r\nElectrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often [latex]\\pm 1\\%,\\pm5\\%,[\/latex] or [latex]\\displaystyle\\pm10\\%[\/latex].\r\n\r\nSuppose we have a resistor rated at 680 ohms, [latex]\\pm 5\\%[\/latex]. Use the absolute value function to express the range of possible values of the actual resistance.\r\n\r\n[reveal-answer q=\"845830\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"845830\"]\r\n\r\n5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance [latex]R[\/latex] in ohms,\r\n<p style=\"text-align: center;\">[latex]|R - 680|\\le 34[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nStudents who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.\r\n\r\n[reveal-answer q=\"584159\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"584159\"]\r\n\r\nUsing the variable [latex]p[\/latex] for passing, [latex]|p - 80|\\le 20[\/latex].[\/hidden-answer]\r\n\r\n<\/div>\r\nThe most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the <strong>origin<\/strong>.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015143\/CNX_Precalc_Figure_01_06_0032.jpg\" alt=\"Graph of an absolute function\" width=\"487\" height=\"251\" \/>\r\n\r\nThe graph below is of [latex]y=2\\left|x - 3\\right|+4[\/latex]. The graph of [latex]y=|x|[\/latex] has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at [latex]\\left(3,4\\right)[\/latex] for this transformed function.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015145\/CNX_Precalc_Figure_01_06_0042.jpg\" alt=\"Graph of the different types of transformations for an absolute function.\" width=\"487\" height=\"486\" \/>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Writing an Equation for an Absolute Value Function<\/h3>\r\nWrite an equation for the function 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\/19015148\/CNX_Precalc_Figure_01_06_0052.jpg\" alt=\"Graph of an absolute function. Two rays stem from the point 3, negative 2. One ray crosses the point 0, 4. The other ray crosses the point 5, 2.\" width=\"487\" height=\"363\" \/>\r\n\r\n[reveal-answer q=\"203899\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"203899\"]\r\n\r\nThe basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015150\/CNX_Precalc_Figure_01_06_0062.jpg\" alt=\"Graph of two transformations for an absolute function at (3, -2).\" width=\"487\" height=\"363\" \/>\r\n\r\nWe also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance.\r\n\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015152\/CNX_Precalc_Figure_01_06_0072.jpg\" alt=\"Graph of two transformations for an absolute function at (3, -2) and describes the ratios between the two different transformations.\" width=\"487\" height=\"363\" \/>\r\n\r\nFrom this information we can write the equation\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(x\\right)=2\\left|x - 3\\right|-2,\\hfill &amp; \\text{treating the stretch as a vertical stretch, or}\\hfill \\\\ f\\left(x\\right)=\\left|2\\left(x - 3\\right)\\right|-2,\\hfill &amp; \\text{treating the stretch as a horizontal compression}.\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<h4>Analysis of the Solution<\/h4>\r\nNote that these equations are algebraically the same\u2014the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>If we couldn\u2019t observe the stretch of the function from the graphs, could we algebraically determine it?<\/strong>\r\n\r\n<em>Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for [latex]x[\/latex] and [latex]f\\left(x\\right)[\/latex].<\/em>\r\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a|x - 3|-2[\/latex]<\/p>\r\n<em>Now substituting in the point <\/em>(1, 2)\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2=a|1 - 3|-2\\hfill \\\\ 4=2a\\hfill \\\\ a=2\\hfill \\end{array}[\/latex]<\/p>\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nWrite the equation for the absolute value function that is horizontally shifted left 2 units, vertically flipped, and vertically shifted up 3 units.\r\n\r\n[reveal-answer q=\"364286\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"364286\"]\r\n\r\n[latex]f\\left(x\\right)=-|x+2|+3\\\\[\/latex][\/hidden-answer]\r\n<iframe id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=60791&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"200\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\n<strong>Q &amp; A<\/strong>\r\n\r\n<strong>Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?\r\n<\/strong>\r\n\r\n<em>Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.\r\n<\/em>\r\n\r\n<em>No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points.<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015154\/CNX_Precalc_Figure_01_06_008abc2.jpg\" alt=\"Graph of the different types of transformations for an absolute function.\" width=\"975\" height=\"415\" \/> (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.[\/caption]\r\n\r\n<\/div>\r\n<h3>Find the Intercepts of an Absolute Value Function<\/h3>\r\nKnowing how to solve problems involving <strong>absolute value functions<\/strong> is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.\r\n<div class=\"textbox\">\r\n<h3>How To: Given the formula for an absolute value function, find the horizontal intercepts of its graph.<\/h3>\r\n<ol>\r\n \t<li>Isolate the absolute value term.<\/li>\r\n \t<li>Use [latex]|A|=B[\/latex] to write [latex]A=B[\/latex] or [latex]\\mathrm{-A}=B[\/latex], assuming [latex]B&gt;0[\/latex].<\/li>\r\n \t<li>Solve for [latex]x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: Finding the Zeros of an Absolute Value Function<\/h3>\r\nFor the function [latex]f\\left(x\\right)=|4x+1|-7[\/latex] , find the values of\u00a0[latex]x[\/latex] such that\u00a0[latex]\\text{ }f\\left(x\\right)=0[\/latex] .\r\n\r\n[reveal-answer q=\"594739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"594739\"]\r\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=|4x+1|-7\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Substitute 0 for }f\\left(x\\right).\\hfill \\\\ 7=|4x+1|\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{Isolate the absolute value on one side of the equation}.\\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ 7=4x+1\\hfill &amp; \\text{or}\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; -7=4x+1\\hfill &amp; \\text{Break into two separate equations and solve}.\\hfill \\\\ 6=4x\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; -8=4x\\hfill &amp; \\hfill \\\\ \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill \\\\ x=\\frac{6}{4}=1.5\\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\hfill &amp; \\text{ }x=\\frac{-8}{4}=-2\\hfill &amp; \\hfill \\end{array}[\/latex]<\/p>\r\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/19015157\/CNX_Precalc_Figure_01_06_011F2.jpg\" alt=\"Graph an absolute function with x-intercepts at -2 and 1.5.\" width=\"731\" height=\"476\" \/>\r\n\r\nThe function's output is 0 when [latex]x=1.5[\/latex] or [latex]x=-2[\/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\nFor the function [latex]f\\left(x\\right)=|2x - 1|-3[\/latex], find the values of [latex]x[\/latex] such that [latex]f\\left(x\\right)=0[\/latex].\r\n\r\n[reveal-answer q=\"206693\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"206693\"]\r\n\r\n[latex]x=-1[\/latex] or [latex]x=2[\/latex][\/hidden-answer]\r\n<iframe id=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=40657&amp;theme=oea&amp;iframe_resize_id=mom900\" width=\"100%\" height=\"500\"><\/iframe>\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 \t<li>Parallel lines have the same slope.<\/li>\r\n \t<li>Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.<\/li>\r\n \t<li>A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the <em>x<\/em>- and <em>y<\/em>-values of the given point into the equation [latex]f\\left(x\\right)=mx+b[\/latex] and using the <em>b<\/em>\u00a0that results. Similarly, point-slope form of an equation can also be used.<\/li>\r\n \t<li>A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.<\/li>\r\n \t<li>The absolute value function is commonly used to measure distances between points.<\/li>\r\n \t<li>Applied problems, such as ranges of possible values, can also be solved using the absolute value function.<\/li>\r\n \t<li>The graph of the absolute value function resembles the letter V. It has a corner point at which the graph changes direction.<\/li>\r\n \t<li>In an absolute value equation, an unknown variable is the input of an absolute value function.<\/li>\r\n \t<li>If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable.<\/li>\r\n \t<li>An absolute value equation may have one solution, two solutions, or no solutions.<\/li>\r\n \t<li>An absolute value inequality is similar to an absolute value equation but takes the form [latex]|A|&lt;B,|A|\\le B,|A|&gt;B,\\text{ or }|A|\\ge B[\/latex]. It can be solved by determining the boundaries of the solution set and then testing which segments are in the set.<\/li>\r\n \t<li>Absolute value inequalities can also be solved graphically.<\/li>\r\n<\/ul>\r\n<h2>Glossary<\/h2>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt><strong>absolute value equation<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex]; it will have solutions when [latex]A=B[\/latex] or [latex]-A=B[\/latex]<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt><strong>absolute value inequality<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">a relationship in the form [latex]|{ A }|&lt;{ B },|{ A }|\\le { B },|{ A }|&gt;{ B },\\text{or }|{ A }|\\ge{ B }[\/latex]<\/dd>\r\n<\/dl>\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>parallel lines<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">two or more lines with the same slope<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-id1165135429388\" class=\"definition\">\r\n \t<dt><strong>perpendicular lines<\/strong><\/dt>\r\n \t<dd id=\"fs-id1165135429394\">two lines that intersect at right angles and have slopes that are negative reciprocals of each other<\/dd>\r\n<\/dl>\r\n<dl id=\"fs-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.<\/li>\n<li>Write the equation of a linear function given its graph.<\/li>\n<li>Match linear functions with their graphs.<\/li>\n<li>Find the x-intercept of a function given its equation.<\/li>\n<li>Find the equations of vertical and horizontal lines.<\/li>\n<li>Determine whether lines are parallel or perpendicular given their equations.<\/li>\n<li>Find equations of lines that are parallel or perpendicular to a given line.<\/li>\n<li>Graph an absolute value function.<\/li>\n<li>Find the intercepts of an absolute value function.<\/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\/multiembedq2.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\/multiembedq2.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<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\/multiembedq2.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\/multiembedq2.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<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\/multiembedq2.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\/multiembedq2.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\/multiembedq2.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\/multiembedq2.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<\/div>\n<h2>Parallel and Perpendicular Lines<\/h2>\n<p>The two lines in the graph below\u00a0are <strong>parallel lines<\/strong>: they will never intersect. Notice that they have exactly the same steepness which means their slopes are identical. The only difference between the two lines is the <em>y<\/em>-intercept. If we shifted one line vertically toward the <em>y<\/em>-intercept of the other, they would become the same line.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201110\/CNX_Precalc_Figure_02_02_019n2.jpg\" alt=\"Graph of two functions where the blue line is y = -2\/3x + 1, and the baby blue line is y = -2\/3x +7. Notice that they are parallel lines.\" width=\"487\" height=\"410\" \/><\/p>\n<p class=\"wp-caption-text\">Parallel lines.<\/p>\n<\/div>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201111\/CNX_Precalc_EQ_02_02_001n2.jpg\" alt=\"The functions 2x plus 6 and negative 2x minus 4 are parallel. The functions 3x plus 2 and 2x plus 2 are not parallel.\" width=\"535\" height=\"71\" \/><\/p>\n<p>We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the <em>y<\/em>-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel.<\/p>\n<p>Unlike parallel lines,<strong> perpendicular lines<\/strong> do intersect. Their intersection forms a right or 90-degree angle. The two lines below are perpendicular.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201113\/CNX_Precalc_Figure_02_02_020n2.jpg\" alt=\"Graph of two functions where the blue line is perpendicular to the orange line.\" width=\"487\" height=\"441\" \/><\/p>\n<p class=\"wp-caption-text\">Perpendicular lines.<\/p>\n<\/div>\n<p>Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. If [latex]{m}_{1}\\text{ and }{m}_{2}[\/latex] are negative reciprocals of one another, they can be multiplied together to yield [latex]-1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]{m}_{1}*{m}_{2}=-1[\/latex]<\/p>\n<p>To find the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is [latex]\\frac{1}{8}[\/latex], and the reciprocal of [latex]\\frac{1}{8}[\/latex] is 8. To find the negative reciprocal, first find the reciprocal and then change the sign.<\/p>\n<p>As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{ll}f\\left(x\\right)=\\frac{1}{4}x+2\\hfill & \\text{negative reciprocal of }\\frac{1}{4}\\text{ is }-4\\hfill \\\\ f\\left(x\\right)=-4x+3\\hfill & \\text{negative reciprocal of }-4\\text{ is }\\frac{1}{4}\\hfill \\end{array}[\/latex]<\/p>\n<p>The product of the slopes is \u20131.<\/p>\n<p style=\"text-align: center;\">[latex]-4\\left(\\frac{1}{4}\\right)=-1[\/latex]<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Parallel and Perpendicular Lines<\/h3>\n<p>Two lines are <strong>parallel lines<\/strong> if they do not intersect. The slopes of the lines are the same.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are parallel if }{m}_{1}={m}_{2}[\/latex].<\/p>\n<p>If and only if [latex]{b}_{1}={b}_{2}[\/latex] and [latex]{m}_{1}={m}_{2}[\/latex], we say the lines coincide. Coincident lines are the same line.<\/p>\n<p>Two lines are <strong>perpendicular lines<\/strong> if they intersect at right angles.<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)={m}_{1}x+{b}_{1}\\text{ and }g\\left(x\\right)={m}_{2}x+{b}_{2}\\text{ are perpendicular if }{m}_{1}*{m}_{2}=-1,\\text{ and }{m}_{2}=-\\frac{1}{{m}_{1}}[\/latex].<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Parallel and Perpendicular Lines<\/h3>\n<p>Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(x\\right)=2x+3\\hfill & \\hfill & h\\left(x\\right)=-2x+2\\hfill \\\\ g\\left(x\\right)=\\frac{1}{2}x - 4\\hfill & \\hfill & j\\left(x\\right)=2x - 6\\hfill \\end{array}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q391905\">Show Solution<\/span><\/p>\n<div id=\"q391905\" class=\"hidden-answer\" style=\"display: none\">\n<p>Parallel lines have the same slope. Because the functions [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] each have a slope of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because \u22122 and [latex]\\frac{1}{2}[\/latex] are negative reciprocals, the equations, [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] represent perpendicular lines.<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the lines is shown below.<\/p>\n<div style=\"width: 497px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201114\/CNX_Precalc_Figure_02_02_0212.jpg\" alt=\"Graph of four functions where the blue line is h(x) = -2x + 2, the orange line is f(x) = 2x + 3, the green line is j(x) = 2x - 6, and the red line is g(x) = 1\/2x - 4.\" width=\"487\" height=\"428\" \/><\/p>\n<p class=\"wp-caption-text\">The graph shows that the lines [latex]f\\left(x\\right)=2x+3[\/latex] and [latex]j\\left(x\\right)=2x - 6[\/latex] are parallel, and the lines [latex]g\\left(x\\right)=\\frac{1}{2}x - 4[\/latex] and [latex]h\\left(x\\right)=-2x+2[\/latex] are perpendicular.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<h3>Writing Equations of Parallel Lines<\/h3>\n<p>If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line.<\/p>\n<p>Suppose we are given the following function:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=3x+1[\/latex]<\/p>\n<p>We know that the slope of the line is 3. We also know that the <em>y-<\/em>intercept is (0, 1). Any other line with a slope of 3 will be parallel to <em>f<\/em>(<em>x<\/em>). The lines formed by all of the following functions will be parallel to <em>f<\/em>(<em>x<\/em>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=3x+6\\hfill \\\\ h\\left(x\\right)=3x+1\\hfill \\\\ p\\left(x\\right)=3x+\\frac{2}{3}\\hfill \\end{array}[\/latex]<\/p>\n<p>Suppose then we want to write the equation of a line that is parallel to <em>f<\/em>\u00a0and passes through the point (1, 7). We already know that the slope is 3. We just need to determine which value for <em>b<\/em>\u00a0will give the correct line. We can begin by using point-slope form of an equation for a line. We can then rewrite it in slope-intercept form.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}y-{y}_{1}=m\\left(x-{x}_{1}\\right)\\hfill \\\\ y - 7=3\\left(x - 1\\right)\\hfill \\\\ y - 7=3x - 3\\hfill \\\\ \\text{}y=3x+4\\hfill \\end{array}[\/latex]<\/p>\n<p>So [latex]g\\left(x\\right)=3x+4[\/latex] is parallel to [latex]f\\left(x\\right)=3x+1[\/latex] and passes through the point (1, 7).<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is parallel to the given line.<\/h3>\n<ol>\n<li>Find the slope of the function.<\/li>\n<li>Substitute the slope and given point into point-slope or slope-intercept form.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding a Line Parallel to a Given Line<\/h3>\n<p>Find a line parallel to the graph of [latex]f\\left(x\\right)=3x+6[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q961163\">Show Solution<\/span><\/p>\n<div id=\"q961163\" class=\"hidden-answer\" style=\"display: none\">\n<p>The slope of the given line is 3. If we choose slope-intercept form, we can substitute [latex]m=3[\/latex],\u00a0[latex]x=3[\/latex], and [latex]f(x)=0[\/latex] into slope-intercept form to find the <em>y-<\/em>intercept.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=3x+b\\hfill \\\\ \\text{}0=3\\left(3\\right)+b\\hfill \\\\ \\text{}b=-9\\hfill \\end{array}[\/latex]<\/p>\n<p>The line parallel to\u00a0<em>f<\/em>(<em>x<\/em>) that passes through (3, 0) is [latex]g\\left(x\\right)=3x - 9[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>We can confirm that the two lines are parallel by graphing them. The figure below shows that the two lines will never intersect.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201115\/CNX_Precalc_Figure_02_02_022n2.jpg\" alt=\"Graph of two functions where the blue line is y = 3x + 6, and the orange line is y = 3x - 9.\" width=\"731\" height=\"619\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Writing Equations of Perpendicular Lines<\/h3>\n<p>We can use a very similar process to write the equation of a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function:<\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=2x+4[\/latex]<\/p>\n<p>The slope of the line is 2, and its negative reciprocal is [latex]-\\frac{1}{2}[\/latex]. Any function with a slope of [latex]-\\frac{1}{2}[\/latex] will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>). The lines formed by all of the following functions will be perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=-\\frac{1}{2}x+4\\hfill \\\\ h\\left(x\\right)=-\\frac{1}{2}x+2\\hfill \\\\ p\\left(x\\right)=-\\frac{1}{2}x-\\frac{1}{2}\\hfill \\end{array}[\/latex]<\/p>\n<p>As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose that we want to write the equation of a line that is perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>) and passes through the point (4, 0). We already know that the slope is [latex]-\\frac{1}{2}[\/latex]. Now we can use the point to find the <em>y<\/em>-intercept by substituting the given values into the slope-intercept form of a line and solving for <em>b<\/em>.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=mx+b\\hfill \\\\ 0=-\\frac{1}{2}\\left(4\\right)+b\\hfill \\\\ 0=-2+b\\hfill \\\\ 2=b\\hfill \\\\ b=2\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation for the function with a slope of [latex]-\\frac{1}{2}[\/latex] and a <em>y-<\/em>intercept of 2 is<\/p>\n<p style=\"text-align: center;\">[latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex].<\/p>\n<p>So [latex]g\\left(x\\right)=-\\frac{1}{2}x+2[\/latex] is perpendicular to [latex]f\\left(x\\right)=2x+4[\/latex] and passes through the point (4, 0). Be aware that perpendicular lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature.<\/p>\n<div class=\"textbox\">\n<h3>Q &amp; A<\/h3>\n<p><strong>A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not \u20131. Doesn\u2019t this fact contradict the definition of perpendicular lines?<\/strong><\/p>\n<p><em>No. For two perpendicular linear functions, the product of their slopes is \u20131. However, a vertical line is not a function so the definition is not contradicted.<\/em><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given the equation of a linear function, write the equation of a line WHICH passes through a given point and is Perpendicular to the given line.<\/h3>\n<ol>\n<li>Find the slope of the given function.<\/li>\n<li>Determine the negative reciprocal of the slope.<\/li>\n<li>Substitute the new slope and the values for <em>x<\/em>\u00a0and <em>y<\/em>\u00a0from given point into [latex]g\\left(x\\right)=mx+b[\/latex].<\/li>\n<li>Solve for <em>b<\/em>.<\/li>\n<li>Write the equation of the line.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Perpendicular Line<\/h3>\n<p>Find the equation of a line perpendicular to [latex]f\\left(x\\right)=3x+3[\/latex] that passes through the point (3, 0).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q668867\">Show Solution<\/span><\/p>\n<div id=\"q668867\" class=\"hidden-answer\" style=\"display: none\">\n<p>The original line has slope [latex]m=3[\/latex], so the slope of the perpendicular line will be its negative reciprocal, or [latex]-\\frac{1}{3}[\/latex]. Using this slope and the given point, we can find the equation for the line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=-\\frac{1}{3}x+b\\hfill \\\\ 0=-\\frac{1}{3}\\left(3\\right)+b\\hfill \\\\ \\text{ }1=b\\hfill \\\\ b=1\\hfill \\end{array}[\/latex]<\/p>\n<p>The line perpendicular to\u00a0<em>f<\/em>(<em>x<\/em>)\u00a0that passes through (3, 0) is [latex]g\\left(x\\right)=-\\frac{1}{3}x+1[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>A graph of the two lines is shown below.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images-archive-read-only\/wp-content\/uploads\/sites\/924\/2015\/11\/25201117\/CNX_Precalc_Figure_02_02_023n2.jpg\" alt=\"Graph of two functions where the blue line is g(x) = -1\/3x + 1, and the orange line is f(x) = 3x + 6.\" width=\"487\" height=\"504\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<ol>\n<li>For what y-intercept will the graph of[latex]f(x)[\/latex] pass through the point [latex](-2,5)[\/latex]?<\/li>\n<li>Add a new function that uses the slope <em>m<\/em> that will create a line that is perpendicular to the function [latex]f(x)=mx-2[\/latex].<\/li>\n<li>For what y-intercept will the new function pass through the point [latex](4,1)[\/latex], and still be perpendicular to [latex]f(x)[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q201131\">Show Solution<\/span><\/p>\n<div id=\"q201131\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>When the y-intercept is [latex](0,3)[\/latex] the function will be [latex]f(x) = mx+3[\/latex] and the function will pass through the point [latex](-2,5)[\/latex].<\/li>\n<li>[latex]f(x)=\\frac{-1}{m}x-2[\/latex] for example. Any value for the y-intercept will work.<\/li>\n<li>The y-intercept [latex](0,-3)[\/latex] will give a line perpendicular to [latex]f(x)[\/latex] that passes through the point [latex](4,1)[\/latex].<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>How To: Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point.<\/h3>\n<ol>\n<li>Determine the slope of the line passing through the points.<\/li>\n<li>Find the negative reciprocal of the slope.<\/li>\n<li>Use slope-intercept form or point-slope form to write the equation by substituting the known values.<\/li>\n<li>Simplify.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Equation of a Line going through a point and Perpendicular to a Given Line<\/h3>\n<p>A line passes through the points (\u20132, 6) and (4, 5). Find the equation of a line that is perpendicular and passes through the point (4, 5).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q289053\">Show Solution<\/span><\/p>\n<div id=\"q289053\" class=\"hidden-answer\" style=\"display: none\">\n<p>From the two points of the given line, we can calculate the slope of that line.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{1}=\\frac{5 - 6}{4-\\left(-2\\right)}\\hfill \\\\ {m}_{1}=\\frac{-1}{6}\\hfill \\\\ {m}_{1}=-\\frac{1}{6}\\hfill \\end{array}[\/latex]<\/p>\n<p>Find the negative reciprocal of the slope.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}{m}_{2}=\\frac{-1}{-\\frac{1}{6}}\\hfill \\\\ {m}_{2}=-1\\left(-\\frac{6}{1}\\right)\\hfill \\\\ {m}_{2}=6\\hfill \\end{array}[\/latex]<\/p>\n<p>We can then solve for the <em>y-<\/em>intercept of the line passing through the point (4, 5).<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}g\\left(x\\right)=6x+b\\hfill \\\\ 5=6\\left(4\\right)+b\\hfill \\\\ 5=24+b\\hfill \\\\ -19=b\\hfill \\\\ b=-19\\hfill \\end{array}[\/latex]<\/p>\n<p>The equation of the line that passes through the point (4, 5) and is perpendicular to the line passing through the two given points is [latex]y=6x - 19[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>A line passes through the points, (\u20132, \u201315) and (2, \u20133). Find the equation of a perpendicular line that passes through the point, (6, 4).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q678591\">Show Solution<\/span><\/p>\n<div id=\"q678591\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]y=-\\frac{1}{3}x+6[\/latex]<\/p><\/div>\n<\/div>\n<\/div>\n<h2>Absolute Value Functions<\/h2>\n<p>&nbsp;<\/p>\n<div style=\"width: 498px\" 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\/19015138\/CNX_Precalc_Figure_01_06_001n2.jpg\" alt=\"The Milky Way.\" width=\"488\" height=\"338\" \/><\/p>\n<p class=\"wp-caption-text\">Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: &#8220;s58y&#8221;\/Flickr)<\/p>\n<\/div>\n<p>Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate <strong>absolute value functions<\/strong>.<\/p>\n<h3>Understanding Absolute Value<\/h3>\n<p>Recall that in its basic form [latex]\\displaystyle{f}\\left({x}\\right)={|x|}[\/latex], the absolute value function, is one of our toolkit functions. The <strong>absolute value<\/strong> function is commonly thought of as providing the distance a number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.<\/p>\n<div class=\"textbox\">\n<h3>A General Note: Absolute Value Function<\/h3>\n<p>The absolute value function can be defined as a piecewise function<\/p>\n<p style=\"text-align: center;\">[latex]f(x) =\\begin{cases}x ,\\ x \\geq 0 \\\\ -x , x < 0\\\\ \\end{cases}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Determine a Number within a Prescribed Distance<\/h3>\n<p>Describe all values [latex]x[\/latex] within or including a distance of 4 from the number 5.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q68130\">Show Solution<\/span><\/p>\n<div id=\"q68130\" class=\"hidden-answer\" style=\"display: none\">\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\/19015140\/CNX_Precalc_Figure_01_06_0022.jpg\" alt=\"Number line describing the difference of the distance of 4 away from 5.\" width=\"487\" height=\"81\" \/><\/p>\n<p>We want the distance between [latex]x[\/latex] and 5 to be less than or equal to 4. We can draw a number line\u00a0to represent the condition to be satisfied.<\/p>\n<p>The distance from [latex]x[\/latex] to 5 can be represented using [latex]|x - 5|[\/latex]. We want the values of [latex]x[\/latex] that satisfy the condition [latex]|x - 5|\\le 4[\/latex].<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that<\/p>\n<p>[latex]\\displaystyle{-4}\\le{x - 5}[\/latex]<br \/>\n[latex]\\displaystyle{1}\\le{x}[\/latex]<br \/>\nAnd:<br \/>\n[latex]\\displaystyle{x-5}\\le{4}[\/latex]<br \/>\n[latex]\\displaystyle{x}\\le{9}[\/latex]<\/p>\n<p>So [latex]|x - 5|\\le 4[\/latex] is equal to [latex]1\\le x\\le 9[\/latex].<\/p>\n<p>However, mathematicians generally prefer absolute value notation.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Describe all values [latex]x[\/latex] within a distance of 3 from the number 2.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q98953\">Show Solution<\/span><\/p>\n<div id=\"q98953\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]|x - 2|\\le 3[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom200\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=469&amp;theme=oea&amp;iframe_resize_id=mom200\" width=\"100%\" height=\"400\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Resistance of a Resistor<\/h3>\n<p>Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often [latex]\\pm 1\\%,\\pm5\\%,[\/latex] or [latex]\\displaystyle\\pm10\\%[\/latex].<\/p>\n<p>Suppose we have a resistor rated at 680 ohms, [latex]\\pm 5\\%[\/latex]. Use the absolute value function to express the range of possible values of the actual resistance.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q845830\">Show Solution<\/span><\/p>\n<div id=\"q845830\" class=\"hidden-answer\" style=\"display: none\">\n<p>5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance [latex]R[\/latex] in ohms,<\/p>\n<p style=\"text-align: center;\">[latex]|R - 680|\\le 34[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q584159\">Show Solution<\/span><\/p>\n<div id=\"q584159\" class=\"hidden-answer\" style=\"display: none\">\n<p>Using the variable [latex]p[\/latex] for passing, [latex]|p - 80|\\le 20[\/latex].<\/p><\/div>\n<\/div>\n<\/div>\n<p>The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the <strong>origin<\/strong>.<\/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\/19015143\/CNX_Precalc_Figure_01_06_0032.jpg\" alt=\"Graph of an absolute function\" width=\"487\" height=\"251\" \/><\/p>\n<p>The graph below is of [latex]y=2\\left|x - 3\\right|+4[\/latex]. The graph of [latex]y=|x|[\/latex] has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at [latex]\\left(3,4\\right)[\/latex] for this transformed function.<\/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\/19015145\/CNX_Precalc_Figure_01_06_0042.jpg\" alt=\"Graph of the different types of transformations for an absolute function.\" width=\"487\" height=\"486\" \/><\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Writing an Equation for an Absolute Value Function<\/h3>\n<p>Write an equation for the function 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\/19015148\/CNX_Precalc_Figure_01_06_0052.jpg\" alt=\"Graph of an absolute function. Two rays stem from the point 3, negative 2. One ray crosses the point 0, 4. The other ray crosses the point 5, 2.\" width=\"487\" height=\"363\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q203899\">Show Solution<\/span><\/p>\n<div id=\"q203899\" class=\"hidden-answer\" style=\"display: none\">\n<p>The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function.<\/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\/19015150\/CNX_Precalc_Figure_01_06_0062.jpg\" alt=\"Graph of two transformations for an absolute function at (3, -2).\" width=\"487\" height=\"363\" \/><\/p>\n<p>We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance.<\/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\/19015152\/CNX_Precalc_Figure_01_06_0072.jpg\" alt=\"Graph of two transformations for an absolute function at (3, -2) and describes the ratios between the two different transformations.\" width=\"487\" height=\"363\" \/><\/p>\n<p>From this information we can write the equation<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}f\\left(x\\right)=2\\left|x - 3\\right|-2,\\hfill & \\text{treating the stretch as a vertical stretch, or}\\hfill \\\\ f\\left(x\\right)=\\left|2\\left(x - 3\\right)\\right|-2,\\hfill & \\text{treating the stretch as a horizontal compression}.\\hfill \\end{array}[\/latex]<\/p>\n<h4>Analysis of the Solution<\/h4>\n<p>Note that these equations are algebraically the same\u2014the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>If we couldn\u2019t observe the stretch of the function from the graphs, could we algebraically determine it?<\/strong><\/p>\n<p><em>Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for [latex]x[\/latex] and [latex]f\\left(x\\right)[\/latex].<\/em><\/p>\n<p style=\"text-align: center;\">[latex]f\\left(x\\right)=a|x - 3|-2[\/latex]<\/p>\n<p><em>Now substituting in the point <\/em>(1, 2)<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}2=a|1 - 3|-2\\hfill \\\\ 4=2a\\hfill \\\\ a=2\\hfill \\end{array}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>Write the equation for the absolute value function that is horizontally shifted left 2 units, vertically flipped, and vertically shifted up 3 units.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q364286\">Show Solution<\/span><\/p>\n<div id=\"q364286\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f\\left(x\\right)=-|x+2|+3\\\\[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom500\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=60791&amp;theme=oea&amp;iframe_resize_id=mom500\" width=\"100%\" height=\"200\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<p><strong>Q &amp; A<\/strong><\/p>\n<p><strong>Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?<br \/>\n<\/strong><\/p>\n<p><em>Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.<br \/>\n<\/em><\/p>\n<p><em>No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points.<\/em><\/p>\n<div style=\"width: 985px\" 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\/19015154\/CNX_Precalc_Figure_01_06_008abc2.jpg\" alt=\"Graph of the different types of transformations for an absolute function.\" width=\"975\" height=\"415\" \/><\/p>\n<p class=\"wp-caption-text\">(a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points.<\/p>\n<\/div>\n<\/div>\n<h3>Find the Intercepts of an Absolute Value Function<\/h3>\n<p>Knowing how to solve problems involving <strong>absolute value functions<\/strong> is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given the formula for an absolute value function, find the horizontal intercepts of its graph.<\/h3>\n<ol>\n<li>Isolate the absolute value term.<\/li>\n<li>Use [latex]|A|=B[\/latex] to write [latex]A=B[\/latex] or [latex]\\mathrm{-A}=B[\/latex], assuming [latex]B>0[\/latex].<\/li>\n<li>Solve for [latex]x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Finding the Zeros of an Absolute Value Function<\/h3>\n<p>For the function [latex]f\\left(x\\right)=|4x+1|-7[\/latex] , find the values of\u00a0[latex]x[\/latex] such that\u00a0[latex]\\text{ }f\\left(x\\right)=0[\/latex] .<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q594739\">Show Solution<\/span><\/p>\n<div id=\"q594739\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]\\begin{array}{l}0=|4x+1|-7\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Substitute 0 for }f\\left(x\\right).\\hfill \\\\ 7=|4x+1|\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{Isolate the absolute value on one side of the equation}.\\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ 7=4x+1\\hfill & \\text{or}\\hfill & \\hfill & \\hfill & \\hfill & -7=4x+1\\hfill & \\text{Break into two separate equations and solve}.\\hfill \\\\ 6=4x\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & -8=4x\\hfill & \\hfill \\\\ \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\hfill \\\\ x=\\frac{6}{4}=1.5\\hfill & \\hfill & \\hfill & \\hfill & \\hfill & \\text{ }x=\\frac{-8}{4}=-2\\hfill & \\hfill \\end{array}[\/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\/19015157\/CNX_Precalc_Figure_01_06_011F2.jpg\" alt=\"Graph an absolute function with x-intercepts at -2 and 1.5.\" width=\"731\" height=\"476\" \/><\/p>\n<p>The function's output is 0 when [latex]x=1.5[\/latex] or [latex]x=-2[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>For the function [latex]f\\left(x\\right)=|2x - 1|-3[\/latex], find the values of [latex]x[\/latex] such that [latex]f\\left(x\\right)=0[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q206693\">Show Solution<\/span><\/p>\n<div id=\"q206693\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]x=-1[\/latex] or [latex]x=2[\/latex]<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom900\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq2.php?id=40657&amp;theme=oea&amp;iframe_resize_id=mom900\" width=\"100%\" height=\"500\"><\/iframe><\/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<li>Parallel lines have the same slope.<\/li>\n<li>Perpendicular lines have negative reciprocal slopes, assuming neither is vertical.<\/li>\n<li>A line parallel to another line, passing through a given point, may be found by substituting the slope value of the line and the <em>x<\/em>- and <em>y<\/em>-values of the given point into the equation [latex]f\\left(x\\right)=mx+b[\/latex] and using the <em>b<\/em>\u00a0that results. Similarly, point-slope form of an equation can also be used.<\/li>\n<li>A line perpendicular to another line, passing through a given point, may be found in the same manner, with the exception of using the negative reciprocal slope.<\/li>\n<li>The absolute value function is commonly used to measure distances between points.<\/li>\n<li>Applied problems, such as ranges of possible values, can also be solved using the absolute value function.<\/li>\n<li>The graph of the absolute value function resembles the letter V. It has a corner point at which the graph changes direction.<\/li>\n<li>In an absolute value equation, an unknown variable is the input of an absolute value function.<\/li>\n<li>If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable.<\/li>\n<li>An absolute value equation may have one solution, two solutions, or no solutions.<\/li>\n<li>An absolute value inequality is similar to an absolute value equation but takes the form [latex]|A|<B,|A|\\le B,|A|>B,\\text{ or }|A|\\ge B[\/latex]. It can be solved by determining the boundaries of the solution set and then testing which segments are in the set.<\/li>\n<li>Absolute value inequalities can also be solved graphically.<\/li>\n<\/ul>\n<h2>Glossary<\/h2>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong>absolute value equation<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">an equation of the form [latex]|A|=B[\/latex], with [latex]B\\ge 0[\/latex]; it will have solutions when [latex]A=B[\/latex] or [latex]-A=B[\/latex]<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong>absolute value inequality<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">a relationship in the form [latex]|{ A }|<{ B },|{ A }|\\le { B },|{ A }|>{ B },\\text{or }|{ A }|\\ge{ B }[\/latex]<\/dd>\n<\/dl>\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>parallel lines<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">two or more lines with the same slope<\/dd>\n<\/dl>\n<dl id=\"fs-id1165135429388\" class=\"definition\">\n<dt><strong>perpendicular lines<\/strong><\/dt>\n<dd id=\"fs-id1165135429394\">two lines that intersect at right angles and have slopes that are negative reciprocals of each other<\/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-1076\">\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":21,"menu_order":3,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Revision and 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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 + 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