{"id":113,"date":"2021-02-03T21:28:37","date_gmt":"2021-02-03T21:28:37","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=113"},"modified":"2022-03-11T21:36:30","modified_gmt":"2022-03-11T21:36:30","slug":"linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/linear-functions\/","title":{"raw":"Linear Functions","rendered":"Linear Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Calculate the slope of a linear function and interpret its meaning<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe easiest type of function to consider is a <strong>linear function<\/strong>. Linear functions have the form [latex]f(x)=ax+b[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are constants. In Figure 1, we see examples of linear functions when [latex]a[\/latex] is positive, negative, and zero. Note that if [latex]a&gt;0[\/latex], the graph of the line rises as [latex]x[\/latex] increases. In other words, [latex]f(x)=ax+b[\/latex] is increasing on [latex](\u2212\\infty, \\infty)[\/latex]. If [latex]a&lt;0[\/latex], the graph of the line falls as [latex]x[\/latex] increases. In this case, [latex]f(x)=ax+b[\/latex] is decreasing on [latex](\u2212\\infty, \\infty)[\/latex]. If [latex]a=0[\/latex], the line is horizontal.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"325\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202222\/CNX_Calc_Figure_01_02_001.jpg\" alt=\"An image of a graph. The y axis runs from -2 to 5 and the x axis runs from -2 to 5. The graph is of the 3 functions. The first function is \u201cf(x) = 3x + 1\u201d, which is an increasing straight line with an x intercept at ((-1\/3), 0) and a y intercept at (0, 1). The second function is \u201cg(x) = 2\u201d, which is a horizontal line with a y intercept at (0, 2) and no x intercept. The third function is \u201ch(x) = (-1\/2)x\u201d, which is a decreasing straight line with an x intercept and y intercept both at the origin. The function f(x) is increasing at a higher rate than the function h(x) is decreasing.\" width=\"325\" height=\"312\" \/> Figure 1. These linear functions are increasing or decreasing on [latex](-\\infty, \\infty)[\/latex] and one function is a horizontal line.[\/caption]\r\n<h2>Slope<\/h2>\r\n<p id=\"fs-id1170573396917\">As suggested by Figure 1, the graph of any linear function is a line. One of the distinguishing features of a line is its slope. The <strong>slope<\/strong> is the change in [latex]y[\/latex] for each unit change in [latex]x[\/latex]. The slope measures both the steepness and the direction of a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we need to determine the ratio of the change in [latex]y[\/latex] versus the change in [latex]x[\/latex]. To do so, we choose any two points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line and calculate [latex]\\dfrac{y_2-y_1}{x_2-x_1}[\/latex]. In Figure 2, we see this ratio is independent of the points chosen.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"465\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202225\/CNX_Calc_Figure_01_02_021.jpg\" alt=\"An image of a graph. The y axis runs from -1 to 10 and the x axis runs from -1 to 6. The graph is of a function that is an increasing straight line. There are four points labeled on the function at (1, 1), (2, 3), (3, 5), and (5, 9). There is a dotted horizontal line from the labeled function point (1, 1) to the unlabeled point (3, 1) which is not on the function, and then dotted vertical line from the unlabeled point (3, 1), which is not on the function, to the labeled function point (3, 5). These two dotted have the label \u201c(y2 - y1)\/(x2 - x1) = (5 -1)\/(3 - 1) = 2\u201d. There is a dotted horizontal line from the labeled function point (2, 3) to the unlabeled point (5, 3) which is not on the function, and then dotted vertical line from the unlabeled point (5, 3), which is not on the function, to the labeled function point (5, 9). These two dotted have the label \u201c(y2 - y1)\/(x2 - x1) = (9 -3)\/(5 - 2) = 2\u201d.\" width=\"465\" height=\"459\" \/> Figure 2. or any linear function, the slope [latex](y_2-y_1)\/(x_2-x_1)[\/latex] is independent of the choice of points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line.[\/caption]&nbsp;\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n\r\nConsider line [latex]L[\/latex] passing through points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex]. Let [latex]\\Delta y=y_2-y_1[\/latex] and [latex]\\Delta x=x_2-x_1[\/latex] denote the changes in [latex]y[\/latex] and [latex]x[\/latex], respectively. The <strong>slope<\/strong> of the line is\r\n<div id=\"fs-id1170573414443\" class=\"equation\" style=\"text-align: center;\">[latex]m=\\dfrac{y_2-y_1}{x_2-x_1}=\\dfrac{\\Delta y}{\\Delta x}[\/latex]<\/div>\r\n<\/div>\r\n<h2>Linear Function Formulas<\/h2>\r\n<p id=\"fs-id1170573356648\"><span style=\"font-size: 1rem; text-align: initial;\">We now examine the relationship between slope and the formula for a linear function. Consider the linear function given by the formula [latex]f(x)=ax+b[\/latex]. As discussed earlier, we know the graph of a linear function is given by a line. We can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating [latex](y_2-y_1)\/(x_2-x_1)[\/latex] for any points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line. Evaluating the function [latex]f[\/latex] at [latex]x=0[\/latex], we see that [latex](0,b)[\/latex] is a point on this line. Evaluating this function at [latex]x=1[\/latex], we see that [latex](1,a+b)[\/latex] is also a point on this line. Therefore, the slope of this line is<\/span><\/p>\r\n\r\n<div id=\"fs-id1170573263457\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{(a+b)-b}{1-0}=a[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573409038\">We have shown that the coefficient [latex]a[\/latex] is the slope of the line. We can conclude that the formula [latex]f(x)=ax+b[\/latex] describes a line with slope [latex]a[\/latex]. Furthermore, because this line intersects the [latex]y[\/latex]-axis at the point [latex](0,b)[\/latex], we see that the [latex]y[\/latex]-intercept for this linear function is [latex](0,b)[\/latex]. We conclude that the formula [latex]f(x)=ax+b[\/latex] tells us the slope, [latex]a[\/latex], and the [latex]y[\/latex]-intercept, [latex](0,b)[\/latex], for this line. Since we often use the symbol [latex]m[\/latex] to denote the slope of a line, we can write<\/p>\r\n\r\n<div id=\"fs-id1170573262665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=mx+b[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573257414\">to denote the <strong>slope-intercept form<\/strong> of a linear function.<\/p>\r\n<p id=\"fs-id1170573381114\">Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function passes through the point [latex](x_1,y_1)[\/latex] and the slope of the line is [latex]m[\/latex]. Since any other point [latex](x,f(x))[\/latex] on the graph of [latex]f[\/latex] must satisfy the equation<\/p>\r\n\r\n<div id=\"fs-id1170573299395\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{f(x)-y_1}{x-x_1}[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573569154\">this linear function can be expressed by writing<\/p>\r\n\r\n<div id=\"fs-id1170573583649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)-y_1=m(x-x_1)[\/latex].<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573309782\">We call this equation the<strong> point-slope equation<\/strong> for that linear function.<\/p>\r\n<p id=\"fs-id1170573309900\">Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the slope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be expressed in either of these forms. Instead, a vertical line is described by the equation [latex]x=k[\/latex] for some constant [latex]k[\/latex]. Since neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation<\/p>\r\n\r\n<div id=\"fs-id1170573418048\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]ax+by=c[\/latex],<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573425137\">where [latex]a,b[\/latex] are both not zero, to denote the <strong>standard form of a line.<\/strong><\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 style=\"text-align: center;\">Definition<\/h3>\r\n\r\n<hr \/>\r\n<p id=\"fs-id1170573328429\">Consider a line passing through the point [latex](x_1,y_1)[\/latex] with slope [latex]m[\/latex]. The equation<\/p>\r\n\r\n<div id=\"fs-id1170573382468\" class=\"equation\" style=\"text-align: center;\">[latex]y-y_1=m(x-x_1)[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170573397934\">is the <strong>point-slope equation<\/strong> for that line.<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170573425678\">Consider a line with slope [latex]m[\/latex] and [latex]y[\/latex]-intercept [latex](0,b)[\/latex]. The equation<\/p>\r\n\r\n<div id=\"fs-id1170573240220\" class=\"equation\" style=\"text-align: center;\">[latex]y=mx+b[\/latex]<\/div>\r\n<div><\/div>\r\n<p id=\"fs-id1170573336985\">is an equation for that line in <strong>slope-intercept form<\/strong>.<\/p>\r\n&nbsp;\r\n<p id=\"fs-id1170573334038\">The <strong>standard form of a line<\/strong> is given by the equation<\/p>\r\n\r\n<div id=\"fs-id1170573399196\" class=\"equation\" style=\"text-align: center;\">[latex]ax+by=c[\/latex]<\/div>\r\n&nbsp;\r\n<p id=\"fs-id1170573392539\">where [latex]a[\/latex] and [latex]b[\/latex] are both not zero. This form is more general because it allows for a vertical line, [latex]x=k[\/latex].<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573262524\" class=\"textbox exercises\">\r\n<h3>Example: Finding the Slope and Equations of Lines<\/h3>\r\n<p id=\"fs-id1170573402450\">Consider the line passing through the points [latex](11,-4)[\/latex] and [latex](-4,5)[\/latex], as shown in Figure 3.<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"694\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202228\/CNX_Calc_Figure_01_02_002.jpg\" alt=\"An image of a graph. The x axis runs from -5 to 12 and the y axis runs from -5 to 6. The graph is of the function that is a decreasing straight line. The function has two points plotted, at (-4, 5) and (11, 4).\" width=\"694\" height=\"459\" \/> Figure 3. Finding the equation of a linear function with a graph that is a line between two given points.[\/caption]\r\n\r\n<div class=\"wp-caption-text\"><\/div>\r\n<ol id=\"fs-id1170573248716\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Find the slope of the line.<\/li>\r\n \t<li>Find an equation for this linear function in point-slope form.<\/li>\r\n \t<li>Find an equation for this linear function in slope-intercept form.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170573411739\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573411739\"]\r\n<ol id=\"fs-id1170573411739\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>The slope of the line is\r\n<div id=\"fs-id1170573334269\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{y_2-y_1}{x_2-x_1}=\\dfrac{5-(-4)}{-4-11}=-\\dfrac{9}{15}=-\\dfrac{3}{5}[\/latex].<\/div><\/li>\r\n \t<li>To find an equation for the linear function in point-slope form, use the slope [latex]m=-\\frac{3}{5}[\/latex] and choose any point on the line. If we choose the point [latex](11,-4)[\/latex], we get the equation\r\n<div id=\"fs-id1170573268080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)+4=-\\frac{3}{5}(x-11)[\/latex].<\/div><\/li>\r\n \t<li>To find an equation for the linear function in slope-intercept form, solve the equation in part (b) for [latex]f(x)[\/latex]. When we do this, we get the equation\r\n<div id=\"fs-id1170573274197\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=-\\frac{3}{5}x+\\frac{13}{5}[\/latex].<\/div><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: Finding the Slope and Equations of Lines[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/iiBBHtVIk9U?controls=0&amp;start=61&amp;end=163&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266833\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266833\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.2BasicClassesOfFunctions61to163_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.2 Basic Classes of Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1170573230955\" class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1170573210343\">Consider the line passing through points [latex](-3,2)[\/latex] and [latex](1,4)[\/latex]. Find the slope of the line.<\/p>\r\n<p id=\"fs-id1170573570490\">Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.<\/p>\r\n&nbsp;\r\n\r\n[reveal-answer q=\"227675\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"227675\"]\r\n\r\n&nbsp;\r\n\r\nThe slope [latex]m=\\dfrac{\\Delta y}{\\Delta x}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"fs-id1170573368474\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573368474\"]\r\n\r\n&nbsp;\r\n<p id=\"fs-id1170573368474\">[latex]m=\\dfrac{1}{2}[\/latex]<\/p>\r\nThe point-slope form is [latex]y-4=\\frac{1}{2}(x-1)[\/latex].\r\n<p id=\"fs-id1170573287470\">The slope-intercept form is [latex]y=\\frac{1}{2}x+\\frac{7}{2}[\/latex].<\/p>\r\n&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div id=\"fs-id1170573575076\" class=\"textbox exercises\">\r\n<h3>Example: A Linear Distance Function<\/h3>\r\n<p id=\"fs-id1170573413936\">Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.<\/p>\r\n\r\n<ol id=\"fs-id1170573534412\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>Describe the distance [latex]D[\/latex] (in miles) Jessica runs as a linear function of her run time [latex]t[\/latex] (in minutes).<\/li>\r\n \t<li>Sketch a graph of [latex]D[\/latex].<\/li>\r\n \t<li>Interpret the meaning of the slope.<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1170573413689\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1170573413689\"]\r\n<ol id=\"fs-id1170573413689\" style=\"list-style-type: lower-alpha;\">\r\n \t<li>At time [latex]t=0[\/latex], Jessica is at her house, so [latex]D(0)=0[\/latex]. At time [latex]t=78[\/latex] minutes, Jessica has finished running 9 mi, so [latex]D(78)=9[\/latex]. The slope of the linear function is\r\n<div id=\"fs-id1170573404013\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{9-0}{78-0}=\\dfrac{3}{26}[\/latex]<\/div>\r\nThe [latex]y[\/latex]-intercept is [latex](0,0)[\/latex], so the equation for this linear function is\r\n<div id=\"fs-id1170573573845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]D(t)=\\frac{3}{26}t[\/latex]<\/div><\/li>\r\n \t<li>To graph [latex]D[\/latex], use the fact that the graph passes through the origin and has slope [latex]m=\\frac{3}{26}[\/latex].[caption id=\"\" align=\"aligncenter\" width=\"731\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202230\/CNX_Calc_Figure_01_02_003.jpg\" alt=\"An image of a graph. The y axis is labeled \u201cy, distance in miles\u201d. The x axis is labeled \u201ct, time in minutes\u201d. The graph is of the function \u201cD(t) = 3t\/26\u201d, which is an increasing straight line that starts at the origin. The function ends at the plotted point (78, 9).\" width=\"731\" height=\"190\" \/> Figure 4. Graph of function [latex]D[\/latex] \u2013\u00a0Jessica's distance from home in miles vs. minutes spent running.[\/caption]<\/li>\r\n \t<li>The slope [latex]m=\\dfrac{3}{26} \\approx 0.115[\/latex] describes the distance (in miles) Jessica runs per minute, or her average velocity.<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n\r\n[caption]Watch the following video to see the worked solution to Example: A Linear Distance Function[\/caption]\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/iiBBHtVIk9U?controls=0&amp;start=166&amp;end=355&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.2BasicClassesOfFunctions166to355_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"1.2 Basic Classes of Functions\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]196593[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Calculate the slope of a linear function and interpret its meaning<\/li>\n<\/ul>\n<\/div>\n<p>The easiest type of function to consider is a <strong>linear function<\/strong>. Linear functions have the form [latex]f(x)=ax+b[\/latex], where [latex]a[\/latex] and [latex]b[\/latex] are constants. In Figure 1, we see examples of linear functions when [latex]a[\/latex] is positive, negative, and zero. Note that if [latex]a>0[\/latex], the graph of the line rises as [latex]x[\/latex] increases. In other words, [latex]f(x)=ax+b[\/latex] is increasing on [latex](\u2212\\infty, \\infty)[\/latex]. If [latex]a<0[\/latex], the graph of the line falls as [latex]x[\/latex] increases. In this case, [latex]f(x)=ax+b[\/latex] is decreasing on [latex](\u2212\\infty, \\infty)[\/latex]. If [latex]a=0[\/latex], the line is horizontal.\n\n\n\n<div style=\"width: 335px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202222\/CNX_Calc_Figure_01_02_001.jpg\" alt=\"An image of a graph. The y axis runs from -2 to 5 and the x axis runs from -2 to 5. The graph is of the 3 functions. The first function is \u201cf(x) = 3x + 1\u201d, which is an increasing straight line with an x intercept at ((-1\/3), 0) and a y intercept at (0, 1). The second function is \u201cg(x) = 2\u201d, which is a horizontal line with a y intercept at (0, 2) and no x intercept. The third function is \u201ch(x) = (-1\/2)x\u201d, which is a decreasing straight line with an x intercept and y intercept both at the origin. The function f(x) is increasing at a higher rate than the function h(x) is decreasing.\" width=\"325\" height=\"312\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 1. These linear functions are increasing or decreasing on [latex](-\\infty, \\infty)[\/latex] and one function is a horizontal line.<\/p>\n<\/div>\n<h2>Slope<\/h2>\n<p id=\"fs-id1170573396917\">As suggested by Figure 1, the graph of any linear function is a line. One of the distinguishing features of a line is its slope. The <strong>slope<\/strong> is the change in [latex]y[\/latex] for each unit change in [latex]x[\/latex]. The slope measures both the steepness and the direction of a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we need to determine the ratio of the change in [latex]y[\/latex] versus the change in [latex]x[\/latex]. To do so, we choose any two points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line and calculate [latex]\\dfrac{y_2-y_1}{x_2-x_1}[\/latex]. In Figure 2, we see this ratio is independent of the points chosen.<\/p>\n<div style=\"width: 475px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202225\/CNX_Calc_Figure_01_02_021.jpg\" alt=\"An image of a graph. The y axis runs from -1 to 10 and the x axis runs from -1 to 6. The graph is of a function that is an increasing straight line. There are four points labeled on the function at (1, 1), (2, 3), (3, 5), and (5, 9). There is a dotted horizontal line from the labeled function point (1, 1) to the unlabeled point (3, 1) which is not on the function, and then dotted vertical line from the unlabeled point (3, 1), which is not on the function, to the labeled function point (3, 5). These two dotted have the label \u201c(y2 - y1)\/(x2 - x1) = (5 -1)\/(3 - 1) = 2\u201d. There is a dotted horizontal line from the labeled function point (2, 3) to the unlabeled point (5, 3) which is not on the function, and then dotted vertical line from the unlabeled point (5, 3), which is not on the function, to the labeled function point (5, 9). These two dotted have the label \u201c(y2 - y1)\/(x2 - x1) = (9 -3)\/(5 - 2) = 2\u201d.\" width=\"465\" height=\"459\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 2. or any linear function, the slope [latex](y_2-y_1)\/(x_2-x_1)[\/latex] is independent of the choice of points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line.<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p>Consider line [latex]L[\/latex] passing through points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex]. Let [latex]\\Delta y=y_2-y_1[\/latex] and [latex]\\Delta x=x_2-x_1[\/latex] denote the changes in [latex]y[\/latex] and [latex]x[\/latex], respectively. The <strong>slope<\/strong> of the line is<\/p>\n<div id=\"fs-id1170573414443\" class=\"equation\" style=\"text-align: center;\">[latex]m=\\dfrac{y_2-y_1}{x_2-x_1}=\\dfrac{\\Delta y}{\\Delta x}[\/latex]<\/div>\n<\/div>\n<h2>Linear Function Formulas<\/h2>\n<p id=\"fs-id1170573356648\"><span style=\"font-size: 1rem; text-align: initial;\">We now examine the relationship between slope and the formula for a linear function. Consider the linear function given by the formula [latex]f(x)=ax+b[\/latex]. As discussed earlier, we know the graph of a linear function is given by a line. We can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating [latex](y_2-y_1)\/(x_2-x_1)[\/latex] for any points [latex](x_1,y_1)[\/latex] and [latex](x_2,y_2)[\/latex] on the line. Evaluating the function [latex]f[\/latex] at [latex]x=0[\/latex], we see that [latex](0,b)[\/latex] is a point on this line. Evaluating this function at [latex]x=1[\/latex], we see that [latex](1,a+b)[\/latex] is also a point on this line. Therefore, the slope of this line is<\/span><\/p>\n<div id=\"fs-id1170573263457\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{(a+b)-b}{1-0}=a[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573409038\">We have shown that the coefficient [latex]a[\/latex] is the slope of the line. We can conclude that the formula [latex]f(x)=ax+b[\/latex] describes a line with slope [latex]a[\/latex]. Furthermore, because this line intersects the [latex]y[\/latex]-axis at the point [latex](0,b)[\/latex], we see that the [latex]y[\/latex]-intercept for this linear function is [latex](0,b)[\/latex]. We conclude that the formula [latex]f(x)=ax+b[\/latex] tells us the slope, [latex]a[\/latex], and the [latex]y[\/latex]-intercept, [latex](0,b)[\/latex], for this line. Since we often use the symbol [latex]m[\/latex] to denote the slope of a line, we can write<\/p>\n<div id=\"fs-id1170573262665\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=mx+b[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573257414\">to denote the <strong>slope-intercept form<\/strong> of a linear function.<\/p>\n<p id=\"fs-id1170573381114\">Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function passes through the point [latex](x_1,y_1)[\/latex] and the slope of the line is [latex]m[\/latex]. Since any other point [latex](x,f(x))[\/latex] on the graph of [latex]f[\/latex] must satisfy the equation<\/p>\n<div id=\"fs-id1170573299395\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{f(x)-y_1}{x-x_1}[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573569154\">this linear function can be expressed by writing<\/p>\n<div id=\"fs-id1170573583649\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)-y_1=m(x-x_1)[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573309782\">We call this equation the<strong> point-slope equation<\/strong> for that linear function.<\/p>\n<p id=\"fs-id1170573309900\">Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the slope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be expressed in either of these forms. Instead, a vertical line is described by the equation [latex]x=k[\/latex] for some constant [latex]k[\/latex]. Since neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation<\/p>\n<div id=\"fs-id1170573418048\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]ax+by=c[\/latex],<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573425137\">where [latex]a,b[\/latex] are both not zero, to denote the <strong>standard form of a line.<\/strong><\/p>\n<div class=\"textbox shaded\">\n<h3 style=\"text-align: center;\">Definition<\/h3>\n<hr \/>\n<p id=\"fs-id1170573328429\">Consider a line passing through the point [latex](x_1,y_1)[\/latex] with slope [latex]m[\/latex]. The equation<\/p>\n<div id=\"fs-id1170573382468\" class=\"equation\" style=\"text-align: center;\">[latex]y-y_1=m(x-x_1)[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170573397934\">is the <strong>point-slope equation<\/strong> for that line.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573425678\">Consider a line with slope [latex]m[\/latex] and [latex]y[\/latex]-intercept [latex](0,b)[\/latex]. The equation<\/p>\n<div id=\"fs-id1170573240220\" class=\"equation\" style=\"text-align: center;\">[latex]y=mx+b[\/latex]<\/div>\n<div><\/div>\n<p id=\"fs-id1170573336985\">is an equation for that line in <strong>slope-intercept form<\/strong>.<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573334038\">The <strong>standard form of a line<\/strong> is given by the equation<\/p>\n<div id=\"fs-id1170573399196\" class=\"equation\" style=\"text-align: center;\">[latex]ax+by=c[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573392539\">where [latex]a[\/latex] and [latex]b[\/latex] are both not zero. This form is more general because it allows for a vertical line, [latex]x=k[\/latex].<\/p>\n<\/div>\n<div id=\"fs-id1170573262524\" class=\"textbox exercises\">\n<h3>Example: Finding the Slope and Equations of Lines<\/h3>\n<p id=\"fs-id1170573402450\">Consider the line passing through the points [latex](11,-4)[\/latex] and [latex](-4,5)[\/latex], as shown in Figure 3.<\/p>\n<div style=\"width: 704px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202228\/CNX_Calc_Figure_01_02_002.jpg\" alt=\"An image of a graph. The x axis runs from -5 to 12 and the y axis runs from -5 to 6. The graph is of the function that is a decreasing straight line. The function has two points plotted, at (-4, 5) and (11, 4).\" width=\"694\" height=\"459\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Finding the equation of a linear function with a graph that is a line between two given points.<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<ol id=\"fs-id1170573248716\" style=\"list-style-type: lower-alpha;\">\n<li>Find the slope of the line.<\/li>\n<li>Find an equation for this linear function in point-slope form.<\/li>\n<li>Find an equation for this linear function in slope-intercept form.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573411739\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573411739\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170573411739\" style=\"list-style-type: lower-alpha;\">\n<li>The slope of the line is\n<div id=\"fs-id1170573334269\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{y_2-y_1}{x_2-x_1}=\\dfrac{5-(-4)}{-4-11}=-\\dfrac{9}{15}=-\\dfrac{3}{5}[\/latex].<\/div>\n<\/li>\n<li>To find an equation for the linear function in point-slope form, use the slope [latex]m=-\\frac{3}{5}[\/latex] and choose any point on the line. If we choose the point [latex](11,-4)[\/latex], we get the equation\n<div id=\"fs-id1170573268080\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)+4=-\\frac{3}{5}(x-11)[\/latex].<\/div>\n<\/li>\n<li>To find an equation for the linear function in slope-intercept form, solve the equation in part (b) for [latex]f(x)[\/latex]. When we do this, we get the equation\n<div id=\"fs-id1170573274197\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=-\\frac{3}{5}x+\\frac{13}{5}[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Finding the Slope and Equations of Lines<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/iiBBHtVIk9U?controls=0&amp;start=61&amp;end=163&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266833\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266833\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.2BasicClassesOfFunctions61to163_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.2 Basic Classes of Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1170573230955\" class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1170573210343\">Consider the line passing through points [latex](-3,2)[\/latex] and [latex](1,4)[\/latex]. Find the slope of the line.<\/p>\n<p id=\"fs-id1170573570490\">Find an equation of that line in point-slope form. Find an equation of that line in slope-intercept form.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q227675\">Hint<\/span><\/p>\n<div id=\"q227675\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p>The slope [latex]m=\\dfrac{\\Delta y}{\\Delta x}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573368474\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573368474\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<p id=\"fs-id1170573368474\">[latex]m=\\dfrac{1}{2}[\/latex]<\/p>\n<p>The point-slope form is [latex]y-4=\\frac{1}{2}(x-1)[\/latex].<\/p>\n<p id=\"fs-id1170573287470\">The slope-intercept form is [latex]y=\\frac{1}{2}x+\\frac{7}{2}[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div id=\"fs-id1170573575076\" class=\"textbox exercises\">\n<h3>Example: A Linear Distance Function<\/h3>\n<p id=\"fs-id1170573413936\">Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.<\/p>\n<ol id=\"fs-id1170573534412\" style=\"list-style-type: lower-alpha;\">\n<li>Describe the distance [latex]D[\/latex] (in miles) Jessica runs as a linear function of her run time [latex]t[\/latex] (in minutes).<\/li>\n<li>Sketch a graph of [latex]D[\/latex].<\/li>\n<li>Interpret the meaning of the slope.<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1170573413689\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1170573413689\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1170573413689\" style=\"list-style-type: lower-alpha;\">\n<li>At time [latex]t=0[\/latex], Jessica is at her house, so [latex]D(0)=0[\/latex]. At time [latex]t=78[\/latex] minutes, Jessica has finished running 9 mi, so [latex]D(78)=9[\/latex]. The slope of the linear function is\n<div id=\"fs-id1170573404013\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]m=\\dfrac{9-0}{78-0}=\\dfrac{3}{26}[\/latex]<\/div>\n<p>The [latex]y[\/latex]-intercept is [latex](0,0)[\/latex], so the equation for this linear function is<\/p>\n<div id=\"fs-id1170573573845\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]D(t)=\\frac{3}{26}t[\/latex]<\/div>\n<\/li>\n<li>To graph [latex]D[\/latex], use the fact that the graph passes through the origin and has slope [latex]m=\\frac{3}{26}[\/latex].\n<div style=\"width: 741px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11202230\/CNX_Calc_Figure_01_02_003.jpg\" alt=\"An image of a graph. The y axis is labeled \u201cy, distance in miles\u201d. The x axis is labeled \u201ct, time in minutes\u201d. The graph is of the function \u201cD(t) = 3t\/26\u201d, which is an increasing straight line that starts at the origin. The function ends at the plotted point (78, 9).\" width=\"731\" height=\"190\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. Graph of function [latex]D[\/latex] \u2013\u00a0Jessica&#8217;s distance from home in miles vs. minutes spent running.<\/p>\n<\/div>\n<\/li>\n<li>The slope [latex]m=\\dfrac{3}{26} \\approx 0.115[\/latex] describes the distance (in miles) Jessica runs per minute, or her average velocity.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: A Linear Distance Function<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/iiBBHtVIk9U?controls=0&amp;start=166&amp;end=355&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3-us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/1.2BasicClassesOfFunctions166to355_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;1.2 Basic Classes of Functions&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm196593\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=196593&theme=oea&iframe_resize_id=ohm196593&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-113\">\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>1.2 Basic Classes of Functions. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/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":17533,"menu_order":8,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"1.2 Basic Classes of Functions\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-113","chapter","type-chapter","status-publish","hentry"],"part":21,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/113","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":43,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/113\/revisions"}],"predecessor-version":[{"id":4742,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/113\/revisions\/4742"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/21"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/113\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=113"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=113"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=113"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=113"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}