{"id":939,"date":"2022-03-02T23:06:20","date_gmt":"2022-03-02T23:06:20","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/?post_type=chapter&#038;p=939"},"modified":"2026-01-17T01:55:57","modified_gmt":"2026-01-17T01:55:57","slug":"2-3-algebraic-analysis-of-linear-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/chapter\/2-3-algebraic-analysis-of-linear-functions\/","title":{"raw":"2.3: Algebraic Analysis of Linear Functions","rendered":"2.3: Algebraic Analysis of Linear Functions"},"content":{"raw":"<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>\r\n<div class=\"textbox learning-objectives\">\r\n<h1>Learning Objectives<\/h1>\r\nDetermine slope of a line using the slope formula.\r\n\r\nDetermine the linear function given:\r\n<ul>\r\n \t<li>a slope and the [latex]y[\/latex]-intercept<\/li>\r\n \t<li>a slope and a point<\/li>\r\n \t<li>two points<\/li>\r\n \t<li>a graph<\/li>\r\n \t<li>a point and a parallel linear function<\/li>\r\n \t<li>a point and a perpendicular linear function<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Slope<\/h2>\r\nThe <em><strong>slope<\/strong><\/em> of a line, [latex]m[\/latex], is a measure of the steepness of the line, and since the steepness is the same all along the line, the slope is constant.\u00a0 The slope of a line is defined as the ratio of [latex]\\frac{\\text{rise}}{\\text{run}}[\/latex] where rise is the change in [latex]y[\/latex]-values, [latex]\\Delta y[\/latex], and run is the corresponding change in [latex]x[\/latex]-values, [latex]\\Delta x[\/latex]. If [latex]y=f(x)[\/latex], then slope = [latex]\\frac{\\Delta y}{\\Delta x}=\\frac{\\text{change in function values}}{\\text{change in }x\\text{-values}}[\/latex].\r\n\r\nConsider the line in figure 1. It passes through the points (\u20134, \u20134) and (10, 3). The run is the distance between [latex]x=-4[\/latex] and [latex]x=10[\/latex]; a total of 14 units. Rather than counting units on the graph, we can subtract the two [latex]x[\/latex]-coordinates: 10 \u2013 (\u20134) = 14. The rise is the distance between [latex]y=-4[\/latex] and [latex]y=3[\/latex]; a total of 7 units. We can find this number by subtracting the [latex]y[\/latex]-coordinates: 3 \u2013 (\u20134) = 7. Consequently, the slope of the line is\u00a0 [latex]m=\\frac{7}{14}=\\frac{1}{2}[\/latex].\r\n\r\n[caption id=\"attachment_1260\" align=\"aligncenter\" width=\"603\"]<img class=\"wp-image-1260\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/02\/31203020\/Slope-between-2-points-300x190.png\" alt=\"Line with two fixed points, (-4,-4) and (10,3) showing the horizontal change of 14 and the vertical change of 7.\" width=\"603\" height=\"382\" \/> Figure 1. Slope between 2 points.[\/caption]\r\n\r\nLet's now consider two arbitrary points on a line. Suppose the first point is designated [latex](x_1, y_1)[\/latex] and the second point is [latex](x_2,y_2)[\/latex]. The run is the difference in [latex]x[\/latex]-values, [latex]x_2-x_1[\/latex], and the rise is the difference in [latex]y[\/latex]-values, [latex]y_2-y_1[\/latex]. So the slope of the line between these two points is [latex]m=\\frac{y_2-y_1}{x_2-x_1}[\/latex] (see figure 2). It does not matter which point we designate as the first point or the second point. We will always get the same slope as long as we subtract from one point to the other in the same order.\r\n\r\n[caption id=\"attachment_1292\" align=\"aligncenter\" width=\"386\"]<img class=\"wp-image-1292\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/31235741\/slope--300x300.png\" alt=\"Line with two arbitrary points, (x_2,y_2) and (x_1,y_1) showing the horizontal change x2-x1and the vertical change y2-y1.\" width=\"386\" height=\"386\" \/> Figure 2. Determining the slope between 2 points on a line.[\/caption]\r\n\r\nThe notation\u00a0[latex]\\Delta[\/latex] is often used in mathematics to represent \"change in\", so \"change in y\" can be denoted by\u00a0[latex]\\Delta y[\/latex], and \"change in [latex]x[\/latex]\" can be denoted\u00a0[latex]\\Delta x[\/latex].\u00a0 So, slope = [latex]m=\\frac{\\Delta y}{\\Delta x}[\/latex].\r\n<div class=\"textbox\">\r\n<h3>SLOPE<\/h3>\r\n<p style=\"text-align: center;\">The slope between two points on a line [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex] is found by the equation\u00a0 [latex]m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{\\Delta y}{\\Delta x}[\/latex].<\/p>\r\n<p style=\"text-align: center;\">If [latex]y=f(x)[\/latex] is a linear function, the rate of change of the function = the slope of the line representing the function on a graph.<\/p>\r\n<p style=\"text-align: center;\"><\/p>\r\n\r\n<\/div>\r\n<div id=\"post-91\" class=\"standard post-91 chapter type-chapter status-publish hentry\">\r\n<div class=\"entry-content\">\r\n<div class=\"textbox examples\">\r\n<h3>Example 1<\/h3>\r\nIf [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function and [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]\u00a0are points on the graph of the function, find the slope of the graphed line. Is the graph of this function increasing, decreasing, or neither?\r\n<h4>Solution<\/h4>\r\nThe coordinate pairs are [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]. To find the slope we find the rate of change of the function between the two points.\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{1-\\left(-2\\right)}{8 - 3}=\\dfrac{3}{5}[\/latex]<\/p>\r\nThe function is increasing because [latex]m&gt;0[\/latex].\r\n\r\n&nbsp;\r\n\r\nAlternatively, changing the order in which we use the points:\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{-2-1}{3-8}=\\dfrac{-3}{-5}=\\dfrac{3}{5}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nNotice that we get the same solution since the order the points are referenced does not matter.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nThe following video shows examples of how to find the slope of a line passing through two points and then determines whether the line is increasing, decreasing or neither.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/in3NTcx11I8?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.3-Video-1.odt\">Transcript 2.3 Video 1<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 1<\/h3>\r\nThe graph of a linear function passes through the points (\u20132, 4) and (3, 1). FInd the slope of the graphed line and determine whether it is increasing, decreasing, or neither.\r\n\r\n[reveal-answer q=\"hjm578\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm578\"]\r\n\r\n[latex]m=-\\frac{3}{5}[\/latex]\r\n\r\nThe line is decreasing because [latex]m&lt;0[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nThe next video shows an example where the increase in cost for producing solar panels is determined by two data points.\r\n\r\n<iframe src=\"https:\/\/www.youtube.com\/embed\/4RbniDgEGE4?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe>\r\n\r\n<a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.3-Video-2.odt\">Transcript 2.3 Video 2<\/a>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 2<\/h3>\r\nA gym charges a membership fee and a fixed monthly amount.\u00a0 For 3 months the total cost is $110. For 8 months, the total cost is $210. Calculate the monthly amount.\r\n\r\n[reveal-answer q=\"hjm810\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm810\"]\r\n\r\n(3, $110) and (8, $210) are data points. Since the monthly fee is fixed, the total cost over time is linear, and the monthly fee = the change in the total cost per month = the slope.\r\n\r\nMonthly fee = $20.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2>Determining Linear Functions<\/h2>\r\nA linear function may be described by the slope-intercept form [latex]f(x) = mx + b[\/latex], where [latex]m[\/latex] is the rate of change of the function or the slope of the graphed function, and [latex]b[\/latex] is the initial value or\u00a0[latex]y[\/latex]-coordinate of the [latex]y[\/latex]-intercept. There are several different situations and conditions we may be given to determine the equation of the linear function.\r\n<h3>Given a Slope and the y-intercept<\/h3>\r\nIf we are given the slope and the [latex]y[\/latex]-intercept, we are given the value of [latex]m[\/latex] and [latex]b[\/latex]. This allows us to jump right to slope-intercept form to find the function. For example, if we are told that the slope of a line is 2 and the [latex]y[\/latex]-intercept is (0, 1), then using slope-intercept form, [latex]f(x)=x+b[\/latex], the linear function will be [latex]f(x) = 2x + 1[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 2<\/h3>\r\nFind the function whose graph is a line with:\r\n<ol>\r\n \t<li>slope = 4 and [latex]y[\/latex]-intercept = (0, 8).<\/li>\r\n \t<li>slope = [latex]\\frac{4}{3}[\/latex] and [latex]y[\/latex]-intercept = (0, \u20135).<\/li>\r\n \t<li>slope = 0 and [latex]y[\/latex]-intercept = (0, 3).<\/li>\r\n \t<li>slope = [latex]-\\frac{5}{6}[\/latex] and [latex]y[\/latex]-intercept = (0, 0).<\/li>\r\n<\/ol>\r\n<h4>Solution<\/h4>\r\nWe are given\u00a0[latex]m[\/latex] and\u00a0[latex]b[\/latex] so can use slope-intercept form [latex]f(x)=mx+b[\/latex] to find the function.\r\n<ol>\r\n \t<li>[latex]m=4[\/latex] and [latex]b=8[\/latex]. Therefore, the function is [latex]f(x)=4x+8[\/latex].<\/li>\r\n \t<li>[latex]m=\\frac{4}{3}[\/latex] and [latex]b=-5[\/latex]. Therefore, the function is [latex]f(x)=\\frac{4}{3}x-5[\/latex].<\/li>\r\n \t<li>[latex]m=0[\/latex] and [latex]b=3[\/latex]. Therefore, the function is [latex]f(x)=0x+3[\/latex], which simplifies to [latex]f(x)=3[\/latex].<\/li>\r\n \t<li>[latex]m=-\\frac{5}{6}[\/latex] and [latex]b=0[\/latex]. Therefore, the function is [latex]f(x)=-\\frac{5}{6}x+0[\/latex], which simplifies to[latex]f(x)=-\\frac{5}{6}x[\/latex].<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 3<\/h3>\r\nFind the function whose graph is a line with:\r\n<ol>\r\n \t<li>slope = \u20132 and [latex]y[\/latex]-intercept = (0, 3).<\/li>\r\n \t<li>slope = [latex]\\frac{2}{3}[\/latex] and [latex]y[\/latex]-intercept = (0, 0).<\/li>\r\n \t<li>slope = 0 and [latex]y[\/latex]-intercept = (0, -7).<\/li>\r\n \t<li>slope = [latex]\\frac{5}{4}[\/latex] and [latex]y[\/latex]-intercept = [latex]-\\frac{7}{2}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm916\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm916\"]\r\n<ol>\r\n \t<li>[latex]f(x)=-2x+3[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\frac{2}{3}x[\/latex]<\/li>\r\n \t<li>[latex]f(x)=-7[\/latex]<\/li>\r\n \t<li>[latex]f(x)=\\frac{5}{4}x-\\frac{7}{2}[\/latex]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Given a Slope and a Point<\/h3>\r\nIf we are given a slope and a point, then we need to find the [latex]y[\/latex]-intercept in order to determine the linear function [latex]f(x)=mx+b[\/latex]. Since the given point [latex](x_1,y_1)[\/latex] lies on the line, it is a solution of the linear function. We say that [latex](x_1,y_1)[\/latex] satisfies the function. We can use this piece of information to solve for [latex]b[\/latex] and determine the function.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">For example, suppose we are told that the slope of a line is [latex]\\dfrac{1}{3}[\/latex]\u00a0and the point (6, \u20131) lies on the line. We know that the function can be found using [latex]f(x)=mx+b[\/latex]. Because [latex]m=\\frac{1}{3}[\/latex] the function must be:<\/span>\r\n<p style=\"text-align: center;\">[latex]f(x) = \\frac{1}{3}x + b[\/latex]<\/p>\r\nWe now need to find the value of [latex]b[\/latex] and we can use the given point (6, \u20131) to do this. Because (6, \u20131) satisfies the equation, [latex]f(6)=-1[\/latex].\r\n\r\nThis means that:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned} -1 &amp;= \\frac{1}{3} \\cdot 6 + b\\\\-1 &amp;= 2 + b\\\\-3 &amp;= b \\end{aligned}\\end{equation}[\/latex]<\/p>\r\nNow that we know [latex]m=\\frac{1}{3}[\/latex] and [latex]b=-3[\/latex], we can write the linear function as [latex] f(x) = \\frac{1}{3}x - 3[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 3<\/h3>\r\nFind the function whose graph is a line with slope [latex]\\frac{3}{4}[\/latex] that passes through the point (\u20133, 1).\r\n\r\nSolution\r\n\r\nFirst we are told that [latex]m=\\frac{3}{4}[\/latex], so the function is [latex]f(x)=\\large\\frac{3}{4}x+b[\/latex].\r\n\r\nTo find the value of [latex]b[\/latex], we use the given point (\u20133, 1) which tells us that [latex]f(-3)=1[\/latex].\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=\\frac{3}{4}x+b\\\\1&amp;=\\frac{3}{4}\\cdot (-3) + b\\\\1&amp;=-\\frac{9}{4}+b\\\\1+\\frac{9}{4}&amp;=b\\\\ \\frac{4}{4}+\\frac{9}{4}&amp;=b\\\\ \\frac{13}{4}&amp;=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n&nbsp;\r\n\r\nNow we know [latex]m[\/latex] and [latex]b[\/latex], the function is [latex]f(x)=\\large\\frac{3}{4}x+\\frac{13}{4}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 4<\/h3>\r\nFind the function whose graph is a line with slope \u20133 that passes through the point (4, \u20135).\r\n\r\n[reveal-answer q=\"hjm411\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm411\"]\r\n\r\n[latex]f(x)=-3x+7[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Given Two Points<\/h3>\r\nIf we are given two points that lie on the graph of the function, then we have to find both the slope and the [latex]y[\/latex]-intercept in order to determine the linear function. We can find the slope by finding the rate of change\u00a0 between the two points:\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{y_2 - y_1}{x_2 - x_1}[\/latex]<\/p>\r\nAfter we find the slope, then we can use either one of the two points on the line to find the value of [latex]b[\/latex] because both of the two points satisfy the linear function.\r\n\r\nFor example, given the two points (\u20131, 3) and (4, \u20137) that satisfy the function , the slope of the line representing the linear function will be:\r\n<p style=\"text-align: center;\">Slope = [latex]m=\\dfrac{-7-3}{4-(-1)} =\\dfrac{-10}{5} = -2 [\/latex]<\/p>\r\nSince both point (\u20131, 3) and (4, \u20137) satisfy the function, we may use either of the two points. Let's use the point (4, \u20137):\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=-2x+b\\\\-7&amp;=-2\\cdot 4+b\\\\-7&amp;=-8+b\\\\1&amp;=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nWe now know that [latex]m=-2[\/latex] and [latex]b=1[\/latex] so the linear function is\u00a0[latex] f(x) = -2x + 1[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 4<\/h3>\r\nFind the function whose graph is a line that passes through the points (\u20133, 5) and (4, \u20132).\r\n<h4>Solution<\/h4>\r\nFIrst we must find the slope of the line:\u00a0 [latex]m=\\frac{\\Delta y}{\\Delta x}=\\frac{-2-5}{4-(-3)}=\\frac{-7}{7}=-1[\/latex]\r\n\r\nThis means that the function is: [latex]f(x)=-x+b[\/latex]\r\n\r\nTo find the value of [latex]b[\/latex], we can use either point on the line that satisfies the function.\r\n\r\nLet's use (4, \u20132), which means [latex]f(4)=-2[\/latex]:\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=-x+b\\\\-2&amp;=-(4)+b\\\\2&amp;=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nNow we know that [latex]m=-1[\/latex] and [latex]b=2[\/latex] so we can write the function as [latex]f(x)=-x+2[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 5<\/h3>\r\nFind the function whose graph is a line that passes through the points (0, 5) and (\u20133, 8).\r\n\r\n[reveal-answer q=\"hjm769\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm769\"]\r\n\r\n[latex]f(x)=-x+5[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Given a Graph<\/h3>\r\nIf we are given a line on the coordinate plane, we may use any of the methods discussed above to find the function of the line by identifying the slope and the [latex]y[\/latex]-intercept of the line, or the slope and a point on the line, or two points on the line.\r\n<div class=\"textbox examples\">\r\n<h3>Example 5<\/h3>\r\nFind the function represented by the graph:\r\n<p style=\"text-align: center;\"><img class=\"aligncenter wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01050649\/y2x-5-227x300.png\" alt=\"A line with y-intercept (0,-5) passing through (5,5).\" width=\"280\" height=\"370\" \/><\/p>\r\n\r\n<h4>Solution<\/h4>\r\nThe [latex]y[\/latex]-intercept is (0, \u20135) so we know that [latex]b=-5[\/latex] in the function [latex]f(x)=mx+b[\/latex].\r\n\r\nTo find the slope [latex]m[\/latex], we use any other point on the line. Let's use (5, 5):\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{5-(-5)}{5-0}=\\dfrac{10}{5}=2[\/latex]<\/p>\r\nTherefore, the function is [latex]f(x)=2x-5[\/latex].\r\n\r\n<\/div>\r\n<div class=\"textbox examples\">\r\n<h3>Example 6<\/h3>\r\nFind the function represented by the graph:\r\n\r\n<img class=\"aligncenter wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01051608\/y-12x32-300x223.png\" alt=\"A line with indeterminant y-intercept passing through the points (3,0) and (-5,4).\" width=\"412\" height=\"306\" \/>\r\n<h4>Solution<\/h4>\r\nThe [latex]y[\/latex]-intercept is fractional between [latex]y=[\/latex] 1 and 2 and therefore not obvious, so we will use two points on the graph to find the function: (3, 0) and (\u20135, 4).\r\n\r\nTo find the slope:\r\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{4-0}{-5-3}=\\dfrac{4}{-8}=-\\dfrac{1}{2}[\/latex].<\/p>\r\nThe function can be written: [latex]f(x)=-\\frac{1}{2}x+b[\/latex]\r\n\r\nTo find [latex]b[\/latex] we can use any point on the line. Let's use (3, 0):\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=-\\frac{1}{2}x+b\\\\0&amp;=-\\frac{1}{2}\\cdot 3+b\\\\0&amp;=-\\frac{3}{2}+b\\\\\\frac{3}{2}=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nNow we know that [latex]m=-\\frac{1}{2}[\/latex] and [latex]b=\\frac{3}{2}[\/latex], so the function is [latex]f(x)=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 6<\/h3>\r\nFind the function represented by each graph:\r\n\r\n[caption id=\"attachment_1302\" align=\"alignleft\" width=\"260\"]<img class=\"wp-image-1302\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01053014\/y4x-7-150x300.png\" alt=\"Line with y-intercept (0,-7) with slope m=4.\" width=\"260\" height=\"520\" \/> Graph 1[\/caption]\r\n\r\n&nbsp;\r\n\r\n[caption id=\"attachment_1303\" align=\"alignright\" width=\"481\"]<img class=\"wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01053119\/2x3y5-300x242.png\" alt=\"A line with indeterminant y-intercept passing through (-2,3) and (1,1).\" width=\"481\" height=\"388\" \/> Graph 2[\/caption]\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n&nbsp;\r\n\r\n[reveal-answer q=\"hjm049\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm049\"]\r\n\r\nGraph 1: [latex]f(x)=4x-7[\/latex]\r\n\r\nGraph 2: [latex]f(x)=-\\frac{2}{3}x+\\frac{5}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf we are given a horizontal line, its slope is 0 because the rise is 0 (i.e., [latex]\\dfrac{0}{run} = 0[\/latex]). Sometimes we can identify the [latex]y[\/latex]-intercept on the graph, and the function is [latex]f(x)=0x +b[\/latex] or\u00a0[latex]f(x)=b[\/latex]. For example, for the function of the red horizontal line in figure 3, the [latex]y[\/latex]-intercept is (0, 3) and therefore the function is [latex]f(x)=3[\/latex].\r\n\r\n[caption id=\"attachment_1138\" align=\"aligncenter\" width=\"300\"]<img class=\"wp-image-1138 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-300x300.png\" alt=\"A horizontal line with y-intercept (0,3), and a vertical line with x-intercept (-2,0).\" width=\"300\" height=\"300\" \/> Figure 3. A horizontal line and a vertical line.[\/caption]\r\n\r\nIf we are given a vertical line, like the blue line in figure 3, it is not a function because a vertical line is a one-to-many mapping (i.e., one input with many different outputs). Therefore, we are not able to use the function form [latex]f(x)=mx+b[\/latex] to express a vertical line. Also, the\u00a0slope of a vertical line is undefined because the run is 0, and a number divided by 0 is undefined (i.e.,\u00a0[latex]\\dfrac{rise}{0}[\/latex] is undefined). The equation for a vertical line is [latex]x = a[\/latex] where [latex]a[\/latex] is the\u00a0[latex]x[\/latex]-coordinate of the [latex]x[\/latex]-intercept. Indeed, for all of the points on a vertical line, the [latex]x[\/latex]-coordinates are identical and the [latex]y[\/latex]-coordinates are any real number on the vertical line. For example, the equation of the blue vertical line in figure 3 is\u00a0[latex]x = -2[\/latex].\r\n<h3>Given a Point and a Parallel Linear Function<\/h3>\r\nFinding the equation of a function whose graph is parallel to another line, still requires finding the slope and [latex]y[\/latex]-intercept. Fortunately, the slope can be found from the parallel line since parallel lines have the same slope. Knowing the slope, we can then use another point on the line to determine the [latex]y[\/latex]-intercept and consequently the function.\r\n\r\nFor example, suppose we are asked to find the linear function that passes through the point (8, \u20132), whose graph runs parallel to the graph of the function [latex]g(x)=\\frac{3}{4}x+2[\/latex]. The slope of the line that represents [latex]g(x)[\/latex] is [latex]\\frac{3}{4}[\/latex]. This means that the line representing [latex]f(x)[\/latex] also has a slope of [latex]\\frac{3}{4}[\/latex] since they are parallel. this means that [latex]f(x)=\\frac{3}{4}x + b[\/latex]. Now, we can find the value of [latex]b[\/latex] by using the given point (8, \u20132):\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=\\frac{3}{4}x+b\\\\-2&amp;=\\frac{3}{4}\\cdot 8+b\\\\-2&amp;=6+b\\\\-8=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\n<p style=\"text-align: left;\">Now we know that [latex]m=\\frac{3}{4}[\/latex] and [latex]b=-8[\/latex].\u00a0Therefore, the function is\u00a0[latex]f(x) = \\frac{3}{4}x - 8[\/latex].<\/p>\r\n\r\n<div class=\"textbox examples\">\r\n<h3>Example 7<\/h3>\r\nFind the function that is represented by a line that passes through the point (4, \u20132) and runs parallel to the line representing the function [latex]f(x)=3x-4[\/latex].\r\n<h4>Solution<\/h4>\r\nThe function\u00a0[latex]f(x)=3x-4[\/latex] is of the form [latex]f(x)=mx+b[\/latex], so [latex]m=3[\/latex].\r\n\r\nThe function we are trying to find is parallel to this so shares the same value of [latex]m[\/latex].\r\n\r\nWe know that the line passes through (4, \u20132) with a slope of 2 so we can substitute this information into the function [latex]f(x)=mx+b[\/latex] to find the value of [latex]b[\/latex]:\r\n\r\n[latex]f(x)=mx+b\\\\-2=3(4)+b\\\\-2=12+b\\\\-14=b[\/latex]\r\n\r\nTherefore, the function is [latex]f(x)=3x-14[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 7<\/h3>\r\nFind the function that is represented by a line that passes through the point (\u20135, 3) and runs parallel to the line representing the function [latex]f(x)=-4x+7[\/latex].\r\n\r\n[reveal-answer q=\"hjm521\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm521\"]\r\n\r\n[latex]f(x)=-4x-17[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h3>Given a Point and a Perpendicular Linear Function<\/h3>\r\nDetermining the function whose graph passes through a given point and is perpendicular to the another function is found in a similar way to that of a given point and a parallel function. We must still determine the slope and [latex]y[\/latex]-intercept of the function, Since the function is perpendicular to the given function, we know that the slopes are the negative reciprocals of each other.\r\n\r\n<span style=\"font-size: 1rem; text-align: initial;\">For example, suppose we are asked to find the linear function whose graph passes through the point (\u20136, 1) and is perpendicular to the graph of the function [latex]g(x)=\\frac{3}{4}x - 5[\/latex]. The line represented by [latex]g(x)[\/latex] has a slope of [latex]\\frac{3}{4}[\/latex]. This means that the line represented by [latex]f(x)[\/latex] must have a slope of [latex]-\\dfrac{4}{3}[\/latex] because the lines are perpendicular.<\/span>\r\n\r\nKnowing the slope, we now have [latex]f(x) = -\\frac{4}{3}x + b[\/latex]. To find the value of [latex]b[\/latex], we can use the point (\u20136, 1):\r\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&amp;=-\\frac{4}{3}x+b\\\\1&amp;=-\\frac{4}{3}\\cdot (-6)+b\\\\1&amp;=8+b\\\\-7&amp;=b\\end{aligned}\\end{equation}[\/latex]<\/p>\r\nNow that we know [latex]m=-\\frac{4}{3}[\/latex] and [latex]b=-7[\/latex], we can write the linear function as\u00a0[latex]f(x) = -\\frac{4}{3}x - 7[\/latex].\r\n<div class=\"textbox examples\">\r\n<h3>Example 8<\/h3>\r\nDetermine the function that is represented by a line that passes through the point (\u20131, \u20133) and runs perpendicular to the line representing the function [latex]f(x)=4x-2[\/latex].\r\n<h4>Solution<\/h4>\r\nThe function\u00a0[latex]f(x)=4x-2[\/latex] is of the form [latex]f(x)=mx+b[\/latex], so [latex]m=4[\/latex].\r\n\r\nThe function we are trying to find is perpendicular to this so has an [latex]m[\/latex]-value that is the negative reciprocal of 4:\u00a0 [latex]m=-\\dfrac{1}{4}[\/latex].\r\n\r\nWe now know that the line passes through (\u20131, \u20133) with a slope of[latex]-\\frac{1}{4}[\/latex] so we can substitute this information into the function [latex]f(x)=mx+b[\/latex] to find the value of [latex]b[\/latex]:\r\n\r\n[latex]f(x)=mx+b\\\\-3=-\\dfrac{1}{4}(-1)+b\\\\-3=\\dfrac{1}{4}+b\\\\-\\dfrac{12}{4}-\\dfrac{1}{4}=b\\\\-\\dfrac{13}{4}=b[\/latex]\r\n\r\nTherefore, the function is [latex]f(x)=-\\dfrac{1}{4}x-\\dfrac{13}{4}[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox tryit\">\r\n<h3>Try It 8<\/h3>\r\nDetermine the function that is represented by a line that passes through the point (4, \u20136) and runs perpendicular to the line representing the function [latex]f(x)=\\dfrac{3}{2}x-1[\/latex].\r\n\r\n[reveal-answer q=\"hjm018\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm018\"]\r\n\r\n[latex]f(x)=-\\dfrac{2}{3}x-\\dfrac{10}{3}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<script>\r\nwindow.embeddedChatbotConfig = {\r\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\r\ndomain: \"www.chatbase.co\"\r\n}\r\n<\/script>\r\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" chatbotId=\"ejVb5sgc1-w972OOCgl5x\" domain=\"www.chatbase.co\" defer>\r\n<\/script>\r\n\r\n<iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe>","rendered":"<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\n<div class=\"textbox learning-objectives\">\n<h1>Learning Objectives<\/h1>\n<p>Determine slope of a line using the slope formula.<\/p>\n<p>Determine the linear function given:<\/p>\n<ul>\n<li>a slope and the [latex]y[\/latex]-intercept<\/li>\n<li>a slope and a point<\/li>\n<li>two points<\/li>\n<li>a graph<\/li>\n<li>a point and a parallel linear function<\/li>\n<li>a point and a perpendicular linear function<\/li>\n<\/ul>\n<\/div>\n<h2>Slope<\/h2>\n<p>The <em><strong>slope<\/strong><\/em> of a line, [latex]m[\/latex], is a measure of the steepness of the line, and since the steepness is the same all along the line, the slope is constant.\u00a0 The slope of a line is defined as the ratio of [latex]\\frac{\\text{rise}}{\\text{run}}[\/latex] where rise is the change in [latex]y[\/latex]-values, [latex]\\Delta y[\/latex], and run is the corresponding change in [latex]x[\/latex]-values, [latex]\\Delta x[\/latex]. If [latex]y=f(x)[\/latex], then slope = [latex]\\frac{\\Delta y}{\\Delta x}=\\frac{\\text{change in function values}}{\\text{change in }x\\text{-values}}[\/latex].<\/p>\n<p>Consider the line in figure 1. It passes through the points (\u20134, \u20134) and (10, 3). The run is the distance between [latex]x=-4[\/latex] and [latex]x=10[\/latex]; a total of 14 units. Rather than counting units on the graph, we can subtract the two [latex]x[\/latex]-coordinates: 10 \u2013 (\u20134) = 14. The rise is the distance between [latex]y=-4[\/latex] and [latex]y=3[\/latex]; a total of 7 units. We can find this number by subtracting the [latex]y[\/latex]-coordinates: 3 \u2013 (\u20134) = 7. Consequently, the slope of the line is\u00a0 [latex]m=\\frac{7}{14}=\\frac{1}{2}[\/latex].<\/p>\n<div id=\"attachment_1260\" style=\"width: 613px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1260\" class=\"wp-image-1260\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/02\/31203020\/Slope-between-2-points-300x190.png\" alt=\"Line with two fixed points, (-4,-4) and (10,3) showing the horizontal change of 14 and the vertical change of 7.\" width=\"603\" height=\"382\" \/><\/p>\n<p id=\"caption-attachment-1260\" class=\"wp-caption-text\">Figure 1. Slope between 2 points.<\/p>\n<\/div>\n<p>Let&#8217;s now consider two arbitrary points on a line. Suppose the first point is designated [latex](x_1, y_1)[\/latex] and the second point is [latex](x_2,y_2)[\/latex]. The run is the difference in [latex]x[\/latex]-values, [latex]x_2-x_1[\/latex], and the rise is the difference in [latex]y[\/latex]-values, [latex]y_2-y_1[\/latex]. So the slope of the line between these two points is [latex]m=\\frac{y_2-y_1}{x_2-x_1}[\/latex] (see figure 2). It does not matter which point we designate as the first point or the second point. We will always get the same slope as long as we subtract from one point to the other in the same order.<\/p>\n<div id=\"attachment_1292\" style=\"width: 396px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1292\" class=\"wp-image-1292\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/31235741\/slope--300x300.png\" alt=\"Line with two arbitrary points, (x_2,y_2) and (x_1,y_1) showing the horizontal change x2-x1and the vertical change y2-y1.\" width=\"386\" height=\"386\" \/><\/p>\n<p id=\"caption-attachment-1292\" class=\"wp-caption-text\">Figure 2. Determining the slope between 2 points on a line.<\/p>\n<\/div>\n<p>The notation\u00a0[latex]\\Delta[\/latex] is often used in mathematics to represent &#8220;change in&#8221;, so &#8220;change in y&#8221; can be denoted by\u00a0[latex]\\Delta y[\/latex], and &#8220;change in [latex]x[\/latex]&#8221; can be denoted\u00a0[latex]\\Delta x[\/latex].\u00a0 So, slope = [latex]m=\\frac{\\Delta y}{\\Delta x}[\/latex].<\/p>\n<div class=\"textbox\">\n<h3>SLOPE<\/h3>\n<p style=\"text-align: center;\">The slope between two points on a line [latex](x_1, y_1)[\/latex] and [latex](x_2, y_2)[\/latex] is found by the equation\u00a0 [latex]m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{\\Delta y}{\\Delta x}[\/latex].<\/p>\n<p style=\"text-align: center;\">If [latex]y=f(x)[\/latex] is a linear function, the rate of change of the function = the slope of the line representing the function on a graph.<\/p>\n<p style=\"text-align: center;\">\n<\/div>\n<div id=\"post-91\" class=\"standard post-91 chapter type-chapter status-publish hentry\">\n<div class=\"entry-content\">\n<div class=\"textbox examples\">\n<h3>Example 1<\/h3>\n<p>If [latex]f\\left(x\\right)[\/latex]\u00a0is a linear function and [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]\u00a0are points on the graph of the function, find the slope of the graphed line. Is the graph of this function increasing, decreasing, or neither?<\/p>\n<h4>Solution<\/h4>\n<p>The coordinate pairs are [latex]\\left(3,-2\\right)[\/latex]\u00a0and [latex]\\left(8,1\\right)[\/latex]. To find the slope we find the rate of change of the function between the two points.<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{1-\\left(-2\\right)}{8 - 3}=\\dfrac{3}{5}[\/latex]<\/p>\n<p>The function is increasing because [latex]m>0[\/latex].<\/p>\n<p>&nbsp;<\/p>\n<p>Alternatively, changing the order in which we use the points:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{-2-1}{3-8}=\\dfrac{-3}{-5}=\\dfrac{3}{5}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Notice that we get the same solution since the order the points are referenced does not matter.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>The following video shows examples of how to find the slope of a line passing through two points and then determines whether the line is increasing, decreasing or neither.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/in3NTcx11I8?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.3-Video-1.odt\">Transcript 2.3 Video 1<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 1<\/h3>\n<p>The graph of a linear function passes through the points (\u20132, 4) and (3, 1). FInd the slope of the graphed line and determine whether it is increasing, decreasing, or neither.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm578\">Show Answer<\/span><\/p>\n<div id=\"qhjm578\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]m=-\\frac{3}{5}[\/latex]<\/p>\n<p>The line is decreasing because [latex]m<0[\/latex].\n\n<\/div>\n<\/div>\n<\/div>\n<p>The next video shows an example where the increase in cost for producing solar panels is determined by two data points.<\/p>\n<p><iframe loading=\"lazy\" src=\"https:\/\/www.youtube.com\/embed\/4RbniDgEGE4?feature=oembed&amp;rel=0\" width=\"500\" height=\"281\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/p>\n<p><a href=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/03\/Transcript-2.3-Video-2.odt\">Transcript 2.3 Video 2<\/a><\/p>\n<div class=\"textbox tryit\">\n<h3>Try It 2<\/h3>\n<p>A gym charges a membership fee and a fixed monthly amount.\u00a0 For 3 months the total cost is $110. For 8 months, the total cost is $210. Calculate the monthly amount.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm810\">Show Answer<\/span><\/p>\n<div id=\"qhjm810\" class=\"hidden-answer\" style=\"display: none\">\n<p>(3, $110) and (8, $210) are data points. Since the monthly fee is fixed, the total cost over time is linear, and the monthly fee = the change in the total cost per month = the slope.<\/p>\n<p>Monthly fee = $20.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Determining Linear Functions<\/h2>\n<p>A linear function may be described by the slope-intercept form [latex]f(x) = mx + b[\/latex], where [latex]m[\/latex] is the rate of change of the function or the slope of the graphed function, and [latex]b[\/latex] is the initial value or\u00a0[latex]y[\/latex]-coordinate of the [latex]y[\/latex]-intercept. There are several different situations and conditions we may be given to determine the equation of the linear function.<\/p>\n<h3>Given a Slope and the y-intercept<\/h3>\n<p>If we are given the slope and the [latex]y[\/latex]-intercept, we are given the value of [latex]m[\/latex] and [latex]b[\/latex]. This allows us to jump right to slope-intercept form to find the function. For example, if we are told that the slope of a line is 2 and the [latex]y[\/latex]-intercept is (0, 1), then using slope-intercept form, [latex]f(x)=x+b[\/latex], the linear function will be [latex]f(x) = 2x + 1[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 2<\/h3>\n<p>Find the function whose graph is a line with:<\/p>\n<ol>\n<li>slope = 4 and [latex]y[\/latex]-intercept = (0, 8).<\/li>\n<li>slope = [latex]\\frac{4}{3}[\/latex] and [latex]y[\/latex]-intercept = (0, \u20135).<\/li>\n<li>slope = 0 and [latex]y[\/latex]-intercept = (0, 3).<\/li>\n<li>slope = [latex]-\\frac{5}{6}[\/latex] and [latex]y[\/latex]-intercept = (0, 0).<\/li>\n<\/ol>\n<h4>Solution<\/h4>\n<p>We are given\u00a0[latex]m[\/latex] and\u00a0[latex]b[\/latex] so can use slope-intercept form [latex]f(x)=mx+b[\/latex] to find the function.<\/p>\n<ol>\n<li>[latex]m=4[\/latex] and [latex]b=8[\/latex]. Therefore, the function is [latex]f(x)=4x+8[\/latex].<\/li>\n<li>[latex]m=\\frac{4}{3}[\/latex] and [latex]b=-5[\/latex]. Therefore, the function is [latex]f(x)=\\frac{4}{3}x-5[\/latex].<\/li>\n<li>[latex]m=0[\/latex] and [latex]b=3[\/latex]. Therefore, the function is [latex]f(x)=0x+3[\/latex], which simplifies to [latex]f(x)=3[\/latex].<\/li>\n<li>[latex]m=-\\frac{5}{6}[\/latex] and [latex]b=0[\/latex]. Therefore, the function is [latex]f(x)=-\\frac{5}{6}x+0[\/latex], which simplifies to[latex]f(x)=-\\frac{5}{6}x[\/latex].<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 3<\/h3>\n<p>Find the function whose graph is a line with:<\/p>\n<ol>\n<li>slope = \u20132 and [latex]y[\/latex]-intercept = (0, 3).<\/li>\n<li>slope = [latex]\\frac{2}{3}[\/latex] and [latex]y[\/latex]-intercept = (0, 0).<\/li>\n<li>slope = 0 and [latex]y[\/latex]-intercept = (0, -7).<\/li>\n<li>slope = [latex]\\frac{5}{4}[\/latex] and [latex]y[\/latex]-intercept = [latex]-\\frac{7}{2}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm916\">Show Answer<\/span><\/p>\n<div id=\"qhjm916\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>[latex]f(x)=-2x+3[\/latex]<\/li>\n<li>[latex]f(x)=\\frac{2}{3}x[\/latex]<\/li>\n<li>[latex]f(x)=-7[\/latex]<\/li>\n<li>[latex]f(x)=\\frac{5}{4}x-\\frac{7}{2}[\/latex]<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<h3>Given a Slope and a Point<\/h3>\n<p>If we are given a slope and a point, then we need to find the [latex]y[\/latex]-intercept in order to determine the linear function [latex]f(x)=mx+b[\/latex]. Since the given point [latex](x_1,y_1)[\/latex] lies on the line, it is a solution of the linear function. We say that [latex](x_1,y_1)[\/latex] satisfies the function. We can use this piece of information to solve for [latex]b[\/latex] and determine the function.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">For example, suppose we are told that the slope of a line is [latex]\\dfrac{1}{3}[\/latex]\u00a0and the point (6, \u20131) lies on the line. We know that the function can be found using [latex]f(x)=mx+b[\/latex]. Because [latex]m=\\frac{1}{3}[\/latex] the function must be:<\/span><\/p>\n<p style=\"text-align: center;\">[latex]f(x) = \\frac{1}{3}x + b[\/latex]<\/p>\n<p>We now need to find the value of [latex]b[\/latex] and we can use the given point (6, \u20131) to do this. Because (6, \u20131) satisfies the equation, [latex]f(6)=-1[\/latex].<\/p>\n<p>This means that:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned} -1 &= \\frac{1}{3} \\cdot 6 + b\\\\-1 &= 2 + b\\\\-3 &= b \\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Now that we know [latex]m=\\frac{1}{3}[\/latex] and [latex]b=-3[\/latex], we can write the linear function as [latex]f(x) = \\frac{1}{3}x - 3[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 3<\/h3>\n<p>Find the function whose graph is a line with slope [latex]\\frac{3}{4}[\/latex] that passes through the point (\u20133, 1).<\/p>\n<p>Solution<\/p>\n<p>First we are told that [latex]m=\\frac{3}{4}[\/latex], so the function is [latex]f(x)=\\large\\frac{3}{4}x+b[\/latex].<\/p>\n<p>To find the value of [latex]b[\/latex], we use the given point (\u20133, 1) which tells us that [latex]f(-3)=1[\/latex].<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=\\frac{3}{4}x+b\\\\1&=\\frac{3}{4}\\cdot (-3) + b\\\\1&=-\\frac{9}{4}+b\\\\1+\\frac{9}{4}&=b\\\\ \\frac{4}{4}+\\frac{9}{4}&=b\\\\ \\frac{13}{4}&=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>Now we know [latex]m[\/latex] and [latex]b[\/latex], the function is [latex]f(x)=\\large\\frac{3}{4}x+\\frac{13}{4}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 4<\/h3>\n<p>Find the function whose graph is a line with slope \u20133 that passes through the point (4, \u20135).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm411\">Show Answer<\/span><\/p>\n<div id=\"qhjm411\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)=-3x+7[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Given Two Points<\/h3>\n<p>If we are given two points that lie on the graph of the function, then we have to find both the slope and the [latex]y[\/latex]-intercept in order to determine the linear function. We can find the slope by finding the rate of change\u00a0 between the two points:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{y_2 - y_1}{x_2 - x_1}[\/latex]<\/p>\n<p>After we find the slope, then we can use either one of the two points on the line to find the value of [latex]b[\/latex] because both of the two points satisfy the linear function.<\/p>\n<p>For example, given the two points (\u20131, 3) and (4, \u20137) that satisfy the function , the slope of the line representing the linear function will be:<\/p>\n<p style=\"text-align: center;\">Slope = [latex]m=\\dfrac{-7-3}{4-(-1)} =\\dfrac{-10}{5} = -2[\/latex]<\/p>\n<p>Since both point (\u20131, 3) and (4, \u20137) satisfy the function, we may use either of the two points. Let&#8217;s use the point (4, \u20137):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=-2x+b\\\\-7&=-2\\cdot 4+b\\\\-7&=-8+b\\\\1&=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>We now know that [latex]m=-2[\/latex] and [latex]b=1[\/latex] so the linear function is\u00a0[latex]f(x) = -2x + 1[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 4<\/h3>\n<p>Find the function whose graph is a line that passes through the points (\u20133, 5) and (4, \u20132).<\/p>\n<h4>Solution<\/h4>\n<p>FIrst we must find the slope of the line:\u00a0 [latex]m=\\frac{\\Delta y}{\\Delta x}=\\frac{-2-5}{4-(-3)}=\\frac{-7}{7}=-1[\/latex]<\/p>\n<p>This means that the function is: [latex]f(x)=-x+b[\/latex]<\/p>\n<p>To find the value of [latex]b[\/latex], we can use either point on the line that satisfies the function.<\/p>\n<p>Let&#8217;s use (4, \u20132), which means [latex]f(4)=-2[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=-x+b\\\\-2&=-(4)+b\\\\2&=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Now we know that [latex]m=-1[\/latex] and [latex]b=2[\/latex] so we can write the function as [latex]f(x)=-x+2[\/latex].<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 5<\/h3>\n<p>Find the function whose graph is a line that passes through the points (0, 5) and (\u20133, 8).<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm769\">Show Answer<\/span><\/p>\n<div id=\"qhjm769\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)=-x+5[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Given a Graph<\/h3>\n<p>If we are given a line on the coordinate plane, we may use any of the methods discussed above to find the function of the line by identifying the slope and the [latex]y[\/latex]-intercept of the line, or the slope and a point on the line, or two points on the line.<\/p>\n<div class=\"textbox examples\">\n<h3>Example 5<\/h3>\n<p>Find the function represented by the graph:<\/p>\n<p style=\"text-align: center;\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1300\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01050649\/y2x-5-227x300.png\" alt=\"A line with y-intercept (0,-5) passing through (5,5).\" width=\"280\" height=\"370\" \/><\/p>\n<h4>Solution<\/h4>\n<p>The [latex]y[\/latex]-intercept is (0, \u20135) so we know that [latex]b=-5[\/latex] in the function [latex]f(x)=mx+b[\/latex].<\/p>\n<p>To find the slope [latex]m[\/latex], we use any other point on the line. Let&#8217;s use (5, 5):<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{5-(-5)}{5-0}=\\dfrac{10}{5}=2[\/latex]<\/p>\n<p>Therefore, the function is [latex]f(x)=2x-5[\/latex].<\/p>\n<\/div>\n<div class=\"textbox examples\">\n<h3>Example 6<\/h3>\n<p>Find the function represented by the graph:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-1301\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01051608\/y-12x32-300x223.png\" alt=\"A line with indeterminant y-intercept passing through the points (3,0) and (-5,4).\" width=\"412\" height=\"306\" \/><\/p>\n<h4>Solution<\/h4>\n<p>The [latex]y[\/latex]-intercept is fractional between [latex]y=[\/latex] 1 and 2 and therefore not obvious, so we will use two points on the graph to find the function: (3, 0) and (\u20135, 4).<\/p>\n<p>To find the slope:<\/p>\n<p style=\"text-align: center;\">[latex]m=\\dfrac{\\Delta y}{\\Delta x}=\\dfrac{4-0}{-5-3}=\\dfrac{4}{-8}=-\\dfrac{1}{2}[\/latex].<\/p>\n<p>The function can be written: [latex]f(x)=-\\frac{1}{2}x+b[\/latex]<\/p>\n<p>To find [latex]b[\/latex] we can use any point on the line. Let&#8217;s use (3, 0):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=-\\frac{1}{2}x+b\\\\0&=-\\frac{1}{2}\\cdot 3+b\\\\0&=-\\frac{3}{2}+b\\\\\\frac{3}{2}=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Now we know that [latex]m=-\\frac{1}{2}[\/latex] and [latex]b=\\frac{3}{2}[\/latex], so the function is [latex]f(x)=-\\frac{1}{2}x+\\frac{3}{2}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 6<\/h3>\n<p>Find the function represented by each graph:<\/p>\n<div id=\"attachment_1302\" style=\"width: 270px\" class=\"wp-caption alignleft\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1302\" class=\"wp-image-1302\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01053014\/y4x-7-150x300.png\" alt=\"Line with y-intercept (0,-7) with slope m=4.\" width=\"260\" height=\"520\" \/><\/p>\n<p id=\"caption-attachment-1302\" class=\"wp-caption-text\">Graph 1<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div id=\"attachment_1303\" style=\"width: 491px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1303\" class=\"wp-image-1303\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5774\/2022\/03\/01053119\/2x3y5-300x242.png\" alt=\"A line with indeterminant y-intercept passing through (-2,3) and (1,1).\" width=\"481\" height=\"388\" \/><\/p>\n<p id=\"caption-attachment-1303\" class=\"wp-caption-text\">Graph 2<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<p>&nbsp;<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm049\">Show Answer<\/span><\/p>\n<div id=\"qhjm049\" class=\"hidden-answer\" style=\"display: none\">\n<p>Graph 1: [latex]f(x)=4x-7[\/latex]<\/p>\n<p>Graph 2: [latex]f(x)=-\\frac{2}{3}x+\\frac{5}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>If we are given a horizontal line, its slope is 0 because the rise is 0 (i.e., [latex]\\dfrac{0}{run} = 0[\/latex]). Sometimes we can identify the [latex]y[\/latex]-intercept on the graph, and the function is [latex]f(x)=0x +b[\/latex] or\u00a0[latex]f(x)=b[\/latex]. For example, for the function of the red horizontal line in figure 3, the [latex]y[\/latex]-intercept is (0, 3) and therefore the function is [latex]f(x)=3[\/latex].<\/p>\n<div id=\"attachment_1138\" style=\"width: 310px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1138\" class=\"wp-image-1138 size-medium\" src=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-300x300.png\" alt=\"A horizontal line with y-intercept (0,3), and a vertical line with x-intercept (-2,0).\" width=\"300\" height=\"300\" srcset=\"https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-300x300.png 300w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-150x150.png 150w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-768x768.png 768w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-65x65.png 65w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-225x225.png 225w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine-350x350.png 350w, https:\/\/courses.lumenlearning.com\/uvu-combinedalgebra\/wp-content\/uploads\/sites\/5774\/2022\/02\/2-2-1-HorVerLine.png 800w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p id=\"caption-attachment-1138\" class=\"wp-caption-text\">Figure 3. A horizontal line and a vertical line.<\/p>\n<\/div>\n<p>If we are given a vertical line, like the blue line in figure 3, it is not a function because a vertical line is a one-to-many mapping (i.e., one input with many different outputs). Therefore, we are not able to use the function form [latex]f(x)=mx+b[\/latex] to express a vertical line. Also, the\u00a0slope of a vertical line is undefined because the run is 0, and a number divided by 0 is undefined (i.e.,\u00a0[latex]\\dfrac{rise}{0}[\/latex] is undefined). The equation for a vertical line is [latex]x = a[\/latex] where [latex]a[\/latex] is the\u00a0[latex]x[\/latex]-coordinate of the [latex]x[\/latex]-intercept. Indeed, for all of the points on a vertical line, the [latex]x[\/latex]-coordinates are identical and the [latex]y[\/latex]-coordinates are any real number on the vertical line. For example, the equation of the blue vertical line in figure 3 is\u00a0[latex]x = -2[\/latex].<\/p>\n<h3>Given a Point and a Parallel Linear Function<\/h3>\n<p>Finding the equation of a function whose graph is parallel to another line, still requires finding the slope and [latex]y[\/latex]-intercept. Fortunately, the slope can be found from the parallel line since parallel lines have the same slope. Knowing the slope, we can then use another point on the line to determine the [latex]y[\/latex]-intercept and consequently the function.<\/p>\n<p>For example, suppose we are asked to find the linear function that passes through the point (8, \u20132), whose graph runs parallel to the graph of the function [latex]g(x)=\\frac{3}{4}x+2[\/latex]. The slope of the line that represents [latex]g(x)[\/latex] is [latex]\\frac{3}{4}[\/latex]. This means that the line representing [latex]f(x)[\/latex] also has a slope of [latex]\\frac{3}{4}[\/latex] since they are parallel. this means that [latex]f(x)=\\frac{3}{4}x + b[\/latex]. Now, we can find the value of [latex]b[\/latex] by using the given point (8, \u20132):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=\\frac{3}{4}x+b\\\\-2&=\\frac{3}{4}\\cdot 8+b\\\\-2&=6+b\\\\-8=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p style=\"text-align: left;\">Now we know that [latex]m=\\frac{3}{4}[\/latex] and [latex]b=-8[\/latex].\u00a0Therefore, the function is\u00a0[latex]f(x) = \\frac{3}{4}x - 8[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 7<\/h3>\n<p>Find the function that is represented by a line that passes through the point (4, \u20132) and runs parallel to the line representing the function [latex]f(x)=3x-4[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>The function\u00a0[latex]f(x)=3x-4[\/latex] is of the form [latex]f(x)=mx+b[\/latex], so [latex]m=3[\/latex].<\/p>\n<p>The function we are trying to find is parallel to this so shares the same value of [latex]m[\/latex].<\/p>\n<p>We know that the line passes through (4, \u20132) with a slope of 2 so we can substitute this information into the function [latex]f(x)=mx+b[\/latex] to find the value of [latex]b[\/latex]:<\/p>\n<p>[latex]f(x)=mx+b\\\\-2=3(4)+b\\\\-2=12+b\\\\-14=b[\/latex]<\/p>\n<p>Therefore, the function is [latex]f(x)=3x-14[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 7<\/h3>\n<p>Find the function that is represented by a line that passes through the point (\u20135, 3) and runs parallel to the line representing the function [latex]f(x)=-4x+7[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm521\">Show Answer<\/span><\/p>\n<div id=\"qhjm521\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)=-4x-17[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h3>Given a Point and a Perpendicular Linear Function<\/h3>\n<p>Determining the function whose graph passes through a given point and is perpendicular to the another function is found in a similar way to that of a given point and a parallel function. We must still determine the slope and [latex]y[\/latex]-intercept of the function, Since the function is perpendicular to the given function, we know that the slopes are the negative reciprocals of each other.<\/p>\n<p><span style=\"font-size: 1rem; text-align: initial;\">For example, suppose we are asked to find the linear function whose graph passes through the point (\u20136, 1) and is perpendicular to the graph of the function [latex]g(x)=\\frac{3}{4}x - 5[\/latex]. The line represented by [latex]g(x)[\/latex] has a slope of [latex]\\frac{3}{4}[\/latex]. This means that the line represented by [latex]f(x)[\/latex] must have a slope of [latex]-\\dfrac{4}{3}[\/latex] because the lines are perpendicular.<\/span><\/p>\n<p>Knowing the slope, we now have [latex]f(x) = -\\frac{4}{3}x + b[\/latex]. To find the value of [latex]b[\/latex], we can use the point (\u20136, 1):<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{equation}\\begin{aligned}f(x)&=-\\frac{4}{3}x+b\\\\1&=-\\frac{4}{3}\\cdot (-6)+b\\\\1&=8+b\\\\-7&=b\\end{aligned}\\end{equation}[\/latex]<\/p>\n<p>Now that we know [latex]m=-\\frac{4}{3}[\/latex] and [latex]b=-7[\/latex], we can write the linear function as\u00a0[latex]f(x) = -\\frac{4}{3}x - 7[\/latex].<\/p>\n<div class=\"textbox examples\">\n<h3>Example 8<\/h3>\n<p>Determine the function that is represented by a line that passes through the point (\u20131, \u20133) and runs perpendicular to the line representing the function [latex]f(x)=4x-2[\/latex].<\/p>\n<h4>Solution<\/h4>\n<p>The function\u00a0[latex]f(x)=4x-2[\/latex] is of the form [latex]f(x)=mx+b[\/latex], so [latex]m=4[\/latex].<\/p>\n<p>The function we are trying to find is perpendicular to this so has an [latex]m[\/latex]-value that is the negative reciprocal of 4:\u00a0 [latex]m=-\\dfrac{1}{4}[\/latex].<\/p>\n<p>We now know that the line passes through (\u20131, \u20133) with a slope of[latex]-\\frac{1}{4}[\/latex] so we can substitute this information into the function [latex]f(x)=mx+b[\/latex] to find the value of [latex]b[\/latex]:<\/p>\n<p>[latex]f(x)=mx+b\\\\-3=-\\dfrac{1}{4}(-1)+b\\\\-3=\\dfrac{1}{4}+b\\\\-\\dfrac{12}{4}-\\dfrac{1}{4}=b\\\\-\\dfrac{13}{4}=b[\/latex]<\/p>\n<p>Therefore, the function is [latex]f(x)=-\\dfrac{1}{4}x-\\dfrac{13}{4}[\/latex]<\/p>\n<\/div>\n<div class=\"textbox tryit\">\n<h3>Try It 8<\/h3>\n<p>Determine the function that is represented by a line that passes through the point (4, \u20136) and runs perpendicular to the line representing the function [latex]f(x)=\\dfrac{3}{2}x-1[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qhjm018\">Show Answer<\/span><\/p>\n<div id=\"qhjm018\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]f(x)=-\\dfrac{2}{3}x-\\dfrac{10}{3}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p><script>\nwindow.embeddedChatbotConfig = {\nchatbotId: \"ejVb5sgc1-w972OOCgl5x\",\ndomain: \"www.chatbase.co\"\n}\n<\/script><br \/>\n<script src=\"https:\/\/www.chatbase.co\/embed.min.js\" defer=\"defer\">\n<\/script><\/p>\n<p><iframe style=\"height: 100%; min-height: 700px;\" src=\"https:\/\/www.chatbase.co\/chatbot-iframe\/ejVb5sgc1-w972OOCgl5x\" width=\"100%\" frameborder=\"0\"><\/iframe><\/p>\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-939\">\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>Algebraic Analysis of Linear Functions. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.desmos.com\/calculator\">http:\/\/www.desmos.com\/calculator<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All graphs created using desmos graphing calculator. <strong>Authored by<\/strong>: Hazel McKenna and Leo Chang. <strong>Provided by<\/strong>: Utah Valley University. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>All Examples and Try Its. <strong>Authored by<\/strong>: Hazel McKenna. <strong>Provided by<\/strong>: Utah Valley University. <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>Ex: Find the Slope Given Two Points and Describe the Line. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) . <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/in3NTcx11I8\">https:\/\/youtu.be\/in3NTcx11I8<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Ex: Slope Application Involving Production Costs. <strong>Authored by<\/strong>: James Sousa (Mathispower4u.com) for Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4RbniDgEGE4\">https:\/\/youtu.be\/4RbniDgEGE4<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":422608,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Algebraic Analysis of Linear Functions\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"www.desmos.com\/calculator\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"All graphs created using desmos graphing calculator\",\"author\":\"Hazel McKenna and Leo Chang\",\"organization\":\"Utah Valley University\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Find the Slope Given Two Points and Describe the Line\",\"author\":\"James Sousa (Mathispower4u.com) \",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/in3NTcx11I8\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Ex: Slope Application Involving Production 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