{"id":759,"date":"2025-07-15T15:50:35","date_gmt":"2025-07-15T15:50:35","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=759"},"modified":"2026-01-09T20:56:52","modified_gmt":"2026-01-09T20:56:52","slug":"graphs-of-linear-functions-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/graphs-of-linear-functions-learn-it-6\/","title":{"raw":"Graphs of Linear Functions: Learn It 6","rendered":"Graphs of Linear Functions: Learn It 6"},"content":{"raw":"

Graphing a Linear Function Using [latex]y[\/latex]-intercept and Slope<\/h3>\r\nWe can graph linear functions by using specific characteristics of the function rather than plotting points. The first characteristic is its [latex]y[\/latex]-<\/em>intercept which is the point at which the input value is zero. To find the [latex]y[\/latex]-<\/em>intercept<\/strong>, we can set [latex]x=0[\/latex] in the equation. The other characteristic of the linear function is its slope [latex]m[\/latex].\r\n\r\n
Remember that if a function has a [latex]y[\/latex]-intercept, we can always find it by setting [latex]x=0[\/latex] and then solving for [latex]y[\/latex].<\/section>
Let\u2019s consider the following function:\r\n

[latex]f\\left(x\\right)=\\frac{1}{2}x+1[\/latex]<\/p>\r\n\r\n

    \r\n \t
  • The slope is [latex]\\frac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right.<\/li>\r\n \t
  • The [latex]y[\/latex]-<\/em>intercept is the point on the graph when [latex]x\u00a0= 0[\/latex]. The graph crosses the [latex]y[\/latex]-axis at [latex](0, 1)[\/latex].<\/li>\r\n<\/ul>\r\nNow we know the slope and the [latex]y[\/latex]-intercept. We can begin graphing by plotting the point [latex](0, 1)[\/latex] We know that the slope is rise over run, [latex]m=\\frac{\\text{rise}}{\\text{run}}[\/latex].\r\n\r\nFrom our example, we have [latex]m=\\frac{1}{2}[\/latex], which means that the rise is [latex]1[\/latex] and the run is [latex]2[\/latex]. Starting from our [latex]y[\/latex]-intercept [latex](0, 1)[\/latex], we can rise [latex]1[\/latex] and then run [latex]2[\/latex] or run [latex]2[\/latex] and then rise [latex]1[\/latex]. We repeat until we have multiple points, and then we draw a line through the points as shown below.\r\n\r\n\"graph\r\n\r\n<\/section>
    \r\n

    graphical interpretation of a linear function<\/h3>\r\nIn the equation [latex]f\\left(x\\right)=mx+b[\/latex]\r\n
      \r\n \t
    • [latex]b[\/latex]\u00a0is the [latex]y[\/latex]-intercept of the graph and indicates the point [latex](0, b)[\/latex] at which the graph crosses the [latex]y[\/latex]-axis.<\/li>\r\n \t
    • [latex]m[\/latex]\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\r\n<\/ul>\r\n

      [latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\r\n\r\n<\/section>

      \r\n

      Match each equation of the linear functions with one of the lines on the graph.<\/p>\r\n\r\n

        \r\n \t
      1. [latex]f\\left(x\\right)=2x+3[\/latex]<\/li>\r\n \t
      2. [latex]g\\left(x\\right)=2x - 3[\/latex]<\/li>\r\n \t
      3. [latex]h\\left(x\\right)=-2x+3[\/latex]<\/li>\r\n \t
      4. [latex]j\\left(x\\right)=\\frac{1}{2}x+3[\/latex]<\/li>\r\n<\/ol>\r\n\"Graph\r\n\r\n[reveal-answer q=\"35922\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"35922\"]\r\n

        Analyze the information for each function.<\/p>\r\n\r\n

          \r\n \t
        1. This function has a slope of 2 and a y<\/em>-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g<\/em>\u00a0has the same slope, but a different y-<\/em>intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f<\/em>\u00a0must be represented by line I.<\/li>\r\n \t
        2. This function also has a slope of 2, but a y<\/em>-intercept of \u20133. It must pass through the point (0, \u20133) and slant upward from left to right. It must be represented by line III.<\/li>\r\n \t
        3. This function has a slope of \u20132 and a y-<\/em>intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right.<\/li>\r\n \t
        4. This function has a slope of [latex]\\frac{1}{2}[\/latex] and a y-<\/em>intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. Lines I and II pass through (0, 3), but the slope of j<\/em>\u00a0is less than the slope of f<\/em>\u00a0so the line for j<\/em>\u00a0must be flatter. This function is represented by Line II.<\/li>\r\n<\/ol>\r\nNow we can re-label the lines.\r\n\r\n\"Graph\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
          How To: Given the equation for a linear function, graph the function using the [latex]y[\/latex]-intercept and slope.<\/strong>\r\n
            \r\n \t
          1. Evaluate the function at an input value of zero to find the [latex]y[\/latex]-<\/em>intercept.<\/li>\r\n \t
          2. Identify the slope.<\/li>\r\n \t
          3. Plot the point represented by the y-<\/em>intercept.<\/li>\r\n \t
          4. Use [latex]\\frac{\\text{rise}}{\\text{run}}[\/latex] to determine at least two more points on the line.<\/li>\r\n \t
          5. Draw a line which passes through the points.<\/li>\r\n<\/ol>\r\n<\/section>
            Graph [latex]f\\left(x\\right)=-\\frac{2}{3}x+5[\/latex] using the [latex]y[\/latex]-<\/em>intercept and slope.[reveal-answer q=\"736682\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"736682\"]Evaluate the function at [latex]x\u00a0= 0[\/latex] to find the [latex]y-[\/latex]intercept. The output value when [latex]x\u00a0= 0[\/latex] is [latex]5[\/latex], so the graph will cross the [latex]y[\/latex]-axis at [latex](0, 5)[\/latex].\r\n[latex]\\\\[\/latex]\r\nAccording to the equation for the function, the slope of the line is [latex]-\\frac{2}{3}[\/latex]. This tells us that for each vertical decrease in the \"rise\" of [latex]\u20132[\/latex] units, the \"run\" increases by [latex]3[\/latex] units in the horizontal direction.We can now graph the function by first plotting the [latex]y[\/latex]-intercept. From the initial value [latex](0, 5)[\/latex] we move down [latex]2[\/latex] units and to the right [latex]3[\/latex] units. We can extend the line to the left and right by repeating, and then draw a line through the points.\"graph\r\nAnalysis of the Solution<\/strong>\r\nThe graph slants downward from left to right which means it has a negative slope as expected.[\/hidden-answer]<\/section>Look at the graph of the function [latex]f[\/latex] given below:\r\n\"This\r\n\r\nWe are not given the slope of the line, but we can choose any two points on the line to find the slope. Let\u2019s choose [latex](0, 7)[\/latex] and [latex](4, 4)[\/latex].\r\n

            [latex] \\begin{array}{rcl} m & = & \\frac{y_2 - y_1}{x_2 - x_1} \\\\ m & = & \\frac{4 - 7}{4 - 0} \\\\ m & = & -\\frac{3}{4} \\end{array} [\/latex]<\/p>\r\nNow we can substitute the slope and the coordinates of one of the points into the point-slope form.\r\n

            [latex] \\begin{array}{rcl} y - y_1 & = & m(x - x_1) \\\\ y - 4 & = & -\\frac{3}{4}(x - 4) \\end{array} [\/latex]<\/p>\r\nIf we want to rewrite the equation in the slope-intercept form, we would find\r\n

            [latex] \\begin{array}{rcl} y - 4 & = & -\\frac{3}{4}(x - 4) \\\\ y - 4 & = & -\\frac{3}{4}x + 3 \\\\ y & = & -\\frac{3}{4}x + 7 \\end{array} [\/latex]<\/p>\r\nIf we want to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, [latex]b = 7[\/latex]. We now have the initial value [latex]b[\/latex] and the slope [latex]m[\/latex], so we can substitute [latex]m[\/latex] and [latex]b[\/latex] into the slope-intercept form of a line.\r\n\r\n\"This\r\n\r\nSo the function is [latex]f(x) = -\\frac{3}{4}x + 7[\/latex], and the linear equation would be [latex]y = -\\frac{3}{4}x + 7[\/latex].\r\n\r\n

            \r\n

            How to: Given the graph of a linear function, write an equation to represent the function.<\/strong><\/p>\r\n\r\n

              \r\n \t
            1. Identify two points on the line.<\/li>\r\n \t
            2. Use the two points to calculate the slope.<\/li>\r\n \t
            3. Determine where the line crosses the\u00a0y<\/em>-axis to identify the\u00a0y<\/em>-intercept by visual inspection.<\/li>\r\n \t
            4. Substitute the slope and\u00a0y<\/em>-intercept into the slope-intercept form of a line equation.<\/li>\r\n<\/ol>\r\n<\/section>
              [ohm_question hide_question_numbers=1]318708[\/ohm_question]<\/section>","rendered":"

              Graphing a Linear Function Using [latex]y[\/latex]-intercept and Slope<\/h3>\n

              We can graph linear functions by using specific characteristics of the function rather than plotting points. The first characteristic is its [latex]y[\/latex]–<\/em>intercept which is the point at which the input value is zero. To find the [latex]y[\/latex]–<\/em>intercept<\/strong>, we can set [latex]x=0[\/latex] in the equation. The other characteristic of the linear function is its slope [latex]m[\/latex].<\/p>\n

              Remember that if a function has a [latex]y[\/latex]-intercept, we can always find it by setting [latex]x=0[\/latex] and then solving for [latex]y[\/latex].<\/section>\n
              Let\u2019s consider the following function:<\/p>\n

              [latex]f\\left(x\\right)=\\frac{1}{2}x+1[\/latex]<\/p>\n

                \n
              • The slope is [latex]\\frac{1}{2}[\/latex]. Because the slope is positive, we know the graph will slant upward from left to right.<\/li>\n
              • The [latex]y[\/latex]–<\/em>intercept is the point on the graph when [latex]x\u00a0= 0[\/latex]. The graph crosses the [latex]y[\/latex]-axis at [latex](0, 1)[\/latex].<\/li>\n<\/ul>\n

                Now we know the slope and the [latex]y[\/latex]-intercept. We can begin graphing by plotting the point [latex](0, 1)[\/latex] We know that the slope is rise over run, [latex]m=\\frac{\\text{rise}}{\\text{run}}[\/latex].<\/p>\n

                From our example, we have [latex]m=\\frac{1}{2}[\/latex], which means that the rise is [latex]1[\/latex] and the run is [latex]2[\/latex]. Starting from our [latex]y[\/latex]-intercept [latex](0, 1)[\/latex], we can rise [latex]1[\/latex] and then run [latex]2[\/latex] or run [latex]2[\/latex] and then rise [latex]1[\/latex]. We repeat until we have multiple points, and then we draw a line through the points as shown below.<\/p>\n

                \"graph<\/p>\n<\/section>\n

                \n

                graphical interpretation of a linear function<\/h3>\n

                In the equation [latex]f\\left(x\\right)=mx+b[\/latex]<\/p>\n

                  \n
                • [latex]b[\/latex]\u00a0is the [latex]y[\/latex]-intercept of the graph and indicates the point [latex](0, b)[\/latex] at which the graph crosses the [latex]y[\/latex]-axis.<\/li>\n
                • [latex]m[\/latex]\u00a0is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope:<\/li>\n<\/ul>\n

                  [latex]m=\\frac{\\text{change in output (rise)}}{\\text{change in input (run)}}=\\frac{\\Delta y}{\\Delta x}=\\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}[\/latex]<\/p>\n<\/section>\n

                  \n

                  Match each equation of the linear functions with one of the lines on the graph.<\/p>\n

                    \n
                  1. [latex]f\\left(x\\right)=2x+3[\/latex]<\/li>\n
                  2. [latex]g\\left(x\\right)=2x - 3[\/latex]<\/li>\n
                  3. [latex]h\\left(x\\right)=-2x+3[\/latex]<\/li>\n
                  4. [latex]j\\left(x\\right)=\\frac{1}{2}x+3[\/latex]<\/li>\n<\/ol>\n

                    \"Graph<\/p>\n