{"id":1291,"date":"2021-12-02T19:37:21","date_gmt":"2021-12-03T00:37:21","guid":{"rendered":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/4-4-2d-inequality-and-absolute-value-graphs\/"},"modified":"2023-08-30T12:54:30","modified_gmt":"2023-08-30T16:54:30","slug":"2d-inequality-and-absolute-value-graphs","status":"publish","type":"chapter","link":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/chapter\/2d-inequality-and-absolute-value-graphs\/","title":{"raw":"4.4 2D Inequality and Absolute Value Graphs","rendered":"4.4 2D Inequality and Absolute Value Graphs"},"content":{"raw":"

Graphing a 2D Inequality<\/h1>\r\nTo graph an inequality, borrow the strategy used to draw a line graph in 2D. To do this, replace the inequality with an equal sign.\r\n
\r\n

Example 4.4.1<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nConsider the following inequalities:\r\n

[latex]\\begin{array}{rrrrr}2x&+&2y&<&12 \\\\ 2x&+&2y&\\le&12 \\\\2x&+&2y&>&12 \\\\2x&+&2y&\\ge&12 \\end{array}[\/latex]<\/p>\r\nAll can be changed to [latex]3x + 2y = 12[\/latex] by replacing the inequality sign with =.\r\n\r\nIt is then possible to create a data table using the new equation.\r\n\r\nCreate a data table of values for the equation [latex]3x + 2y = 12.[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/th>\r\n[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n
0<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n3<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
6<\/td>\r\n\u22123<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nUsing these values, plot the data points on a graph.\r\n\r\n\"Graph.\r\n\r\nOnce the data points are\u00a0plotted, draw a line that\u00a0connects them all. The\u00a0type of line drawn depends on the original inequality that was replaced.\r\n\r\nIf the inequality had \u2264 or \u2265, then draw a solid line to represent data points that are on the line.\r\n\r\n\"Solid\r\n\r\nIf the inequality had < or >, then draw a dashed line instead to indicate that some data points are excluded.\r\n\r\n\"Dashed\r\n\r\nIf the inequality is either [latex]3x + 2y \\le 12[\/latex] or [latex]3x + 2y \\ge 12[\/latex], then draw its graph using a solid line and solid dots.\r\n\r\n\"Solid\r\n\r\nIf the inequality is either [latex]3x + 2y < 12[\/latex] or [latex]3x + 2y[\/latex] > [latex]12[\/latex], then draw its graph using a dashed line and hollow dots.\r\n\r\n\"Dashed\r\n\r\nThere remains only one step to complete this graph: finding which side of the line makes the inequality true and shading it. The easiest way to do this is to choose the data point [latex](0, 0)[\/latex].\r\n\r\nIt is evident that, for [latex]3(0) + 2(0) \\le 12[\/latex] and [latex]3(0) + 2(0) < 12[\/latex], the data point [latex](0, 0)[\/latex] is true for the inequality. In this case, shade the side of the line that contains the data point [latex](0, 0)[\/latex].\r\n\r\n\"Completed\r\n\r\nIt is also clear that, for [latex]3(0) + 2(0) \\ge 12[\/latex] and [latex]3(0) + 2(0)[\/latex] > [latex]12[\/latex], the data point [latex](0, 0)[\/latex] is false for the inequality. In this case, shade the side of the line that does not contain the data point [latex](0, 0)[\/latex].\r\n\r\n\"Completed\r\n\r\n<\/div>\r\n<\/div>\r\n

Graphing an Absolute Value Function<\/h1>\r\nTo graph an absolute value function, first create a data table using the absolute value part of the equation.\r\n\r\nThe data point that is started with is the one that makes the absolute value equal to 0 (this is the [latex]x[\/latex]-value of the vertex). Calculating the value of this point is quite simple.\r\n\r\nFor example, for [latex]| x - 3 |[\/latex], the value [latex]x = 3[\/latex] makes the absolute value equal to 0.\r\n\r\nExamples of others are:\r\n\r\n[latex]\\begin{array}{rrrrrrrrr}\r\n|x&+&2|&=&0&\\text{when}&x&=&-2 \\\\\r\n|x&-&11|&=&0&\\text{when}&x&=&11 \\\\\r\n|x&+&9|&=&0&\\text{when}&x&=&-9 \\\\\r\n\\end{array}[\/latex]\r\n\r\nThe graph of an absolute value equation will be a V-shape that opens upward for any positive absolute function and opens downward for any negative absolute value function.\r\n
\r\n

Example 4.4.2<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nPlot the graph of [latex]y = | x + 2 | - 3.[\/latex]\r\n\r\nThe data point that gives the [latex]x[\/latex]-value of the vertex is [latex]| x + 2 | = 0,[\/latex] in which [latex]x = -2.[\/latex] This is the first value.\r\n\r\nFor [latex]x = -2, y = | -2 + 2 | - 3,[\/latex] which yields [latex]y = -3.[\/latex]\r\n\r\nNow pick [latex]x[\/latex]-values that are larger and less than \u22122 to get three data points on both sides of the vertex, [latex](-2, -3).[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/th>\r\n[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n
1<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
0<\/td>\r\n\u22121<\/td>\r\n<\/tr>\r\n
\u22121<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n
\u22122<\/td>\r\n\u22123<\/td>\r\n<\/tr>\r\n
\u22123<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n
\u22124<\/td>\r\n\u22121<\/td>\r\n<\/tr>\r\n
\u22125<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOnce there are three data points on either side of the vertex, plot and connect them in a line. The graph is complete.\r\n\r\n\"Positive<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n
\r\n

Example 4.4.3<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nPlot the graph of [latex]y = -| x - 2 | + 1.[\/latex]\r\n\r\nThe data point that gives the [latex]x[\/latex]-value of the vertex is [latex]| x - 2 | = 0,[\/latex] in which [latex]x = 2.[\/latex] This is the first value.\r\n\r\nFor [latex]x = 2, y = -| 2 - 2 | + 1,[\/latex] which yields [latex]y = 1.[\/latex]\r\n\r\nNow pick [latex]x[\/latex]-values that are larger and less than 2 to get three data points on both sides of the vertex, [latex](2, 1).[\/latex]\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
[latex]x[\/latex]<\/th>\r\n[latex]y[\/latex]<\/th>\r\n<\/tr>\r\n
5<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n
4<\/td>\r\n\u22121<\/td>\r\n<\/tr>\r\n
3<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
2<\/td>\r\n1<\/td>\r\n<\/tr>\r\n
1<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
0<\/td>\r\n\u22121<\/td>\r\n<\/tr>\r\n
\u22121<\/td>\r\n\u22122<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nOnce there are three data points on either side of the vertex, plot and connect them in a line. The graph is complete.\r\n\r\n\"Negative<\/span>\r\n\r\n<\/div>\r\n<\/div>\r\n

Questions<\/h1>\r\nFor questions 1 to 8, graph each linear inequality.\r\n
    \r\n \t
  1. [latex]y[\/latex] > [latex]3x + 2[\/latex]<\/li>\r\n \t
  2. [latex]3x - 4y[\/latex] > [latex]12[\/latex]<\/li>\r\n \t
  3. [latex]2y \\ge 3x + 6[\/latex]<\/li>\r\n \t
  4. [latex]3x - 2y \\ge 6[\/latex]<\/li>\r\n \t
  5. [latex]2y[\/latex] > [latex]5x + 10[\/latex]<\/li>\r\n \t
  6. [latex]5x + 4y[\/latex] > [latex]-20[\/latex]<\/li>\r\n \t
  7. [latex]4y \\ge 5x + 20[\/latex]<\/li>\r\n \t
  8. [latex]5x + 2y \\ge -10[\/latex]<\/li>\r\n<\/ol>\r\nFor questions 9 to 16, graph each absolute value equation.\r\n
      \r\n \t
    1. [latex]y=|x-4|[\/latex]<\/li>\r\n \t
    2. [latex]y=|x-3|+3[\/latex]<\/li>\r\n \t
    3. [latex]y=|x-2|[\/latex]<\/li>\r\n \t
    4. [latex]y=|x-2|+2[\/latex]<\/li>\r\n \t
    5. [latex]y=-|x-2|[\/latex]<\/li>\r\n \t
    6. [latex]y=-|x-2|+2[\/latex]<\/li>\r\n \t
    7. [latex]y=-|x+2|[\/latex]<\/li>\r\n \t
    8. [latex]y=-|x+2|+2[\/latex]<\/li>\r\n<\/ol>\r\nAnswer Key 4.4<\/a>","rendered":"

      Graphing a 2D Inequality<\/h1>\n

      To graph an inequality, borrow the strategy used to draw a line graph in 2D. To do this, replace the inequality with an equal sign.<\/p>\n

      \n
      \n

      Example 4.4.1<\/p>\n<\/header>\n

      \n

      Consider the following inequalities:<\/p>\n

      [latex]\\begin{array}{rrrrr}2x&+&2y&<&12 \\\\ 2x&+&2y&\\le&12 \\\\2x&+&2y&>&12 \\\\2x&+&2y&\\ge&12 \\end{array}[\/latex]<\/p>\n

      All can be changed to [latex]3x + 2y = 12[\/latex] by replacing the inequality sign with =.<\/p>\n

      It is then possible to create a data table using the new equation.<\/p>\n

      Create a data table of values for the equation [latex]3x + 2y = 12.[\/latex]<\/p>\n\n\n\n\n\n\n\n
      [latex]x[\/latex]<\/th>\n[latex]y[\/latex]<\/th>\n<\/tr>\n
      0<\/td>\n6<\/td>\n<\/tr>\n
      2<\/td>\n3<\/td>\n<\/tr>\n
      4<\/td>\n0<\/td>\n<\/tr>\n
      6<\/td>\n\u22123<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Using these values, plot the data points on a graph.<\/p>\n

      \"Graph.<\/p>\n

      Once the data points are\u00a0plotted, draw a line that\u00a0connects them all. The\u00a0type of line drawn depends on the original inequality that was replaced.<\/p>\n

      If the inequality had \u2264 or \u2265, then draw a solid line to represent data points that are on the line.<\/p>\n

      \"Solid<\/p>\n

      If the inequality had < or >, then draw a dashed line instead to indicate that some data points are excluded.<\/p>\n

      \"Dashed<\/p>\n

      If the inequality is either [latex]3x + 2y \\le 12[\/latex] or [latex]3x + 2y \\ge 12[\/latex], then draw its graph using a solid line and solid dots.<\/p>\n

      \"Solid<\/p>\n

      If the inequality is either [latex]3x + 2y < 12[\/latex] or [latex]3x + 2y[\/latex] > [latex]12[\/latex], then draw its graph using a dashed line and hollow dots.\n\n\"Dashed<\/p>\n

      There remains only one step to complete this graph: finding which side of the line makes the inequality true and shading it. The easiest way to do this is to choose the data point [latex](0, 0)[\/latex].<\/p>\n

      It is evident that, for [latex]3(0) + 2(0) \\le 12[\/latex] and [latex]3(0) + 2(0) < 12[\/latex], the data point [latex](0, 0)[\/latex] is true for the inequality. In this case, shade the side of the line that contains the data point [latex](0, 0)[\/latex].\n\n\"Completed<\/p>\n

      It is also clear that, for [latex]3(0) + 2(0) \\ge 12[\/latex] and [latex]3(0) + 2(0)[\/latex] > [latex]12[\/latex], the data point [latex](0, 0)[\/latex] is false for the inequality. In this case, shade the side of the line that does not contain the data point [latex](0, 0)[\/latex].<\/p>\n

      \"Completed<\/p>\n<\/div>\n<\/div>\n

      Graphing an Absolute Value Function<\/h1>\n

      To graph an absolute value function, first create a data table using the absolute value part of the equation.<\/p>\n

      The data point that is started with is the one that makes the absolute value equal to 0 (this is the [latex]x[\/latex]-value of the vertex). Calculating the value of this point is quite simple.<\/p>\n

      For example, for [latex]| x - 3 |[\/latex], the value [latex]x = 3[\/latex] makes the absolute value equal to 0.<\/p>\n

      Examples of others are:<\/p>\n

      [latex]\\begin{array}{rrrrrrrrr} |x&+&2|&=&0&\\text{when}&x&=&-2 \\\\ |x&-&11|&=&0&\\text{when}&x&=&11 \\\\ |x&+&9|&=&0&\\text{when}&x&=&-9 \\\\ \\end{array}[\/latex]<\/p>\n

      The graph of an absolute value equation will be a V-shape that opens upward for any positive absolute function and opens downward for any negative absolute value function.<\/p>\n

      \n
      \n

      Example 4.4.2<\/p>\n<\/header>\n

      \n

      Plot the graph of [latex]y = | x + 2 | - 3.[\/latex]<\/p>\n

      The data point that gives the [latex]x[\/latex]-value of the vertex is [latex]| x + 2 | = 0,[\/latex] in which [latex]x = -2.[\/latex] This is the first value.<\/p>\n

      For [latex]x = -2, y = | -2 + 2 | - 3,[\/latex] which yields [latex]y = -3.[\/latex]<\/p>\n

      Now pick [latex]x[\/latex]-values that are larger and less than \u22122 to get three data points on both sides of the vertex, [latex](-2, -3).[\/latex]<\/p>\n\n\n\n\n\n\n\n\n\n\n
      [latex]x[\/latex]<\/th>\n[latex]y[\/latex]<\/th>\n<\/tr>\n
      1<\/td>\n0<\/td>\n<\/tr>\n
      0<\/td>\n\u22121<\/td>\n<\/tr>\n
      \u22121<\/td>\n\u22122<\/td>\n<\/tr>\n
      \u22122<\/td>\n\u22123<\/td>\n<\/tr>\n
      \u22123<\/td>\n\u22122<\/td>\n<\/tr>\n
      \u22124<\/td>\n\u22121<\/td>\n<\/tr>\n
      \u22125<\/td>\n0<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Once there are three data points on either side of the vertex, plot and connect them in a line. The graph is complete.<\/p>\n

      \"Positive<\/span><\/p>\n<\/div>\n<\/div>\n

      \n
      \n

      Example 4.4.3<\/p>\n<\/header>\n

      \n

      Plot the graph of [latex]y = -| x - 2 | + 1.[\/latex]<\/p>\n

      The data point that gives the [latex]x[\/latex]-value of the vertex is [latex]| x - 2 | = 0,[\/latex] in which [latex]x = 2.[\/latex] This is the first value.<\/p>\n

      For [latex]x = 2, y = -| 2 - 2 | + 1,[\/latex] which yields [latex]y = 1.[\/latex]<\/p>\n

      Now pick [latex]x[\/latex]-values that are larger and less than 2 to get three data points on both sides of the vertex, [latex](2, 1).[\/latex]<\/p>\n\n\n\n\n\n\n\n\n\n\n
      [latex]x[\/latex]<\/th>\n[latex]y[\/latex]<\/th>\n<\/tr>\n
      5<\/td>\n\u22122<\/td>\n<\/tr>\n
      4<\/td>\n\u22121<\/td>\n<\/tr>\n
      3<\/td>\n0<\/td>\n<\/tr>\n
      2<\/td>\n1<\/td>\n<\/tr>\n
      1<\/td>\n0<\/td>\n<\/tr>\n
      0<\/td>\n\u22121<\/td>\n<\/tr>\n
      \u22121<\/td>\n\u22122<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

      Once there are three data points on either side of the vertex, plot and connect them in a line. The graph is complete.<\/p>\n

      \"Negative<\/span><\/p>\n<\/div>\n<\/div>\n

      Questions<\/h1>\n

      For questions 1 to 8, graph each linear inequality.<\/p>\n

        \n
      1. [latex]y[\/latex] > [latex]3x + 2[\/latex]<\/li>\n
      2. [latex]3x - 4y[\/latex] > [latex]12[\/latex]<\/li>\n
      3. [latex]2y \\ge 3x + 6[\/latex]<\/li>\n
      4. [latex]3x - 2y \\ge 6[\/latex]<\/li>\n
      5. [latex]2y[\/latex] > [latex]5x + 10[\/latex]<\/li>\n
      6. [latex]5x + 4y[\/latex] > [latex]-20[\/latex]<\/li>\n
      7. [latex]4y \\ge 5x + 20[\/latex]<\/li>\n
      8. [latex]5x + 2y \\ge -10[\/latex]<\/li>\n<\/ol>\n

        For questions 9 to 16, graph each absolute value equation.<\/p>\n

          \n
        1. [latex]y=|x-4|[\/latex]<\/li>\n
        2. [latex]y=|x-3|+3[\/latex]<\/li>\n
        3. [latex]y=|x-2|[\/latex]<\/li>\n
        4. [latex]y=|x-2|+2[\/latex]<\/li>\n
        5. [latex]y=-|x-2|[\/latex]<\/li>\n
        6. [latex]y=-|x-2|+2[\/latex]<\/li>\n
        7. [latex]y=-|x+2|[\/latex]<\/li>\n
        8. [latex]y=-|x+2|+2[\/latex]<\/li>\n<\/ol>\n

          Answer Key 4.4<\/a><\/p>\n","protected":false},"author":90,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":"cc-by-nc-sa"},"chapter-type":[],"contributor":[],"license":[56],"class_list":["post-1291","chapter","type-chapter","status-publish","hentry","license-cc-by-nc-sa"],"part":1241,"_links":{"self":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1291","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/users\/90"}],"version-history":[{"count":2,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1291\/revisions"}],"predecessor-version":[{"id":2093,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1291\/revisions\/2093"}],"part":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/parts\/1241"}],"metadata":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapters\/1291\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/media?parent=1291"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/pressbooks\/v2\/chapter-type?post=1291"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/contributor?post=1291"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/opentextbc.ca\/intermediatealgebraberg\/wp-json\/wp\/v2\/license?post=1291"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}