{"id":6680,"date":"2020-10-06T15:02:07","date_gmt":"2020-10-06T15:02:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/slcc-beginalgebra\/?post_type=chapter&p=6680"},"modified":"2026-02-05T01:57:12","modified_gmt":"2026-02-05T01:57:12","slug":"8-5-graphing-absolute-value-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-elementaryalgebra\/chapter\/8-5-graphing-absolute-value-functions\/","title":{"raw":"8.3: Graphing Basic Absolute Value and Quadratic Functions","rendered":"8.3: Graphing Basic Absolute Value and Quadratic Functions"},"content":{"raw":"
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Section 8.3 Learning Objectives<\/h3>\r\n8.3: Graphing Absolute Value and Quadratic Functions<\/strong>\r\n
    \r\n \t
  • Graph basic absolute value functions<\/li>\r\n \t
  • Graph basic quadratic functions<\/li>\r\n<\/ul>\r\n<\/div>\r\n \r\n

    The Basic Absolute Value Function<\/h1>\r\nIn Chapter 2, we reviewed the idea of absolute value and introduced strategies to solve absolute value equations and inequalities. In this section, our goal is to graph absolute value functions.\r\n\r\nLet us begin with the most basic absolute value function, [latex]f(x)=|x|.[\/latex]\r\n\r\nRecall that the absolute value of a number is its distance from zero on the number line. As with any function, we can use a table to assist us in graphing it.\r\n\r\nSince we are unsure of the shape, we want to pick several points, including positives and negatives. This leads to the table given below.\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]<\/td>\r\n[latex]f(x)=|x|[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-3[\/latex]<\/td>\r\n[latex]|-3|=3[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-2[\/latex]<\/td>\r\n[latex]|-2|=2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-1[\/latex]<\/td>\r\n[latex]|-1|=1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]|0|=0[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex]|1|=1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex]|2|=2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3[\/latex]<\/td>\r\n[latex]|3|=3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNext we plot the points.\r\n\r\n\"Seven\r\n\r\nHad we not chosen enough points, we may be tempted to draw something more like a smooth curve through the points. However, we can see that this will require a graph that is more linear, except that it seems to suddenly change directions at the origin. Had we only chosen points to the right of 0, we may have thought the graph would be a simple straight line instead of the \"V\" shape that is ends up being. The final graph of [latex]f(x)=|x|[\/latex] is shown below.\r\n\r\n\"Upward\r\n\r\nThis \"V\" shape will appear in all of the simple absolute value functions we will consider in this section. However, the \"V\" might be shifted, made narrower or wider, or even flipped.\r\n

    Graphing Absolute Value Functions of the Form [latex]f(x)=a|x|+b[\/latex]<\/h2>\r\nKnowing the basic shape of an absolute value graph will help us as we continue our exploration of absolute values.\r\n\r\nConsider the function [latex]f(x)=|x|-3[\/latex]. Let us try plugging in the same values for [latex]x[\/latex] that we use for the basic absolute value function.\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]<\/td>\r\n[latex]f(x)=|x|-3[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-3[\/latex]<\/td>\r\n[latex]|-3|-3=3-3=0[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-2[\/latex]<\/td>\r\n[latex]|-2|-3=2-3=-1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-1[\/latex]<\/td>\r\n[latex]|-1|-3=1-3=-2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]|0|-3=0-3=-3[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex]|1|-3=1-3=-2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex]|2|-3=2-3=-1[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]3[\/latex]<\/td>\r\n[latex]|3|-3=3-3=0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting these points gives the graph shown below.\r\n

    \"Upward<\/p>\r\nWe see that same \"V\" shape, which we will come to expect with this type of absolute value function.\u00a0 However, not coincidentally, notice that the curve has moved down 3 units.\r\n

    \r\n

    Example 1<\/h3>\r\nGraph [latex]f(x)=|x|+2[\/latex].\r\n\r\n[reveal-answer q=\"335188\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"335188\"]\r\n\r\nNow that we are starting to recognize what shape to expect, we could plug in fewer values. Let us plug in integer value for [latex]x[\/latex] from [latex]-2[\/latex] to [latex]2[\/latex]. However, if you are more comfortable plugging in more values, it never hurts!\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
    [latex]x[\/latex]<\/td>\r\n[latex]f(x)=|x|+2[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-2[\/latex]<\/td>\r\n[latex]|-2|+2=2+2=4[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]-1[\/latex]<\/td>\r\n[latex]|-1|+2=1+2=3[\/latex]<\/td>\r\n[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]0[\/latex]<\/td>\r\n[latex]|0|+2=0+2=2[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]1[\/latex]<\/td>\r\n[latex]|1|+2=1+2=3[\/latex]<\/td>\r\n[latex]3[\/latex]<\/td>\r\n<\/tr>\r\n
    [latex]2[\/latex]<\/td>\r\n[latex]|2|+2=2+2=4[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting these points, we arrive at the following graph:\r\n

    \"Upward<\/p>\r\nSo, we again end up with the same shape, but this time it is shifted up 2 units.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nStrategy for Graphing Absolute Value Functions of the Form [latex]f(x)=a|x|+b[\/latex]<\/strong>\r\n

      \r\n \t
    1. Make an [latex]xy[\/latex]-table. We recommend (at least) five values for [latex]x[\/latex], [latex]x=-2,-1,0,1,2[\/latex].<\/li>\r\n \t
    2. Graph. The resulting curve should have a \"V\" shape.<\/li>\r\n<\/ol>\r\nIn the next problem, we will see what happens when there is a coefficient other than 1 in front of the absolute value.\r\n
      \r\n

      Example 2<\/h3>\r\nGraph [latex]f(x)=2|x|[\/latex].\r\n\r\n[reveal-answer q=\"60724\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"60724\"]\r\n\r\nWe proceed by plugging in the same set of [latex]x[\/latex]-values.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=2|x|[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-2[\/latex]<\/td>\r\n[latex]2|-2|=2(2)=4[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1[\/latex]<\/td>\r\n[latex]2|-1|=2(1)=2[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex]2|0|=2(0)=0[\/latex]<\/td>\r\n[latex]0[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex]2|1|=2(1)=2[\/latex]<\/td>\r\n[latex]2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex]2|2|=2(2)=4[\/latex]<\/td>\r\n[latex]4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nThis gives us the graph below.\r\n

      \"Upward<\/p>\r\nAnd there is our \"V\" shape again. If we are exploring this for relationships and patterns, you may notice that the [latex]y[\/latex]-values were all doubled as compared to the basic absolute value function. In a way, this made the curve \"twice as tall,\" or equivalently, narrower.\r\n\r\n[\/hidden-answer]<\/span>\r\n\r\n<\/div>\r\nIn our last example, we combine some of the ideas we have seen.\u00a0 Moreover, we look at the effect of a negative coefficient.\r\n

      \r\n

      Example 3<\/h3>\r\nGraph [latex]f(x)=-2|x|+1[\/latex].\r\n\r\n[reveal-answer q=\"486270\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"486270\"]\r\n\r\nDespite everything going on with this function, we utilize the same approach. Just be careful with the computations as we follow the order of operations.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=-2|x|+1[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-2[\/latex]<\/td>\r\n[latex]-2|-2|+1=-2(2)+1=-4+1=-3[\/latex]<\/td>\r\n[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1[\/latex]<\/td>\r\n[latex]-2|-1|+1=-2(1)+1=-2+1=-1[\/latex]<\/td>\r\n[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex]-2|0|+1=-2(0)+1=0+1=1[\/latex]<\/td>\r\n[latex]1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex]-2|1|+1=-2(1)+1=-2+1=-1[\/latex]<\/td>\r\n[latex]-1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex]-2|2|+1=-2(2)+1=-4+1=-3[\/latex]<\/td>\r\n[latex]-3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe obtain the following graph:\r\n

      \"Downward<\/p>\r\nOnce again, we may find it interesting to explore how the different pieces of the function affected the graph. How do you think the negative exponent affected the graph? If you were thinking that it caused our \"V\" shape to \"flip over,\" you are right! Putting everything together, the curve flipped (vertically), is narrower, and shifted up 1 unit.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIf you would like to investigate further into absolute value graphs, check out the \"Think About It\" below.\r\n

      \r\n

      think about it<\/h3>\r\nGraph [latex]f(x)=|x-3|[\/latex].\r\n\r\n[reveal-answer q=\"799499\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"799499\"]\r\n\r\nWe could simply plug in the same values for [latex]x[\/latex] that we selected for the basic graph, but we will see that this leads to a slight problem.\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]<\/td>\r\n[latex]f(x)=|x-3|[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-3[\/latex]<\/td>\r\n[latex]|-3-3|=|-6|=6[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-2[\/latex]<\/td>\r\n[latex]|-2-3|=|-5|=5[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1[\/latex]<\/td>\r\n[latex]|-1-3|=|-4|=4[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex]|0-3|=|-3|=3[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex]|1-3|=|-2|=2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex]|2-3|=|-1|=1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]3[\/latex]<\/td>\r\n[latex]|3-3|=|0|=0[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nIf you plot these points (try it), it looks like a line and does not change directions. Yet, we know we are anticipating that \"V\" shape. So maybe we have not picked enough points or the right points. Let's continue adding to our table.\r\n\r\n\r\n\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]<\/td>\r\n[latex]f(x)=|x-3|[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-3[\/latex]<\/td>\r\n[latex]|-3-3|=|-6|=6[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-2[\/latex]<\/td>\r\n[latex]|-2-3|=|-5|=5[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]-1[\/latex]<\/td>\r\n[latex]|-1-3|=|-4|=4[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex]|0-3|=|-3|=3[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex]|1-3|=|-2|=2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex]|2-3|=|-1|=1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]3[\/latex]<\/td>\r\n[latex]|3-3|=|0|=0[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]4[\/latex]<\/td>\r\n[latex]|4-3|=|1|=1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]5[\/latex]<\/td>\r\n[latex]|5-3|=|2|=2[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]6[\/latex]<\/td>\r\n[latex]|6-3|=|3|=3[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow that traditional absolute value shape begins to emerge, and we obtain the graph for [latex]f(x)=|x-3|[\/latex] below.\r\n\r\n\"Upward\r\n\r\nThis example teaches us an important and extremely helpful lesson. Notice where the graph changed directions and where we begin to see the \"V\" shape...it occurred at [latex]x=3[\/latex]. This was not coincidence, as we saw a 3 in the original function. However, it was \"minus 3\" in the original. The key is to focus on the value for [latex]x[\/latex] that will make the quantity inside the absolute value equal to zero. In this case, with\u00a0[latex]f(x)=|x-3|[\/latex], when [latex]x=3[\/latex], we get 0 inside the absolute value. So, it is wise to choose values for [latex]x[\/latex] that are around [latex]x=3[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n

      <\/h2>\r\n

      Graphing Basic Quadratic Functions<\/h1>\r\nIn Modules 5 and 6 we learned about polynomials. An equation containing a second-degree polynomial is called a quadratic equation<\/strong>. For example, equations such as [latex]{x}^{2}-3x-4=0[\/latex] and [latex]{x}^{2}-16=0[\/latex] are in the \"family\" of quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.\r\n\r\nJust like the functions in other \"families\" we've learned about (linear, exponential, absolute value), quadratic\u00a0functions can also be graphed. It is helpful to have an idea about what the shape should be so you can be sure that you have chosen enough points to plot as a guide. Let us start with the most basic quadratic function,\u00a0[latex]f(x)=x^{2}[\/latex].\r\n\r\nGraph [latex]f(x)=x^{2}[\/latex].\r\nWe'll start with a table of values. Then think of each row of the table as an ordered pair.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
      [latex]\u22122[\/latex]<\/td>\r\n[latex](-2)^2=4[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]\u22121[\/latex]<\/td>\r\n[latex](-1)^2=1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex](0)^2=0[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex](1)^2=1[\/latex]<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex](2)^2=4[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nNow we'll plot the points [latex](-2,4), (-1,1), (0,0), (1,1), (2,4)[\/latex]\r\n\r\n\"Coordinate\r\n\r\nAt first glance, it might appear that these points make a \"V\" shape like the absolute value functions we have graphed previously. But, the quadratic family of functions is not drawn using straight lines. Since the points are not<\/i> on a line, you cannot use a straight edge<\/strong>. Connect the points as best you can using a smooth curve<\/i> (not a series of straight lines). You may want to find and plot additional points (such as the ones in blue below). Placing arrows on the tips of the lines implies that they continue in that direction forever.\r\n\r\n\"Graph\r\n\r\nNotice that the shape is similar to the letter U. This is called a parabola<\/em>. One-half of the parabola is a mirror image of the other half. The lowest point on this graph is called the vertex<\/em>. The vertical line that goes through the vertex is called the line of symmetry<\/em>. In this case, that line is the\u00a0y<\/i>-axis, which is the line [latex]x=0[\/latex].\r\n\r\nIn the following video, we show an example of graphing another quadratic function,\u00a0[latex]f(x)=\\frac{1}{2}x^2[\/latex], using a table of values.\r\n\r\nhttps:\/\/youtu.be\/wYfEzOJugS8\r\n\r\nThe equations for quadratic functions can be written in the form [latex]f(x)=ax^{2}+bx+c[\/latex] (where [latex] a\\ne 0[\/latex]). In the two basic quadratic functions we graphed above, notice there was only the [latex]ax^2[\/latex] term and that both [latex]b=0[\/latex], and [latex]c=0[\/latex] for those functions.\r\n\r\nAlthough there are many forms that quadratic functions can be written in, this Elementary Algebra course will focus only on graphing basic<\/em> quadratic functions of the form [latex]f(x)=ax^{2}+c[\/latex].\r\n

      Quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] will always be centered around the y-axis which is the line [latex]x=0[\/latex].\u00a0<\/strong><\/p>\r\nWe have shared two examples above of graphing a basic quadratic function of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], where [latex]c=0[\/latex], so let's now explore how to graph a quadratic function of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] where\u00a0[latex]c\\ne 0[\/latex] .\r\n\r\nGraph [latex]f(x)=x^{2}-4[\/latex].\r\nSimilar to the problems above, lets start with a table of values.\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2-4[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n<\/thead>\r\n
      [latex]\u22122[\/latex]<\/td>\r\n[latex](-2)^2-4 = 4-4 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      [latex]\u22121[\/latex]<\/td>\r\n[latex](-1)^2-4 = 1-4 = -3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
      [latex]0[\/latex]<\/td>\r\n[latex](0)^2-4 = 0-4 = -4[\/latex]<\/td>\r\n-4<\/td>\r\n<\/tr>\r\n
      [latex]1[\/latex]<\/td>\r\n[latex](1)^2-4 = 1-4 = -3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
      [latex]2[\/latex]<\/td>\r\n[latex](2)^2-4 = 4-4 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nWe'll now plot the points [latex](-2,0), (-1,-3), (0,-4), (1,-3), (2,0)[\/latex] on the graph.\r\n\r\n\"Coordinate\r\n\r\nIf we connect all the points with a smooth curve we will have the graph of our function [latex]f(x)=x^{2}-4[\/latex].\r\n\r\n\"Coordinate\r\n\r\nLet's have you now try one on your own:\r\n
      \r\n

      Example 4<\/h3>\r\nGraph the function: [latex]f(x)=x^2-16[\/latex]\r\n\r\n[reveal-answer q=\"184639\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"184639\"]\r\n\r\nNotice our function\u00a0[latex]f(x)=x^2-16[\/latex] is of the form [latex]f(x)=ax^{2}+c[\/latex]. This means it will be centered around [latex]x=0[\/latex].\r\n\r\nWe can start our table with any values we want, but will want to make sure to choose values on both sides of 0, as well as the value of [latex]x=0[\/latex].\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2-16[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      4<\/td>\r\n[latex](4)^2-16=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      -4<\/td>\r\n[latex](-4)^2-16=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      0<\/td>\r\n[latex](0)^2-16=-16[\/latex]<\/td>\r\n-16<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n[latex](2)^2-16=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n
      -2<\/td>\r\n[latex](-2)^2-16=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting the ordered pairs from the table above gives us:\r\n\r\n\"Coordinate\r\n\r\nConnecting the points with a smooth curve gives our final graph:\r\n\r\n\"Coordinate\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nLets try a few more examples.\r\n
      \r\n

      Example 5<\/h3>\r\nGraph the function: [latex]f(x)=x^2+5[\/latex]\r\n\r\n[reveal-answer q=\"613009\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"613009\"]\r\n\r\nNotice our function [latex]f(x)=x^2+5[\/latex] is of the form [latex]f(x)=ax^{2}+c[\/latex]. This means it will be centered around [latex]x=0[\/latex].\r\n\r\nCreate a table of values\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=x^2+5[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      -2<\/td>\r\n[latex](-2)^2+5=9[\/latex]<\/td>\r\n9<\/td>\r\n<\/tr>\r\n
      -1<\/td>\r\n[latex](-1)^2+5=6[\/latex]<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
      0<\/td>\r\n[latex](0)^2+5=5[\/latex]<\/td>\r\n5<\/td>\r\n<\/tr>\r\n
      1<\/td>\r\n[latex](1)^2+5=6[\/latex]<\/td>\r\n6<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n[latex](2)^2+5=9[\/latex]<\/td>\r\n9<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlot the points (-2,9), (-1,6), (0,5), (1,6) and (2,9) and connect them with a smooth curve.\r\n\r\n\"Coordinate\r\n\r\nExtension:\u00a0<\/b>\r\n\r\nNotice some quadratic functions never cross the x-axis. The function [latex]f(x)=x^2 +5[\/latex] is one example. If we try to solve for the x-intercepts in this function by setting the function equal to 0, we would have the following:\r\n

      [latex]x^2 +5=0[\/latex]<\/p>\r\n

      [latex]x^2 =-5[\/latex]<\/p>\r\n

      [latex]\\sqrt{x^2}=\u00b1\\sqrt{-5}[\/latex]<\/p>\r\nIf we try and find the square root of -5 in our calculator we will find that it gives us an error message. There are no real number answers to the square root of a negative value. Thus this function does not have any x-intercepts on our graph.\r\n\r\nAs we can see with the graph we made above, the parabola falls above the x-axis.\r\n\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nIn the next example we will explore how parabolas don't always open upwards.\r\n

      \r\n

      Example 6<\/h3>\r\nGraph the function: [latex]f(x)=-3x^2[\/latex]\r\n\r\n[reveal-answer q=\"824856\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"824856\"]\r\n\r\nNotice our function\u00a0[latex]f(x)=-3x^2[\/latex], is of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], where [latex] a=-3[\/latex] and [latex] c=0[\/latex]. We will start to graph this by choosing values surrounding 0.\r\n\r\n*It is important that you understand how to evaluate a function like this correctly. For this reason, we will show the steps for the input value of -2 in detail:\r\n

      [latex]f(-2)= -3(-2)^2[\/latex]<\/p>\r\nWe must make sure we follow the correct \"Order of Operations\" here. This means that\u00a0before\u00a0<\/em>we multiply the [latex]-3[\/latex] with anything, we must\u00a0first\u00a0<\/em>raise the\u00a0 [latex]-2[\/latex] to the exponent of 2. When we raise\u00a0[latex](-2)^2[\/latex], we are multiplying\u00a0[latex](-2)\\cdot(-2)[\/latex], which is equal to\u00a0[latex]4[\/latex].\r\n

      [latex]f(-2)= -3(4)[\/latex]<\/p>\r\nWe can now multiply the -3 with the 4 to find the final output.\r\n

      [latex]f(-2)= -12[\/latex]<\/p>\r\nThis process will be repeated with all the x values input in the table of values below:\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      [latex]x[\/latex]<\/td>\r\n[latex]f(x)=-3x^2[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
      -2<\/td>\r\n[latex]-3(-2)^2=(-3)(4)=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n
      -1<\/td>\r\n[latex]-3(-1)^2=(-3)(1)=-3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
      0<\/td>\r\n[latex]-3(0)^2=(-3)(0)=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
      1<\/td>\r\n[latex]-3(1)^2=(-3)(1)=-3[\/latex]<\/td>\r\n-3<\/td>\r\n<\/tr>\r\n
      2<\/td>\r\n[latex]-3(2)^2=(-3)(4)=-12[\/latex]<\/td>\r\n-12<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nPlotting the ordered pairs from the table above gives us:\r\n\r\n\"Coordinate\r\n\r\nConnecting the points with a smooth curve gives our final graph below. Notice that this is an upside down U-shape this time.\r\n\r\n\"Coordinate\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs we saw with the last example, with quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex], changing the value of [latex]a[\/latex]\u00a0can change the width of the parabola and whether it opens up ([latex]a>0[\/latex]) or down ([latex]a<0[\/latex]). If [latex]a[\/latex] is positive, the vertex is the lowest point, and the parabola opens up. If [latex]a[\/latex] is negative, the vertex is the highest point, and the parabola opens down.\r\n
      \r\n\r\nWhen graphing quadratic functions of the form\u00a0[latex]f(x)=ax^{2}+c[\/latex] follow the steps below:<\/span><\/strong>\r\n
        \r\n \t
      1. Recognize the form of the quadratic function and that it will be a parabola centered around x=0.<\/span><\/strong><\/li>\r\n \t
      2. Make a table of values, making sure to choose some values on either side of x=0, as well as the value of x=0<\/span><\/strong><\/li>\r\n \t
      3. Plot your points and connect them with a smooth curve into a parabola shape.\u00a0<\/span><\/strong><\/li>\r\n<\/ol>\r\n<\/div>\r\n
        \r\n

        Think About it<\/h3>\r\nGraph the equation: [latex]f(x)={x}^{2}+x - 6[\/latex]. (It may be helpful to factor it, and set it equal to 0 to find the [latex]x[\/latex]-intercepts.)\r\n[reveal-answer q=\"609710\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"609710\"]\r\n\r\nTo factor [latex]f(x)={x}^{2}+x - 6[\/latex], we look for two numbers whose product equals [latex]-6[\/latex] and whose sum equals\u00a0[latex]1[\/latex]. Begin by looking at the possible factors of [latex]-6[\/latex].\r\n
        [latex]\\begin{array}{l}1\\cdot \\left(-6\\right)\\hfill \\\\ \\left(-6\\right)\\cdot 1\\hfill \\\\ 2\\cdot \\left(-3\\right)\\hfill \\\\ 3\\cdot \\left(-2\\right)\\hfill \\end{array}[\/latex]<\/div>\r\nThe last pair, [latex]3\\cdot \\left(-2\\right)[\/latex] sums to\u00a0[latex]1[\/latex], so these are the numbers. Note that only one pair of numbers will work. Then, write the factors.\r\n
        [latex]f(x)=\\left(x - 2\\right)\\left(x+3\\right)[\/latex]<\/div>\r\nWe can now set this equation equal to 0 and use the zero-product property to find the\u00a0[latex]x[\/latex]-intercepts. To do this, we will set each factor equal to zero and solve.<\/span>\r\n

        [latex]\\left(x - 2\\right)\\left(x+3\\right)=0[\/latex]<\/p>\r\n\r\n

        [latex]\\begin{array}{lll}& \\left(x - 2\\right)\\left(x+3\\right)=0 & \\hfill \\\\ \\left(x - 2\\right)\\hfill=0 & & \\left(x+3\\right)=0 \\hfill \\\\ x=2 & & x=-3 \\hfill \\end{array}[\/latex]<\/div>\r\nRecall that x-intercepts are where the outputs, or y values are zero, therefore the points\u00a0[latex](-3,0)[\/latex] and\u00a0[latex](2,0)[\/latex] represent the places where the parabola crosses the x axis.\r\n\r\nTo find other points on the graph, we could make a table of values\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
        [latex]x[\/latex]<\/td>\r\n[latex]f(x)={x}^{2}+x - 6[\/latex]<\/td>\r\n[latex]f(x)[\/latex]<\/td>\r\n<\/tr>\r\n
        2<\/td>\r\n[latex](2)^2+(2) - 6 = 0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
        -3<\/td>\r\n[latex](-3)^2+(-3)-6=0[\/latex]<\/td>\r\n0<\/td>\r\n<\/tr>\r\n
        0<\/td>\r\n[latex](0)^2+(0)-6=-6[\/latex]<\/td>\r\n-6<\/td>\r\n<\/tr>\r\n
        1<\/td>\r\n[latex](1)^2+(1)-6=-4[\/latex]<\/td>\r\n-4<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n \r\n\r\nIf we plot the points found above and connect them with a smooth curve, we will get the parabola in the figure below.\r\n\r\n\"Coordinate\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

        Graphing Quadratics Summary<\/h2>\r\nCreating a graph of a function is one way to understand the relationship between the inputs and outputs of that function. Creating a graph can be done by choosing values for x<\/em>, finding the corresponding y<\/em> values, and plotting them. However, it helps to understand the basic shape of the function. Knowing how changes to the basic function equation affect the graph is also helpful.\r\n\r\nThe shape of a quadratic function is a parabola. Parabolas have the equation [latex]f(x)=ax^{2}+bx+c[\/latex], where [latex]a, b[\/latex] and [latex]c[\/latex] are real numbers and [latex]a\\ne0[\/latex]. The value of [latex]a[\/latex] determines the width and the direction of the parabola, while the vertex depends on the values of [latex]a, b[\/latex] and [latex]c[\/latex].","rendered":"
        \n

        Section 8.3 Learning Objectives<\/h3>\n

        8.3: Graphing Absolute Value and Quadratic Functions<\/strong><\/p>\n

          \n
        • Graph basic absolute value functions<\/li>\n
        • Graph basic quadratic functions<\/li>\n<\/ul>\n<\/div>\n

           <\/p>\n

          The Basic Absolute Value Function<\/h1>\n

          In Chapter 2, we reviewed the idea of absolute value and introduced strategies to solve absolute value equations and inequalities. In this section, our goal is to graph absolute value functions.<\/p>\n

          Let us begin with the most basic absolute value function, [latex]f(x)=|x|.[\/latex]<\/p>\n

          Recall that the absolute value of a number is its distance from zero on the number line. As with any function, we can use a table to assist us in graphing it.<\/p>\n

          Since we are unsure of the shape, we want to pick several points, including positives and negatives. This leads to the table given below.<\/p>\n\n\n\n\n\n\n\n\n\n\n
          [latex]x[\/latex]<\/td>\n[latex]f(x)=|x|[\/latex]<\/td>\n<\/tr>\n
          [latex]-3[\/latex]<\/td>\n[latex]|-3|=3[\/latex]<\/td>\n<\/tr>\n
          [latex]-2[\/latex]<\/td>\n[latex]|-2|=2[\/latex]<\/td>\n<\/tr>\n
          [latex]-1[\/latex]<\/td>\n[latex]|-1|=1[\/latex]<\/td>\n<\/tr>\n
          [latex]0[\/latex]<\/td>\n[latex]|0|=0[\/latex]<\/td>\n<\/tr>\n
          [latex]1[\/latex]<\/td>\n[latex]|1|=1[\/latex]<\/td>\n<\/tr>\n
          [latex]2[\/latex]<\/td>\n[latex]|2|=2[\/latex]<\/td>\n<\/tr>\n
          [latex]3[\/latex]<\/td>\n[latex]|3|=3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Next we plot the points.<\/p>\n

          \"Seven<\/p>\n

          Had we not chosen enough points, we may be tempted to draw something more like a smooth curve through the points. However, we can see that this will require a graph that is more linear, except that it seems to suddenly change directions at the origin. Had we only chosen points to the right of 0, we may have thought the graph would be a simple straight line instead of the “V” shape that is ends up being. The final graph of [latex]f(x)=|x|[\/latex] is shown below.<\/p>\n

          \"Upward<\/p>\n

          This “V” shape will appear in all of the simple absolute value functions we will consider in this section. However, the “V” might be shifted, made narrower or wider, or even flipped.<\/p>\n

          Graphing Absolute Value Functions of the Form [latex]f(x)=a|x|+b[\/latex]<\/h2>\n

          Knowing the basic shape of an absolute value graph will help us as we continue our exploration of absolute values.<\/p>\n

          Consider the function [latex]f(x)=|x|-3[\/latex]. Let us try plugging in the same values for [latex]x[\/latex] that we use for the basic absolute value function.<\/p>\n\n\n\n\n\n\n\n\n\n\n
          [latex]x[\/latex]<\/td>\n[latex]f(x)=|x|-3[\/latex]<\/td>\n<\/tr>\n
          [latex]-3[\/latex]<\/td>\n[latex]|-3|-3=3-3=0[\/latex]<\/td>\n<\/tr>\n
          [latex]-2[\/latex]<\/td>\n[latex]|-2|-3=2-3=-1[\/latex]<\/td>\n<\/tr>\n
          [latex]-1[\/latex]<\/td>\n[latex]|-1|-3=1-3=-2[\/latex]<\/td>\n<\/tr>\n
          [latex]0[\/latex]<\/td>\n[latex]|0|-3=0-3=-3[\/latex]<\/td>\n<\/tr>\n
          [latex]1[\/latex]<\/td>\n[latex]|1|-3=1-3=-2[\/latex]<\/td>\n<\/tr>\n
          [latex]2[\/latex]<\/td>\n[latex]|2|-3=2-3=-1[\/latex]<\/td>\n<\/tr>\n
          [latex]3[\/latex]<\/td>\n[latex]|3|-3=3-3=0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Plotting these points gives the graph shown below.<\/p>\n

          \"Upward<\/p>\n

          We see that same “V” shape, which we will come to expect with this type of absolute value function.\u00a0 However, not coincidentally, notice that the curve has moved down 3 units.<\/p>\n

          \n

          Example 1<\/h3>\n

          Graph [latex]f(x)=|x|+2[\/latex].<\/p>\n

          Show Solution<\/span><\/p>\n
          \n

          Now that we are starting to recognize what shape to expect, we could plug in fewer values. Let us plug in integer value for [latex]x[\/latex] from [latex]-2[\/latex] to [latex]2[\/latex]. However, if you are more comfortable plugging in more values, it never hurts!<\/p>\n\n\n\n\n\n\n\n\n
          [latex]x[\/latex]<\/td>\n[latex]f(x)=|x|+2[\/latex]<\/td>\n[latex]f(x)[\/latex]<\/td>\n<\/tr>\n
          [latex]-2[\/latex]<\/td>\n[latex]|-2|+2=2+2=4[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n<\/tr>\n
          [latex]-1[\/latex]<\/td>\n[latex]|-1|+2=1+2=3[\/latex]<\/td>\n[latex]3[\/latex]<\/td>\n<\/tr>\n
          [latex]0[\/latex]<\/td>\n[latex]|0|+2=0+2=2[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr>\n
          [latex]1[\/latex]<\/td>\n[latex]|1|+2=1+2=3[\/latex]<\/td>\n[latex]3[\/latex]<\/td>\n<\/tr>\n
          [latex]2[\/latex]<\/td>\n[latex]|2|+2=2+2=4[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

          Plotting these points, we arrive at the following graph:<\/p>\n

          \"Upward<\/p>\n

          So, we again end up with the same shape, but this time it is shifted up 2 units.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

          Strategy for Graphing Absolute Value Functions of the Form [latex]f(x)=a|x|+b[\/latex]<\/strong><\/p>\n

            \n
          1. Make an [latex]xy[\/latex]-table. We recommend (at least) five values for [latex]x[\/latex], [latex]x=-2,-1,0,1,2[\/latex].<\/li>\n
          2. Graph. The resulting curve should have a “V” shape.<\/li>\n<\/ol>\n

            In the next problem, we will see what happens when there is a coefficient other than 1 in front of the absolute value.<\/p>\n

            \n

            Example 2<\/h3>\n

            Graph [latex]f(x)=2|x|[\/latex].<\/p>\n

            Show Solution<\/span><\/p>\n
            \n

            We proceed by plugging in the same set of [latex]x[\/latex]-values.<\/p>\n\n\n\n\n\n\n\n\n
            [latex]x[\/latex]<\/td>\n[latex]f(x)=2|x|[\/latex]<\/td>\n[latex]f(x)[\/latex]<\/td>\n<\/tr>\n
            [latex]-2[\/latex]<\/td>\n[latex]2|-2|=2(2)=4[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n<\/tr>\n
            [latex]-1[\/latex]<\/td>\n[latex]2|-1|=2(1)=2[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr>\n
            [latex]0[\/latex]<\/td>\n[latex]2|0|=2(0)=0[\/latex]<\/td>\n[latex]0[\/latex]<\/td>\n<\/tr>\n
            [latex]1[\/latex]<\/td>\n[latex]2|1|=2(1)=2[\/latex]<\/td>\n[latex]2[\/latex]<\/td>\n<\/tr>\n
            [latex]2[\/latex]<\/td>\n[latex]2|2|=2(2)=4[\/latex]<\/td>\n[latex]4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            This gives us the graph below.<\/p>\n

            \"Upward<\/p>\n

            And there is our “V” shape again. If we are exploring this for relationships and patterns, you may notice that the [latex]y[\/latex]-values were all doubled as compared to the basic absolute value function. In a way, this made the curve “twice as tall,” or equivalently, narrower.<\/p>\n

            <\/div>\n<\/div>\n

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

            In our last example, we combine some of the ideas we have seen.\u00a0 Moreover, we look at the effect of a negative coefficient.<\/p>\n

            \n

            Example 3<\/h3>\n

            Graph [latex]f(x)=-2|x|+1[\/latex].<\/p>\n

            Show Solution<\/span><\/p>\n
            \n

            Despite everything going on with this function, we utilize the same approach. Just be careful with the computations as we follow the order of operations.<\/p>\n\n\n\n\n\n\n\n\n
            [latex]x[\/latex]<\/td>\n[latex]f(x)=-2|x|+1[\/latex]<\/td>\n[latex]f(x)[\/latex]<\/td>\n<\/tr>\n
            [latex]-2[\/latex]<\/td>\n[latex]-2|-2|+1=-2(2)+1=-4+1=-3[\/latex]<\/td>\n[latex]-3[\/latex]<\/td>\n<\/tr>\n
            [latex]-1[\/latex]<\/td>\n[latex]-2|-1|+1=-2(1)+1=-2+1=-1[\/latex]<\/td>\n[latex]-1[\/latex]<\/td>\n<\/tr>\n
            [latex]0[\/latex]<\/td>\n[latex]-2|0|+1=-2(0)+1=0+1=1[\/latex]<\/td>\n[latex]1[\/latex]<\/td>\n<\/tr>\n
            [latex]1[\/latex]<\/td>\n[latex]-2|1|+1=-2(1)+1=-2+1=-1[\/latex]<\/td>\n[latex]-1[\/latex]<\/td>\n<\/tr>\n
            [latex]2[\/latex]<\/td>\n[latex]-2|2|+1=-2(2)+1=-4+1=-3[\/latex]<\/td>\n[latex]-3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            We obtain the following graph:<\/p>\n

            \"Downward<\/p>\n

            Once again, we may find it interesting to explore how the different pieces of the function affected the graph. How do you think the negative exponent affected the graph? If you were thinking that it caused our “V” shape to “flip over,” you are right! Putting everything together, the curve flipped (vertically), is narrower, and shifted up 1 unit.<\/p>\n<\/div>\n<\/div>\n<\/div>\n

            If you would like to investigate further into absolute value graphs, check out the “Think About It” below.<\/p>\n

            \n

            think about it<\/h3>\n

            Graph [latex]f(x)=|x-3|[\/latex].<\/p>\n

            Show Solution<\/span><\/p>\n
            \n

            We could simply plug in the same values for [latex]x[\/latex] that we selected for the basic graph, but we will see that this leads to a slight problem.<\/p>\n\n\n\n\n\n\n\n\n\n\n
            [latex]x[\/latex]<\/td>\n[latex]f(x)=|x-3|[\/latex]<\/td>\n<\/tr>\n
            [latex]-3[\/latex]<\/td>\n[latex]|-3-3|=|-6|=6[\/latex]<\/td>\n<\/tr>\n
            [latex]-2[\/latex]<\/td>\n[latex]|-2-3|=|-5|=5[\/latex]<\/td>\n<\/tr>\n
            [latex]-1[\/latex]<\/td>\n[latex]|-1-3|=|-4|=4[\/latex]<\/td>\n<\/tr>\n
            [latex]0[\/latex]<\/td>\n[latex]|0-3|=|-3|=3[\/latex]<\/td>\n<\/tr>\n
            [latex]1[\/latex]<\/td>\n[latex]|1-3|=|-2|=2[\/latex]<\/td>\n<\/tr>\n
            [latex]2[\/latex]<\/td>\n[latex]|2-3|=|-1|=1[\/latex]<\/td>\n<\/tr>\n
            [latex]3[\/latex]<\/td>\n[latex]|3-3|=|0|=0[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            If you plot these points (try it), it looks like a line and does not change directions. Yet, we know we are anticipating that “V” shape. So maybe we have not picked enough points or the right points. Let’s continue adding to our table.<\/p>\n\n\n\n\n\n\n\n\n\n\n\n\n\n
            [latex]x[\/latex]<\/td>\n[latex]f(x)=|x-3|[\/latex]<\/td>\n<\/tr>\n
            [latex]-3[\/latex]<\/td>\n[latex]|-3-3|=|-6|=6[\/latex]<\/td>\n<\/tr>\n
            [latex]-2[\/latex]<\/td>\n[latex]|-2-3|=|-5|=5[\/latex]<\/td>\n<\/tr>\n
            [latex]-1[\/latex]<\/td>\n[latex]|-1-3|=|-4|=4[\/latex]<\/td>\n<\/tr>\n
            [latex]0[\/latex]<\/td>\n[latex]|0-3|=|-3|=3[\/latex]<\/td>\n<\/tr>\n
            [latex]1[\/latex]<\/td>\n[latex]|1-3|=|-2|=2[\/latex]<\/td>\n<\/tr>\n
            [latex]2[\/latex]<\/td>\n[latex]|2-3|=|-1|=1[\/latex]<\/td>\n<\/tr>\n
            [latex]3[\/latex]<\/td>\n[latex]|3-3|=|0|=0[\/latex]<\/td>\n<\/tr>\n
            [latex]4[\/latex]<\/td>\n[latex]|4-3|=|1|=1[\/latex]<\/td>\n<\/tr>\n
            [latex]5[\/latex]<\/td>\n[latex]|5-3|=|2|=2[\/latex]<\/td>\n<\/tr>\n
            [latex]6[\/latex]<\/td>\n[latex]|6-3|=|3|=3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            Now that traditional absolute value shape begins to emerge, and we obtain the graph for [latex]f(x)=|x-3|[\/latex] below.<\/p>\n

            \"Upward<\/p>\n

            This example teaches us an important and extremely helpful lesson. Notice where the graph changed directions and where we begin to see the “V” shape…it occurred at [latex]x=3[\/latex]. This was not coincidence, as we saw a 3 in the original function. However, it was “minus 3” in the original. The key is to focus on the value for [latex]x[\/latex] that will make the quantity inside the absolute value equal to zero. In this case, with\u00a0[latex]f(x)=|x-3|[\/latex], when [latex]x=3[\/latex], we get 0 inside the absolute value. So, it is wise to choose values for [latex]x[\/latex] that are around [latex]x=3[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n

            <\/h2>\n

            Graphing Basic Quadratic Functions<\/h1>\n

            In Modules 5 and 6 we learned about polynomials. An equation containing a second-degree polynomial is called a quadratic equation<\/strong>. For example, equations such as [latex]{x}^{2}-3x-4=0[\/latex] and [latex]{x}^{2}-16=0[\/latex] are in the “family” of quadratic equations. They are used in countless ways in the fields of engineering, architecture, finance, biological science, and, of course, mathematics.<\/p>\n

            Just like the functions in other “families” we’ve learned about (linear, exponential, absolute value), quadratic\u00a0functions can also be graphed. It is helpful to have an idea about what the shape should be so you can be sure that you have chosen enough points to plot as a guide. Let us start with the most basic quadratic function,\u00a0[latex]f(x)=x^{2}[\/latex].<\/p>\n

            Graph [latex]f(x)=x^{2}[\/latex].
            \nWe’ll start with a table of values. Then think of each row of the table as an ordered pair.<\/p>\n\n\n\n\n\n\n\n\n\n
            [latex]x[\/latex]<\/td>\n[latex]f(x)=x^2[\/latex]<\/td>\n<\/tr>\n<\/thead>\n
            [latex]\u22122[\/latex]<\/td>\n[latex](-2)^2=4[\/latex]<\/td>\n<\/tr>\n
            [latex]\u22121[\/latex]<\/td>\n[latex](-1)^2=1[\/latex]<\/td>\n<\/tr>\n
            [latex]0[\/latex]<\/td>\n[latex](0)^2=0[\/latex]<\/td>\n<\/tr>\n
            [latex]1[\/latex]<\/td>\n[latex](1)^2=1[\/latex]<\/td>\n<\/tr>\n
            [latex]2[\/latex]<\/td>\n[latex](2)^2=4[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

            Now we’ll plot the points [latex](-2,4), (-1,1), (0,0), (1,1), (2,4)[\/latex]<\/p>\n

            \"Coordinate<\/p>\n

            At first glance, it might appear that these points make a “V” shape like the absolute value functions we have graphed previously. But, the quadratic family of functions is not drawn using straight lines. Since the points are not<\/i> on a line, you cannot use a straight edge<\/strong>. Connect the points as best you can using a smooth curve<\/i> (not a series of straight lines). You may want to find and plot additional points (such as the ones in blue below). Placing arrows on the tips of the lines implies that they continue in that direction forever.<\/p>\n

            \"Graph<\/p>\n

            Notice that the shape is similar to the letter U. This is called a parabola<\/em>. One-half of the parabola is a mirror image of the other half. The lowest point on this graph is called the vertex<\/em>. The vertical line that goes through the vertex is called the line of symmetry<\/em>. In this case, that line is the\u00a0y<\/i>-axis, which is the line [latex]x=0[\/latex].<\/p>\n

            In the following video, we show an example of graphing another quadratic function,\u00a0[latex]f(x)=\\frac{1}{2}x^2[\/latex], using a table of values.<\/p>\n