{"id":2550,"date":"2025-08-13T18:02:34","date_gmt":"2025-08-13T18:02:34","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=2550"},"modified":"2025-10-22T23:15:31","modified_gmt":"2025-10-22T23:15:31","slug":"parabolas-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/parabolas-learn-it-2\/","title":{"raw":"Parabolas: Learn It 2","rendered":"Parabolas: Learn It 2"},"content":{"raw":"

Graphing Parabolas with Vertices Not at the Origin<\/h2>\r\nLike other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].\r\n\r\nTo graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.\r\n\r\n
\r\n

parabolas with vertex (h,k)<\/h3>\r\n\r\n\r\n\r\n\r\n\r\n
Axis of Symmetry<\/strong><\/td>\r\nEquation<\/strong><\/td>\r\nFocus<\/strong><\/td>\r\nDirectrix<\/strong><\/td>\r\nEndpoints of Latus Rectum<\/strong><\/td>\r\n<\/tr>\r\n
[latex]y=k[\/latex]<\/td>\r\n[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\r\n[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\r\n[latex]x=h-p[\/latex]<\/td>\r\n[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n
[latex]x=h[\/latex]<\/td>\r\n[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\r\n[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\r\n[latex]y=k-p[\/latex]<\/td>\r\n[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\"\"\r\n\r\n<\/section>
How To: Given a standard form equation for a parabola centered at (h<\/em>, k<\/em>), sketch the graph.<\/strong>\r\n
    \r\n \t
  1. Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\r\n \t
  2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.\r\n
      \r\n \t
    1. If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\r\n
        \r\n \t
      • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
      • use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\r\n \t
      • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\r\n \t
      • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\r\n \t
      • use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\r\n \t
      • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n \t
      • If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\r\n
          \r\n \t
        • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\r\n \t
        • use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\r\n \t
        • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\r\n \t
        • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\r\n \t
        • use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\r\n \t
        • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n \t
        • Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\r\n<\/ol>\r\n<\/section>
          Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the vertex<\/strong>, axis of symmetry<\/strong>, focus<\/strong>, directrix<\/strong>, and endpoints of the latus rectum<\/strong>.[reveal-answer q=\"260087\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"260087\"]\r\n\r\nThe standard form that applies to the given equation is [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]. Thus, the axis of symmetry is parallel to the x<\/em>-axis. It follows that:\r\n
          \r\n
            \r\n \t
          • the vertex is [latex]\\left(h,k\\right)=\\left(-3,1\\right)[\/latex]<\/li>\r\n \t
          • the axis of symmetry is [latex]y=k=1[\/latex]<\/li>\r\n \t
          • [latex]-16=4p[\/latex], so [latex]p=-4[\/latex]. Since [latex]p<0[\/latex], the parabola opens left.<\/li>\r\n \t
          • the coordinates of the focus are [latex]\\left(h+p,k\\right)=\\left(-3+\\left(-4\\right),1\\right)=\\left(-7,1\\right)[\/latex]<\/li>\r\n \t
          • the equation of the directrix is [latex]x=h-p=-3-\\left(-4\\right)=1[\/latex]<\/li>\r\n \t
          • the endpoints of the latus rectum are [latex]\\left(h+p,k\\pm 2p\\right)=\\left(-3+\\left(-4\\right),1\\pm 2\\left(-4\\right)\\right)[\/latex], or [latex]\\left(-7,-7\\right)[\/latex] and [latex]\\left(-7,9\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
            \r\n
            \r\n\r\nGraph [latex]{\\left(y+1\\right)}^{2}=4\\left(x - 8\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.\r\n\r\n[reveal-answer q=\"252238\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"252238\"]\r\n\r\nVertex: [latex]\\left(8,-1\\right)[\/latex]; Axis of symmetry: [latex]y=-1[\/latex]; Focus: [latex]\\left(9,-1\\right)[\/latex]; Directrix: [latex]x=7[\/latex]; Endpoints of the latus rectum: [latex]\\left(9,-3\\right)[\/latex] and [latex]\\left(9,1\\right)[\/latex].\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<\/section>\r\n
            Graph [latex]{x}^{2}-8x - 28y - 208=0[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.[reveal-answer q=\"974028\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"974028\"]\r\n\r\nStart by writing the equation of the parabola<\/strong> in standard form. The standard form that applies to the given equation is [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]. Thus, the axis of symmetry is parallel to the y<\/em>-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable [latex]x[\/latex] in order to complete the square.\r\n

            [latex]\\begin{gathered}{x}^{2}-8x - 28y - 208=0 \\\\ {x}^{2}-8x=28y+208 \\\\ {x}^{2}-8x+16=28y+208+16 \\\\ {\\left(x - 4\\right)}^{2}=28y+224 \\\\ {\\left(x - 4\\right)}^{2}=28\\left(y+8\\right) \\\\ {\\left(x - 4\\right)}^{2}=4\\cdot 7\\cdot \\left(y+8\\right) \\end{gathered}[\/latex]<\/p>\r\nIt follows that:\r\n

            \r\n
              \r\n \t
            • the vertex is [latex]\\left(h,k\\right)=\\left(4,-8\\right)[\/latex]<\/li>\r\n \t
            • the axis of symmetry is [latex]x=h=4[\/latex]<\/li>\r\n \t
            • since [latex]p=7,p>0[\/latex] and so the parabola opens up<\/li>\r\n \t
            • the coordinates of the focus are [latex]\\left(h,k+p\\right)=\\left(4,-8+7\\right)=\\left(4,-1\\right)[\/latex]<\/li>\r\n \t
            • the equation of the directrix is [latex]y=k-p=-8 - 7=-15[\/latex]<\/li>\r\n \t
            • the endpoints of the latus rectum are [latex]\\left(h\\pm 2p,k+p\\right)=\\left(4\\pm 2\\left(7\\right),-8+7\\right)[\/latex], or [latex]\\left(-10,-1\\right)[\/latex] and [latex]\\left(18,-1\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\nNext we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.\r\n\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
              Graph [latex]{\\left(x+2\\right)}^{2}=-20\\left(y - 3\\right)[\/latex]. Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.[reveal-answer q=\"84557\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"84557\"]Vertex: [latex]\\left(-2,3\\right)[\/latex]; Axis of symmetry: [latex]x=-2[\/latex]; Focus: [latex]\\left(-2,-2\\right)[\/latex]; Directrix: [latex]y=8[\/latex]; Endpoints of the latus rectum: [latex]\\left(-12,-2\\right)[\/latex] and [latex]\\left(8,-2\\right)[\/latex].\r\n\"\"\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
              [ohm_question hide_question_numbers=1]174066[\/ohm_question]<\/section><\/div>","rendered":"

              Graphing Parabolas with Vertices Not at the Origin<\/h2>\n

              Like other graphs we\u2019ve worked with, the graph of a parabola can be translated. If a parabola is translated [latex]h[\/latex] units horizontally and [latex]k[\/latex] units vertically, the vertex will be [latex]\\left(h,k\\right)[\/latex]. This translation results in the standard form of the equation we saw previously with [latex]x[\/latex] replaced by [latex]\\left(x-h\\right)[\/latex] and [latex]y[\/latex] replaced by [latex]\\left(y-k\\right)[\/latex].<\/p>\n

              To graph parabolas with a vertex [latex]\\left(h,k\\right)[\/latex] other than the origin, we use the standard form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the x<\/em>-axis, and [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex] for parabolas that have an axis of symmetry parallel to the y<\/em>-axis. These standard forms are given below, along with their general graphs and key features.<\/p>\n

              \n

              parabolas with vertex (h,k)<\/h3>\n\n\n\n\n\n
              Axis of Symmetry<\/strong><\/td>\nEquation<\/strong><\/td>\nFocus<\/strong><\/td>\nDirectrix<\/strong><\/td>\nEndpoints of Latus Rectum<\/strong><\/td>\n<\/tr>\n
              [latex]y=k[\/latex]<\/td>\n[latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex]<\/td>\n[latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/td>\n[latex]x=h-p[\/latex]<\/td>\n[latex]\\left(h+p,\\text{ }k\\pm 2p\\right)[\/latex]<\/td>\n<\/tr>\n
              [latex]x=h[\/latex]<\/td>\n[latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex]<\/td>\n[latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/td>\n[latex]y=k-p[\/latex]<\/td>\n[latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

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

              How To: Given a standard form equation for a parabola centered at (h<\/em>, k<\/em>), sketch the graph.<\/strong><\/p>\n
                \n
              1. Determine which of the standard forms applies to the given equation: [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex] or [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex].<\/li>\n
              2. Use the standard form identified in Step 1 to determine the vertex, axis of symmetry, focus, equation of the directrix, and endpoints of the latus rectum.\n
                  \n
                1. If the equation is in the form [latex]{\\left(y-k\\right)}^{2}=4p\\left(x-h\\right)[\/latex], then:\n
                    \n
                  • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                  • use the value of [latex]k[\/latex] to determine the axis of symmetry, [latex]y=k[\/latex]<\/li>\n
                  • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(x-h\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens right. If [latex]p<0[\/latex], the parabola opens left.<\/li>\n
                  • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h+p,\\text{ }k\\right)[\/latex]<\/li>\n
                  • use [latex]h[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]x=h-p[\/latex]<\/li>\n
                  • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h+p,k\\pm 2p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n
                  • If the equation is in the form [latex]{\\left(x-h\\right)}^{2}=4p\\left(y-k\\right)[\/latex], then:\n
                      \n
                    • use the given equation to identify [latex]h[\/latex] and [latex]k[\/latex] for the vertex, [latex]\\left(h,k\\right)[\/latex]<\/li>\n
                    • use the value of [latex]h[\/latex] to determine the axis of symmetry, [latex]x=h[\/latex]<\/li>\n
                    • set [latex]4p[\/latex] equal to the coefficient of [latex]\\left(y-k\\right)[\/latex] in the given equation to solve for [latex]p[\/latex]. If [latex]p>0[\/latex], the parabola opens up. If [latex]p<0[\/latex], the parabola opens down.<\/li>\n
                    • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the coordinates of the focus, [latex]\\left(h,\\text{ }k+p\\right)[\/latex]<\/li>\n
                    • use [latex]k[\/latex] and [latex]p[\/latex] to find the equation of the directrix, [latex]y=k-p[\/latex]<\/li>\n
                    • use [latex]h,k[\/latex], and [latex]p[\/latex] to find the endpoints of the latus rectum, [latex]\\left(h\\pm 2p,\\text{ }k+p\\right)[\/latex]<\/li>\n<\/ul>\n<\/li>\n<\/ol>\n<\/li>\n
                    • Plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola.<\/li>\n<\/ol>\n<\/section>\n
                      Graph [latex]{\\left(y - 1\\right)}^{2}=-16\\left(x+3\\right)[\/latex]. Identify and label the vertex<\/strong>, axis of symmetry<\/strong>, focus<\/strong>, directrix<\/strong>, and endpoints of the latus rectum<\/strong>.<\/p>\n