{"id":843,"date":"2025-07-15T20:51:38","date_gmt":"2025-07-15T20:51:38","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=843"},"modified":"2026-01-12T20:48:48","modified_gmt":"2026-01-12T20:48:48","slug":"quadratic-functions-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/quadratic-functions-learn-it-2\/","title":{"raw":"Quadratic Functions: Learn It 2","rendered":"Quadratic Functions: Learn It 2"},"content":{"raw":"

Equations of Quadratic Functions<\/h2>\r\nLet's start by examining the general form of a quadratic function, which is the most basic way to express these equations mathematically.\r\n\r\n
\r\n

general form of a quadratic function<\/h3>\r\nThe general form of a quadratic function<\/strong> presents the function in the form\r\n\r\n \r\n

[latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex]<\/p>\r\n \r\n\r\nwhere [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]\u00a0are real numbers and [latex]a\\ne 0[\/latex].\r\n\r\n \r\n\r\nIf [latex]a>0[\/latex], the parabola opens upward. If [latex]a<0[\/latex], the parabola opens downward.\r\n\r\n<\/section>We can use the general form of a parabola to find the equation for the axis of symmetry<\/strong>.\r\n\r\n

\r\n

axis of symmetry<\/h3>\r\nThe axis of symmetry is the vertical line that passes through the vertex of the parabola. The axis of symmetry can be found with [latex]x=-\\dfrac{b}{2a}[\/latex] for a quadratic in the form [latex]ax^2+bx+c[\/latex].\r\n\r\n<\/section>The figure below shows\u00a0the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[\/latex]. In this form, [latex]a=1,\\text{ }b=4[\/latex], and [latex]c=3[\/latex]. Because [latex]a>0[\/latex], the parabola opens upward. The axis of symmetry is [latex]x=-\\dfrac{4}{2\\left(1\\right)}=-2[\/latex]. This also makes sense because we can see from the graph that the vertical line [latex]x=-2[\/latex] divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, [latex]\\left(-2,-1\\right)[\/latex]. The [latex]x[\/latex]-intercepts, those points where the parabola crosses the [latex]x[\/latex]-axis, occur at [latex]\\left(-3,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex].\r\n\r\n\"Graph\r\n

Given a quadratic function in general form, find the vertex of the parabola.<\/h3>\r\nOne reason we may want to identify the vertex <\/strong>of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, [latex]k[\/latex], and where it occurs, [latex]h[\/latex].\r\n\r\n
\r\n

vertex of the parabola<\/h3>\r\nIf we are given the general form of a quadratic function:\r\n\r\n \r\n

[latex]f(x)=ax^2+bx+c[\/latex]<\/p>\r\n \r\n\r\nWe can define the vertex, [latex](h,k)[\/latex], by doing the following:\r\n\r\n \r\n

    \r\n \t
  • Identify [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex].<\/li>\r\n \t
  • Find [latex]h[\/latex], the [latex]x[\/latex]-coordinate of the vertex, by substituting [latex]a[\/latex] and [latex]b[\/latex]\u00a0into [latex]h=-\\dfrac{b}{2a}[\/latex].<\/li>\r\n \t
  • Find [latex]k[\/latex], the [latex]y[\/latex]-coordinate of the vertex, by evaluating [latex]k=f\\left(h\\right)=f\\left(-\\dfrac{b}{2a}\\right)[\/latex]<\/li>\r\n<\/ul>\r\n<\/section>
    Find the vertex of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}-6x+7[\/latex]. Rewrite the quadratic in standard form (vertex form).[reveal-answer q=\"466886\"]Show Solution[\/reveal-answer][hidden-answer a=\"466886\"]The horizontal coordinate of the vertex will be at\r\n

    [latex]\\begin{align}h&=-\\dfrac{b}{2a}\\ \\\\[2mm] &=-\\dfrac{-6}{2\\left(2\\right)} \\\\[2mm]&=\\dfrac{6}{4} \\\\[2mm]&=\\dfrac{3}{2} \\end{align}[\/latex]<\/p>\r\nThe vertical coordinate of the vertex will be at\r\n

    [latex]\\begin{align}k&=f\\left(h\\right) \\\\[2mm]&=f\\left(\\dfrac{3}{2}\\right) \\\\[2mm]&=2{\\left(\\dfrac{3}{2}\\right)}^{2}-6\\left(\\dfrac{3}{2}\\right)+7 \\\\[2mm]&=\\dfrac{5}{2}\\end{align}[\/latex]<\/p>\r\nSo the vertex is [latex]\\left(\\dfrac{3}{2},\\dfrac{5}{2}\\right)[\/latex]\r\n\r\nRewriting into standard form, the stretch factor will be the same as the [latex]a[\/latex] in the original quadratic.\r\n

    [latex]f\\left(x\\right)=2{\\left(x-\\frac{3}{2}\\right)}^{2}+\\frac{5}{2}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]318785[\/ohm_question]<\/section>
    [ohm_question hide_question_numbers=1]318786[\/ohm_question]<\/section>
    \r\n

    standard form of a quadratic function<\/h3>\r\nThe standard form of a quadratic function<\/strong> presents the function in the form\r\n\r\n \r\n

    [latex]f\\left(x\\right)=a{\\left(x-h\\right)}^{2}+k[\/latex]<\/p>\r\n \r\n\r\nwhere [latex]\\left(h,\\text{ }k\\right)[\/latex] is the vertex<\/strong>.\r\n\r\n \r\n\r\nBecause the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function<\/strong>.\r\n\r\n<\/section><\/section>

    [ohm_question hide_question_numbers=1]317667[\/ohm_question]<\/section><\/section>","rendered":"

    Equations of Quadratic Functions<\/h2>\n

    Let’s start by examining the general form of a quadratic function, which is the most basic way to express these equations mathematically.<\/p>\n

    \n

    general form of a quadratic function<\/h3>\n

    The general form of a quadratic function<\/strong> presents the function in the form<\/p>\n

     <\/p>\n

    [latex]f\\left(x\\right)=a{x}^{2}+bx+c[\/latex]<\/p>\n

     <\/p>\n

    where [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex]\u00a0are real numbers and [latex]a\\ne 0[\/latex].<\/p>\n

     <\/p>\n

    If [latex]a>0[\/latex], the parabola opens upward. If [latex]a<0[\/latex], the parabola opens downward.\n\n<\/section>\n

    We can use the general form of a parabola to find the equation for the axis of symmetry<\/strong>.<\/p>\n

    \n

    axis of symmetry<\/h3>\n

    The axis of symmetry is the vertical line that passes through the vertex of the parabola. The axis of symmetry can be found with [latex]x=-\\dfrac{b}{2a}[\/latex] for a quadratic in the form [latex]ax^2+bx+c[\/latex].<\/p>\n<\/section>\n

    The figure below shows\u00a0the graph of the quadratic function written in general form as [latex]y={x}^{2}+4x+3[\/latex]. In this form, [latex]a=1,\\text{ }b=4[\/latex], and [latex]c=3[\/latex]. Because [latex]a>0[\/latex], the parabola opens upward. The axis of symmetry is [latex]x=-\\dfrac{4}{2\\left(1\\right)}=-2[\/latex]. This also makes sense because we can see from the graph that the vertical line [latex]x=-2[\/latex] divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, [latex]\\left(-2,-1\\right)[\/latex]. The [latex]x[\/latex]-intercepts, those points where the parabola crosses the [latex]x[\/latex]-axis, occur at [latex]\\left(-3,0\\right)[\/latex] and [latex]\\left(-1,0\\right)[\/latex].<\/p>\n

    \"Graph<\/p>\n

    Given a quadratic function in general form, find the vertex of the parabola.<\/h3>\n

    One reason we may want to identify the vertex <\/strong>of the parabola is that this point will inform us where the maximum or minimum value of the output occurs, [latex]k[\/latex], and where it occurs, [latex]h[\/latex].<\/p>\n

    \n

    vertex of the parabola<\/h3>\n

    If we are given the general form of a quadratic function:<\/p>\n

     <\/p>\n

    [latex]f(x)=ax^2+bx+c[\/latex]<\/p>\n

     <\/p>\n

    We can define the vertex, [latex](h,k)[\/latex], by doing the following:<\/p>\n

     <\/p>\n

      \n
    • Identify [latex]a[\/latex], [latex]b[\/latex], and [latex]c[\/latex].<\/li>\n
    • Find [latex]h[\/latex], the [latex]x[\/latex]-coordinate of the vertex, by substituting [latex]a[\/latex] and [latex]b[\/latex]\u00a0into [latex]h=-\\dfrac{b}{2a}[\/latex].<\/li>\n
    • Find [latex]k[\/latex], the [latex]y[\/latex]-coordinate of the vertex, by evaluating [latex]k=f\\left(h\\right)=f\\left(-\\dfrac{b}{2a}\\right)[\/latex]<\/li>\n<\/ul>\n<\/section>\n
      Find the vertex of the quadratic function [latex]f\\left(x\\right)=2{x}^{2}-6x+7[\/latex]. Rewrite the quadratic in standard form (vertex form).<\/p>\n