{"id":3017,"date":"2025-08-15T20:09:19","date_gmt":"2025-08-15T20:09:19","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=3017"},"modified":"2025-08-15T20:29:46","modified_gmt":"2025-08-15T20:29:46","slug":"derivatives-learn-it-5","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/derivatives-learn-it-5\/","title":{"raw":"Derivatives: Learn It 5","rendered":"Derivatives: Learn It 5"},"content":{"raw":"

Finding an Equation of a Line Tangent to the Graph of a Function<\/h2>\r\nThe equation of a tangent line to a curve of the function [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] is derived from the point-slope form of a line, [latex]y=m\\left(x-{x}_{1}\\right)+{y}_{1}[\/latex]. The slope of the line is the slope of the curve at [latex]x=a[\/latex] and is therefore equal to [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex], the derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex]. The coordinate pair of the point on the line at [latex]x=a[\/latex] is [latex]\\left(a,f\\left(a\\right)\\right)[\/latex].\r\n\r\nIf we substitute into the point-slope form, we have\r\n\r\n\"The\r\n\r\nThe equation of the tangent line is\r\n
[latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex]<\/div>\r\n
\r\n

equation of the tangent line<\/h3>\r\nThe equation of a line tangent to the curve of a function [latex]f[\/latex] at a point [latex]x=a[\/latex] is\r\n

[latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex]<\/p>\r\n\r\n<\/section><\/div>\r\n

How To: Given a function [latex]f[\/latex], find the equation of a line tangent to the function at [latex]x=a[\/latex].<\/strong>\r\n
    \r\n \t
  1. Find the derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] using [latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\r\n \t
  2. Evaluate the function at [latex]x=a[\/latex]. This is [latex]f\\left(a\\right)[\/latex].<\/li>\r\n \t
  3. Substitute [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] into [latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex].<\/li>\r\n \t
  4. Write the equation of the tangent line in the form [latex]y=mx+b[\/latex].<\/li>\r\n<\/ol>\r\n<\/section>
    Find the equation of a line tangent to the curve [latex]f\\left(x\\right)={x}^{2}-4x[\/latex] at [latex]x=3[\/latex].[reveal-answer q=\"363951\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"363951\"]\r\n\r\nUsing:\r\n

    [latex]f^{\\prime}\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex]<\/p>\r\nSubstitute [latex]f\\left(a+h\\right)={\\left(a+h\\right)}^{2}-4\\left(a+h\\right)[\/latex] and [latex]f\\left(a\\right)={a}^{2}-4a[\/latex].\r\n

    [latex]\\begin{align} {f}^{\\prime }\\left(a\\right)&=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\left(a+h\\right)\\left(a+h\\right)-4\\left(a+h\\right)-\\left({a}^{2}-4a\\right)}{h} \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{{a}^{2}+2ah+{h}^{2}-4a - 4h-{a}^{2}+4a}{h}&& \\text{Remove parentheses}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{{a}^{2}+2ah+{h}^{2}-4a-4h-{a}^{2}+4a}{h}&& \\text{Combine like terms}. \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{2ah+{h}^{2}-4h}{h} \\\\ &=\\underset{h\\to 0}{\\mathrm{lim}}\\frac{\\cancel{h}\\left(2a+h - 4\\right)}{\\cancel{h}}&& \\text{Factor out }h. \\\\ &=2a+0 - 4 \\\\ {f}^{\\prime }\\left(a\\right)&=2a - 4&& \\text{Evaluate the limit}. \\\\ {f}^{\\prime }\\left(3\\right)&=2\\left(3\\right)-4=2 \\end{align}[\/latex]<\/p>\r\nEquation of tangent line at [latex]x=3:[\/latex]\r\n

    [latex]\\begin{align}&y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right) \\\\ &y=f^{\\prime}\\left(3\\right)\\left(x - 3\\right)+f\\left(3\\right) \\\\ &y=2\\left(x - 3\\right)+\\left(-3\\right) \\\\ &y=2x - 9 \\end{align}[\/latex]<\/p>\r\n\r\n

    Analysis of the Solution<\/h4>\r\nWe can use a graphing utility to graph the function and the tangent line. In so doing, we can observe the point of tangency at [latex]x=3[\/latex] as shown in Figure 19.\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"Graph Graph confirms the point of tangency at [latex]x=3[\/latex].[\/caption][\/hidden-answer]<\/section>
    Find the equation of a tangent line to the curve of the function [latex]f\\left(x\\right)=5{x}^{2}-x+4[\/latex] at [latex]x=2[\/latex].[reveal-answer q=\"952946\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"952946\"][latex]y=19x - 16[\/latex][\/hidden-answer]<\/section>","rendered":"

    Finding an Equation of a Line Tangent to the Graph of a Function<\/h2>\n

    The equation of a tangent line to a curve of the function [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] is derived from the point-slope form of a line, [latex]y=m\\left(x-{x}_{1}\\right)+{y}_{1}[\/latex]. The slope of the line is the slope of the curve at [latex]x=a[\/latex] and is therefore equal to [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex], the derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex]. The coordinate pair of the point on the line at [latex]x=a[\/latex] is [latex]\\left(a,f\\left(a\\right)\\right)[\/latex].<\/p>\n

    If we substitute into the point-slope form, we have<\/p>\n

    \"The<\/p>\n

    The equation of the tangent line is<\/p>\n

    [latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex]<\/div>\n
    \n
    \n

    equation of the tangent line<\/h3>\n

    The equation of a line tangent to the curve of a function [latex]f[\/latex] at a point [latex]x=a[\/latex] is<\/p>\n

    [latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex]<\/p>\n<\/section>\n<\/div>\n

    How To: Given a function [latex]f[\/latex], find the equation of a line tangent to the function at [latex]x=a[\/latex].<\/strong><\/p>\n
      \n
    1. Find the derivative of [latex]f\\left(x\\right)[\/latex] at [latex]x=a[\/latex] using [latex]{f}^{\\prime }\\left(a\\right)=\\underset{h\\to 0}{\\mathrm{lim}}\\dfrac{f\\left(a+h\\right)-f\\left(a\\right)}{h}[\/latex].<\/li>\n
    2. Evaluate the function at [latex]x=a[\/latex]. This is [latex]f\\left(a\\right)[\/latex].<\/li>\n
    3. Substitute [latex]\\left(a,f\\left(a\\right)\\right)[\/latex] and [latex]\\begin{align}{f}^{\\prime }\\left(a\\right)\\end{align}[\/latex] into [latex]\\begin{align}y=f^{\\prime}\\left(a\\right)\\left(x-a\\right)+f\\left(a\\right)\\end{align}[\/latex].<\/li>\n
    4. Write the equation of the tangent line in the form [latex]y=mx+b[\/latex].<\/li>\n<\/ol>\n<\/section>\n
      Find the equation of a line tangent to the curve [latex]f\\left(x\\right)={x}^{2}-4x[\/latex] at [latex]x=3[\/latex].<\/p>\n