{"id":2901,"date":"2024-02-09T20:13:07","date_gmt":"2024-02-09T20:13:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/?post_type=chapter&p=2901"},"modified":"2026-03-10T20:10:45","modified_gmt":"2026-03-10T20:10:45","slug":"7-6-1-slope-intercept-form-of-a-linear-equation","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/uvu-introductoryalgebra\/chapter\/7-6-1-slope-intercept-form-of-a-linear-equation\/","title":{"raw":"7.6.1 Slope-Intercept Form of a Linear Equation","rendered":"7.6.1 Slope-Intercept Form of a Linear Equation"},"content":{"raw":"
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Learning Objectives<\/h1>\r\n
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  • Write a linear equation in slope-intercept form.<\/li>\r\n \t
  • Identify the slope and [latex]y[\/latex]-intercept given a linear equation.<\/li>\r\n \t
  • Write the linear equation of a graphed line.<\/li>\r\n \t
  • Graph a linear equation in slope-intercept form.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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    Key words<\/h1>\r\n
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    • Slope-intercept form<\/strong>: a linear equation in the form [latex]y=mx+b[\/latex]<\/li>\r\n<\/ul>\r\n<\/div>\r\n

      Slope-Intercept Form<\/h2>\r\nWhen an equation is written in the form [latex]y = mx+b[\/latex] it is said to be in slope-intercept form. Let's see why it has that name. Consider the equation [latex]y=2x+3[\/latex].\u00a0 If we set [latex]x=0[\/latex], then [latex]y=2(0)+3=3[\/latex]. This means that the point [latex](0, 3)[\/latex] is the [latex]y[\/latex]-intercept of the line formed by the equation.\r\n\r\n[caption id=\"attachment_1968\" align=\"aligncenter\" width=\"300\"]\"The Figure 1. Graph of [latex]y=2x+3[\/latex][\/caption]The graph in figure 1, also shows that the point [latex](-2 -1)[\/latex] lies on the graph. Consequently, the slope of the line is [latex]m=\\frac{y_2-y_1}{x_2-x_1}=\\frac{3-(-1)}{0-(-2)}=\\frac{4}{2}=2[\/latex]. This is the coefficient of [latex]x[\/latex] in\u00a0the equation [latex]y=2x+3[\/latex].\u00a0 So with slope = [latex]m[\/latex] and the [latex]y[\/latex]-intercept [latex]= (0, b)[\/latex], the equation written in the form [latex]y=mx+b[\/latex] immediately tells us the slope,\u00a0[latex]m[\/latex], and the\u00a0[latex]y[\/latex]-intercept,\u00a0[latex](0, b)[\/latex].\r\n
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      Slope-Intercept Form of a Linear Equation<\/h3>\r\nIn the equation [latex]y=mx+b[\/latex],\r\n
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      • [latex]m[\/latex] is the slope of the graph.<\/li>\r\n \t
      • [latex]b[\/latex] is the [latex]y[\/latex]-coordinate of the [latex]y[\/latex]-intercept [latex](0, b)[\/latex] of the graph.<\/li>\r\n<\/ul>\r\n<\/div>\r\n
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        Examples<\/h3>\r\nFind the slope and [latex]y[\/latex]-intercept of the line with the given equation:\r\n
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        1. [latex]y=-3x+7[\/latex]<\/li>\r\n \t
        2. [latex]y=\\frac{2}{3}x-\\frac{1}{5}[\/latex]<\/li>\r\n \t
        3. [latex]2x-3y=6[\/latex]<\/li>\r\n<\/ol>\r\n

          Solution<\/h4>\r\n
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          1. [latex]m=-3\\text{ and }b=7\\text{ so the slope is }-3\\text{ and the }y\\text{-intercept is the point } (0, 7)[\/latex]<\/li>\r\n \t
          2. [latex]m=\\frac{2}{3}\\text{ and }b=-\\frac{1}{5}\\text{ so the slope is }\\frac{2}{3}\\text{ and the }y\\text{-intercept is the point } \\left (0,-\\frac{1}{5}\\right )[\/latex]<\/li>\r\n \t
          3. We first have to rearrange the equation to solve for [latex]y[\/latex]:<\/li>\r\n<\/ol>\r\n[latex]\\begin{equation}\\begin{aligned}2x-3y & = 6 \\\\ -3y & = -2x+6 \\\\ y & = \\frac{2}{3}x-2\\end{aligned}\\end{equation}[\/latex]\r\n\r\n[latex]\\text{So, }m=\\frac{2}{3}\\text{ and }b=-2\\text{ so the slope is }\\frac{2}{3}\\text{ and the }y\\text{-intercept is the point } \\left (0,2\\right )[\/latex]\r\n\r\n<\/div>\r\n
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            Try It<\/h3>\r\nFind the slope and [latex]y[\/latex]-intercept of the line with the given equation:\r\n
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            1. [latex]y=6x-1[\/latex]<\/li>\r\n \t
            2. [latex]y=\\frac{4}{7}x-\\frac{2}{3}[\/latex]<\/li>\r\n \t
            3. [latex]5x-y=10[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"hjm415\"]Show Answer[\/reveal-answer]\r\n[hidden-answer a=\"hjm415\"]\r\n
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              1. [latex]m=6\\text{ and }b=-1\\text{ so the slope is }6\\text{ and the }y\\text{-intercept is the point } (0, -1)[\/latex]<\/li>\r\n \t
              2. [latex]m=\\frac{4}{7}\\text{ and }b=-\\frac{2}{3}\\text{ so the slope is }\\frac{4}{7}\\text{ and the }y\\text{-intercept is the point } \\left (0,-\\frac{2}{3}\\right )[\/latex]<\/li>\r\n \t
              3. We first have to rearrange the equation to solve for [latex]y[\/latex]:<\/li>\r\n<\/ol>\r\n[latex]\\begin{equation}\\begin{aligned}5x-y & = 10 \\\\ y & = 5x-10\\end{aligned}\\end{equation}[\/latex]\r\n\r\n[latex]\\text{So, }m=-5\\text{ and }b=10\\text{ so the slope is }-5\\text{ and the }y\\text{-intercept is the point } \\left (0,10\\right )[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n \r\n

                We can also find the linear equation that represents te graph of a line If we know the slope and the [latex]y[\/latex]-intercept.<\/p>\r\n\r\n

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                Example<\/h3>\r\nWrite the equation of the line that has a slope of [latex] \\displaystyle \\frac{1}{2}[\/latex] and a\u00a0y<\/i>-intercept of [latex](0, \u22125)[\/latex].\r\n

                Solution<\/h4>\r\nSubstitute the slope, [latex]m[\/latex],\u00a0into [latex]y=mx+b[\/latex].\r\n

                [latex] \\displaystyle y=\\frac{1}{2}x+b[\/latex]<\/p>\r\nSubstitute [latex]b[\/latex], into the equation.\r\n

                [latex] \\displaystyle y=\\frac{1}{2}x-5[\/latex]<\/p>\r\n\r\n

                Answer<\/h4>\r\n[latex]y=\\frac{1}{2}x-5[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n \r\n
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                Example<\/h3>\r\nWrite the equation of the line in the graph by identifying the slope and [latex]y[\/latex]-intercept.\r\n

                \"ASolution<\/h4>\r\nIdentify the point where the graph crosses the [latex]y[\/latex]-axis [latex](0,3)[\/latex]. This means that [latex]b=3[\/latex]. Identify one other point and draw a slope triangle to find the slope.\r\n\r\nSlope: [latex]m=\\frac{-2}{3}[\/latex].\r\n\r\nSubstitute [latex]m[\/latex] and [latex]b[\/latex] into the slope-intercept equation.\r\n

                [latex]y=mx+b\\\\y=\\frac{-2}{3}x+b\\\\y=\\frac{-2}{3}x+3[\/latex]<\/p>\r\n\r\n

                Answer<\/h4>\r\n[latex]y=\\frac{-2}{3}x+3[\/latex]\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\nWhen we are given an equation in the slope-intercept form of [latex]y=mx+b[\/latex], we can identify the slope\u00a0and\u00a0y<\/em>-intercept and use these to graph the equation.\u00a0 When we have an equation in slope-intercept form we can graph it by first plotting\u00a0the [latex]y[\/latex]-intercept, then using the slope to find a second point, and connecting the dots.\r\n
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                Example<\/h3>\r\nGraph [latex]y=\\frac{1}{2}x-4[\/latex] using the slope-intercept equation.\r\n

                Solution<\/h4>\r\nFirst, plot the [latex]y[\/latex]-intercept.\r\n
                \"The<\/div>\r\n \r\n\r\nNow use the slope to count up or down and over left or right to the next point. This slope is [latex]\\frac{1}{2}[\/latex], so we can count up one and right two\u2014both positive because both parts of the slope are positive.\r\n\r\nConnect the dots.\r\n\r\n\"A\r\n\r\n<\/div>\r\n<\/div>\r\nNOTE: it is\u00a0important for the equation to first be in slope-intercept form. If it is not, we have to solve it for [latex]y[\/latex] so we can identify the slope and the [latex]y[\/latex]-intercept.\r\n
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                Try It<\/h3>\r\n