{"id":812,"date":"2025-07-15T19:37:30","date_gmt":"2025-07-15T19:37:30","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/precalculus\/?post_type=chapter&p=812"},"modified":"2026-01-12T17:50:48","modified_gmt":"2026-01-12T17:50:48","slug":"linear-models-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/precalculus\/chapter\/linear-models-learn-it-2\/","title":{"raw":"Linear Models: Learn It 2","rendered":"Linear Models: Learn It 2"},"content":{"raw":"

Using a Given Intercept to Build a Model<\/h2>\r\nSome real-world problems provide the [latex]y[\/latex]-intercept, which is the constant or initial value. Once the [latex]y[\/latex]-intercept is known, the [latex]x[\/latex]-intercept can be calculated.\r\n\r\nLet's explore this through an example.\r\n\r\n
Hannah plans to pay off a no-interest loan from her parents. Her loan balance is [latex]$1,000[\/latex]. She plans to pay [latex]$250[\/latex] per month until her balance is [latex]$0[\/latex].\r\n[latex]\\\\[\/latex]\r\nThe [latex]y[\/latex]-intercept is the initial amount of her debt, or [latex]$1,000[\/latex]. The rate of change, or slope, is [latex]\u2013$250[\/latex] per month. We can then use slope-intercept form and the given information to develop a linear model.\r\n

[latex]\\begin{array}{l}f\\left(x\\right)=mx+b\\hfill \\\\ f\\left(x\\right)=-250x+1000\\hfill \\end{array}[\/latex]<\/p>\r\nNow we can set the function equal to [latex]0[\/latex] and solve for [latex]x[\/latex]\u00a0to find the [latex]x[\/latex]-intercept.\r\n

[latex]\\begin{array}{l}0=-250x+1000\\hfill \\\\ 1000=250x\\hfill \\\\ 4=x\\hfill \\\\ x=4\\hfill \\end{array}[\/latex]<\/p>\r\nThe [latex]x[\/latex]-intercept is the number of months it takes her to reach a balance of [latex]$0[\/latex]. The [latex]x[\/latex]-intercept is [latex]4[\/latex] months, so it will take Hannah four months to pay off her loan.\r\n\r\n<\/section>\r\n

Using a Given Input and Output to Build a Model<\/h2>\r\nMany real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.\r\n\r\n
How to: Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.<\/strong>\r\n
    \r\n \t
  1. Identify the input and output values.<\/li>\r\n \t
  2. Convert the data to two coordinate pairs.<\/li>\r\n \t
  3. Find the slope.<\/li>\r\n \t
  4. Write the linear model.<\/li>\r\n \t
  5. Use the model to make a prediction by evaluating the function at a given [latex]x[\/latex]\u00a0value.<\/li>\r\n \t
  6. Use the model to identify an [latex]x[\/latex]\u00a0value that results in a given [latex]y[\/latex]\u00a0value.<\/li>\r\n \t
  7. Answer the question posed.<\/li>\r\n<\/ol>\r\n<\/section>
    A town\u2019s population has been growing linearly. In 2004 the population was [latex]6,200[\/latex]. By 2009 the population had grown to [latex]8,100[\/latex]. Assume this trend continues.\r\n
      \r\n \t
    1. Predict the population in 2013.<\/li>\r\n \t
    2. Identify the year in which the population will reach [latex]15,000[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"413748\"]Show Solution[\/reveal-answer] [hidden-answer a=\"413748\"]The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the [latex]y[\/latex]-intercept would correspond to the year [latex]0[\/latex], more than [latex]2000[\/latex] years ago!\r\n\r\nTo make computation a little nicer, we will define our input as the number of years since 2004:\r\n
        \r\n \t
      • Input: [latex]t[\/latex], years since 2004<\/li>\r\n \t
      • Output: [latex]P(t)[\/latex], the town\u2019s population<\/li>\r\n<\/ul>\r\nTo predict the population in 2013 ([latex]t= 9[\/latex]), we would first need an equation for the population. Likewise, to find when the population would reach [latex]15,000[\/latex], we would need to solve for the input that would provide an output of [latex]15,000[\/latex]. To write an equation, we need the initial value and the rate of change, or slope.\r\n\r\nTo determine the rate of change, we will use the change in output per change in input.\r\n

        [latex]m=\\frac{\\text{change in output}}{\\text{change in input}}[\/latex]<\/p>\r\nThe problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to [latex]t=0[\/latex], giving the point [latex]\\left(0,\\text{6200}\\right)[\/latex]. Notice that through our clever choice of variable definition, we have \"given\" ourselves the [latex]y[\/latex]-intercept of the function. The year 2009 would correspond to [latex]t=\\text{5}[\/latex], giving the point [latex]\\left(5,\\text{8100}\\right)[\/latex].\r\n\r\nThe two coordinate pairs are [latex]\\left(0,\\text{6200}\\right)[\/latex] and [latex]\\left(5,\\text{8100}\\right)[\/latex]. Recall that we encountered examples in which we were provided two points earlier in the module. We can use these values to calculate the slope.\r\n

        [latex]\\begin{array}{l} m=\\frac{8100 - 6200}{5 - 0}\\hfill \\\\ \\text{}m=\\frac{1900}{5}\\hfill \\\\ \\text{}m=380\\text{ people per year}\\hfill \\end{array}[\/latex]<\/p>\r\n

        We already know the [latex]y[\/latex]-intercept of the line, so we can immediately write the equation: [latex]\\begin{array}{l}P\\left(t\\right)=380t+6200 \\\\ \\hfill \\end{array}[\/latex]<\/p>\r\nTo predict the population in 2013, we evaluate our function at [latex]t= 9[\/latex].\r\n

        [latex]\\begin{array}{l}P\\left(9\\right)=380\\left(9\\right)+6,200\\hfill \\\\ \\text{}P\\left(9\\right)=9,620\\hfill \\end{array}[\/latex]<\/p>\r\nIf the trend continues, our model predicts a population of [latex]9,620[\/latex] in 2013.\r\n\r\nTo find when the population will reach [latex]15,000[\/latex], we can set [latex]P\\left(t\\right)=15000[\/latex] and solve for [latex]t[\/latex].\r\n

        [latex]\\begin{array}{l}15000=380t+6200\\hfill \\\\ \\text{ }8800=380t\\hfill \\\\ \\text{ }t\\approx 23.158\\hfill \\end{array}[\/latex]<\/p>\r\nOur model predicts the population will reach [latex]15,000[\/latex] in a little more than [latex]23[\/latex] years after 2004, or somewhere around the year 2027.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>

        [ohm_question hide_question_numbers=1]318762[\/ohm_question]<\/section>
        [ohm_question hide_question_numbers=1]318763[\/ohm_question]<\/section>\r\n

        Using a Diagram to Model a Problem<\/h2>\r\nIt is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometric shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.\r\n\r\n
        Sketching a diagram or trying a few sample equations is an essential part of doing mathematics. Paths that don't pan out and wrong turns are common and natural. We need to make mistakes in math in order to rule out possibilities that don't lead to success. Don't be afraid to sketch a diagram or take a wrong turn. It's all part of the mathematical process.<\/section>
        Anna and Emanuel start at the same intersection. Anna walks east at [latex]4[\/latex] miles per hour while Emanuel walks south at [latex]3[\/latex] miles per hour. They are communicating with a two-way radio that has a range of [latex]2[\/latex] miles.\r\n[latex]\\\\[\/latex]\r\nHow long after they start walking will they fall out of radio contact? [reveal-answer q=\"960486\"]Show Solution[\/reveal-answer] [hidden-answer a=\"960486\"] In essence, we can partially answer this question by saying they will fall out of radio contact when they are [latex]2[\/latex] miles apart, which leads us to ask a new question:\u00a0\"How long will it take them to be [latex]2[\/latex] miles apart?\"\r\n[latex]\\\\[\/latex]\r\nIn this problem, our changing quantities are time and position, but ultimately we need to know how long will it take for them to be [latex]2[\/latex] miles apart. We can see that time will be our input variable, so we\u2019ll define our input and output variables.\r\n
          \r\n \t
        • Input: [latex]t[\/latex], time in hours.<\/li>\r\n \t
        • Output: [latex]A\\left(t\\right)[\/latex], distance in miles, and [latex]E\\left(t\\right)[\/latex], distance in miles<\/li>\r\n<\/ul>\r\nBecause it is not obvious how to define our output variable, we\u2019ll start by drawing a picture.\r\n\r\n
          \"Picture<\/center> \r\n\r\nInitial Value: They both start at the same intersection so when [latex]t=0[\/latex], the distance traveled by each person should also be [latex]0[\/latex]. Thus the initial value for each is [latex]0[\/latex].\r\n[latex]\\\\[\/latex]\r\nRate of Change: Anna is walking [latex]4[\/latex] miles per hour and Emanuel is walking [latex]3[\/latex] miles per hour, which are both rates of change. The slope for [latex]A[\/latex]\u00a0is [latex]4[\/latex] and the slope for [latex]E[\/latex]\u00a0is [latex]3[\/latex].\r\n[latex]\\\\[\/latex]\r\nUsing those values, we can write formulas for the distance each person has walked.\r\n

          [latex]\\begin{array}{l}A\\left(t\\right)=4t\\\\ E\\left(t\\right)=3t\\end{array}[\/latex]<\/p>\r\nFor this problem, the distances from the starting point are important. To notate these, we can define a coordinate system, identifying the \"starting point\" at the intersection where they both started. Then we can use the variable, [latex]A[\/latex], which we introduced above, to represent Anna\u2019s position and define it to be a measurement from the starting point in the eastward direction. Likewise, can use the variable, [latex]E[\/latex], to represent Emanuel\u2019s position measured from the starting point in the southward direction.\r\n[latex]\\\\[\/latex]\r\n Note that in defining the coordinate system, we specified both the starting point of the measurement and the direction of measure.<\/em>\r\n[latex]\\\\[\/latex]\r\nWe can then define a third variable, [latex]D[\/latex], to be the measurement of the distance between Anna and Emanuel. Showing the variables on the diagram is often helpful. Recall that we need to know how long it takes for [latex]D[\/latex], the distance between them, to equal [latex]2[\/latex] miles. Notice that for any given input [latex]t[\/latex], the outputs [latex]A(t)[\/latex], [latex]E(t)[\/latex], and [latex]D(t)[\/latex] represent distances.\r\n\r\n

          \"Picture<\/center> \r\n\r\nThis picture\u00a0shows us that we can use the Pythagorean Theorem because we have drawn a right triangle. Using the Pythagorean Theorem, we get:\r\n

          [latex]\\begin{array}{lllllll}D{\\left(t\\right)}^{2}=A{\\left(t\\right)}^{2}+E{\\left(t\\right)}^{2}\\hfill & \\hfill \\\\ D{\\left(t\\right)}^{2}={\\left(4t\\right)}^{2}+{\\left(3t\\right)}^{2}\\hfill & \\hfill \\\\ D{\\left(t\\right)}^{2}=16{t}^{2}+9{t}^{2}\\hfill & \\hfill \\\\ D{\\left(t\\right)}^{2}=25{t}^{2}\\hfill & \\hfill \\\\ \\text{}D\\left(t\\right)=\\pm \\sqrt{25{t}^{2}}\\hfill & \\text{Solve for }D\\left(t\\right)\\text{by taking the square root of each side of the equation}\\hfill \\\\ D{\\left(t\\right)}=\\pm 5t\\hfill & \\hfill \\end{array}[\/latex]<\/p>\r\nIn this scenario we are considering only positive values of [latex]t[\/latex], so our distance [latex]D(t)[\/latex] will always be positive. We can simplify this answer to [latex]D(t) = 5t[\/latex]. This means that the distance between Anna and Emanuel is also a linear function. Because [latex]D[\/latex]\u00a0is a linear function, we can now answer the question of when the distance between them will reach [latex]2[\/latex] miles. We will set the output [latex]D(t) = 2[\/latex] and solve for [latex]t[\/latex].\r\n

          [latex]\\begin{array}{l}D\\left(t\\right)=2\\hfill \\\\ \\text{ }5t=2\\hfill \\\\ \\text{ }t=\\frac{2}{5}=0.4\\hfill \\end{array}[\/latex]<\/p>\r\nThey will fall out of radio contact in [latex]0.4[\/latex] hours, or [latex]24[\/latex] minutes. [\/hidden-answer]\r\n\r\n<\/section>

          Q & A<\/strong> Should I draw diagrams when given information based on a geometric shape?<\/strong>\r\n\r\n
          \r\n\r\nYes. Sketch the figure and label the quantities and unknowns on the sketch.\r\n\r\n<\/section>
          There is a straight road leading from the town of Westborough to Agritown [latex]30[\/latex] miles east and [latex]10[\/latex] miles north. A certain distance down this road, it junctions with a second road, perpendicular to the first, leading to the town of Eastborough.\r\n[latex]\\\\[\/latex]\r\nIf the town of Eastborough is located [latex]20[\/latex] miles directly east of the town of Westborough, how far is the road junction from Westborough? [reveal-answer q=\"156610\"]Show Solution[\/reveal-answer] [hidden-answer a=\"156610\"] It might help here to draw a picture of the situation.\u00a0It would then be helpful to introduce a coordinate system. While we could place the origin anywhere, placing it at Westborough seems convenient. This puts Agritown at coordinates [latex](30, 10)[\/latex], and Eastborough at [latex](20, 0)[\/latex].
          \"Picture<\/center>Using this point along with the origin, we can find the slope of the line from Westborough to Agritown:\r\n

          [latex]m=\\frac{10 - 0}{30 - 0}=\\frac{1}{3}[\/latex]<\/p>\r\nThe equation of the road from Westborough to Agritown would be\r\n

          [latex]W\\left(x\\right)=\\frac{1}{3}x[\/latex]<\/p>\r\nFrom this, we can determine the perpendicular road to Eastborough will have slope [latex]m=-3[\/latex]. Because the town of Eastborough is at the point [latex](20, 0)[\/latex], we can find the equation:\r\n

          [latex]\\begin{array}{l}E\\left(x\\right)=-3x+b\\hfill & \\hfill \\\\ 0=-3\\left(20\\right)+b\\hfill & \\text{Substitute in (20, 0)}\\hfill \\\\ b=60\\hfill & \\hfill \\\\ E\\left(x\\right)=-3x+60\\hfill & \\hfill \\end{array}[\/latex]<\/p>\r\nWe can now find the coordinates of the junction of the roads by finding the intersection of these lines. Setting them equal,\r\n

          [latex]\\begin{array}{lllllll}\\text{ }\\frac{1}{3}x=-3x+60\\hfill & \\hfill \\\\ \\frac{10}{3}x=60\\hfill & \\hfill \\\\ 10x=180\\hfill & \\hfill \\\\ \\text{ }x=18\\hfill & \\text{Substituting this back into }W\\left(x\\right)\\hfill \\\\ \\text{ }y=W\\left(18\\right)\\hfill & \\hfill \\\\ \\text{ }y=\\frac{1}{3}\\left(18\\right)\\hfill & \\hfill \\\\ \\text{ }y=6\\hfill & \\hfill \\end{array}[\/latex]<\/p>\r\nThe roads intersect at the point [latex](18, 6)[\/latex]. Using the distance formula, we can now find the distance from Westborough to the junction.\r\n

          [latex]\\begin{array}{l}\\text{distance}=\\sqrt{{\\left({x}_{2}-{x}_{1}\\right)}^{2}+{\\left({y}_{2}-{y}_{1}\\right)}^{2}}\\hfill \\\\ \\text{ }=\\sqrt{{\\left(18 - 0\\right)}^{2}+{\\left(6 - 0\\right)}^{2}}\\hfill \\\\ \\text{ }\\approx 18.974\\text{ miles}\\hfill \\end{array}[\/latex]<\/p>\r\n[latex]\\\\[\/latex]\r\nAnalysis of the Solution<\/strong>\r\n[latex]\\\\[\/latex]\r\nOne nice use of linear models is to take advantage of the fact that the graphs of these functions are lines. This means real-world applications discussing maps need linear functions to model the distances between reference points. [\/hidden-answer]\r\n\r\n<\/section>","rendered":"

          Using a Given Intercept to Build a Model<\/h2>\n

          Some real-world problems provide the [latex]y[\/latex]-intercept, which is the constant or initial value. Once the [latex]y[\/latex]-intercept is known, the [latex]x[\/latex]-intercept can be calculated.<\/p>\n

          Let’s explore this through an example.<\/p>\n

          Hannah plans to pay off a no-interest loan from her parents. Her loan balance is [latex]$1,000[\/latex]. She plans to pay [latex]$250[\/latex] per month until her balance is [latex]$0[\/latex].
          \n[latex]\\\\[\/latex]
          \nThe [latex]y[\/latex]-intercept is the initial amount of her debt, or [latex]$1,000[\/latex]. The rate of change, or slope, is [latex]\u2013$250[\/latex] per month. We can then use slope-intercept form and the given information to develop a linear model.<\/p>\n

          [latex]\\begin{array}{l}f\\left(x\\right)=mx+b\\hfill \\\\ f\\left(x\\right)=-250x+1000\\hfill \\end{array}[\/latex]<\/p>\n

          Now we can set the function equal to [latex]0[\/latex] and solve for [latex]x[\/latex]\u00a0to find the [latex]x[\/latex]-intercept.<\/p>\n

          [latex]\\begin{array}{l}0=-250x+1000\\hfill \\\\ 1000=250x\\hfill \\\\ 4=x\\hfill \\\\ x=4\\hfill \\end{array}[\/latex]<\/p>\n

          The [latex]x[\/latex]-intercept is the number of months it takes her to reach a balance of [latex]$0[\/latex]. The [latex]x[\/latex]-intercept is [latex]4[\/latex] months, so it will take Hannah four months to pay off her loan.<\/p>\n<\/section>\n

          Using a Given Input and Output to Build a Model<\/h2>\n

          Many real-world applications are not as direct as the ones we just considered. Instead they require us to identify some aspect of a linear function. We might sometimes be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.<\/p>\n

          How to: Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.<\/strong><\/p>\n
            \n
          1. Identify the input and output values.<\/li>\n
          2. Convert the data to two coordinate pairs.<\/li>\n
          3. Find the slope.<\/li>\n
          4. Write the linear model.<\/li>\n
          5. Use the model to make a prediction by evaluating the function at a given [latex]x[\/latex]\u00a0value.<\/li>\n
          6. Use the model to identify an [latex]x[\/latex]\u00a0value that results in a given [latex]y[\/latex]\u00a0value.<\/li>\n
          7. Answer the question posed.<\/li>\n<\/ol>\n<\/section>\n
            A town\u2019s population has been growing linearly. In 2004 the population was [latex]6,200[\/latex]. By 2009 the population had grown to [latex]8,100[\/latex]. Assume this trend continues.<\/p>\n
              \n
            1. Predict the population in 2013.<\/li>\n
            2. Identify the year in which the population will reach [latex]15,000[\/latex].<\/li>\n<\/ol>\n