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How to Apply Linear Functions to Life

Linear functions, which represent a constant proportional relationship between two different variables, may seem like they only belong on pages of an algebra textbook, but they have their foundation in real-world application. In functions, one variable is dependent on the other, which means linear functions can be used to calculate everything from salaries to phone calls costs.

Instructions

  1. Calculate Phone Call Costs

    • 1

      Read the example problem: Judy calls a special hotline, which costs $2 for the first minute. Each additional minute costs $1.50. Write the function of this equation. How much will a 10 minute phone call cost her?

    • 2

      Check to see if the two variables have a dependent relationship. In this case, the cost is contingent on time.

    • 3

      Write the relationship of the function: The total cost is equal to the initial cost ($2) and $1.50 times each additional minute: f(x)= $2 + $1.50(x-1).

    • 4

      Calculate how much 10 minutes will cost by substituting 10 for x. f(10)= $2 + $1.50(9). f(10)= $15.50. The total cost of a 10 minute phone call is $15.50.

    Calculate Paychecks and Commission

    • 5

      Read the following problem: Jane is making a base of $12.50/hr and 10 percent commission on her sales. If her goal is to make $3,500 a month, what is the total amount of sales she will need to make? Assume, she works 40 hours a week for four weeks.

    • 6

      Check to see if the two variables have a dependent relationship. In this case, the amount of her total income is contingent on her sales.

    • 7

      Write the relationship of the function. The total salary= $12.50 (160 or total number of hours for the month) + 10 percent (x), where x represents the total amount of sales; f(x)=$12.50(160)+.10x

    • 8

      Solve the equation by substituting $3,500 for the total salary or f(x): $3500=$12.50(160)+.10x

      $3500= $2000 +.10x

      $1500=.10x

      x=$15,000

      Jane will have to make $15,000 to make her monthly goal of $3,500.

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