• Post a relationship from a real-world situation or concept that illustrates the idea of a mathematical function. For example, we use the function F =9/5(C + 32) to convert from degrees Celsius to degrees Fahrenheit.
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    Another application of mathematical function from a real-world situation is profit, revenue and cost functions.

    Note that profit = revenue – cost.

    P(x) = R(x) – C(x) where x= number of items sold.

    Note: P(x) = positive value means it earns a profit and R(x) > C(x)

            P(x) = negative value means loss and R(x) < C(x)

            P(x) =0 or break-even where there is no loss or gain of profit. R(x)=C(x).

    Suppose a store franchised a toy for $5 each with a fixed cost of $230.

    Cost function = total variable cost + total fixed cost.

    C(x) = 5x + 230.

    The toys are sold for a price $8.

    Revenue function = price * items

    R(x) = 8*x  or R(x) = 8x.

    Profit function:

     P(x) = 8x – (5x+230)

            = 8x – 5x -230

            = 3x -230

    *Break-even when R(x) = C(x).

     8x = 5x +230

    -5x   -5x

    ------------------------

    3x  =  0 +230

    3x = 230

    ---    ------

    3        3

    x = 76.7

    Round off to 77 toys since 0.70 toy is not possible.

    The number of items can be counted as 0,1,2,3,…

    Interpretation:

    The store should at least had sold 77 toys to earn a profit.

    When the sold items is less than 77 toys, there will be a loss.

     Let x = 77 toys then profit is:

    P(x) = 3(x) -230

    P(77) = 3(77) -230

              = 231 -230

              = $1

    The store earns $1 profit for 77 toys sold.

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    Function relates input values (independent values) to exactly one possible output value (dependent value).

    Usually, it follows f(x) =y where x = input and y = output.

    One common application is projectile motion: f(t) = `at^2 +bt+c`

    where a <0 and f(t) as height of the object at time  "t".

    The graph of parabola opens downward (projectile path) when a <0.

    Example: `f(t) = -4t^2 +14t+6 ` where f(t) is height in feet and t is time in seconds.

    a) Finding initial height:

    Let t=0  (starting time) then

    f(0) =` -4(0)^2+14(0)+6`  = 6 ft

     Interpretation: An object is launched from the height of 6 feet.

    b) Finding maximum height.

    Hint: Vertex is at the peak of the parabola in projectile trajectory.

    `Vertex t = -b/(2a) = -(14/(2*(-4))) =1.75 seconds`

     maximum height: `f(3.5) = -4(3.5)^2 +14(3.5) +6 =18.25 feet.`

    Interpretation: The object will reach 18.25 feet after 1.75 seconds before it falls down along its projectile trajectory.

    c)  Time it hits the ground.

    At the ground level, the height is 0 or f(t)=0

      f(t) = -4t^2 +14t +6

      0 = -4t^2 +14t +6

       0 = -(t-2)(4t+3)    Factoring

       Zero-factor property:

    t-2 =0

     +2  +2

    ---------------

        t =2 seconds

    4t +3 =0

        -3   -3

    ----------------

        4t = -3

        ---  ----

         4     4

         t= -0.75 seconds (negative time value does not exists).

    Interpretation: The object hits the ground after 2 seconds.

    Approved by eNotes Editorial Team