- Write the equation for a linear function given an application
In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know or can figure out the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems.
How To: Given a linear function [latex]f[/latex] and the initial value and rate of change, evaluate [latex]f(c)[/latex]
- Determine the initial value and the rate of change (slope).
- Substitute the values into [latex]f\left(x\right)=mx+b[/latex].
- Evaluate the function at [latex]x=c[/latex].
Initial value is a term that is typically used in applications of functions. It can be represented as the starting point of the relationship we are describing with a function. In the case of linear functions, the initial value is typically the y-intercept. Here are some characteristics of the initial value:
- The point [latex](0,y)[/latex] is often the initial value of a linear function
- The y-value of the initial value comes from b in slope-intercept form of a linear function, [latex]f\left(x\right)=mx+b[/latex]
- The initial value can be found by solving for b or substituting [latex]0[/latex] in for x in a linear function.
In our first example, we are given a scenario where Marcus wants to increase the number of songs in his music collection by a fixed amount each month. This is a perfect candidate for a linear function because the increase in the number of songs stays the same each month. We will identify the initial value for the music collection and write an equation that represents the number of songs in the collection for any number of months t.
Marcus currently has [latex]200[/latex] songs in his music collection. Every month he adds [latex]15[/latex] new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year?
The initial value for this function is [latex]200[/latex] because he currently owns [latex]200[/latex] songs, so N(0) = [latex]200[/latex]. This means that b =[latex]200[/latex].
The number of songs increases by [latex]15[/latex] songs per month, so the rate of change is [latex]15[/latex] songs per month. Therefore, we know that [latex]m=15[/latex]. We can substitute the initial value and the rate of change into slope-intercept form.
We can write the formula [latex]N\left(t\right)=15t+200[/latex].
With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at [latex]t = 12 [/latex] .
Marcus will have [latex]380[/latex] songs in [latex]12[/latex] months.
In the example we just completed, notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well.
In our next example, we will show that you can write the equation for a linear function given two data points. In this case, Ilya’s weekly income depends on the number of insurance policies he sells. We are given his income for two different weeks and the number of policies sold. We first find the rate of change and then solve for the initial value.
Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week, he sold [latex]3[/latex] new policies and earned [latex]$760[/latex] for the week. The week before, he sold [latex]5[/latex] new policies and earned [latex]$920[/latex]. Find an equation for I(n) and interpret the meaning of the components of the equation.
The given information gives us two input-output pairs: [latex](3, 760)[/latex] and [latex](5, 920)[/latex]. We start by finding the rate of change.
Keeping track of units can help us interpret this quantity. Income increased by [latex]$160[/latex] when the number of policies increased by [latex]2[/latex], so the rate of change is [latex]$80[/latex] per policy. Therefore, Ilya earns a commission of [latex]$80[/latex] for each policy sold during the week.
We can then solve for the initial value.
The value of b is the starting value for the function and represents Ilya’s income when [latex]n = 0[/latex], or when no new policies are sold. We can interpret this as Ilya’s base salary for the week which does not depend upon the number of policies sold.
We can now write the final equation.
Our final interpretation is that Ilya’s base salary is [latex]$520[/latex] per week and he earns an additional [latex]$80[/latex] commission for each policy sold.
Notice that we used units to help us verify that we were calculating the rate correctly. It makes sense to speak in terms of the price per policy. To calculate the initial value, we solved for b by substituting values from one of the points we were given for n and I.
In the following video example, we show how to identify the initial value, slope, and equation for a linear function.
We will show one more example of how to write a linear function that represents the monthly cost to run a company given monthly fixed costs and production costs per item.
Suppose Ben starts a company in which he incurs a fixed cost of [latex]$1,250[/latex] per month for the overhead which includes his office rent. His production costs are [latex]$37.50[/latex] per item. Write a linear function C where C(x) is the cost for x items produced in a given month.
The fixed cost is present every month, [latex]$1,250[/latex]. The costs that can vary include the cost to produce each item, which is [latex]$37.50[/latex] for Ben. The variable cost, called the marginal cost, is represented by [latex]37.5[/latex]. The cost Ben incurs is the sum of these two costs represented by [latex]C\left(x\right)=1250+37.5x[/latex].
It is important to note that we are writing a function based on monthly costs, so the initial cost will be [latex]$1,250[/latex], because Ben has to pay that amount monthly for rent. If Ben produces [latex]100[/latex] items in a month, his monthly cost is represented by
So his monthly cost would be [latex]$5,000[/latex].
The following video example shows how to write a linear function that represents how many miles you can travel in a rental car given a fixed amount of money.
In the next example, we will take data that is in tabular (table) form to write an equation that describes the rate of change of a rat population.
The table below relates the number of rats in a population to time (in weeks). Use the table to write a linear equation.
|w, number of weeks||[latex]0[/latex]||[latex]2[/latex]||[latex]4[/latex]||[latex]6[/latex]|
|P(w), number of rats||[latex]1000[/latex]||[latex]1080[/latex]||[latex]1160[/latex]||[latex]1240[/latex]|
We can see from the table that the initial value for the number of rats is [latex]1000[/latex], so b =[latex]1000[/latex].
Rather than solving for m, we can tell from looking at the table that the population increases by [latex]80[/latex] for every [latex]2[/latex] weeks that pass. This means that the rate of change is [latex]80[/latex] rats per [latex]2[/latex] weeks, which can be simplified to [latex]40[/latex] rats per week.
If we did not notice the rate of change from the table, we could still solve for the slope using any two points from the table. For example, using [latex](2, 1080)[/latex] and [latex](6, 1240)[/latex],
Think About It
Is the initial value always provided in a table of values like the table in the previous example? Write your ideas in the textbox below before you look at the answer.
If your answer is no, give a description of how you would find the initial value.
No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of [latex]0[/latex], then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to [latex]0[/latex], find the slope, substitute one coordinate pair and the slope into [latex]f\left(x\right)=mx+b[/latex], and solve for b.
- Sometimes we are given an initial value and sometimes we have to solve for it.
- Using units can help you verify that you have calculated slope correctly.
- We can write the equation for a line given a slope and a data point or from a table of data.