2.3: Modeling with Linear Functions
Identifying Steps to Model and Solve Problems
When modeling scenarios with linear functions and solving problems involving quantities with a constant rate of change , we typically follow the same problem strategies that we would use for any type of function. Let’s briefly review them:
Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
Carefully read the problem to identify important information. Look for information that provides values for the variables or values for parts of the functional model, such as slope and initial value.
Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
Identify a solution pathway from the provided information to what we are trying to find. Often this will involve checking and tracking units, building a table, or even finding a formula for the function being used to model the problem.
When needed, write a formula for the function.
Solve or evaluate the function using the formula.
Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
Clearly convey your result using appropriate units, and answer in full sentences when necessary.
Building Linear Models
Now let’s take a look at the student in Seattle. In her situation, there are two changing quantities: time and money. The amount of money she has remaining while on vacation depends on how long she stays. We can use this information to define our variables, including units.
- Output: \(M\), money remaining, in dollars
- Input: \(t\), time, in weeks
So, the amount of money remaining depends on the number of weeks: \(M(t)\)
We can also identify the initial value and the rate of change.
- Initial Value: She saved $3,500, so $3,500 is the initial value for M.
- Rate of Change: She anticipates spending $400 each week, so –$400 per week is the rate of change, or slope.
Notice that the unit of dollars per week matches the unit of our output variable divided by our input variable. Also, because the slope is negative, the linear function is decreasing. This should make sense because she is spending money each week.
The rate of change is constant , so we can start with the linear model \(M(t)=mt+b\). Then we can substitute the intercept and slope provided.
To find the x-intercept, we set the output to zero, and solve for the input.
\[\begin{align*} 0&=−400t+3500 \\ t&=\dfrac{3500}{400} \\ &=8.75 \end{align*}\]
The x-intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Emily will have no money left after 8.75 weeks.
When modeling any real-life scenario with functions, there is typically a limited domain over which that model will be valid—almost no trend continues indefinitely. Here the domain refers to the number of weeks. In this case, it doesn’t make sense to talk about input values less than zero. A negative input value could refer to a number of weeks before she saved $3,500, but the scenario discussed poses the question once she saved $3,500 because this is when her trip and subsequent spending starts. It is also likely that this model is not valid after the x-intercept, unless Emily will use a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is \(0{\leq}t{\leq}8.75\).
In the above example, we were given a written description of the situation. We followed the steps of modeling a problem to analyze the information. However, the information provided may not always be the same. Sometimes we might be provided with an intercept. Other times we might be provided with an output value. We must be careful to analyze the information we are given, and use it appropriately to build a linear model.
Using a Given Intercept to Build a Model
Some real-world problems provide the y-intercept, which is the constant or initial value. Once the y-intercept is known, the x-intercept can be calculated. Suppose, for example, that Hannah plans to pay off a no-interest loan from her parents. Her loan balance is $1,000. She plans to pay $250 per month until her balance is $0. The y-intercept is the initial amount of her debt, or $1,000. The rate of change, or slope, is -$250 per month. We can then use the slope-intercept form and the given information to develop a linear model.
\[\begin{align*} f(x)&=mx+b \\ &=-250x+1000 \end{align*}\]
Now we can set the function equal to 0, and solve for \(x\) to find the x-intercept.
\[\begin{align*} 0&=-250+1000 \\ 1000&=250x \\ 4&=x \\ x&=4 \end{align*}\]
The x-intercept is the number of months it takes her to reach a balance of $0. The x-intercept is 4 months, so it will take Hannah four months to pay off her loan.
Using a Given Input and Output to Build a Model
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 instead be asked to evaluate the linear model at a given input or set the equation of the linear model equal to a specified output.
Building Systems of Linear Models
Real-world situations including two or more linear functions may be modeled with a system of linear equations . Remember, when solving a system of linear equations, we are looking for points the two lines have in common. Typically, there are three types of answers possible, as shown in Figure \(\PageIndex{6}\).