# Linear Functions

## Modeling with Linear Functions

Elan is a college student who plans to spend a summer in Seattle. Elan has saved $3,500 for the trip and anticipates spending $400 each week on rent, food, and activities. How can we write a linear model to represent the situation? What would be the x-intercept, and what can Elan learn from it? To answer these and related questions, we can create a model using a linear function. Models such as this one can be extremely useful for analyzing relationships and making predictions based on those relationships. In this section, we will explore examples of linear function models.

### 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:
1. Identify changing quantities, and then define descriptive variables to represent those quantities. When appropriate, sketch a picture or define a coordinate system.
2. 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.
3. Carefully read the problem to determine what we are trying to find, identify, solve, or interpret.
4. 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.
5. When needed, write a formula for the function.
6. Solve or evaluate the function using the formula.
7. Reflect on whether your answer is reasonable for the given situation and whether it makes sense mathematically.
8. 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 Elan's situation, there are two changing quantities: time and money. The amount of money they have remaining while on vacation depends on how long they stay. We can use this information to define our variables, including units.
1. Output: 
 money remaining, in dollars
2. Input: 
 time, in weeks
So, the amount of money remaining depends on the number of weeks: 
 
We can also identify the initial value and the rate of change.
1. Initial Value: They saved $3,500, so $3,500 is the initial value for 
 
2. Rate of Change: They anticipate 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 they are spending money each week.
The rate of change is constant, so we can start with the linear model 
 Then we can substitute the intercept and slope provided.
To find the 
 intercept, we set the output to zero, and solve for the input.
The 
 intercept is 8.75 weeks. Because this represents the input value when the output will be zero, we could say that Elan 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 Elan saved $3,500, but the scenario discussed poses the question once they saved $3,500 because this is when the trip and subsequent spending starts. It is also likely that this model is not valid after the 
 intercept, unless Elan will use a credit card and go into debt. The domain represents the set of input values, so the reasonable domain for this function is 
 
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 
 intercept, which is the constant or initial value. Once the 
 intercept is known, the 
 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 
 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.
Now we can set the function equal to 0, and solve for 
 to find the 
 intercept.
The 
  intercept is the number of months it takes her to reach a balance of $0. The 
 -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.

### Using a Diagram to Model a Problem

It 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, geometrical 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.

### 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 .
1. We can use the same problem strategies that we would use for any type of function.
2. When modeling and solving a problem, identify the variables and look for key values, including the slope and y-intercept. See .
3. Draw a diagram, where appropriate. See  and .
4. Check for reasonableness of the answer.
5. Linear models may be built by identifying or calculating the slope and using the y-intercept.
6. The x-intercept may be found by setting 
  which is setting the expression 
  equal to 0.
7. The point of intersection of a system of linear equations is the point where the x- and y-values are the same. See .
8. A graph of the system may be used to identify the points where one line falls below (or above) the other line.

### Verbal


### Algebraic

For the following exercises, consider this scenario: A town’s population has been decreasing at a constant rate. In 2010 the population was 5,900. By 2012 the population had dropped to 4,700. Assume this trend continues.
For the following exercises, consider this scenario: A town’s population has been increased at a constant rate. In 2010 the population was 46,020. By 2012 the population had increased to 52,070. Assume this trend continues.
For the following exercises, consider this scenario: A town has an initial population of 75,000. It grows at a constant rate of 2,500 per year for 5 years.
For the following exercises, consider this scenario: The weight of a newborn is 7.5 pounds. The baby gained one-half pound a month for its first year.
For the following exercises, consider this scenario: The number of people afflicted with the common cold in the winter months steadily decreased by 205 each year from 2005 until 2010. In 2005, 12,025 people were afflicted.

### Graphical

For the following exercises, use the graph in , which shows the profit, 
  in thousands of dollars, of a company in a given year, 
  where 
  represents the number of years since 1980.
For the following exercises, use the graph in , which shows the profit, 
  in thousands of dollars, of a company in a given year, 
  where 
  represents the number of years since 1980.

### Numeric

For the following exercises, use the median home values in Mississippi and Hawaii (adjusted for inflation) shown in . Assume that the house values are changing linearly.
For the following exercises, use the median home values in Indiana and Alabama (adjusted for inflation) shown in . Assume that the house values are changing linearly.

### Real-World Applications
