# 4.2 Modeling with linear functions  (Page 2/9)

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To find the x- intercept, we set the output to zero, and solve for the input.

$\begin{array}{ccc}\hfill 0& =& -400t+3500\hfill \\ \hfill t& =& \frac{3500}{400}\hfill \\ & =& 8.75\hfill \end{array}$

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 uses a credit card and goes into debt. The domain represents the set of input values, so the reasonable domain for this function is $\text{\hspace{0.17em}}0\le t\le 8.75.$

In this 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. $\begin{array}{ccc}\hfill f\left(x\right)& =& mx+b\hfill \\ & =& -250x+1000\hfill \end{array}$ Now we can set the function equal to 0, and solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to find the x -intercept. $\begin{array}{ccc}\hfill 0& =& -250x+1000\hfill \\ \hfill 1000& =& 250x\hfill \\ \hfill 4& =& x\hfill \\ \hfill x& =& 4\hfill \end{array}$ 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.

Given a word problem that includes two pairs of input and output values, use the linear function to solve a problem.

1. Identify the input and output values.
2. Convert the data to two coordinate pairs.
3. Find the slope.
4. Write the linear model.
5. Use the model to make a prediction by evaluating the function at a given x -value.
6. Use the model to identify an x -value that results in a given y -value.

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