# Problem Solving With Linear Functions Key

Rate of Change: She anticipates spending 0 each week, so –0 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.

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.

$\begin 0&=−400t 3500 \ t&=\dfrac \ &=8.75 \end$ The x-intercept is 8.75 weeks.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.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.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.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.The domain represents the set of input values, so the reasonable domain for this function is $$0t8.75$$.In the above example, we were given a written description of the situation.

## One thought on “Problem Solving With Linear Functions Key”

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