This is the first example of many in which an expression, , only approximately models a value—tuition, in this case. In reality, the tuition would probably be rounded and would not have a fraction of a cent. 

If students are confused by the complicated number that results when they use a calculator to find the tuition from 5 years ago, consider asking:

• “Can you explain how you determined the tuition value for 2007.”

• “What is the same and what is different about the tuition value you found and the values given from 2012, 2013, and 2014?”

Here are some possible questions for discussion:

• “How can we tell what the percent of increase is?” (The 1.04 from the function can be interpreted as 104%, which is 100% of the tuition plus an increase of 4%.”)
• “If the equation weren’t given, can we still tell that the tuition is growing by 4% or by a factor of 1.04? How?” (Yes. We can find the quotient of the tuition costs in 2012 to 2013, or from 2013 to 2014. The changes from 30,000 to 31,200 in the first year and from 31,200 to 32,448 are each an increase of 4% because and .)
• “Why does -5 make sense as the input needed to calculate the cost of tuition in 2012?” (Since the input of is years since 2012, finding the tuition in 2007, which is 5 years before 2012, means using an input value of -5.)
• “Are the tuition costs changing exponentially? How do we know?” (Yes. Each year it is increasing by the same percentage or by the same factor.)

If students are unsure how to start calculating the value of the van after 8 years, consider asking:

• “How did you decide which graph correctly represents the value of the van?”

• “How could you use a table to find the value of the car after 8 years?”

The goal of this discussion is for students to make a connection between a constant percentage change and exponential growth or decay.

Display 2–3 strategies from previously selected students. Use Compare and Connect to help students compare, contrast, and connect the different strategies. Here are some questions for discussion:

• “What do the strategies have in common? How are they different?”
• “How does the initial value of the van show up in each method?”
• “How does the percentage decrease show up in each method?”

Make sure students see that:

• The $40,000 price of the van appears as the vertical intercept of the graph.
• The 15% rate of decay tells you that after 1 year, the van loses 15% of its value, retaining 85% of its original value. , so the point must be on the correct graph.

If students haven’t already shown (in their partner discussions) that they understood why Graphs A, C, and D cannot be the right representations, clarify the reasons.

If no students wrote expressions, invite them to do so now. Highlight that:

• Calculating the decay factor means subtracting the depreciation from 100% to find the remaining value of the vehicle. For example, the van loses 15% of its value each year, which means each year, it is worth 85% of the amount of the previous year.
• To find the value after 8 years, we could write an expression for the value years after it was purchased (  for the van and   for the car, where represents time in years since purchase), and calculate its value when the input is 8.