A car's value V, in dollars, m months after purchase is modeled by \(\mathrm{V(m) = 36,000(0.97)^m}\). Which function W best...

1. TRANSLATE the problem information

• Given information:
  • Original function: \(\mathrm{V(m) = 36,000(0.97)^m}\) (value after m months)
  • Need: \(\mathrm{W(y)}\) = function for value after y years
• What this tells us: We need to convert the time variable from months to years

2. INFER the relationship between time units

• Since 1 year = 12 months, then y years = 12y months
• This means: \(\mathrm{m = 12y}\)
• Strategy: Substitute this relationship into the original function

3. SIMPLIFY by direct substitution

• Start with: \(\mathrm{V(m) = 36,000(0.97)^m}\)
• Replace m with 12y: \(\mathrm{W(y) = 36,000(0.97)^{12y}}\)
• This gives us the value after y years, where the monthly decay factor 0.97 is applied 12y times

Answer: B. \(\mathrm{W(y) = 36,000(0.97)^{12y}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students incorrectly think converting from years to months means dividing in the exponent.

They reason: "If y is in years and m is in months, then \(\mathrm{m = y/12}\)" (backwards relationship). This leads them to \(\mathrm{W(y) = 36,000(0.97)^{y/12}}\), thinking this converts years to the "monthly scale."

This may lead them to select Choice A (\(\mathrm{36,000(0.97)^{y/12}}\))

Second Most Common Error:

Conceptual confusion about exponential decay: Students think the initial value should be adjusted for the time unit change.

They reason: "Since we're going from monthly to yearly, maybe divide the starting value by 12" while also misunderstanding how many times the decay factor should be applied.

This may lead them to select Choice C (\(\mathrm{(36,000/12)(0.97)^y}\))

The Bottom Line:

The key insight is that when converting time units in exponential functions, you substitute the equivalent time expression. If you want yearly input but have a monthly model, substitute the number of months (12y) for the monthly variable (m). The decay factor gets applied more times over longer periods, which is exactly what \(\mathrm{(0.97)^{12y}}\) represents.