A financial analyst estimates that a new car had a value of $24,000 in 2020. The analyst's model predicts that...

1. TRANSLATE the decay information

• Given information:
  • Initial value: \(\$24,000\)
  • Each year: value decreases by \(15\%\)
  • Need equation after t years

• Key insight: "Decreases by \(15\%\)" means the car keeps \(100\% - 15\% = 85\%\) of its value each year

2. INFER the mathematical pattern

• Since the car keeps \(85\% = 0.85\) of its value each year, we multiply by \(0.85\) repeatedly
• This creates an exponential decay pattern:
  • After 1 year: \(24,000 \times 0.85\)
  • After 2 years: \(24,000 \times 0.85 \times 0.85 = 24,000 \times (0.85)^2\)
  • After t years: \(24,000 \times (0.85)^\mathrm{t}\)

3. TRANSLATE to equation form

• The pattern gives us: \(\mathrm{v} = 24,000(0.85)^\mathrm{t}\)
• This matches the standard exponential decay form: \(\mathrm{y} = \mathrm{a}(1-\mathrm{r})^\mathrm{t}\) where:
  • \(\mathrm{a} = 24,000\) (initial value)
  • \(\mathrm{r} = 0.15\) (decay rate)
  • So \((1-\mathrm{r}) = 0.85\) (retention factor)

Answer: Choice C: \(\mathrm{v} = 24,000(0.85)^\mathrm{t}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "decreases by \(15\%\)" and incorrectly think this means multiplying by \(1.15\) instead of \(0.85\).

Their reasoning: "\(15\%\) decrease means I use \(1.15\) in the equation" without recognizing that decrease means the value becomes smaller, not larger. They don't connect that decreasing by \(15\%\) means keeping only \(85\%\) of the original.

This may lead them to select Choice D (\(\mathrm{v} = 24,000(1.15)^\mathrm{t}\)).

Second Most Common Error:

Poor INFER reasoning about exponential structure: Students understand that \(0.85\) is the correct multiplier but get confused about whether it goes with the initial value or gets raised to the power t.

They might think the \(24,000\) should be raised to the power t instead of \(0.85\), not recognizing that in exponential decay, the initial amount stays as a coefficient while the decay factor gets raised to the power.

This may lead them to select Choice A (\(\mathrm{v} = 0.85(24,000)^\mathrm{t}\)).

The Bottom Line:

The key challenge is correctly translating percentage decrease language into mathematical operations, then recognizing which part of the exponential model gets the exponent.