\(\mathrm{A(t) = 1200(1.02)^{(t/0.5)}}\) The function A models the value, in dollars, of an investment account t years after the account...

1. TRANSLATE the problem information

• Given function: \(\mathrm{A(t) = 1200(1.02)^{(t/0.5)}}\) where t is in years
• Need to find: percentage increase every 6 months (0.5 years)
• The question asks for n% increase every 6 months

2. INFER the relationship between time periods and exponents

• Every time t increases by 0.5 years (6 months), the exponent \(\mathrm{t/0.5}\) increases by 1
• When the exponent increases by 1, the entire expression gets multiplied by the base: \(\mathrm{1.02}\)

3. INFER what this multiplication factor means

• Starting value at some time: \(\mathrm{A = 1200(1.02)^k}\)
• Value 6 months later: \(\mathrm{A = 1200(1.02)^{(k+1)} = [1200(1.02)^k] \times 1.02}\)
• The account value gets multiplied by \(\mathrm{1.02}\) every 6 months

4. TRANSLATE multiplication factor to percentage increase

• Multiplying by \(\mathrm{1.02}\) means the new value = original value + 2% of original value
• This represents a 2% increase every 6 months
• Therefore: \(\mathrm{n = 2}\)

Answer: A (2)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students see \(\mathrm{1.02}\) in the function and think this directly represents 1.02% increase rather than recognizing it as a multiplication factor.

They reason: "The base is \(\mathrm{1.02}\), so the increase must be 1.02%." This misunderstands that \(\mathrm{1.02}\) as a multiplier means \(\mathrm{100\% + 2\% = 102\%}\) of the original, which is a 2% increase.

This may lead them to select Choice B (4) if they double \(\mathrm{1.02}\) to get close to an answer, or cause confusion and guessing.

Second Most Common Error:

Poor INFER reasoning about time periods: Students might not recognize that the exponent structure \(\mathrm{t/0.5}\) specifically makes the base (\(\mathrm{1.02}\)) apply to each 6-month period.

They might think the 2% applies to each year instead of each 6-month period, leading to calculations involving annual rates rather than semi-annual rates.

This leads to confusion and guessing among the given choices.

The Bottom Line:

This problem tests whether students can connect the mechanical structure of an exponential function to its real-world interpretation. The key insight is recognizing that the denominator in the exponent (\(\mathrm{0.5}\)) determines the time period for the growth rate represented by the base.