For the function p, the value of \(\mathrm{p(x)}\) decreases by 75% each time x increases by 3. If \(\mathrm{p(0) =...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{p(x)}\) decreases by 75% each time \(\mathrm{x}\) increases by 3
  • \(\mathrm{p(0) = 32}\)
• What this tells us: If something decreases by 75%, it retains 25% of its original value
  • So \(\mathrm{p(x+3) = 0.25 \times p(x)}\)

2. INFER the pattern by testing specific values

• Let's see what happens at regular intervals:
  • At \(\mathrm{x = 0}\): \(\mathrm{p(0) = 32}\) (given)
  • At \(\mathrm{x = 3}\): \(\mathrm{p(3) = 0.25 \times 32 = 8}\)
  • At \(\mathrm{x = 6}\): \(\mathrm{p(6) = 0.25 \times 8 = 2}\)
  • At \(\mathrm{x = 9}\): \(\mathrm{p(9) = 0.25 \times 2 = 0.5}\)

3. INFER the general relationship

• After \(\mathrm{k}\) intervals of size 3, we're at \(\mathrm{x = 3k}\)
• Each interval multiplies by 0.25, so after \(\mathrm{k}\) intervals: \(\mathrm{p(3k) = 32 \times (0.25)^k}\)
• For any \(\mathrm{x}\) value: \(\mathrm{k = \frac{x}{3}}\), so \(\mathrm{p(x) = 32 \times (0.25)^{\frac{x}{3}}}\)

4. APPLY CONSTRAINTS to select the correct choice

• Looking at the options, only choice (C) matches our derived formula: \(\mathrm{p(x) = 32(0.25)^{\frac{x}{3}}}\)

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students often misinterpret "decreases by 75%" as meaning the value becomes 0.75 times the original, rather than 0.25 times the original.

This leads them to think \(\mathrm{p(x+3) = 0.75 \times p(x)}\), which would point toward Choice D (\(\mathrm{p(x) = 32(0.75)^{\frac{x}{3}}}\)).

Second Most Common Error:

Poor INFER reasoning: Students correctly identify that each step multiplies by 0.25, but fail to account for the interval size of 3. They assume the pattern applies for every unit increase in \(\mathrm{x}\).

This leads them to conclude \(\mathrm{p(x) = 32(0.25)^x}\), making them select Choice B (\(\mathrm{p(x) = 32(0.25)^x}\)).

The Bottom Line:

This problem requires careful attention to both the percentage calculation and the interval over which the change occurs. The key insight is recognizing that the exponent must be \(\mathrm{\frac{x}{3}}\), not just \(\mathrm{x}\), because the decay happens every 3 units, not every 1 unit.