For the function q, the value of \(\mathrm{q(x)}\) decreases by 75% for every increase in the value of x by...

1. TRANSLATE the problem information

• Given information:
  • q(x) decreases by 75% for every increase in x by 2
  • \(\mathrm{q(0) = 16}\)

• What this tells us: When x increases by 2, the function value becomes 25% of what it was (since 100% - 75% = 25%)

2. INFER the mathematical relationship

• Since this is exponential decay, q(x) has the form \(\mathrm{q(x) = 16 \times r^x}\)
• The key insight: if the function is multiplied by some factor every 2 units, and that factor is 0.25, then \(\mathrm{r^2 = 0.25}\)
• We need to find the per-unit multiplier r

3. SIMPLIFY to find the base

• From \(\mathrm{r^2 = 0.25}\), take the square root of both sides
• \(\mathrm{r = \sqrt{0.25} = \sqrt{1/4} = 1/2 = 0.5}\)
• Since we expect decay, r should be less than 1, so \(\mathrm{r = 0.5}\) makes sense

4. APPLY CONSTRAINTS to verify our answer

• Our function: \(\mathrm{q(x) = 16 \times (0.5)^x}\)
• Check: \(\mathrm{q(0) = 16 \times (0.5)^0 = 16 \times 1 = 16}\) ✓
• Check: \(\mathrm{q(2) = 16 \times (0.5)^2 = 16 \times 0.25 = 4}\) ✓
• This confirms a 75% decrease from 16 to 4

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor TRANSLATE reasoning: Students misinterpret "decreases by 75%" as meaning the function is multiplied by 0.75 instead of 0.25.

They think: "75% decrease means 75% remains" instead of recognizing that a 75% decrease leaves only 25%. This leads them to set up \(\mathrm{r^2 = 0.75}\), giving \(\mathrm{r = \sqrt{0.75} \approx 0.87}\). Looking at the answer choices, they might approximate this and select Choice (A) \(\mathrm{q(x) = 0.25(16)^x}\) because they see 0.25 and think it relates to their 75% somehow.

Second Most Common Error:

Weak INFER skill: Students correctly identify that each 2-unit increase multiplies by 0.25, but fail to connect this to finding the per-unit multiplier.

They might think the function should be \(\mathrm{q(x) = 16 \times (0.25)^x}\), not realizing they need r where \(\mathrm{r^2 = 0.25}\). This leads them to select Choice (D) \(\mathrm{q(x) = 16(0.25)^x}\) because it has the right initial value and the 0.25 factor they identified.

The Bottom Line:

This problem requires students to translate percentage language precisely and then work backwards from a multi-unit change to find the fundamental per-unit rate. The trap answers exploit both translation errors and incomplete analysis of the exponential relationship.