For the function f, \(\mathrm{f(0) = 86}\), and for each increase in x by 1, the value of \(\mathrm{f(x)}\) decreases...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(0) = 86}\) (initial value)
  • For each increase in x by 1, \(\mathrm{f(x)}\) decreases by 80%

• What "decreases by 80%" means:
  • If something decreases by 80%, you keep \(\mathrm{100\% - 80\% = 20\%}\) of the original
  • So each step multiplies the previous value by \(\mathrm{0.2}\)

2. INFER the function type and form

• Since the function decreases by a constant percentage at each step, this describes exponential decay
• The general form is \(\mathrm{f(x) = a \cdot r^x}\) where:
  • \(\mathrm{a}\) = initial value = \(\mathrm{86}\)
  • \(\mathrm{r}\) = decay factor = \(\mathrm{0.2}\) (the decimal form of 20%)

• Therefore: \(\mathrm{f(x) = 86 \cdot (0.2)^x}\)

3. SIMPLIFY to find f(2)

• Substitute \(\mathrm{x = 2}\) into the function:

\(\mathrm{f(2) = 86 \cdot (0.2)^2}\)

• Calculate the exponent first:

\(\mathrm{(0.2)^2 = 0.2 \times 0.2 = 0.04}\)

• Complete the multiplication:

\(\mathrm{f(2) = 86 \times 0.04 = 3.44}\)

• As a fraction: \(\mathrm{(0.2)^2 = (1/5)^2 = 1/25}\), so \(\mathrm{f(2) = 86/25}\)

Answer: 3.44 or 86/25

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret "decreases by 80%" as meaning the function value becomes 80% of the original (instead of 20% of the original).

This leads them to use \(\mathrm{r = 0.8}\) instead of \(\mathrm{r = 0.2}\), giving \(\mathrm{f(x) = 86 \cdot (0.8)^x}\). Then \(\mathrm{f(2) = 86 \cdot (0.8)^2 = 86 \times 0.64 = 55.04}\). This causes them to get stuck since this value doesn't match the expected answer format.

Second Most Common Error:

Missing conceptual knowledge about exponential functions: Students don't recognize that "constant percentage decrease" indicates exponential decay, and instead try to solve this as a linear function.

They might think \(\mathrm{f(x)}\) decreases by 80% means subtracting 80 each time, leading to \(\mathrm{f(1) = 86 - 80 = 6}\), and \(\mathrm{f(2) = 6 - 80 = -74}\). This negative result should signal an error, but students may proceed with confusion and guessing.

The Bottom Line:

The key challenge is correctly translating percentage language into mathematical operations. "Decreases by 80%" means you multiply by \(\mathrm{(1 - 0.8) = 0.2}\), not by \(\mathrm{0.8}\).