t\(\mathrm{h(t)}\)-1360204-{4}The table above shows three values of t and their corresponding values of \(\mathrm{h(t)}\). The function h is defined b...

1. TRANSLATE the problem information

• Given: \(\mathrm{h(t) = k(t)(2-t)}\) where \(\mathrm{k(t)}\) is linear
• Table shows: \(\mathrm{t = -1}\) gives \(\mathrm{h(-1) = 36}\); \(\mathrm{t = 0}\) gives \(\mathrm{h(0) = 20}\); \(\mathrm{t = 4}\) gives \(\mathrm{h(4) = -4}\)
• Find: t-intercept of \(\mathrm{y = k(t)}\)

2. INFER the approach to find k(t)

• Since we know \(\mathrm{h(t) = k(t)(2-t)}\), we can solve for \(\mathrm{k(t)}\)
• Rearranging: \(\mathrm{k(t) = \frac{h(t)}{(2-t)}}\)
• We can use each table value to find corresponding points on \(\mathrm{k(t)}\)

3. SIMPLIFY to find points on the linear function k(t)

• At \(\mathrm{t = -1}\):
  \(\mathrm{k(-1) = \frac{36}{(2-(-1))}}\)
  \(\mathrm{= \frac{36}{3}}\)
  \(\mathrm{= 12}\)
  → Point \(\mathrm{(-1, 12)}\)
• At \(\mathrm{t = 0}\):
  \(\mathrm{k(0) = \frac{20}{(2-0)}}\)
  \(\mathrm{= \frac{20}{2}}\)
  \(\mathrm{= 10}\)
  → Point \(\mathrm{(0, 10)}\)
• At \(\mathrm{t = 4}\):
  \(\mathrm{k(4) = \frac{-4}{(2-4)}}\)
  \(\mathrm{= \frac{-4}{-2}}\)
  \(\mathrm{= 2}\)
  → Point \(\mathrm{(4, 2)}\)

4. INFER the linear equation from two points

• Since \(\mathrm{k(t)}\) is linear, we can use any two points to find its equation
• Using \(\mathrm{(-1, 12)}\) and \(\mathrm{(0, 10)}\):
• Slope:
  \(\mathrm{m = \frac{(10-12)}{(0-(-1))}}\)
  \(\mathrm{= \frac{-2}{1}}\)
  \(\mathrm{= -2}\)
• From point \(\mathrm{(0, 10)}\), the y-intercept is 10
• Therefore: \(\mathrm{k(t) = -2t + 10}\)

5. SIMPLIFY to find the t-intercept

• Set \(\mathrm{k(t) = 0}\):
  \(\mathrm{0 = -2t + 10}\)
• Solve:
  \(\mathrm{2t = 10}\)
  \(\mathrm{t = 5}\)

Answer: C. 5

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize they need to find \(\mathrm{k(t)}\) first before finding its intercept. They might try to work directly with \(\mathrm{h(t) = k(t)(2-t)}\) without realizing they can solve for \(\mathrm{k(t)}\) using the table values. This leads to confusion about how to proceed, causing them to get stuck and guess randomly.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify that \(\mathrm{k(t) = \frac{h(t)}{(2-t)}}\) but make calculation errors when computing the values. For example, calculating \(\mathrm{k(-1) = \frac{36}{(2-(-1))} = \frac{36}{1} = 36}\) instead of \(\mathrm{\frac{36}{3} = 12}\), or making sign errors with negative values. These errors cascade through the linear equation determination, potentially leading them to select Choice A (-2) or Choice B (2) if their incorrect slope becomes their final answer.

The Bottom Line:

This problem requires students to work backwards from a composite function to find its component, then apply linear function analysis. The key insight is recognizing that the given relationship can be algebraically manipulated to reveal the unknown linear function.