Question: \(\mathrm{h(t) = -2t^2 + 20t - 47}\) The function k is defined by \(\mathrm{k(t) = h(2t)}\). For what value...

1. INFER the problem structure

• Given information:
  • \(\mathrm{h(t) = -2t^2 + 20t - 47}\) (original quadratic function)
  • \(\mathrm{k(t) = h(2t)}\) (transformed function)
  • Need to find where \(\mathrm{k(t)}\) reaches maximum

• Key insight: \(\mathrm{k(t)}\) is a horizontal compression of \(\mathrm{h(t)}\) by factor of 1/2

2. INFER the solution strategy

• Since \(\mathrm{k(t) = h(2t)}\), the maximum of k occurs when h receives its optimal input
• First find where \(\mathrm{h(t)}\) maximizes, then determine what t-value makes 2t equal that optimal input

3. SIMPLIFY to find h(t)'s maximum location

• \(\mathrm{h(t) = -2t^2 + 20t - 47}\) is a downward-opening parabola (\(\mathrm{a = -2 \lt 0}\))
• Maximum occurs at vertex: \(\mathrm{t = -b/(2a)}\)
  \(\mathrm{t = -20/(2 \times (-2))}\)
  \(\mathrm{t = -20/(-4) = 5}\)
• So \(\mathrm{h(t)}\) reaches maximum when \(\mathrm{t = 5}\)

4. INFER the connection to k(t)'s maximum

• For \(\mathrm{k(t) = h(2t)}\) to maximize, need \(\mathrm{h(2t)}\) to maximize
• This happens when \(\mathrm{2t = 5}\) (the input that maximizes h)

5. SIMPLIFY to solve for t

• \(\mathrm{2t = 5}\)
• \(\mathrm{t = 5/2 = 2.5}\)

Answer: B (2.5)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students find that \(\mathrm{h(t)}\) maximizes at \(\mathrm{t = 5}\) and incorrectly conclude that \(\mathrm{k(t)}\) also maximizes at \(\mathrm{t = 5}\), not recognizing the input transformation.

They reason: "If \(\mathrm{h(t)}\) has maximum at \(\mathrm{t = 5}\), then \(\mathrm{k(t)}\) must also have maximum at \(\mathrm{t = 5}\)." This completely ignores that \(\mathrm{k(t) = h(2t)}\) involves a different input variable relationship.

This may lead them to select Choice C (5).

Second Most Common Error:

Conceptual confusion about function transformations: Students attempt to expand \(\mathrm{k(t) = h(2t)}\) but make algebraic errors, particularly when squaring (2t), leading to incorrect coefficients.

For example, they might incorrectly write \(\mathrm{k(t) = -2(2t)^2 + 20(2t) - 47 = -4t^2 + 40t - 47}\) instead of \(\mathrm{-8t^2 + 40t - 47}\), then get \(\mathrm{t = 5}\) using the vertex formula.

This also leads them to select Choice C (5).

The Bottom Line:

The core challenge is understanding that when you have \(\mathrm{k(t) = h(2t)}\), the t-values that produce extrema are scaled by the transformation factor. Students must distinguish between where the original function extremizes versus where the transformed function extremizes.