A quadratic function models the temperature, in degrees Celsius, inside a climate-controlled chamber. The model shows the temperature at time...

1. TRANSLATE the problem information

• Given information:
  • Initial temperature: \(\mathrm{T(0) = 22°C}\)
  • Minimum temperature: \(\mathrm{4°C}\) occurs at \(\mathrm{t = 12}\) hours
  • Need to find: \(\mathrm{T(16)}\)

2. INFER the mathematical approach

• Since we have a quadratic with a known minimum point, this is a vertex form problem
• The vertex (minimum) is at (12, 4)
• We'll use vertex form: \(\mathrm{T(t) = a(t - h)^2 + k}\) where \(\mathrm{(h,k) = (12, 4)}\)

3. Set up the vertex form equation

• \(\mathrm{T(t) = a(t - 12)^2 + 4}\)
• We need to find the value of coefficient 'a'

4. SIMPLIFY to find coefficient 'a'

• Use the initial condition \(\mathrm{T(0) = 22}\):
  \(\mathrm{22 = a(0 - 12)^2 + 4}\)
  \(\mathrm{22 = a(144) + 4}\)
  \(\mathrm{18 = 144a}\)
  \(\mathrm{a = \frac{18}{144} = \frac{1}{8}}\)

5. Write the complete model and evaluate

• \(\mathrm{T(t) = \frac{1}{8}(t - 12)^2 + 4}\)
• SIMPLIFY the calculation at \(\mathrm{t = 16}\):
  \(\mathrm{T(16) = \frac{1}{8}(16 - 12)^2 + 4}\)
  \(\mathrm{T(16) = \frac{1}{8}(4)^2 + 4}\)
  \(\mathrm{T(16) = \frac{1}{8}(16) + 4}\)
  \(\mathrm{T(16) = 2 + 4 = 6}\)

Answer: B. 6

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize this as a vertex form problem and instead try to work with standard form \(\mathrm{ax^2 + bx + c}\), making the algebra much more complex and error-prone. They might attempt to use the quadratic formula or try to set up a system of equations, leading to confusion and potentially guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the vertex form approach but make calculation errors when solving for 'a' or when evaluating \(\mathrm{T(16)}\). Common mistakes include sign errors with \(\mathrm{(-12)^2}\), arithmetic errors in fraction reduction \(\mathrm{(\frac{18}{144})}\), or substitution errors at \(\mathrm{t=16}\). This may lead them to select Choice C (8) if they get \(\mathrm{a = \frac{1}{4}}\) instead of \(\mathrm{a = \frac{1}{8}}\), or other incorrect values.

The Bottom Line:

This problem tests whether students can recognize when vertex form is the most efficient approach for quadratic applications. The key insight is that when given a minimum/maximum point, vertex form immediately provides the structure needed to solve the problem.