The functions f and g are defined for x geq 0. Which of the following equations shows, as a constant...

1. TRANSLATE the question requirements

• We need to determine which function shows its minimum value as a constant or coefficient in the equation
• Both functions are defined for \(\mathrm{x \geq 0}\) only
• We're looking at the equation form, not just calculating minimums

2. INFER the approach for analyzing minimum values

• Since both functions are in vertex form \(\mathrm{a(x-h)^2 + k}\), we know the vertex is at \(\mathrm{(h,k)}\)
• For upward-opening parabolas \(\mathrm{a \gt 0}\), the vertex gives the minimum
• But we must check if the vertex is within our domain \(\mathrm{x \geq 0}\)

3. INFER and analyze function I: \(\mathrm{f(x) = 2(x-3)^2 + 10}\)

• Vertex form shows vertex at \(\mathrm{(3, 10)}\)
• Since \(\mathrm{a = 2 \gt 0}\), parabola opens upward, so vertex is the minimum point
• Since \(\mathrm{x = 3 \geq 0}\), the vertex is within our domain
• Therefore, minimum value is 10, which appears as the constant '+10' in the equation

4. APPLY CONSTRAINTS and analyze function II: \(\mathrm{g(x) = 4(x+2)^2 + 5}\)

• Vertex form shows vertex at \(\mathrm{(-2, 5)}\)
• Since \(\mathrm{a = 4 \gt 0}\), parabola opens upward
• But \(\mathrm{x = -2 \lt 0}\), so vertex is outside our domain \(\mathrm{x \geq 0}\)
• The function is increasing throughout the domain \(\mathrm{x \geq 0}\)
• Minimum in domain occurs at \(\mathrm{x = 0}\):
  \(\mathrm{g(0) = 4(2)^2 + 5}\)
  \(\mathrm{= 21}\)
• The actual minimum value 21 does NOT appear anywhere in the equation

5. INFER the final comparison

• Function I: minimum value 10 appears as constant in equation ✓
• Function II: minimum value 21 does not appear in equation ✗

Answer: A. I only

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students identify that function II has vertex at \(\mathrm{(-2, 5)}\) and incorrectly conclude that 5 is the minimum value shown in the equation, without recognizing that this vertex is outside the given domain.

They see 'minimum at vertex = 5' and notice 5 appears in the equation, leading them to think both functions show their minimum values.

This may lead them to select Choice C (I and II).

Second Most Common Error:

Poor APPLY CONSTRAINTS reasoning: Students correctly find the vertices but fail to consider how the domain restriction \(\mathrm{x \geq 0}\) affects the actual minimum values. They assume the vertex always gives the minimum without checking if it's within the allowed domain.

This leads to confusion about which values are actually the minimums on the restricted domain, causing them to get stuck and guess.

The Bottom Line:

This problem tests whether students can distinguish between mathematical minimum points (vertices) and practical minimum values within domain constraints, then identify which minimum values actually appear in the equation form.