What is the minimum value of the function f defined by \(\mathrm{f(x) = (x - 2)^2 - 4}\)?

1. INFER the form of the function

• Given: \(\mathrm{f(x) = (x - 2)^2 - 4}\)
• This matches vertex form: \(\mathrm{f(x) = a(x - h)^2 + k}\)
• Identify the parameters:
  • \(\mathrm{a = 1}\) (leading coefficient)
  • \(\mathrm{h = 2}\) (from \(\mathrm{(x - 2)}\))
  • \(\mathrm{k = -4}\) (constant term)

2. INFER the vertex location

• In vertex form \(\mathrm{f(x) = a(x - h)^2 + k}\), the vertex is at point \(\mathrm{(h, k)}\)
• Therefore, vertex is at \(\mathrm{(2, -4)}\)

3. INFER whether this vertex gives minimum or maximum

• Since \(\mathrm{a = 1 \gt 0}\), the parabola opens upward
• When parabola opens upward, the vertex represents the lowest point
• Therefore, the vertex gives us the minimum value

4. Extract the minimum value

• The minimum value is the y-coordinate of the vertex
• Minimum value = \(\mathrm{k = -4}\)

Answer: A. -4

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize vertex form or confuse the vertex coordinates

Many students see \(\mathrm{(x - 2)^2}\) and think the vertex has x-coordinate of -2 instead of +2, leading them to incorrectly identify the vertex. Others might not realize this is vertex form at all and try to complete the square or use calculus, making the problem much harder than needed.

This confusion about form recognition leads to guessing among the answer choices.

Second Most Common Error:

Conceptual confusion about minimum vs maximum: Students identify the vertex correctly but confuse whether it represents minimum or maximum

Some students correctly find the vertex at \(\mathrm{(2, -4)}\) but then think that since the y-coordinate is negative, this must be a maximum rather than minimum. They forget to check the leading coefficient to determine the parabola's direction.

This may lead them to select Choice D (4) by taking the absolute value, thinking they need a positive minimum.

The Bottom Line:

This problem tests whether students can recognize standard forms and connect algebraic structure to graphical meaning. The key insight is that vertex form gives you the extremum directly - no complex calculations needed, just pattern recognition and understanding of parabola behavior.