Let \(\mathrm{f(x) = x^2 - 10x + 29}\), where x is a real number. Which of the following is the...

1. INFER the problem strategy

• Given information:
  • \(\mathrm{f(x) = x^2 - 10x + 29}\)
  • Need to find minimum value

• Key insight: This is a quadratic function with positive leading coefficient (\(\mathrm{a = 1 \gt 0}\)), so the parabola opens upward and has a minimum value at its vertex. I can find this using completing the square.

2. SIMPLIFY by completing the square

• Start with: \(\mathrm{f(x) = x^2 - 10x + 29}\)

• To complete the square for \(\mathrm{x^2 - 10x}\):
  • Take half of the x coefficient: \(\mathrm{-10/2 = -5}\)
  • Square it: \(\mathrm{(-5)^2 = 25}\)
  • Add and subtract this value

• \(\mathrm{f(x) = x^2 - 10x + 25 - 25 + 29}\)
• \(\mathrm{f(x) = (x^2 - 10x + 25) - 25 + 29}\)
• \(\mathrm{f(x) = (x - 5)^2 + 4}\)

3. INFER the minimum value

• In vertex form \(\mathrm{f(x) = (x - 5)^2 + 4}\):
  • Since \(\mathrm{(x - 5)^2 \geq 0}\) for all real numbers
  • The minimum value of \(\mathrm{(x - 5)^2}\) is 0
  • This occurs when \(\mathrm{x - 5 = 0}\), so \(\mathrm{x = 5}\)

• Minimum value \(\mathrm{= 0 + 4 = 4}\)

Answer: C (4)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skills: Students make arithmetic errors when completing the square, particularly with the +25/-25 step.

They might write: \(\mathrm{f(x) = x^2 - 10x + 29 = (x - 5)^2 + 29 - 25 = (x - 5)^2 + 54}\), leading to a minimum value of 54. Since 54 isn't an answer choice, this leads to confusion and guessing.

Second Most Common Error:

Poor INFER reasoning: Students don't recognize that they need to find the vertex of the parabola and instead try to solve \(\mathrm{f(x) = 0}\) or substitute the answer choices.

They might think "let me try each answer choice to see which one works" without understanding what "minimum value" means. This leads to random testing and likely results in selecting the wrong answer.

The Bottom Line:

This problem requires students to understand that finding the minimum value means finding the y-coordinate of the vertex, and completing the square is the most reliable method to transform the quadratic into vertex form where the minimum is clearly visible.