\(\mathrm{p(x) + 57 = x²}\) The given equation relates the value of x and its corresponding value of \(\mathrm{p(x)}\) for...

1. TRANSLATE the given equation into standard form

• Given: \(\mathrm{p(x) + 57 = x²}\)
• TRANSLATE this by isolating \(\mathrm{p(x)}\): subtract 57 from both sides
• Result: \(\mathrm{p(x) = x² - 57}\)

2. INFER the type of function and its key properties

• We now have \(\mathrm{p(x) = x² - 57}\), which is a quadratic function
• Since the coefficient of \(\mathrm{x²}\) is positive (\(\mathrm{1 \gt 0}\)), this parabola opens upward
• This means the function has a minimum value, not a maximum

3. INFER the vertex form and minimum value

• The function \(\mathrm{p(x) = x² - 57}\) can be written as \(\mathrm{p(x) = (x - 0)² + (-57)}\)
• This matches vertex form \(\mathrm{p(x) = a(x - h)² + k}\) where \(\mathrm{a = 1, h = 0, k = -57}\)
• For vertex form, the minimum value is always k when \(\mathrm{a \gt 0}\)

Answer: B. -57

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students recognize they need to rearrange the equation but then confuse what the minimum value represents. They might think the minimum occurs at \(\mathrm{x = -57}\) or that they need to find where \(\mathrm{p(x)}\) equals zero.

This confusion about interpreting quadratic functions leads to guessing among the answer choices.

Second Most Common Error:

Conceptual confusion about vertex form: Students correctly find \(\mathrm{p(x) = x² - 57}\) but don't recognize this as vertex form or remember that k represents the minimum value. They might try to complete the square unnecessarily or use the quadratic formula.

This may lead them to select Choice C (57) by taking the absolute value of the constant term.

The Bottom Line:

This problem tests whether students can connect algebraic manipulation with the geometric meaning of quadratic functions. The key insight is recognizing that rearranging to standard form immediately reveals the minimum value.