Let \(\mathrm{q(x) = x^2 - 6x + 3}\).For a neq 1, define \(\mathrm{A(a) = \frac{q(a) - q(1)}{a - 1}}\).What is...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{q(x) = x^2 - 6x + 3}\)
  • \(\mathrm{A(a) = \frac{q(a) - q(1)}{a - 1}}\) for \(\mathrm{a \neq 1}\)
  • Need to find \(\mathrm{A(-1)}\)

• What this tells us: We need to substitute specific values into both the original function \(\mathrm{q(x)}\) and the derived function \(\mathrm{A(a)}\)

2. INFER the solution approach

• To find \(\mathrm{A(-1)}\), I need to use the formula \(\mathrm{A(a) = \frac{q(a) - q(1)}{a - 1}}\) with \(\mathrm{a = -1}\)
• This means I need to compute both \(\mathrm{q(-1)}\) and \(\mathrm{q(1)}\) first
• Strategy: Calculate \(\mathrm{q(1)}\), then \(\mathrm{q(-1)}\), then apply the formula

3. SIMPLIFY by computing q(1)

\(\mathrm{q(1) = 1^2 - 6(1) + 3}\)

\(\mathrm{q(1) = 1 - 6 + 3 = -2}\)

4. SIMPLIFY by computing q(-1)

\(\mathrm{q(-1) = (-1)^2 - 6(-1) + 3}\)

\(\mathrm{q(-1) = 1 + 6 + 3 = 10}\)

• Key: \(\mathrm{(-1)^2 = +1}\), and \(\mathrm{-6(-1) = +6}\)

5. SIMPLIFY by applying the A(a) formula

\(\mathrm{A(-1) = \frac{q(-1) - q(1)}{-1 - 1}}\)

\(\mathrm{A(-1) = \frac{10 - (-2)}{-2}}\)

\(\mathrm{A(-1) = \frac{10 + 2}{-2}}\)

\(\mathrm{A(-1) = \frac{12}{-2} = -6}\)

Answer: -6 (Choice B)

Why Students Usually Falter on This Problem

Most Common Error Path:

Poor SIMPLIFY execution: Sign errors when evaluating \(\mathrm{q(-1)}\)

Students often make mistakes like:

• Writing \(\mathrm{(-1)^2 = -1}\) instead of \(\mathrm{+1}\)
• Computing \(\mathrm{-6(-1) = -6}\) instead of \(\mathrm{+6}\)
• Getting \(\mathrm{q(-1) = -1 - 6 + 3 = -4}\) instead of the correct \(\mathrm{q(-1) = 10}\)

With \(\mathrm{q(-1) = -4}\), they would get \(\mathrm{A(-1) = \frac{-4 - (-2)}{-2} = \frac{-2}{-2} = 1}\), but this isn't among the choices, leading to confusion and guessing.

Second Most Common Error:

Weak TRANSLATE skills: Misunderstanding the composite nature of \(\mathrm{A(a)}\)

Some students try to substitute \(\mathrm{x = -1}\) directly into \(\mathrm{A(x)}\) thinking it's like \(\mathrm{q(x)}\), not recognizing that \(\mathrm{A(a)}\) is defined in terms of \(\mathrm{q(a)}\) and \(\mathrm{q(1)}\). This leads to confusion about what to compute first and often results in abandoning systematic solution and guessing.

The Bottom Line:

This problem tests both function evaluation skills and careful arithmetic with negative numbers. Success requires systematically working through the function definitions while maintaining accuracy with signs throughout multiple calculation steps.