Question:The function g is defined for all real numbers x by \(\mathrm{g(x) = 6(2x + \frac{1}{3})^2 + \frac{9}{8}}\).What is the...

1. TRANSLATE the problem information

• Given: \(\mathrm{g(x) = 6(2x + 1/3)^2 + 9/8}\)
• Find: \(\mathrm{g(-1/6)}\)
• This means substitute \(\mathrm{x = -1/6}\) into the function

2. SIMPLIFY through direct substitution

• Replace every x with -1/6:
  \(\mathrm{g(-1/6) = 6(2(-1/6) + 1/3)^2 + 9/8}\)

3. SIMPLIFY the expression inside parentheses

• Calculate \(\mathrm{2(-1/6) + 1/3}\):
  • \(\mathrm{2(-1/6) = -2/6 = -1/3}\)
  • \(\mathrm{-1/3 + 1/3 = 0}\)
• So we have: \(\mathrm{g(-1/6) = 6(0)^2 + 9/8}\)

4. SIMPLIFY the remaining calculation

• \(\mathrm{0^2 = 0}\)
• \(\mathrm{6(0) = 0}\)
• \(\mathrm{0 + 9/8 = 9/8}\)

Answer: \(\mathrm{9/8}\) (also acceptable as \(\mathrm{1.125}\) or \(\mathrm{1\,1/8}\))

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY skill: Arithmetic errors when calculating \(\mathrm{2(-1/6) + 1/3}\)

Students often make sign errors or fraction mistakes:

• Getting \(\mathrm{2(-1/6) = 1/3}\) instead of \(\mathrm{-1/3}\)
• Calculating \(\mathrm{-1/3 + 1/3 = -2/3}\) instead of \(\mathrm{0}\)
• Converting fractions incorrectly

This leads to getting a non-zero value inside the parentheses, resulting in answers like \(\mathrm{3/2}\) or other incorrect values, causing confusion and potentially guessing.

Second Most Common Error:

Weak SIMPLIFY skill: Order of operations mistakes

Students might multiply by 6 before squaring, or forget to square the expression entirely. For instance, calculating \(\mathrm{6(2(-1/6) + 1/3) + 9/8}\) instead of \(\mathrm{6(2(-1/6) + 1/3)^2 + 9/8}\).

This typically leads to answers that are far from the correct value, causing students to second-guess their work.

The Bottom Line:

This problem tests careful algebraic manipulation and attention to order of operations. The key insight is recognizing that when the expression inside the parentheses equals zero, the squared term disappears entirely, leaving only the constant term.