The function f is defined by \(\mathrm{f(x) = ax^2 + bx + c}\), where a, b, and c are constants....

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = ax^2 + bx + c}\) (quadratic function)
  • Graph passes through \(\mathrm{(7, 0)}\) and \(\mathrm{(-3, 0)}\)
  • \(\mathrm{a}\) is an integer greater than 1
  • Need to find possible value of \(\mathrm{a + b}\)
• What this tells us: The points \(\mathrm{(7, 0)}\) and \(\mathrm{(-3, 0)}\) are zeros of the function

2. INFER the solution strategy

• Since we know the zeros, we can write the quadratic in factored form
• Key insight: If zeros are 7 and -3, then \(\mathrm{f(x) = a(x - 7)(x + 3)}\)
• We can expand this and compare coefficients to find the relationship between \(\mathrm{a}\) and \(\mathrm{b}\)

3. SIMPLIFY by expanding the factored form

• \(\mathrm{f(x) = a(x - 7)(x + 3)}\)
• \(\mathrm{f(x) = a(x^2 + 3x - 7x - 21)}\)
• \(\mathrm{f(x) = a(x^2 - 4x - 21)}\)
• \(\mathrm{f(x) = ax^2 - 4ax - 21a}\)

4. INFER the coefficient relationships

• Comparing \(\mathrm{ax^2 - 4ax - 21a}\) with \(\mathrm{ax^2 + bx + c}\):
  • Coefficient of \(\mathrm{x^2}\): \(\mathrm{a = a}\) ✓
  • Coefficient of \(\mathrm{x}\): \(\mathrm{b = -4a}\)
  • Constant term: \(\mathrm{c = -21a}\)
• Therefore: \(\mathrm{a + b = a + (-4a) = -3a}\)

5. APPLY CONSTRAINTS to find valid answers

• Since \(\mathrm{a}\) is an integer greater than 1: \(\mathrm{a \geq 2}\)
• This means: \(\mathrm{a + b = -3a \leq -3(2) = -6}\)
• Also, since \(\mathrm{a}\) is an integer, \(\mathrm{-3a}\) must be a multiple of 3
• Checking each choice:
  • A. -6: This works when \(\mathrm{a = 2}\) (since \(\mathrm{-3(2) = -6}\)) ✓
  • B. -3: Would need \(\mathrm{a = 1}\), but \(\mathrm{a \gt 1}\) ✗
  • C. 4: Would need \(\mathrm{-3a = 4}\), so \(\mathrm{a = -4/3}\) (not integer) ✗
  • D. 5: Would need \(\mathrm{-3a = 5}\), so \(\mathrm{a = -5/3}\) (not integer) ✗

Answer: A. -6

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that the given points are zeros, so they can't make the connection to factored form. Instead, they try to substitute the points directly into \(\mathrm{f(x) = ax^2 + bx + c}\), creating a system of equations:

• \(\mathrm{49a + 7b + c = 0}\)
• \(\mathrm{9a - 3b + c = 0}\)

This approach is much more complex and often leads to algebraic mistakes or getting stuck completely. This leads to confusion and guessing.

Second Most Common Error:

Inadequate APPLY CONSTRAINTS execution: Students correctly find that \(\mathrm{a + b = -3a}\) but forget that \(\mathrm{a}\) must be an integer greater than 1. They might select Choice B (-3) because it "looks right" without checking that it requires \(\mathrm{a = 1}\), which violates the given constraint.

The Bottom Line:

This problem rewards recognizing the connection between zeros and factored form. Students who miss this key insight face unnecessarily complex algebra, while those who see it can solve it efficiently through strategic reasoning.