In the xy-plane, a parabola has a minimum value of -{7} at x = 4 and intersects the x-axis at...

1. TRANSLATE the problem information

• Given information:
  • Parabola has minimum value of -7 at \(\mathrm{x = 4}\)
  • Intersects x-axis at two distinct points
  • Written as \(\mathrm{y = ax^2 + bx + c}\)
• What this tells us: The vertex is (4, -7) and since there's a minimum (not maximum), the parabola opens upward

2. INFER the constraint on coefficient a

• Since the parabola has a minimum value, it opens upward
• This means \(\mathrm{a \gt 0}\) (positive leading coefficient)
• This constraint will be crucial for eliminating wrong answers

3. TRANSLATE to vertex form

• With vertex (4, -7), the parabola can be written as:
  \(\mathrm{y = a(x - 4)^2 - 7}\)
• This form makes calculations easier than expanding to standard form

4. INFER the strategy to find a + b + c

• We need the value of a + b + c from \(\mathrm{y = ax^2 + bx + c}\)
• Key insight: When we substitute \(\mathrm{x = 1}\) into any quadratic \(\mathrm{y = ax^2 + bx + c}\), we get:
  \(\mathrm{y = a(1)^2 + b(1) + c = a + b + c}\)
• So \(\mathrm{f(1) = a + b + c}\)

5. SIMPLIFY the calculation

• Substitute \(\mathrm{x = 1}\) into vertex form:
  \(\mathrm{f(1) = a(1 - 4)^2 - 7}\)
  \(\mathrm{= a(-3)^2 - 7}\)
  \(\mathrm{= 9a - 7}\)
• Therefore: \(\mathrm{a + b + c = 9a - 7}\)

6. APPLY CONSTRAINTS to eliminate choices

• Since \(\mathrm{a \gt 0}\), we have \(\mathrm{9a \gt 0}\)
• This means \(\mathrm{9a - 7 \gt -7}\)
• Looking at choices: -15, -10, -7, -6
• Only -6 is greater than -7

Answer: D (-6)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students don't recognize that \(\mathrm{f(1) = a + b + c}\)

Many students try to expand the vertex form completely to find individual values of a, b, and c, which creates unnecessary complexity. They might attempt to use the fact that the parabola crosses the x-axis at two points to set up a system of equations, but this approach becomes algebraically messy without additional information about the x-intercepts.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Missing APPLY CONSTRAINTS reasoning: Students find that \(\mathrm{a + b + c = 9a - 7}\) but don't use the constraint \(\mathrm{a \gt 0}\)

Without recognizing that \(\mathrm{a \gt 0}\) (since the parabola opens upward), students can't eliminate the impossible choices. They might randomly select any of the negative values, particularly Choice C (-7) since it matches the minimum value given in the problem.

The Bottom Line:

The key insight is recognizing the elegant shortcut: \(\mathrm{f(1)}\) always equals \(\mathrm{a + b + c}\) for any quadratic in standard form. Combined with the constraint from the parabola's orientation, this eliminates the need for complex algebraic manipulation.