In the xy-plane, a parabola has vertex \((-5, 3)\) and crosses the y-axis at two points. If the equation of...

1. TRANSLATE the problem information

• Given information:
  • Vertex: (-5, 3)
  • Crosses y-axis at two points (meaning when \(\mathrm{x = 0}\), there are two y-values)
  • Form: \(\mathrm{x = ay^2 + by + c}\)
• What this tells us: We're working with a horizontal parabola (x depends on y)

2. INFER the approach

• Since we know the vertex, start with vertex form for horizontal parabolas
• The vertex form is \(\mathrm{x = a(y - k)^2 + h}\) where \(\mathrm{(h, k)}\) is the vertex
• We'll expand this to match the required form \(\mathrm{x = ay^2 + by + c}\)

3. SIMPLIFY using vertex form

• With vertex (-5, 3): \(\mathrm{x = a(y - 3)^2 + (-5)}\)
  \(\mathrm{x = a(y - 3)^2 - 5}\)
• Expand: \(\mathrm{x = a(y^2 - 6y + 9) - 5}\)
  \(\mathrm{x = ay^2 - 6ay + 9a - 5}\)
• Comparing to \(\mathrm{x = ay^2 + by + c}\):
  • Coefficient of \(\mathrm{y^2}\): a (same)
  • Coefficient of y: \(\mathrm{b = -6a}\)
  • Constant term: \(\mathrm{c = 9a - 5}\)

4. INFER the constraint from two y-intercepts

• Two y-axis intersections means when \(\mathrm{x = 0}\), the equation \(\mathrm{ay^2 - 6ay + 9a - 5 = 0}\) has two real solutions
• For two real solutions, we need \(\mathrm{discriminant \gt 0}\)
• Discriminant = \(\mathrm{(-6a)^2 - 4(a)(9a - 5)}\)
  \(\mathrm{= 36a^2 - 36a^2 + 20a}\)
  \(\mathrm{= 20a}\)

5. APPLY CONSTRAINTS to find valid values

• For \(\mathrm{20a \gt 0}\), we need \(\mathrm{a \gt 0}\)
• Since \(\mathrm{a + b + c = a + (-6a) + (9a - 5)}\)
  \(\mathrm{= 4a - 5}\)
• With \(\mathrm{a \gt 0}\), we get \(\mathrm{4a - 5 \gt -5}\)

6. INFER which answer works

• Check each choice by solving \(\mathrm{4a - 5 =}\) [choice value]:
  • (A) -12: \(\mathrm{a = -7/4 \lt 0}\) ✗
  • (B) -9: \(\mathrm{a = -1 \lt 0}\) ✗
  • (C) -5: \(\mathrm{a = 0}\) (no parabola) ✗
  • (D) -3: \(\mathrm{a = 1/2 \gt 0}\) ✓

Answer: D. -3

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students often forget that "crosses the y-axis at two points" creates a constraint on the parameter 'a'. They calculate \(\mathrm{a + b + c = 4a - 5}\) correctly but then think any answer choice could work since they can solve for 'a' from each one. They miss that the discriminant condition requires \(\mathrm{a \gt 0}\), which eliminates most choices. This leads to confusion and guessing among the first few choices.

Second Most Common Error:

Conceptual confusion about parabola orientation: Students might try to use the vertical parabola vertex form \(\mathrm{y = a(x - h)^2 + k}\) instead of the horizontal form \(\mathrm{x = a(y - k)^2 + h}\). This creates completely wrong coefficients and leads them to an incorrect expression for \(\mathrm{a + b + c}\). This causes them to get stuck and randomly select an answer.

The Bottom Line:

This problem tests whether students can work with horizontal parabolas AND apply intersection constraints. The key insight is that the phrase "crosses the y-axis at two points" isn't just descriptive—it's a mathematical requirement that constrains which values of 'a' are actually possible.