In the xy-plane, the parabola y = x^2 + 4x + 5 intersects the horizontal line y = 10. How...

1. TRANSLATE the intersection problem

• Given information:
  • Parabola: \(\mathrm{y = x^2 + 4x + 5}\)
  • Horizontal line: \(\mathrm{y = 10}\)
  • Need to find: number of intersection points

• What this tells us: Intersections occur where both equations have the same y-value for the same x-value

2. INFER the mathematical approach

• Since both expressions equal y, I can set them equal to each other
• This creates a quadratic equation whose solutions give the x-coordinates of intersections
• The number of real solutions equals the number of intersection points

3. TRANSLATE to equation form

Set the equations equal: \(\mathrm{x^2 + 4x + 5 = 10}\)

4. SIMPLIFY to standard quadratic form

• Subtract 10 from both sides: \(\mathrm{x^2 + 4x + 5 - 10 = 0}\)
• Standard form: \(\mathrm{x^2 + 4x - 5 = 0}\)

5. INFER using discriminant analysis

• For \(\mathrm{ax^2 + bx + c = 0}\), the discriminant \(\mathrm{Δ = b^2 - 4ac}\) tells us:
  • \(\mathrm{Δ \gt 0}\): exactly two real solutions
  • \(\mathrm{Δ = 0}\): exactly one real solution
  • \(\mathrm{Δ \lt 0}\): no real solutions

6. SIMPLIFY the discriminant calculation

• With \(\mathrm{a = 1, b = 4, c = -5}\):
• \(\mathrm{Δ = (4)^2 - 4(1)(-5)}\)
  \(\mathrm{= 16 + 20}\)
  \(\mathrm{= 36}\)

7. INFER the final answer

• Since \(\mathrm{Δ = 36 \gt 0}\), there are exactly two real solutions
• Two solutions means exactly two intersection points

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that "intersection points" means setting the equations equal to each other.

Instead, they might try to solve each equation separately or attempt to substitute specific values. This leads to confusion about what the problem is actually asking for, causing them to abandon systematic solution and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up \(\mathrm{x^2 + 4x + 5 = 10}\) but make arithmetic errors when rearranging or calculating the discriminant.

Common mistakes include getting \(\mathrm{x^2 + 4x + 15 = 0}\) (adding instead of subtracting) or calculating \(\mathrm{Δ = 16 - 20 = -4}\) (sign error). A negative discriminant would suggest zero intersections, leading them to select Choice A (Zero).

The Bottom Line:

This problem tests whether students can connect the geometric concept of intersections with the algebraic process of solving equations. The key insight is that intersection problems always reduce to solving equations, and the discriminant immediately tells you how many solutions exist.