\(\mathrm{x - 1 = (y + 2)^2}\)x = 10When the equations above are graphed in the xy-plane, what are the...

1. TRANSLATE the problem information

• Given information:
  • First equation: \(\mathrm{x - 1 = (y + 2)^2}\) (a parabola opening rightward)
  • Second equation: \(\mathrm{x = 10}\) (a vertical line)
  • Need: coordinates where these graphs intersect

2. INFER the solution approach

• Intersection points satisfy both equations simultaneously
• Since the second equation gives us \(\mathrm{x = 10}\) directly, substitute this value into the first equation
• This will give us the y-coordinates of intersection points

3. SIMPLIFY by substituting \(\mathrm{x = 10}\)

• Substitute into \(\mathrm{x - 1 = (y + 2)^2}\):
  \(\mathrm{10 - 1 = (y + 2)^2}\)
  \(\mathrm{9 = (y + 2)^2}\)

4. CONSIDER ALL CASES when solving the square equation

• Take the square root of both sides: \(\mathrm{y + 2 = ±3}\)
• This gives us two cases:
  • \(\mathrm{y + 2 = 3}\) → \(\mathrm{y = 1}\)
  • \(\mathrm{y + 2 = -3}\) → \(\mathrm{y = -5}\)

5. INFER the final coordinates

• Since \(\mathrm{x = 10}\) for both solutions, our intersection points are:
  • \(\mathrm{(10, 1)}\) and \(\mathrm{(10, -5)}\)

Answer: A

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES skill: Students solve \(\mathrm{(y + 2)^2 = 9}\) by taking only the positive square root, getting \(\mathrm{y + 2 = 3}\), so \(\mathrm{y = 1}\). They conclude there's only one intersection point at \(\mathrm{(10, 1)}\) and either select a choice with only one point or get confused when all choices show two points.

This leads to confusion and guessing since all answer choices contain two points.

Second Most Common Error:

Poor SIMPLIFY execution: Students make arithmetic errors when solving \(\mathrm{y + 2 = ±3}\), perhaps getting \(\mathrm{y = 5}\) and \(\mathrm{y = -1}\) instead of \(\mathrm{y = 1}\) and \(\mathrm{y = -5}\).

This may lead them to select Choice B (\(\mathrm{(10, -1)}\) and \(\mathrm{(10, 5)}\)).

The Bottom Line:

This problem tests whether students understand that quadratic equations typically have two solutions. The key insight is recognizing that when you square root both sides of an equation, you must consider both positive and negative possibilities.