y = x^2 - 1 y = 3 When the equations above are graphed in the xy-plane, what are the...

1. INFER the solution strategy

• Given information:
  • First equation: \(\mathrm{y = x^2 - 1}\) (parabola)
  • Second equation: \(\mathrm{y = 3}\) (horizontal line)
  • Need: intersection points of both graphs

• Key insight: Intersection points occur where both equations are satisfied simultaneously - this means we have a system of equations to solve

2. INFER the most efficient approach

• Since both equations are already solved for y, we can use substitution
• The second equation tells us that \(\mathrm{y = 3}\) at any intersection point
• We can substitute this y-value into the first equation

3. SIMPLIFY through substitution

• Substitute \(\mathrm{y = 3}\) into \(\mathrm{y = x^2 - 1}\):
  \(\mathrm{3 = x^2 - 1}\)
• Add 1 to both sides:
  \(\mathrm{4 = x^2}\)
• Take the square root of both sides:
  \(\mathrm{x = \pm 2}\)

4. CONSIDER ALL CASES for complete solution

• From \(\mathrm{x^2 = 4}\), we get two solutions: \(\mathrm{x = 2}\) and \(\mathrm{x = -2}\)
• Since \(\mathrm{y = 3}\) for both values (from the second equation)
• The intersection points are: \(\mathrm{(2, 3)}\) and \(\mathrm{(-2, 3)}\)

Answer: A. (2,3) and (-2,3)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak CONSIDER ALL CASES execution: Students solve \(\mathrm{x^2 = 4}\) and only consider the positive solution \(\mathrm{x = 2}\), missing \(\mathrm{x = -2}\).

They might think "square root of 4 is 2" without remembering that every positive number has both positive and negative square roots. This leads them to find only one intersection point \(\mathrm{(2, 3)}\) and then guess among the remaining choices, or they might get confused about why their single point doesn't match any of the given answer choices that all show two points.

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when rearranging \(\mathrm{3 = x^2 - 1}\), such as subtracting 1 instead of adding 1, leading to \(\mathrm{x^2 = 2}\) and \(\mathrm{x = \pm\sqrt{2}}\).

This calculation error leads them to select Choice D. \(\mathrm{(\sqrt{2}, 3)}\) and \(\mathrm{(-\sqrt{2}, 3)}\).

The Bottom Line:

This problem tests whether students can systematically solve a system of equations and remember that quadratic equations typically have two solutions. The key challenge is maintaining precision through each algebraic step while ensuring all solutions are found.