Question:y = x^2 + 5y = -x^2 + 13When the equations above are graphed in the xy-plane, what are the...

1. TRANSLATE the problem information

• Given information:
  • First equation: \(\mathrm{y = x^2 + 5}\)
  • Second equation: \(\mathrm{y = -x^2 + 13}\)
  • Need to find: coordinates where the graphs intersect

• What this tells us: At intersection points, both equations must have the same x and y values

2. INFER the approach

• Since both expressions equal y, we can set them equal to each other
• This will give us the x-coordinates where the graphs meet
• Then we substitute back to find the y-coordinates

3. SIMPLIFY to solve for x

Set the right sides equal:

\(\mathrm{x^2 + 5 = -x^2 + 13}\)

Add \(\mathrm{x^2}\) to both sides:

\(\mathrm{2x^2 + 5 = 13}\)

Subtract 5 from both sides:

\(\mathrm{2x^2 = 8}\)

Divide by 2:

\(\mathrm{x^2 = 4}\)

4. CONSIDER ALL CASES when taking the square root

Take the square root of both sides:

\(\mathrm{x = ±2}\)

This gives us \(\mathrm{x = 2}\) and \(\mathrm{x = -2}\)

5. SIMPLIFY to find the y-coordinates

Substitute each x-value into either original equation. Using \(\mathrm{y = x^2 + 5}\):

For \(\mathrm{x = 2}\):

\(\mathrm{y = (2)^2 + 5 = 4 + 5 = 9}\)

For \(\mathrm{x = -2}\):

\(\mathrm{y = (-2)^2 + 5 = 4 + 5 = 9}\)

Answer: \(\mathrm{(2, 9)}\) and \(\mathrm{(-2, 9)}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

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

They correctly set up \(\mathrm{x^2 + 5 = -x^2 + 13}\) and solve to get \(\mathrm{x^2 = 4}\), but then only take the positive square root. This leads them to find just one intersection point \(\mathrm{(2, 9)}\) instead of both \(\mathrm{(2, 9)}\) and \(\mathrm{(-2, 9)}\). Since all answer choices show two points, this causes confusion and random guessing.

Second Most Common Error:

Poor TRANSLATE reasoning: Students misunderstand what "intersection points" means and try to solve each equation separately instead of setting them equal.

They might find individual points on each parabola rather than points where the parabolas meet. This approach doesn't lead to any systematic solution method, causing them to abandon the problem and guess among the answer choices.

The Bottom Line:

This problem tests whether students understand that intersection points require setting equations equal AND whether they remember to find both positive and negative square root solutions. The algebraic steps themselves are straightforward once the setup is correct.