y = 3 y = |x - 2| + 5 If the given equations are graphed in the xy-plane, at...

1. TRANSLATE the problem information

• Given information:
  • First equation: \(\mathrm{y = 3}\) (horizontal line)
  • Second equation: \(\mathrm{y = |x - 2| + 5}\) (absolute value function)
• We need to find how many intersection points exist

2. INFER the approach

• Intersection points occur where both graphs have the same y-value for the same x-value
• This means we need to set the right sides of the equations equal: \(\mathrm{3 = |x - 2| + 5}\)

3. SIMPLIFY the equation

• Start with: \(\mathrm{3 = |x - 2| + 5}\)
• Subtract 5 from both sides: \(\mathrm{3 - 5 = |x - 2|}\)
• This gives us: \(\mathrm{-2 = |x - 2|}\)

4. INFER the final conclusion

• We know that absolute value expressions are always non-negative: \(\mathrm{|x - 2| \geq 0}\)
• Since -2 is negative and \(\mathrm{|x - 2|}\) must be non-negative, there's no value of x that satisfies \(\mathrm{-2 = |x - 2|}\)
• Therefore, the equation has no solutions

Answer: (A) Zero

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may not recognize the fundamental property that absolute values are always non-negative. They might attempt to solve \(\mathrm{-2 = |x - 2|}\) by considering cases like "when \(\mathrm{x - 2 \geq 0}\)" and "when \(\mathrm{x - 2 \lt 0}\)," leading to:

• Case 1: \(\mathrm{-2 = x - 2}\), so \(\mathrm{x = 0}\)
• Case 2: \(\mathrm{-2 = -(x - 2)}\), so \(\mathrm{x = 4}\)

They then conclude there are two intersection points, leading them to select Choice (C) (Exactly two).

Second Most Common Error:

Poor SIMPLIFY execution: Students make algebraic errors when isolating the absolute value expression, perhaps getting \(\mathrm{2 = |x - 2|}\) instead of \(\mathrm{-2 = |x - 2|}\). This would actually have two solutions (\(\mathrm{x = 0}\) and \(\mathrm{x = 4}\)), causing them to select Choice (C) (Exactly two).

The Bottom Line:

This problem tests whether students truly understand that absolute value expressions represent distances and are therefore always non-negative. The key insight is recognizing when an equation has no solution rather than trying to force a solution to exist.