A quadratic function f is defined by \(\mathrm{f(x) = ax^2 + bx + c}\). The graph of \(\mathrm{y = f(x)}\)...

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{f(x) = ax^2 + bx + c}\) (quadratic function)
  • Vertex at \(\mathrm{(1, -1)}\)
  • Passes through origin \(\mathrm{(0, 0)}\)
  • Need to find number of distinct real solutions to \(\mathrm{f(x) = 0}\)

2. INFER the most efficient approach

• Since we know the vertex coordinates, vertex form will be much easier to work with than trying to use standard form
• Vertex form: \(\mathrm{f(x) = a(x - h)^2 + k}\) where \(\mathrm{(h, k)}\) is the vertex
• We'll use the additional point to find the value of 'a'

3. TRANSLATE to vertex form

• With vertex at \(\mathrm{(1, -1)}\): \(\mathrm{f(x) = a(x - 1)^2 + (-1) = a(x - 1)^2 - 1}\)

4. SIMPLIFY to find the value of 'a'

• Use the condition that \(\mathrm{f(0) = 0}\) (passes through origin):

\(\mathrm{f(0) = a(0 - 1)^2 - 1 = 0}\)
\(\mathrm{a(1) - 1 = 0}\)
\(\mathrm{a - 1 = 0}\)
\(\mathrm{a = 1}\)

• Therefore: \(\mathrm{f(x) = (x - 1)^2 - 1}\)

5. SIMPLIFY to solve f(x) = 0

• Set up the equation: \(\mathrm{(x - 1)^2 - 1 = 0}\)
• Add 1 to both sides: \(\mathrm{(x - 1)^2 = 1}\)
• Take square root of both sides: \(\mathrm{x - 1 = ±1}\)
• Solve for x: \(\mathrm{x = 1 ± 1}\), so \(\mathrm{x = 2}\) or \(\mathrm{x = 0}\)

6. INFER the final answer

• We found \(\mathrm{x = 0}\) and \(\mathrm{x = 2}\), which are two different values
• Therefore, there are 2 distinct real solutions

Answer: C

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students attempt to expand the vertex form or work backward to standard form unnecessarily, creating complex algebra that leads to calculation errors.

Instead of keeping \(\mathrm{f(x) = a(x - 1)^2 - 1}\) and using the y-intercept directly, they might expand to get \(\mathrm{f(x) = a(x^2 - 2x + 1) - 1 = ax^2 - 2ax + a - 1}\), then try to match this with \(\mathrm{ax^2 + bx + c}\). This creates unnecessary complexity and increases chances for algebraic mistakes.

This may lead them to select Choice A (0) or get confused and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students make errors when taking the square root, either forgetting the ± or making sign errors in the final step.

When solving \(\mathrm{(x - 1)^2 = 1}\), they might only take the positive square root and get \(\mathrm{x - 1 = 1}\), so \(\mathrm{x = 2}\), missing the second solution. Or they might make a sign error and get \(\mathrm{x = 0}\) and \(\mathrm{x = -2}\) instead of \(\mathrm{x = 0}\) and \(\mathrm{x = 2}\).

This may lead them to select Choice B (1) if they find only one solution.

The Bottom Line:

This problem tests whether students can efficiently use vertex form and systematically handle the ± when taking square roots. The key insight is recognizing that vertex form makes the algebra much simpler than working with standard form.