x\(\mathrm{h(x)}\)-{2}-{1/4}-1-11-12-{1/4}For function h, the table shows four values of x and their corresponding values of \(\mathrm{h(x)}\). Which ...

1. TRANSLATE the problem information

• Given information:
  • Table with four x-values and corresponding h(x) values
  • Four possible function equations to choose from
• What we need: Determine which equation produces the exact table values

2. INFER the solution approach

• Since we have specific input-output pairs and multiple choice options, the most efficient strategy is systematic substitution
• Test each option by plugging in the x-values and checking if we get the correct h(x) values
• We only need one mismatch to eliminate an option

3. SIMPLIFY by testing Option A: \(\mathrm{h(x) = -\frac{1}{x^2}}\)

• \(\mathrm{h(-2) = -\frac{1}{(-2)^2} = -\frac{1}{4}}\) ✓ (matches table)
• \(\mathrm{h(-1) = -\frac{1}{(-1)^2} = -\frac{1}{1} = -1}\) ✓ (matches table)
• \(\mathrm{h(1) = -\frac{1}{(1)^2} = -\frac{1}{1} = -1}\) ✓ (matches table)
• \(\mathrm{h(2) = -\frac{1}{(2)^2} = -\frac{1}{4}}\) ✓ (matches table)

All values match! But let's verify by checking at least one other option.

4. SIMPLIFY by testing Option B: \(\mathrm{h(x) = -\frac{1}{x}}\)

• \(\mathrm{h(-2) = -\frac{1}{(-2)} = \frac{1}{2} ≠ -\frac{1}{4}}\) ✗

This doesn't match, so Option B is eliminated.

Answer: A

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students make sign errors when evaluating expressions with negative bases in exponents.

For example, when calculating \(\mathrm{h(-2) = -\frac{1}{(-2)^2}}\), they might incorrectly think \(\mathrm{(-2)^2 = -4}\) instead of \(\mathrm{+4}\), leading to \(\mathrm{h(-2) = -\frac{1}{(-4)} = \frac{1}{4}}\) instead of the correct \(\mathrm{-\frac{1}{4}}\). This creates confusion because their calculated values don't match any of the table entries, causing them to doubt their approach and potentially guess randomly.

Second Most Common Error:

Insufficient INFER reasoning: Students test only one or two x-values instead of systematically checking all given points.

They might test \(\mathrm{x = 1}\) for each option, find that both Options A and C give \(\mathrm{h(1) = -1}\), and then guess between them without testing additional values. This incomplete verification often leads them to select Choice C \(\mathrm{(\frac{1}{x^2})}\) since it seems simpler, missing that it gives positive values for negative x-inputs.

The Bottom Line:

This problem rewards systematic, careful calculation over shortcuts. Success depends on methodical substitution and precise handling of negative numbers in exponential expressions.