In the linear function h, \(\mathrm{h(3) = 7}\) and \(\mathrm{h(7) = 15}\). Which equation defines h?

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{h(3) = 7}\) means when \(\mathrm{x = 3}\), the output is 7
  • \(\mathrm{h(7) = 15}\) means when \(\mathrm{x = 7}\), the output is 15
  • This gives us two points: \(\mathrm{(3, 7)}\) and \(\mathrm{(7, 15)}\)

2. INFER the approach needed

• Since we have two points on a linear function, we need to:
  • First find the slope using the slope formula
  • Then use point-slope form to write the equation
  • Convert to slope-intercept form if needed

3. Calculate the slope

• Using the slope formula: \(\mathrm{m = \frac{y_2 - y_1}{x_2 - x_1}}\)
• \(\mathrm{m = \frac{15 - 7}{7 - 3}}\)
  \(\mathrm{= \frac{8}{4}}\)
  \(\mathrm{= 2}\)

4. SIMPLIFY using point-slope form

• Start with: \(\mathrm{y - y_1 = m(x - x_1)}\)
• Using point \(\mathrm{(3, 7)}\): \(\mathrm{h(x) - 7 = 2(x - 3)}\)
• Expand: \(\mathrm{h(x) - 7 = 2x - 6}\)
• Add 7 to both sides: \(\mathrm{h(x) = 2x - 6 + 7}\)
• Combine: \(\mathrm{h(x) = 2x + 1}\)

5. Verify the solution

• Check \(\mathrm{h(3) = 2(3) + 1 = 7}\) ✓
• Check \(\mathrm{h(7) = 2(7) + 1 = 15}\) ✓

Answer: D

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that \(\mathrm{h(3) = 7}\) means the point \(\mathrm{(3, 7)}\) is on the graph. They might try to work directly with the function notation without converting to coordinate points, leading to confusion about how to proceed.

This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly find the slope as 2 and set up the point-slope form, but make algebraic errors when expanding \(\mathrm{h(x) - 7 = 2(x - 3)}\). Common mistakes include:

• Forgetting to distribute the 2: getting \(\mathrm{h(x) = 2x - 3 + 7 = 2x + 4}\)
• Sign errors when combining terms: getting \(\mathrm{h(x) = 2x - 1}\)

These algebraic mistakes don't match any of the given choices exactly, causing students to guess.

The Bottom Line:

This problem requires students to bridge between function notation and coordinate geometry. The key insight is recognizing that function values give you points, and two points determine a unique line.