In the linear function h, \(\mathrm{h(28) = 15}\) and \(\mathrm{h(26) = 22}\). Which equation defines h?

1. TRANSLATE the problem information

• Given information:
  • \(\mathrm{h(28) = 15}\) means when \(\mathrm{x = 28}\), the function output is 15
  • \(\mathrm{h(26) = 22}\) means when \(\mathrm{x = 26}\), the function output is 22
  • h is a linear function

• What this tells us: We have two coordinate points: \(\mathrm{(28, 15)}\) and \(\mathrm{(26, 22)}\)

2. INFER the solution strategy

• For any linear function \(\mathrm{h(x) = mx + b}\), we need to find m (slope) and b (y-intercept)
• With two points, we can find the slope first, then use either point to find b

3. SIMPLIFY to find the slope

• Using the slope formula: \(\mathrm{m = \frac{y_2 - y_1}{x_2 - x_1}}\)
• \(\mathrm{m = \frac{22 - 15}{26 - 28}}\)
  \(\mathrm{= \frac{7}{-2}}\)
  \(\mathrm{= -\frac{7}{2}}\)

4. SIMPLIFY to find the y-intercept

• Substitute \(\mathrm{m = -\frac{7}{2}}\) and point \(\mathrm{(28, 15)}\) into \(\mathrm{h(x) = mx + b}\):
• \(\mathrm{15 = \left(-\frac{7}{2}\right)(28) + b}\)
  \(\mathrm{15 = -98 + b}\)
  \(\mathrm{b = 113}\)

5. Write the final equation

• \(\mathrm{h(x) = -\frac{7}{2}x + 113}\)

Answer: D. \(\mathrm{h(x) = -\frac{7}{2}x + 113}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students may not recognize that \(\mathrm{h(28) = 15}\) means the point \(\mathrm{(28, 15)}\) is on the graph. Instead, they might try to substitute \(\mathrm{x = 28}\) directly into the answer choices without understanding the coordinate relationship.

This leads to confusion and random guessing among the choices.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the need for slope but make sign errors in the calculation, particularly with \(\mathrm{m = \frac{7}{-2} = -\frac{7}{2}}\). They might get \(\mathrm{+\frac{7}{2}}\) instead, leading them to select Choice A or C.

The Bottom Line:

This problem tests whether students can bridge the gap between function notation and coordinate geometry. Success requires recognizing that function values create coordinate points, then systematically applying the slope formula and point-slope relationship.