The function f is linear. The graph of \(\mathrm{y = f(x)}\) crosses the y-axis at \(\mathrm{(0, 3)}\) and the x-axis...

1. TRANSLATE the problem information

• Given information:
  • Function f is linear
  • Line crosses y-axis at \(\mathrm{(0, 3)}\)
  • Line crosses x-axis at \(\mathrm{(-1.5, 0)}\)

• What this tells us: We have two specific points on the line: \(\mathrm{(0, 3)}\) and \(\mathrm{(-1.5, 0)}\)

2. INFER the approach

• Since we know two points on a linear function, we can find both the slope and y-intercept
• The y-intercept is already given directly: when \(\mathrm{x = 0}\), \(\mathrm{y = 3}\), so \(\mathrm{b = 3}\)
• We need to calculate the slope using the slope formula

3. SIMPLIFY to find the slope

• Using slope formula: \(\mathrm{m = \frac{y_2 - y_1}{x_2 - x_1}}\)
• With points \(\mathrm{(0, 3)}\) and \(\mathrm{(-1.5, 0)}\):
  \(\mathrm{m = \frac{0 - 3}{-1.5 - 0}}\)
  \(\mathrm{m = \frac{-3}{-1.5}}\)
  \(\mathrm{m = 2}\)

• Important: When dividing two negative numbers, the result is positive!

4. Combine slope and y-intercept

• Linear function format: \(\mathrm{f(x) = mx + b}\)
• With \(\mathrm{m = 2}\) and \(\mathrm{b = 3}\): \(\mathrm{f(x) = 2x + 3}\)

5. Verify the solution

• Check y-intercept:
  \(\mathrm{f(0) = 2(0) + 3}\)
  \(\mathrm{f(0) = 3}\) ✓
• Check x-intercept:
  \(\mathrm{f(-1.5) = 2(-1.5) + 3}\)
  \(\mathrm{f(-1.5) = -3 + 3}\)
  \(\mathrm{f(-1.5) = 0}\) ✓

Answer: D. \(\mathrm{f(x) = 2x + 3}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak SIMPLIFY execution: Students correctly set up the slope formula but make a sign error when calculating \(\mathrm{\frac{-3}{-1.5}}\). They might get \(\mathrm{-2}\) instead of \(\mathrm{+2}\), especially if they lose track of the negative signs.

This leads them to think the slope is \(\mathrm{-2}\), giving them \(\mathrm{f(x) = -2x + 3}\), causing them to select Choice A (\(\mathrm{f(x) = -2x + 3}\)).

Second Most Common Error:

Poor INFER reasoning: Students might confuse which intercept gives which value. They might try to use the x-intercept \(\mathrm{(-1.5, 0)}\) as the y-intercept value, thinking \(\mathrm{b = -1.5}\), or get confused about which coordinate represents what.

This conceptual confusion about intercepts leads to incorrect equation setup and causes them to guess among the remaining choices.

The Bottom Line:

This problem tests whether students can systematically extract information from intercept descriptions and execute the slope calculation correctly. The key insight is recognizing that "crosses the y-axis at \(\mathrm{(0, 3)}\)" directly gives you \(\mathrm{b = 3}\), while the slope requires careful arithmetic with the negative coordinates.