For which of the following functions is \(\mathrm{f(x) \gt 0}\) for all real numbers x? \(\mathrm{f(x) = -2x^2 + 6x...

1. INFER the requirements for f(x) > 0 for all real x

For a quadratic \(\mathrm{f(x) = ax^2 + bx + c}\) to always be positive, we need:

• The parabola opens upward: \(\mathrm{a \gt 0}\)
• The parabola never touches/crosses the x-axis: discriminant \(\mathrm{\lt 0}\)

This ensures the parabola sits entirely above the x-axis.

2. CONSIDER ALL CASES by checking each option systematically

We'll test both conditions for every choice:

3. SIMPLIFY the discriminant calculations

Option A: f(x) = -2x² + 6x - 5

• \(\mathrm{a = -2 \lt 0}\) → Opens downward, immediately fails

Option B: f(x) = 2x² + 6x + 5

• \(\mathrm{a = 2 \gt 0}\) → Opens upward ✓
• \(\mathrm{\Delta = 6^2 - 4(2)(5)}\)
• \(\mathrm{= 36 - 40}\)
• \(\mathrm{= -4 \lt 0}\) → No real roots ✓
• Both conditions met!

Option C: f(x) = 3x² - 12x + 12

• \(\mathrm{a = 3 \gt 0}\) → Opens upward ✓
• \(\mathrm{\Delta = (-12)^2 - 4(3)(12)}\)
• \(\mathrm{= 144 - 144}\)
• \(\mathrm{= 0}\) → Touches x-axis once
• Fails because \(\mathrm{f(x) = 0}\) at one point

Option D: f(x) = x² + 5x - 6

• \(\mathrm{a = 1 \gt 0}\) → Opens upward ✓
• \(\mathrm{\Delta = 5^2 - 4(1)(-6)}\)
• \(\mathrm{= 25 + 24}\)
• \(\mathrm{= 49 \gt 0}\) → Two real roots
• Fails because parabola crosses x-axis

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students check only that \(\mathrm{a \gt 0}\) (parabola opens upward) but forget to verify the discriminant condition.

They see that options B, C, and D all have positive leading coefficients and think any of these could work. Without checking discriminants, they might guess or pick the "nicest looking" option. This leads to confusion and random guessing between the upward-opening parabolas.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students attempt discriminant calculations but make arithmetic errors, especially with option C where \(\mathrm{\Delta = 144 - 144 = 0}\).

A student might miscalculate this as \(\mathrm{\Delta = -144}\) or get confused by the zero result, not realizing that \(\mathrm{\Delta = 0}\) means the parabola touches (but doesn't stay above) the x-axis. This may lead them to select Choice C (3x² - 12x + 12).

The Bottom Line:

This problem requires understanding that "always positive" means both structural conditions (upward opening) AND positional conditions (above the x-axis) must be satisfied simultaneously.