What is the range of the function f(x) = -x^2 + 4x - 1, where x is a real number?
Show the answer and explanation
Correct answer
A. (-∞, 3]
Principle or equation
For a quadratic function f(x) = ax^2 + bx + c with a < 0, the range is (-∞, f(h)] where h = -b/(2a) is the x-coordinate of the vertex.
Why this answer is correct
The function is a downward-opening parabola. Complete the square: f(x) = -(x^2 - 4x) - 1 = -(x^2 - 4x + 4) + 4 - 1 = -(x - 2)^2 + 3. The maximum value is 3 at x = 2, so the range is all real numbers less than or equal to 3.
Example
For f(x) = -x^2 + 4x - 1, the vertex is at x = 2, giving f(2) = -4 + 8 - 1 = 3. Since the parabola opens downward, the range is (-∞, 3].
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