Question

The quadratic function f(x) has these characteristics:

The vertex is located at (1, 8).
The range is -[infinity] < f(x) ≤ 8.
Which function could be the f(x)?
A) 2x^2 + 4x + 6
B) 2x^2 + x + 8
C) -2x^2 + 4x + 6
D) -2x^2 + x + 8

Answer 1

The correct quadratic function is D) -2x² + x + 8, as it is the only one with a negative leading coefficient and has a vertex at (1, 8), matching both the vertex position and the range requirement.

Step-by-step explanation:

The quadratic function f(x) has a vertex at (1, 8) and a range of ∞ < f(x) ≤ 8. Because the range has an upper bound of 8, we know that the parabola opens downwards, indicating that the leading coefficient must be negative. The vertex form of a quadratic function is f(x) = a(x-h)² + k, where (h,k) is the vertex. So, with our vertex at (1,8), and knowing a must be negative, the function will appear as f(x) = -a(x-1)² + 8, where a > 0. Looking at the given options, only choice D, -2x² + x + 8, has a negative leading coefficient and matches the y-coordinate of the vertex. Expanding this option, we have f(x) = -(2x² - x - 8), which simplifies to the given form with a vertex at (1, 8) and opens downwards, as required by the range. Therefore, option D is the correct quadratic function.