Final answer:
The equation of the quadratic graph with a focus of (5, -1) and a directrix of y = 1 is given by option C, which is f(x) = -1/4(x - 5)².
Step-by-step explanation:
The question asks for the equation of a quadratic graph with a given focus and directrix. The standard form of a parabolic equation whose axis of symmetry is parallel to the y-axis is f(x) = a(x - h)² + k, where (h, k) is the vertex of the parabola. In this case, the axis of symmetry of the parabola is the line that lies exactly midway between the focus and the directrix, which is y = 0 since the directrix is y = 1 and the focus is at (5, -1). The vertex thus lies at (5, 0), halfway between the focus and directrix.
Given the focus (5, -1) and directrix y = 1, the distance from the vertex to the focus (p) is 1 unit. This distance is the same as the distance from the vertex to the directrix. The value of 'a' in the parabola's equation is 1/(4p). Since the focus is below the directrix, 'a' must be negative. Therefore, 'a' = -1/4. The equation of the parabola then becomes f(x) = -1/4(x - 5)². However, we must account for the vertex's y-coordinate (which is 'k' in the equation), which is 0. Thus, the equation does not need to be shifted vertically, eliminating options A and B. The correct equation is f(x) = -1/4(x - 5)², which corresponds to answer choice C.