The vertex of the parabola is at (4,-1). Knowing the vertex alone will not determine a parabola. Assume the parabola's equation is f(x) = ax2 + bx + c.
The x-coordinate of the vertex is x = -b/2a = 4, so that b = -8a.
(4,-1) is on the parabola, so f(4) = -1 = 16a + 4b + c;
substituting, b = -8a, -1 = 16a -32a + c, so that c = 16a -1.
So f(x) = ax2 -8ax + 16a - 1 = a(x2-8x+16) -1. This is the FAMILY of parabolas with vertex (4,-1).
The limit of the family as a approaches 0 is the horizontal line y = -1. As a increases to infinity the parabola becomes steeper, until a approaches infinity, the parabola approaches the vertical line segment at x = 4 where y is in (-1,infinity). As a decreases to -infinity, the parabola approaches the vertical like at x =4 where y is in the interval (-infinity, -1).
If we specify any point in the x-y plane where x is not equal to 4, there is a unique parabola in this family that passes through that point.