y=f(x)=a(x-1)(x-5)=ax^2-6ax+5a=ax^2-6ax+9a-4a=a(x-3)^2-4a, where a is a constant.
So y+4a=a(x-3)^2. The vertex is at (3,-4a)=(3,-2) so -4a=-2 and a=1/2. y=f(x)=x^2/2-3x+5/2 or (1/2)(x-3)^2-2 and the y intercept is (1/2)(9)-2=5/2, the constant of the quadratic.