One thing we can do is expand this function.

But first, let's just isolate #y# as follows: #y=-(1/16)(x-1)^2+2#

Now, let's expand your product of differences:

#y=-(x^2-2x+1)/16+2=(x^2-2x+33)/16#

#y=-x^2/16+x/8-2.0625#

Using Bhaskara we can find its roots:

#((-1/8)+-sqrt((1/8)^2-4(-1/16)(-33/16)))/-(2/16)#

#((-1/8)+-sqrt(1/64-33/64))/-2/16#

Ends up our #Delta# is negative, so there are no Real roots, which means this function does not cross the axis #x#.

Alternatively, we can find two other pieces of information:

Where the function crosses the axis #y#, when, by definition, #x=0#:

#y=0+0-2.0625#

#y=2.0625# is our intercept.

Now, we can find the coordinates for our vertex #(x_v,y_v)#

#x_v=-b/(2a)=(-1/8)/-(2/16)=1#

#y_v=-Delta/(4a)=1/2/-1/4=-2#

Vertex = (#1,-2#)

Finally, as the coeficient of your #x^2# is negative, you know that this is a parabola where the vertex indicates the point of maximum.

graph{-(x^2)/16+(x/8)-2.065 [-10, 10, -5, 5]}