Question 1090251
None.  No maximum value unless you have a specific interval.  Your parabola has a minimum value.


{{{y=4(x^2-2x+4)}}}
{{{y=4(x^2-2x+1-1+4)}}}
{{{4((x-1)^2+3)}}}
{{{4(x-1)^2+12}}}


The MINIMUM value is {{{f(1)=highlight_green(12)}}}.