Question 1043305
{{{f>=g}}}
{{{x^2-6x+12>=k}}}
{{{x^2-6x+12-k>=0}}}


Maybe better, expect k to be the vertex, minimum value for f(x).


{{{f(x)=x^2-6x+9-9+12}}} to complete-the-square;
{{{(x-3)^2+3}}}
Vertex is  (3,3).
The line g(x)=k must be no higher than y=3.  The MAXIMUM value for k is 3.