Question 330857
You only have 1/2 of the answer.
{{{x^2 - 8x + 16 >= 0}}} 
{{{(x-4)^2>=0}}}
When you take the square root of both sides you get,
{{{x-4>=0}}} and {{{-(x-4)>=0}}}
{{{x>=4}}} and {{{x-4<=0}}}
{{{x>=4}}} and {{{x<=4}}}
which together makes up ({{{-infinity}}}, {{{infinity}}}).
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The function is always greater than or equal to zero as shown in the graph.
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{{{graph(300,300,-10,10,-10,10,x^2-8x+16)}}}