Question 328737
Complete the square to put into vertex form,{{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
{{{f(x)=(x^2-4x)-4}}}
{{{f(x)=(x^2-4x+4)-4-4}}}
{{{f(x)=(x-2)^2-8}}}
Comparing, the vertex is (2,-8).
THe vertex lies on the line of symmetry, {{{x=2}}}.
The parabola opens upwards since the coefficient of the {{{x^2}}} term is positive so the function value at the vertex is a minimum.
{{{ymin=-8}}}
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Verify by graphing,
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{{{drawing(300,300,-10,10,-10,10,grid(1),circle(2,-8,.35),blue(line(2,-10,2,10)),graph(300,300,-10,10,-10,10,x^2-4x-4))}}}