Question 329083
Complete the square to convert to vertex form, {{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
{{{f(x)=4x^2-40x+104}}}
{{{f(x)=4(x^2-10x)+104}}}
{{{f(x)=4(x^2-10x+25)+104-4(25)}}}
{{{f(x)=4(x-5)^2+4}}}
Comparing to the equation above,
(h,k)=(5,4)
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The vertex lies on the axis of symmetry, {{{x=5}}}
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The min or max value occurs at the vertex.
Since the coefficient of the {{{x^2}}} term is positive, the parabola opens upwards and the vertex value is a minimum.
{{{ymin=4}}}
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{{{drawing(300,300,-2,12,-2,12,circle(5,4,.3),blue(line(5,-20,5,20)),grid(1),graph(300,300,-2,12,-2,12,4x^2-40x+104))}}}