Question 323631
Convert to vertex form, {{{y=a(x-h)^2+k}}} where (h,k) is the vertex.
{{{f(x)=4x^2-16x+19}}}
{{{f(x)=4(x^2-4x)+19}}}
{{{f(x)=4(x^2-4x+4)+19-16}}}
{{{f(x)=4(x-2)^2+3}}}
Comparing, the vertex is (2,3).
The vertex lies on the axis of symmetry {{{x=2}}}.

The parabola opens upwards since the coefficient of the {{{x^2}}} term is positive so the vertex value is the minimum value.
{{{ymin=3}}}
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{{{drawing(300,300,-2,8,-2,8,grid(1),circle(2,3,0.2),blue(line(2,-10,2,10)), graph(300,300,-2,8,-2,8,4(x-2)^2+3,4x^2-16x+19))}}}