Question 371932
{{{f(x)=-16x^2+8x+3}}}
{{{f(x)=-16(x^2-(1/2)x)+3}}}
{{{f(x)=-16(x^2-(1/2)x+1/16)+3+16(1/16)}}}
{{{f(x)=-16(x-1/4)^2+3+1}}}
{{{f(x)=-16(x-1/4)^2+4}}}
.
.
.
({{{1/4}}},{{{4}}})
.
.
.
{{{drawing(300,300,-3,3,-1,5,grid(1),circle(1/4,4,0.1),graph(300,300,-3,3,-1,5,0,-16x^2+8x+3,-16(x-1/4)^2+4))}}}