Question 357822
Convert to vertex form, {{{y=a(x-h)^2+k}}}.
{{{x^2-5x+12=0}}}
{{{x^2-5x+25/4+12-25/4=0}}}
{{{(x-5/2)^2+48/4-25/4=0}}}
{{{(x-5/2)^2+23/4=0}}}
Since {{{a>0}}}, then the value at the vertex is a minimum.
{{{y[min]=23/4}}}
.
.
.
{{{drawing(300,300,-2,10,-2,10,grid(1),circle(5/2,23/4,0.2),graph(300,300,-2,10,-2,10,x^2-5x+12))}}}