Question 368460

Convert to vertex form, {{{y=a(x-h)^2+k}}}, by completing the square.
{{{f(x)=8x^2-4x+50}}}
{{{f(x)=8(x^2-(1/2)x)+50}}}
{{{f(x)=8(x^2-(1/2)x+1/16)+50-8/16}}}
{{{f(x)=8(x-1/4)^2+100/2-1/2}}}
{{{f(x)=8(x-1/4)^2+99/2}}}
The maximum/minimum occurs at the vertex.
Since {{{a=8>0}}}, the parabola opens upwards and the vertex value is a minimum.
{{{y[min]=99/2}}}