Question 127804
{{{y = 6x^2 + 7x + 1}}}


The vertex of a parabola expressed in the form {{{f(x)=ax^2+bx+c}}} is at ({{{-b/2a}}},{{{f(-b/2a)}}})


So, the x-coordinate of the vertex of your function is:


{{{-7/12}}},


and the y-coordinate is:


{{{f(-7/12)=6(-7/12)^2+7(-7/12)+1}}}


{{{f(-7/12)=294/144-49/12+1}}}


{{{f(-7/12)=294/144-588/144+144/144}}}


{{{f(-7/12)=-150/144}}}


{{{f(-7/12)=-25/24}}}


Therefore the vertex is at:  ({{{-7/12}}},{{{-25/24}}})


{{{drawing(600,600,-3,3,-3,3,grid(1),
graph(600,600,-3,3,-3,3,6x^2 + 7x + 1,-25/24),
green(line(-7/12,3,-7/12,-3))
)}}}