Vertex form of quadratic equation is #y=a(x-h)^2+k#, with #(h,k)# as vertex.
To convert #y=(3x+1)(x+2)+2#, what we need is to expand and then convert part containing #x# into a complete square and leave remaining constant as #k#. The process is as shown below.
#y=(3x+1)(x+2)+2#
= #3x xx x+3x xx2+1xx x+1xx2+2#
= #3x^2+6x+x+2+2#
= #3x^2+7x+4#
= #3(x^2+7/3x)+4#
= #3(color(blue)(x^2)+2xxcolor(blue)x xxcolor(red)(7/6)+color(red)((7/6)^2))-3xx(7/6)^2+4#
= #3(x+7/6)^2-(cancel3xx49)/(cancel(36)^12)+4#
= #3(x+7/6)^2-49/12+48/12#
= #3(x+7/6)^2-1/12#
i.e. #y=3(x+7/6)^2-1/12# and vertex is #(-7/6,-1/12)#
graph{(3x+1)(x+2)+2 [-2.402, 0.098, -0.54, 0.71]}
