Vertex form of quadratic equation is y=a(x-h)^2+ky=a(x−h)2+k, with (h,k)(h,k) as vertex.
To convert y=(3x+1)(x+2)+2y=(3x+1)(x+2)+2, what we need is to expand and then convert part containing xx into a complete square and leave remaining constant as kk. The process is as shown below.
y=(3x+1)(x+2)+2y=(3x+1)(x+2)+2
= 3x xx x+3x xx2+1xx x+1xx2+23x×x+3x×2+1×x+1×2+2
= 3x^2+6x+x+2+23x2+6x+x+2+2
= 3x^2+7x+43x2+7x+4
= 3(x^2+7/3x)+43(x2+73x)+4
= 3(color(blue)(x^2)+2xxcolor(blue)x xxcolor(red)(7/6)+color(red)((7/6)^2))-3xx(7/6)^2+43(x2+2×x×76+(76)2)−3×(76)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]}
