\(y=ax^2+bx+c\\\\The\ vertex\ form:y=a(x-h)^2+k\ where\ h=\frac{-b}{2a}\ and\ k=f(h)\\========================================\\\\y=3x^2-18x+30\\\\a=3;\ b=-18;\ c=30\\\\h=\frac{-(-18)}{2\cdot3}=\frac{18}{6}=3\\\\k=f(3)=3(3^2)-18(3)+30=3(9)-54+30=27-24=3\\\\therefore:\boxed{y=3(x-3)^2+3}\)
