\(The\ vertex\ form\ of\ y=ax^2+bx+c:y=a(x-h)^2+k\\\\where:h=\dfrac{-b}{2a}\ and\ k=\dfrac{-(b^2-4ac)}{4a}\\-------------------------------\\y=6x^2-12x+1\to a=6;\ b=-12;\ c=1\)
\(h=\dfrac{-(-12)}{2\cdot6}=\dfrac{12}{12}=1\\\\k=\dfrac{-[(-12)^2-4(6)(1)}{4(6)}=\dfrac{-(144-24)}{24}=\dfrac{-120}{24}=-5\\\\vertex:(1-5)\)
\(the\ vertex\ form:\\\boxed{y=6(x-1)^2-5}\)
