Completing the square is also possible when the coefficient of the $x^{2}$ term is not $1$. Work your way through this problem to see how.

$\begin{align*}y&=3x^2 -18x+1\\y-1&=3(x^2 -6x)\end{align*}$

1. What number needs to be added to both sides of the equation so that the expression inside the parentheses is a perfect square?
   Hint: $9$ is the wrong answer.

   Hint (a):

   The outside $3$ will distribute with whatever is inside the ( )'s.

2. Finish completing the square and find the vertex of the parabola.

   Hint (b):

   $y − 1 = 3\left(x^{2} − 6x + 9\right) − 27$