Answer
The vertex form: $y=-3(x-1/6)^2+1/12$
Work Step by Step
The vertex form of a quadratic function is $f(x) = a(x – h)^2 + k$, where $a, h, k$ are constants. The vertex of the parabola is at $(h, k)$.
By completing the square of $y=-3x^2+x$ we get, $y=-3(x-1/6)^2+1/12$
Here $a=-3 , h=1/6, k=1/12$
