Solution:
\( \displaylines{\left(0,1\right)\text{ is the y-intercept.}\\ \text{Using quadratic formula can provide a solution, or we can notice that}\\ 6x^2-5x+1=\left(3x-1\right)\left(2x-1\right)\\ \text{so that }\left(\frac13,0\right)\text{ and }\left(\frac12,0\right)\text{ are x-intercepts.}\\ \text{The vertex is }x=-\frac{b}{2a}=\frac{5}{12}\text{ (directly in between the two roots)}\\ \text{So we have }\left(\frac{5}{12},f\left(\frac{5}{12}\right)\right)=\left(\frac{5}{12},-\frac{1}{24}\right)\text{ as a vertex.}} \)