[tex]y=-x^2+4x+6 \\ \\a =-1, \ b = 4 , \ c=6 \\ \\ The \ vertex \ is \ (x,f(x)) \\ \\x=\frac{-b}{2a} =\frac{-4}{2\cdot (-1)}=\frac{-4}{-2}=2\\ \\ f(2)=-x^2+4x+6 =-2^2+4 \cdot 2+6 =-4+8+6=10 \\ \\ The \ vertex \ is \ (2,10) \\ \\ \\ Answer: \ 2) \ \ (2,10)[/tex]