we have
[tex]f(x) =-3x^{2}+ 6x-2[/tex]
Since the leading coefficient is negative, the function has a maximum
Let
y=f(x)
[tex]y=-3x^{2}+ 6x-2[/tex]
Group terms that contain the same variable, and move the constant to the opposite side of the equation
[tex]y+2=-3x^{2}+ 6x[/tex]
Factor the leading coefficient
[tex]y+2=-3(x^{2}-2x)[/tex]
Complete the square. Remember to balance the equation by adding the same constants to each side
[tex]y+2-3=-3(x^{2}-2x+1)[/tex]
[tex]y-1=-3(x^{2}-2x+1)[/tex]
Rewrite as perfect squares
[tex]y-1=-3(x-1)^{2}[/tex]
[tex]y=-3(x-1)^{2}+1[/tex]
the vertex is the point [tex](1,1)[/tex]
therefore
the answer is the option
A. Maximum at (1, 1)
