Given quadratic equation : [tex]y=-3x^2+12x-33[/tex].
A quadratic equation represents a parabola.
We have leading coefficient is -3, a negative number.
So, it would be a downward parabola and it would have it's vertex point on the top on the parabola.
So, it would obviously have a maximum value.
Let us find the x-coordinate of the parabola first and then y-coordinate of the parabola.
x = -b/2a = -12/2(-3) = -12/-6 = 2.
Plugging x=in given equation, we get
y=-3(2)^2+12(2)-33 = -3(4) +24 -33
= -12 +24 -33 = -21.
Therefore, maximum value is -21.
Therefore, correct option is :
