Answer:
A. The equation has a maximum value with a y-coordinate of -21.
Step-by-step explanation:
From the given equation:
[tex]\mathbf{y = -3x^2 + 12x -33}[/tex]
This parabola is vertical and is goes downward via the negative path
Where the vertex represents the maximum value;
[tex]\mathbf{y = -3 (x^2 + 4x) -33}[/tex]
Using completing the square method;
[tex]\mathbf{y = -3 (x^2 + 4x+2^2) -33+12}[/tex]
[tex]\mathbf{y = -3 (x^2 + 4x+4) -21}[/tex]
To perfect square:
[tex]\mathbf{y = -3 (x-2)^2 -21}[/tex]
The vertex point is (2, -21)
Hence ; the equation has a maximum value with a y-coordinate of -21.