To determine the extreme value of the given quadratic equation [tex]\( y = -3x^2 + 12x - 33 \)[/tex], we need to find the vertex of the parabola described by the equation.
A quadratic equation in the form [tex]\( y = ax^2 + bx + c \)[/tex] has its vertex at:
[tex]\[ x = -\frac{b}{2a} \][/tex]
where [tex]\( a \)[/tex] and [tex]\( b \)[/tex] are the coefficients of [tex]\( x^2 \)[/tex] and [tex]\( x \)[/tex] terms respectively.
For the equation [tex]\( y = -3x^2 + 12x - 33 \)[/tex]:
- The coefficient [tex]\( a = -3 \)[/tex]
- The coefficient [tex]\( b = 12 \)[/tex]
Substitute these values into the vertex formula to find the x-coordinate of the vertex:
[tex]\[ x = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2 \][/tex]
Next, we substitute this x-coordinate back into the quadratic equation to find the y-coordinate of the vertex:
[tex]\[ y = -3(2)^2 + 12(2) - 33 \][/tex]
[tex]\[ y = -3(4) + 24 - 33 \][/tex]
[tex]\[ y = -12 + 24 - 33 \][/tex]
[tex]\[ y = 12 - 33 \][/tex]
[tex]\[ y = -21 \][/tex]
The y-coordinate of the vertex is [tex]\(-21\)[/tex], and since the coefficient of [tex]\( x^2 \)[/tex] ([tex]\( a \)[/tex]) is negative ([tex]\( a = -3 \)[/tex]), the parabola opens downwards. Thus, the vertex represents a maximum value of the function.
Therefore, the correct statement is:
C. The equation has a maximum value with a [tex]\( y \)[/tex]-coordinate of -21.
