Given the function \( f(x) = 3x^2 - 12 \), what is its vertex?

To find the vertex of the quadratic function ( f(x) = 3x^2 - 12 ), we can use the vertex formula for a parabola defined by the equation ( f(x) = ax^2 + bx + c ). In this case, the coefficients are ( a = 3 ), ( b = 0 ), and ( c = -12 ).

The x-coordinate of the vertex can be determined using the formula ( x = -\frac{b}{2a} ). Substituting the values into this formula gives:

[

x = -\frac{0}{2 \times 3} = 0.

]

Now that we have the x-coordinate, we can find the corresponding y-coordinate by substituting ( x = 0 ) back into the function:

[

f(0) = 3(0)^2 - 12 = -12.

]

Thus, the vertex of the function is at the point ( (0, -12) ).

However, the question specifies options that suggest there may have been a miscalculation in assessment. The answer provided as correct (D. (2, -12)) does not represent the vertex derived from