12
36
24
6
Given that the quadratic expression is \( \ge 0 \) for all real \( x \), the parabola opens upwards with a minimum value of zero. The expression passes through \( (2, 0) \) and \( (4, 6) \). Since the expression is 0 at \( x = 2 \), this point is the vertex.
The vertex form of a quadratic is \( y = a(x - h)^2 + k \). With vertex \( (2, 0) \), this simplifies to \( y = a(x - 2)^2 \).
To determine \( a \), we use the point \( (4, 6) \): \( 6 = a(4 - 2)^2 = a \cdot 4 \). Thus, \( a = \frac{6}{4} = \frac{3}{2} \).
The quadratic expression is \( y = \frac{3}{2}(x - 2)^2 \).
Evaluating at \( x = -2 \): \( y = \frac{3}{2}(-2 - 2)^2 = \frac{3}{2} \cdot 16 = 24 \).
The value of the expression at \( x = -2 \) is \( \boxed{24} \).