Please submit your concern

Explanation:

Let f(x) = ax² + bx + c
Since, f(x) ≥ 0, and f(2) = 0, it means the graph of the quadratic lies above x-axis and touches x-axis at x = 2.

Since x = 2 is the only root of the equation (i.e., graph is symmetric about x = 2),
⇒ f(2 + a) = f(2 – a)
⇒ f(2 + 2) = f(2 - 2)
⇒ f(4) = f(0)
⇒ f(0) = 6
⇒ c = 6

f(2) = 0
⇒ 4a + 2b + c = 0   
⇒ 4a + 2b = -6   …(2)

f(4) = 6
⇒ 16a + 4b + c = 6
⇒ 16a + 4b = 0   …(3)

Solving (2) and (4), we get
a = -3/2 and b = -6

⇒ f(-2) = $\frac{3}{2}$(-2)² – 6 × - 2 + 6 = 6 + 12 + 6 = 24

Hence, option (a).