Given, f (x) = ax2 + bx + c.
At x = 0, f(0) = 6 (given)
∴ a × 0 + b × 0 + c = 6
∴ c = 6 (i)
At x = 2, f(2) = 1 (given)
∴ a(2)2 + b(2) + c = 1
⇒ 4a + 2b + 6 = 1 (using (i))
⇒ 4a + 2b = – 5 (ii)
At x = – 3, f(–3) = 6
∴ a(– 3)2 + b(– 3) + c = 6
⇒ 9a – 3b + 6 = 6 (using (i))
⇒ 9a – 3b = 0 (iii)
On solving eqs. (ii) And (iii), we get
a = \(-\frac{1}{2}\) and a = \(-\frac{3}{2}\) ∴ Required quadratic function (f) = \(\frac{-1}{2}\) x2 + \(\frac{-3}{2}\)x + 6