Find a second-degree polynomial

Question

P(x) such that p(3)=16 P'(3)=14 P"(3)=6? 

Answer 1

Let the second degree polynomial is P(x) =  ax² + bx + c

Apply dervative on each side.

P' = 2ax + b

Apply derivative on each side.

P'' = 2a

P(3) =  a(3)² + b(3) + c

16 = 9a + 3b + c ----> (1)

P'(3) = 2a(3) + b

14 = 6a + b ---> (2)

P''(3) = 2a

6 = 2a

a = 3

Substitute the value of a in equation (2).

14 = 6(3) + b

14 = 18 + b

b = - 4

Substitute the values of a, b in equation (1).

16 = 9a + 3b + c

16 = 9(3) + 3(-4) + c

16 = 27 - 12 + c

16 = 15 + c

c = 1

Substitute the values of a, b ,c in P(x) = ax² + bx + c.

P(x) = (3)x² + (-4)x + 1

The second degree polynomial is P(x) = 3x² - 4x + 1.