Welcome :: Homework Help and Answers :: Mathskey.com
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,435 questions

17,804 answers

1,438 comments

774,866 users

Find a second-degree polynomial

0 votes

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

asked Oct 4, 2014 in PRECALCULUS by anonymous

1 Answer

0 votes

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.

answered Oct 4, 2014 by david Expert

Related questions

...