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# Find a second-degree polynomial P such that P(2) = 5, P'(2) = 4, and P''(2) = 4. P(x)=?

### 2 Answers

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- PhilipLv 64 weeks ago
Put P(x) = ax^2+bx+c;

P'(x) = 2ax+b;

P''(x) = 2a;

P(2) = 5 ---> 4a + 2b +c = 5...(1).;

P'(2) = 4 ---> 4a..+..b......= 4...(2).;

P''(2)= 4 ---> 2a..............= 4...(3),;

(3)---> a = 2;

(2)---> 8+b=4, ie., b = -4;

(1)---> 8-8+c = 5, ie., c = 5;

Then P(x) = 2x^2-4x+5.

- stanschimLv 74 weeks ago
P(x) = ax^2 + bx + c

P'(x) = 2ax + b

P''(x) = 2a

P''(2) = 4 = 2a, so a = 2

P'(2) = 4 = 2ax + b = 2(2)(2) + b, 4 = 8 + b, b = -4

P(2) = 5 = ax^2 + bx + c = 2(2^2) + (-4)(2) + c, 8 - 8 + c = 5, c = 5

P(x) = 2x^2 - 4x + 5 <==answer

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