Anonymous
Anonymous asked in Science & MathematicsMathematics · 4 weeks ago

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

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  • Philip
    Lv 6
    4 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.

  • 4 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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