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Math Help - Third degree

  1. #1
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    Post Third degree

    Find the third degree polynomial whose graph is shown in the figure. My answer is f(x)= 1/4x^3 - 1/4x^2 +2x+2, but I'm not sure. Thanks for checking my answer.
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    Last edited by CaptainBlack; July 5th 2006 at 09:19 AM.
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  2. #2
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    Quote Originally Posted by kwtolley
    Find the third degree polynomial whose graph is shown in the figure. My answer is f(x)= 1/4x^3 - 1/4x^2 +2x+2, but I'm not sure. Thanks for checking my answer.
    Can you tell us why you think this is the cubic in question?

    RonL
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  3. #3
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    cubic

    The graph I put together is off some, as for being cubed or not thats why I'm asking. Some of my home work is hard, thats why I ask because I like to see how I messed up my answer, and checking the instructor also. Instructor/teachers can make mistakes just like all of us.
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  4. #4
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    Hello, kwtolley!

    Find the third degree polynomial whose graph is shown in the figure.

    My answer is: f(x) \:= \:\frac{1}{4}x^3 - \frac{1}{4}x^2 +2x+2 , but I'm not sure.

    We have a big problem . . .

    A cubic function, f(x) \;=\;ax^3 + bx^2 + cx + d, has four constants to determine,
    . . but we are given only three facts.

    [1] The point (0,2) is on the graph.
    [2] The point (-2,0) is on the graph.
    [3] There is a critical point at x = 2.

    Since f(0) = 2:\;\;a\cdot0^3 + b\cdot0^2 + c\cdot0 + d\:=\:2\quad\Rightarrow\quad d = 2

    . . The function (so far) is: . f(x)\;=\;ax^3 + bx^2 + cx + 2


    Since f(-2) = 0:\;\;a(-2)^3 + b(-2)^2 + c(-2) + 2\;=\;0

    . . -8a + 4b - 2c + 2\:=\:0\quad\Rightarrow\quad 8a - 4b + 2c\:=\:2 [1]


    The derivative is: . f'(x) \:=\:3ax^2 + 2bx + c

    Since f'(2) = 0:\;\;3a\cdot2^2 + 2b\cdot2 + c\:=\:0\quad\Rightarrow\quad 12a + 4b + c \:=\:0 [2]


    Add [1] and [2]: . 20a + 3c\:=\:2\quad\Rightarrow\quad c\,=\,\frac{2-20a}{3}

    Substitute into [2]: . 12a + 4b + \left(\frac{2 - 20a}{3}\right)\:=\:0\quad\Rightarrow\quad b = \frac{-8a - 1}{6}


    Therefore: . \boxed{f(x) \;= \;ax^3 - \left(\frac{8a + 1}{6}\right)\!x^2 + \left(\frac{2 - 20a}{3}\right)\!x \,+ \,2\,}

    And, if there is no more information, that's the best we can do.


    Edit: corrected that silly "3" . . . Thanks for the heads-up!
    Last edited by Soroban; July 5th 2006 at 03:57 PM.
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  5. #5
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    Quote Originally Posted by Soroban
    . . -8a + 4b - 2c + 2\:=\:0\quad\Rightarrow\quad 8a - 3b + 2c\:=\:2 [1]
    wouldn't -8a + 4b - 2c + 2\:=\:0\quad\Rightarrow\quad8a - 4b + 2c \:=\:2 instead of 8a - 3b + 2c\:=\:2
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  6. #6
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    thanks

    thanks for checking my answer I reworked it and got the right answer.
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