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Math Help - Finding function of graph

  1. #1
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    Post Finding function of graph

    Graph of a 3 degree polynomial is like the figure. What will be the function?
    Finding function of graph-dsc-0000172.jpg

    Thanks for your help.
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  2. #2
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    From the figure we see that -3 is a root, 2 is a double root, and the y-intercept is (0,10).
    So the cubic polynomial has the form f(x)=a(x+3)(x-2)^2 for some constant a.
    See if you can find a.
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  3. #3
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    Thanks but
    1-How did you find that (x+3)(x-2)^2 and not (x-2)(x+3)^2
    2- You see in the figure f(0)=10 but in this solution f(0)=0
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  4. #4
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    Hello, hejmh!

    Graph of a 3 degree polynomial is like the figure.
    What will be the function?
    Click image for larger version. 

Name:	DSC-0000172.jpg 
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    The general cubic function is: . f(x) \:=\:ax^3 + bx^2 + cx + d

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

    We need to determine the four coefficients: . a,b,c,d.


    We know five facts about the function:

    . . \begin{Bmatrix} f(0) &=& 10 \\ f(2) &=& 0 \\ f(\text{-}3) &=& 0 \\ \\[-3mm] f'(0) &=& 0 \\ f'(2) &=& 0 \end{Bmatrix}


    Set up a system of equations and solve.



    Edit: Ah, I see that DrSteve has a nicer form of the cubic.
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  5. #5
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    Quote Originally Posted by hejmh View Post
    Thanks but
    1-How did you find that (x+3)(x-2)^2 and not (x-2)(x+3)^2
    Because, as Dr. Steve said, x= 2 is a double root- the x-axis is tangent to the graph at x= 2.
    Another way of seeing the same thing- moving from one side of 2 to the other, say from x= 1.9 to x= 2.1, changes the sign of x- 2. x- 2 must be to an even power since y does not change sign.

    2- You see in the figure f(0)=10 but in this solution f(0)=0
    NO, it isn't. f(0)= a(0+ 3)(0- 2)^2= 12a, not 0.
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  6. #6
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    Hello Soroban
    If we set f(0)=10 then d=10
    If we set f'(0)=0 then c=0
    If we set f(2)=0 then 8a+4b+10=0
    If we set f(-3)=0 then -27a+9b+10=0
    If we set f'(2)= 0 then 12a+4b=0
    Looking to tree last equations we have not unique value for a and b.
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, hejmh!


    The general cubic function is: . f(x) \:=\:ax^3 + bx^2 + cx + d

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

    We need to determine the four coefficients: . a,b,c,d.


    We know five facts about the function:

    . . \begin{Bmatrix} f(0) &=& 10 \\ f(2) &=& 0 \\ f(\text{-}3) &=& 0 \\ \\[-3mm] f'(0) &=& 0 \\ f'(2) &=& 0 \end{Bmatrix}


    Set up a system of equations and solve.



    Edit: Ah, I see that DrSteve has a nicer form of the cubic.
    I'm not sure if it's clear from this picture that f'(0)=0. Is there a reason we can be certain of this?
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  8. #8
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    Quote Originally Posted by DrSteve View Post
    I'm not sure if it's clear from this picture that f'(0)=0. Is there a reason we can be certain of this?
    It is like a change in direction at 0 so I think f'0) could be0.
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  9. #9
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    According to your graph f'(0) may or may not be 0. It turns out that it is not. If you combine my post above with HallsofIvy's you will see that

    f(x)=\frac{5}{6}(x+3)(x-2)^2.

    The derivative of this function is not 0 at x=0.

    Soroban's method should work if you just leave out the condition f'(0)=0. You will get a system of 3 equations in 3 unknowns which should have a unique solution (I'm pretty sure that the 3 equations in 2 unknowns you have above is inconsistent). You'll probably want to do Gauss-Jordan Reduction to solve this. It's quite a bit more work than the original solution I posted.
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