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Thread: 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 $\displaystyle f(x)=a(x+3)(x-2)^2$ for some constant $\displaystyle a$.
    See if you can find $\displaystyle a$.
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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: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx + d$

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

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


    We know five facts about the function:

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


    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. $\displaystyle f(0)= a(0+ 3)(0- 2)^2= 12a$, 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: .$\displaystyle f(x) \:=\:ax^3 + bx^2 + cx + d$

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

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


    We know five facts about the function:

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


    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 $\displaystyle f'(0)$ may or may not be $\displaystyle 0$. It turns out that it is not. If you combine my post above with HallsofIvy's you will see that

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

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

    Soroban's method should work if you just leave out the condition $\displaystyle f'(0)=0$. 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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