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Math Help - Function diagram

  1. #1
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    Function diagram

    Greetings,

    I am unable to draw the diagram of f(x) = x|x-b|. I know we use the mirror effect to draw the diagrams of functions with modules, however I do not how to do it when there are variables both inside and outside the module signs like x|x|. Let alone that, there is also this b parameter I have not the vaguest idea how to deal with.

    Thank you.
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  2. #2
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    Quote Originally Posted by Logic View Post
    Greetings,

    I am unable to draw the diagram of f(x) = x|x-b|. I know we use the mirror effect to draw the diagrams of functions with modules, however I do not how to do it when there are variables both inside and outside the module signs like x|x|. Let alone that, there is also this b parameter I have not the vaguest idea how to deal with.

    Thank you.
    If x< b, then x- b< 0 so |x- b|= -(x-b)= b-x and f(x)= x|x-b|= x(b-x)= bx- x^2. If x> b, then x- b> 0 so |x- b|= x- b and f(x)= x|x- b|= x(x- b)= x^2- bx. Can you graph those?
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  3. #3
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    That was very helpful - thank you.
    However, I am still unsure as to what to do with the parameter.
    I mean, assume x>b for example, the coordinates of the top of the parabola are something like: (b/2; (b^2-4)/4) and I do not know where this point on the coordinate system would be. I have encountered similar problems several times and all textbooks have proven useless... at least the ones used in school.
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  4. #4
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    Quote Originally Posted by Logic View Post
    That was very helpful - thank you.
    However, I am still unsure as to what to do with the parameter.
    I mean, assume x>b for example, the coordinates of the top of the parabola are something like: (b/2; (b^2-4)/4) and I do not know where this point on the coordinate system would be. I have encountered similar problems several times and all textbooks have proven useless... at least the ones used in school.
    1. With an additional parameter you define a family of curves.

    2. Have a look at the zeros of the given function: f(x)=x \cdot |x-b|
    - all graphs of f pass through the origin.
    - all graphs have a second zero at x = b

    3. At the second zero the graph of f is continuous but not differentiable if b \neq 0

    4. I've attached the graphs of f for b\in \{-4, -3, ..., 3, 4\}

    5. You find the vertex at V\left(\dfrac b2\ ,\ \dfrac{b^2}4\right), b \geq 0 or at V\left(\dfrac b2\ ,\ -\dfrac{b^2}4 \right), b < 0. .
    That means you know

    x = \dfrac {|b|}2~\wedge~y = \dfrac{b^2}4, b \geq 0~\vee~y=- \dfrac{b^2}4, b < 0

    Calcute b from the first equation and plug in this term into the second equation:

    b\geq 0: \ b = 2x~\implies~y = \dfrac{(2x)^2}4~\implies~y=x^2

    b < 0: \ b = 2x~\implies~y = -\dfrac{(2x)^2}4~\implies~y=-x^2

    That means the locus of all vertices is the function with b = 0: f_0(x)=x \cdot |x|
    Attached Thumbnails Attached Thumbnails Function diagram-betrg_fktschar.png  
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