1. ## 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.

2. Originally Posted by Logic
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?

3. 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.

4. Originally Posted by Logic
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 $b = 0: f_0(x)=x \cdot |x|$