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.
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.
2. Have a look at the zeros of the given function:
- 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
4. I've attached the graphs of f for
5. You find the vertex at or at . .
That means you know
Calcute b from the first equation and plug in this term into the second equation:
That means the locus of all vertices is the function with