I tried making the graph lines bolder, but it does not work:

myPlot.addPlot('sin(x)',FOOPLOT_TYPE_FUNCTION,{'color':'#00F', 'width':'5' });

How to do it?

Would this be possible with the recent version of fooplot:

Thanks.

I am astonished that fooplot can "understand" so many formulas, such as sin() etc.

How would I go and use this parsing capability in my own project? What functions do I have to use?

Goal: I have an input field, that holds for example 2*sin(3*x) and I want to throw in my own x values to get the corresponding y values.

For now I am using mathjs.org but I run into some problems, also the math.min.js is 210 KB...

is:issue is:open How do I plot a vertical line such as x=2? You can currently do this by creating a Parametric equation with x=2 and y=s, and setting s to an interval that covers your viewing window. See this for an example. the link 'See this for an example' is 失效了

The graph of (x-4)^(3)x^2(x+5) has also a root at x = -5 but fooplot is not plotting this.

How it should look like: http://www.matheretter.de/tools/funktionsplotter/?%28x-4%29^3*x^2*%28x+5%29

How it looks like: http://fooplot.com/plot/qvb34774yj

How to solve the bug?

Example x(1+x): http://fooplot.com/plot/nr9v8lks2i throws:

TypeError: vars.x is not a function

Maybe we should "scan" the string beforehand and add a * if we find a string like "variable(...)".

I have tried:

pointplot = "1,1";
myPlot.addPlot(pointplot, FOOPLOT_TYPE_POINTS);
myPlot.reDraw();

without success. Could you please give an example?

how can i shade area between two curves using fooplt.js

I would like to draw a graph beginning from x=0 and then into the positive: http://fooplot.com/plot/bqb5rfvg9n

Function's equation is: ((21/20*x+1)^(1/2))*(x>0)

Problem is, even though I define (x>0) the drawing can be seen for negative x, starting by x = -1:

How comes?

Drawing f(x)=x with the limit to x>0 draws a red line onto the x-axis for negative x: http://fooplot.com/plot/s0x03w0mn9

In the FAQ I found:

How do I draw a piecewise function?
You can enter a piecewise function using the comparison operators <, >, <=, >=, and == which return 1 if the comparison is true and 0 if the comparison is false. For example, (x<0) returns 1 if the statement (x<0) is true and 0 otherwise. To plot the piecewise function y={x, x<0; x^2, x>0} you could enter (x<0)*x+(x>0)*x^2.

Is there any way to achieve this:

Example:

Or at least the area between graph and x-axis?

Set of points:

0,0
pi,0
pi/2,1

only 0,0 gets plotted: http://fooplot.com/plot/k10cxaz1q1

I guess this.parseEquation and this.parseConst are missing in:

  case FOOPLOT_TYPE_POINTS:
    if(eq.length===null) return null;
    this.plots.push({'id':this.plotlastid++,'type':type,'eq':eq,'options':options});
  break;

But my attempts to parse the eq there failed unfortunately.

Any plans for bringing it back up?

Seems not to work: |x^3+1| or -2*|x+1|+4

Ah, update: It is this one: abs(x^3+1)

I leave my question and answer here, in case somebody googles the same.