Graph of $|x|=x$

Question

I typed the equation $|x|=x$ on varius graphing calculators waiting to see the part
of the plane where $x\ge0$. I thought that because for every point $A(x,y)$ with $x\ge0$ the above equation is true. Instead only the y'y axis is displayed(as if the equation is only true for $x=0$).

I would like to draw this semiplane without using the inequality $x\ge0$ (which works) but an equation. Is there a problem or I don't understand something? Any help would be appreciated.

In fact I want to graph the semicirle $|x^2+y^2-1|+|x|-x=0$ but noticed that the problem is the $|x|-x=0$ part on every graphing calculator. Is there any other possible way?

Answer 1

The problem is presumably that most graphing calculators work in two dimensions, i.e. they need an $x$ and a $y$ (sometimes also called $f(x)$). Each point on that plane is an ordered pair, $(x,y)$ (or $(x,f(x))$ if you're talking about the graph of a function in that plane).

Consider $|x|=x$. Where is your $y$? I suppose you could graph it in one dimension, in which case it would just be the ray $[0,\infty)$, but most graphing calculators don't care to do something so simple.

So, how do we find $|x|=x$ in a standard two-dimensional graph? Well, move everything to one side, define $f(x)$ by replacing $0$ with it, and then look at the graph: where $y=0$ is where the equality is satisfied.

$$|x| = x \;\;\; \text{becomes} \;\;\; x - |x| = 0 \;\;\; \text{becomes} \;\;\; f(x) = x - |x|$$

Graph of $f(x) = x - |x|$: the solutions to $|x|=x$ will be where $f(x)=0$:

Answer 2

ok...from my aspect,

$1.$ where is your "y" variable? to plot any function you you must have to give a variable representing the function itself.you can write $y=x\text{ or }y=|x|$.But can't simply write $x=|x|$

$2.$ Now coming to your equation.I think you want to have a graph like $y=\sqrt{1-x^2}$ but your given function and this function is not a same thing.there is a problem in your function.that is, given, $$|x^2+y^2-1|+|x|-x=0$$ $$\implies |x^2+y^2-1|=x-|x|$$ $$\text{now this is only possible when $x\ge 0$.and every time for those values of $x$,$x-|x|$ yields $0$.}$$ you must have to add this condition while graphing it.because it is a necessary condition to the function be valid. so,we can say, $$|x^2+y^2-1|=0$$ now it is only possible when $(x^2+y^2-1)$ is $0$ itself.

Now look you can compute it in geogebra 3D .and that is your problem. https://www.geogebra.org/3d/ypk8vzc5

look where $z=0$