# Question about implicit functions of this form

I'm sorry I don't know how to make the title more descriptive. Last year I learned in Calculus something I don't remember quite well, but I need it now for another subject, I just want to be sure this is right.

If we have a function$:\mathbb{R}\rightarrow\mathbb{R}^2$, given in the form: $$h(x,y)\cdot g(x,y)=0$$

Is it true that the plot will be the two independent curves defined by $h=0$ and $g=0$, for example: $(x-1)(y-1)=0$, is that just the two lines $x=1,y=1$?, in that case it's true. Is the general case I wrote up also true, or not?

I only need it for something else I study, so I accept a yes/no answer, although some insight would be appreciated. Thanks.

After some discussion with amWhy, this is clear: What I said happens if and only if the function h and g can be factorised into more functions of the same kind, for example the plot of this function:

$(y-x)(y-x^2)(y-3x^3)(y^2+x^2-1)=0$

will be all together the lines $y=x$, the parabola $y=x^2$, $y=3x^3$ and the circle $x^2+y^2=1$, which if it's factorised in more terms, then those terms will be plotted: $(x^2-1)(y-1)=0$ will plot the line $y=1$ and the factors of the other term: lines $x=1, x=-1.$

Thank you.

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$$h(x,y)\cdot g(x,y)=0 \quad?\implies? \quad h = 0,\;\text{and}\;g = 0$$

No, all that you know is that EITHER $\;h(x, y) = 0\;$ OR $\;g(x, y) = 0\;$ OR both are $0$.

Any of these cases will satisfy $\;h(x,y)\cdot g(x,y)=0.$

The solution to the equation: (product of any number of functions)$= 0$ will be the union of the solutions to the equations (each function)$= 0$.

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And isn't there something like that? I remember something similar, like in my example, how do you deduce that? – MyUserIsThis Feb 7 '13 at 21:57
I'm not clear what you're asking; the product of two function is $0$ iff one, or both, functions evaluate to zero. Put simply $a\cdot b = 0 \iff a = 0 \lor b = 0$. – amWhy Feb 7 '13 at 22:02
It is true that the solution to your example includes, or is given by the two lines $x = 1$, $y = 1$. – amWhy Feb 7 '13 at 22:06
I think you meant what I wrote about "or": either being a possibility. Perhaps I was nitpicking with the use of "and": each possibility. It is true that $h(x,y) = 0$ is a solution, and $g(x, y) = 0$ happens to be a solution. Essentially, I was simply pointing out that they needn't both be zero, so yes, there are two "independent" solutions, in that sense. – amWhy Feb 7 '13 at 22:10
Yes, I know that. I just remember that last year our multivariable calculus teacher told us something similar to what I was saying, like a trick when you see something similar to what I asked, but I don't remember very well what it was. It was that if you found two functions like I wrote, (not exactly like that but I don't remember), the plot would be the two independent functions. I have tried some examples and what I wrote is not correct, but it was similar. I just found some problems that might be usefull for and I would like to remember it. – MyUserIsThis Feb 7 '13 at 22:12