Write an equation whose graph has intercepts x = -2 and x = 2 and is symmetric with respect to the origin.

Question

debbieg · Answer

The intercepts are easy enough to get, if you make your function a polynomial, then each root of c corresponds to an x-intercept, and a factor of x-c.  So use x=-2 and x=2 to cook up the two associated factors and you will have the two x-intercepts.

To make it symmetric wrt the origin, it needs to be "odd". For a polynomial, that will be automatic if every term has odd degree. Can you think of way to do that? (it will also give an additional root, but nothing in the problem says that you can't have that)

debbieg · Answer

In other words, start by cooking up a polynomial that has intercepts (which means, zeros) at x=-2 and x=2. Once you have that, you can easily make it symmetric about the origin, with just one slight tweak. :)

anonymous · Answer

Very simple is a circle, centered in the origin (0,0) and radius=2
$$x^2+y^2=4$$Nothing more symmetric than that. It is symetric vs y-axis, x-axis and origin

debbieg · Answer

That one would work too.... not the one I was trying to help her figure out, but certainly that is symmetric. It's not a function, but then, the problem as stated does not require a function, so that would work.

anonymous · Answer

That is the point, the problem is not asking for a function but for an equation, therefore the circle´s equation is a valid one

anonymous · Answer

Your solution is more "mathematical" if you let me put it that way. f(x)=(x-2)(x+2) is symetric versus y-axis--->f(x)=f(-x). In order to get a function symmetric vs origin we only need a function such g(x)=-g(-x)--->-g(-x)·f(-x)=g(x)·f(x). The simplest is g(x)=x, then the function sought is ???

debbieg · Answer

Yes, I agree that the circle equation you gave her is a valid one. I said it would work. :) I still think it would be a useful exercise though for her to come up with one on her own, that's what I was trying to do.

debbieg · Answer

Right, like I said,  f(x)=(x-2)(x+2) is not the final solution. It isn't symmetric about the origin, but can easily be tweaked to create a new function, that is.

anonymous · Answer

I think that @amelapal can be now left to her own devices