# Testing for symmetry about the origin

• Apr 29th 2014, 08:34 PM
precalc
Testing for symmetry about the origin
Hi everybody!

I'm stumped as to why I can't seem to get the right answer when testing for symmetry in the following equation:

x^2 + xy + y^2 = 0.

Normally to test for symmetry I test to see if f(-x)=-f(x) for symmetry about the origin.

When I do this, I find that f(-x) = x^2 - xy + y^2 which is not equal to -f(x). Therefore, I would write down that it's not symmetric about the origin. But my math book says that it is symmetric about the origin because f(x)=-f(-x). Any help as to clarifying why I'm getting this answer wrong would be much appreciated.

It's the Barrons SAT Math II subject test book, and the way they approach problems is really counter to how I've learned them.

Thank you very much,
• Apr 29th 2014, 08:52 PM
romsek
Re: Testing for symmetry about the origin
Quote:

Originally Posted by precalc
Hi everybody!

I'm stumped as to why I can't seem to get the right answer when testing for symmetry in the following equation:

x^2 + xy + y^2 = 0.

Normally to test for symmetry I test to see if f(-x)=-f(x) for symmetry about the origin.

When I do this, I find that f(-x) = x^2 - xy + y^2 which is not equal to -f(x). Therefore, I would write down that it's not symmetric about the origin. But my math book says that it is symmetric about the origin because f(x)=-f(-x). Any help as to clarifying why I'm getting this answer wrong would be much appreciated.

It's the Barrons SAT Math II subject test book, and the way they approach problems is really counter to how I've learned them.

Thank you very much,

symmetry about the origin means that if (x,y) satisfies some equation then (-x,-y) does as well.

here we have

\$x^2+xy+y^2=0\$

and

\$(-x)^2 + (-x)(-y)+(-y)^2 = x^2+xy+y^2 = 0\$

thus we have symmetry about the origin.
• Apr 30th 2014, 02:58 PM
precalc
Re: Testing for symmetry about the origin
Okay, that definitely makes sense, thank you very much for the help, romsek!
Would the reason why testing if f(-x)=-f(x) does not work in this situation have something to do with the fact that there is both a y2 and an x2 within the equation?
• Apr 30th 2014, 03:15 PM
Plato
Re: Testing for symmetry about the origin
Quote:

Originally Posted by precalc
Okay, that definitely makes sense, thank you very much for the help, romsek!
Would the reason why testing if f(-x)=-f(x) does not work in this situation have something to do with the fact that there is both a y2 and an x2 within the equation?

Actually it does work if one uses the correct notation.
In this case \$f(x,y)=x^2+xy+y^2\$. Does \$f(x,y)=f(-x,-y)~?\$
• Apr 30th 2014, 04:04 PM
HallsofIvy
Re: Testing for symmetry about the origin
Quote:

Originally Posted by precalc
Okay, that definitely makes sense, thank you very much for the help, romsek!
Would the reason why testing if f(-x)=-f(x) does not work in this situation have something to do with the fact that there is both a y2 and an x2 within the equation?

f(x) should depend only on x! You can use "f(-x)= -f(x)" if you recognize that f(x)= y, not the whole formula.
• May 2nd 2014, 01:30 PM
precalc
Re: Testing for symmetry about the origin
Thank you very much, Plato and HallsofIvy! I finally understand both ways to do it, by using f(x,y)=f(−x,−y) as well as that if I want to do f(-x)=-f(x) I need to isolate y on one side. I appreciate everybody's help so much!