How to determine symmetry algebraically?

Question

"Algebraically determine the symmetry of the graph of the relation."

How would I go about doing this? (The symmetries are x-axis, y-axis, origin, and line y=x) - please use my first exercise as an example:

y = x^4 - 2x^2

Thanks!

Answer 1

For axis-symmetry, simply allow the "other" variable to transform from itself to (minus-itself).

In other words:

If you want to test whether a function is symmetric about the y-axis, simply transform x to -x and see if you get the same expression:

ynew = (-x)^4 - 2*(-x)^2 = (x)^4 - 2*(x)^2 = yold

Since "ynew" and yold" are the same, the function is symmetric about the y-axis.

If one wants to find if something is symmetric about the origin, simply change BOTH x and y to -x and -y simultaneously.

Finally, if one wants to find if something is symmetric about the line y = x then simply change each x into a y and each y into an x, again simultaneously.

In all cases, if you can transform it in this way and the resulting expression can be found to be identical to the original, the function is said to be symmetric in that way.

Answer 2

If a graph is symmetric across the x-axis, then for every (x, y) on the graph (x, -y) is also on the graph. If it's a graph where y is some function of x, then for any given x, there can only be single value of y. Therefore y = f(x) = 0 is the only function graph that is x-axis symmetric. In your case, y is indeed a non-zero function of x, so it is NOT x-axis symmetric.

If a graph is symmetric around the origin, then for every (x, y) on the graph (-x, -y) is also on the graph. Let's try one: (2, 8) is on the graph; is (-2, -8)? No. So your graphs is not origin symmetric.

If a graph is symmetric across the line for y=x, then for every (x, y) on the graph (y, x) is also on the graph. Let's try one: (2, 8) is on the graph; is (8, 2)? No. So your graphs is not symmetric across y=x.

Of course, I saved the best for last:
If a graph is symmetric across the y-axis, then for every (x, y) on the graph (-x, y) is also on the graph. Let's try one: (2, 8) is on the graph; is (-2, 8)? Yes! In fact, a graph where y is a function of x and f(x) = f(-x) for all x is symmetric about the y-axis.

In fact, when y is a polynomial function of x...
- if only even powers of x (including constants) have non-zero coefficients, it's symmetric about the y-axis.
- if only odd powers of x have non-zero coefficients, it's symmetric about the origin.

Is it possible for a graph to have all four of these kinds of symmetry? Sure. A circle centered at the origin is just one example.

Answer 3

If f(x)=f(-x) then f(x) is symmetric to the y-axis.
Notice that f(x)=x^4-2x^2 is the same if x is replaced by -x

If -f(x) = f(-x), then f(x) is symmetric with respect to the origin.

If f(x) is reflected across the line y=x then the reflection is the inverse of f(x).

If f(x) remains the same when y is replaced by -y, then f(x) is symmetric to the x-axis.

Answer 4

x-axis :

for any x you will have two values for y only the sign differ.

y-axis:
+x and -x results in the same y-value. ( that is the case with your example )

rest is way too difficult for me.

i think that symetric in the origin implies that you will also have symmetry in x-axis and y-axis. and vice versa symmetry in x and y axis implies symetric in origin.