Graphing Symmetry Question II?

Question

X^2 - 2X - Y^2=4
.
.
.
(1)^2 - 2(1) -y^2=4,,
1 - 2 -y^2 = 4,,
(-1) - y^2 = 4,,
-y^2 = 5,,
y^2=(-5),,
,
,
,
(-1)^2 - 2(-1) -y^2=4,,
1 +2 - y^2 =4..
3 - y^2 = 4..
-y^2 = 1..
y^2 = (-1)..
,
,

,

I know this exhibits x-axis, and y-axis symmetry. however, I believe it is not because I have obtained an equivent equation by replacing a postive value with a negative. I believe it is because of obtaining the irrational solution. Is this correct to think along these lines? Am I choosing incorrect values maybe, to test for symmetry?
Thank you kindly " once again."

Answer 1

It clearly doesn't exhibit y-axis symmetry, because you got different equations for x = 1 and x = -1.

And no, it has nothing to do with whether solutions are rational or irrational. Almost every function has irrational values almost everywhere.

x^2 - 2x - y^2=4
<=> (x-1)^2 - y^2 = 3

So this is symmetrical about x = 1 and about y = 0.

The previous answerer has some issues; they got the wrong equation for y, then assumed that y^2 = a => y = sqrt(a) [which of course is only true for y >= 0] and used this assumption to prove that y had to be non-negative - which of course it does not.

Answer 2

X^2 - 2X - Y^2=4


It is not symmetrical about x axis, because it's shape is that of a half circle because Y cannot be negative (Y = (4-x^2+2*x)^0.5)
they gotta to be positive real numbers.

and yea it is symmetrical about some vertical line: x=2.2361.

I got this from a graphing software but.

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true!!
I forgot to putt negative sign in front of that y=..
>< oops! sorry. I lost marks for this mistake like two years ago in a test, but I made that mistake again
anyways try graphing it, it is the best way!