Note that your expression is only real for
Underneath the radical it's the difference of two squares, which iirc, are always symmetrical in the y axis. Plus your graph is a semi-circle.
No doubt someone can put it more elegantly than I though
Hi!
i am trying to determine if this equation is symmetrical to the x-axis and/or the y-axis. or both. the equation is...
which i changed to...
the book states that this equation is ONLY symmetrical to the y-axis. i agree it is symmetrical to the y-axis, but it also seems to me to be symmetrical to the x-axis.
since...
Because a square root has both a positive and negative value, doesn't this make sense?
what am i missing/misunderstanding?
p.s. is this algebra or pre-calc? i wasn't entirely sure
ok lets say a=.5
then...
is this totally wrong?
my graphing calculator does show that it is a semi-circle.
edit: ok, so you can never have a negative number as the answer for a square root. so is this just something that one has to remember? i guess i'm struggling with the fact that the method i learned from the book i'm reading explained how to determine symmetry using specific rules.
it states that you can substitute in -y or -b in place of y or b respectively when trying to see if a graph or function displays symmetry with the x-axis. (y=b). If the resulting equation is the same as the original, then it is symmetrical with respect to the x-axis.
so when i plug in -y i get
so is it as simple as saying that ?
this inequality isn't always true though, since y could be zero.
You correctly squared the equation, the issue with squaring is that it introduces possible extraneous solutions (solutions not in the original domain). For example if you had and squared it you'd get . Solutions here are
In your case if you wanted to solve for y then you'd have and the negative solution is extraneous.
Your equation will hold for
Using your example of a = 0.5
and
Say if you were only given the second equation then you'd deduce that - but where has that negative solution come from?
Wiki: http://en.wikipedia.org/wiki/Extrane...sing_solutions
Squaring both sides of an equation in general is a "non-reversible" step. When you "undo" the squaring operation you don't get back what you started with. For example, if , and you square both sides you get . Reversing this gives . An extra solution has been generated that wasn't there before.