Pre-Calculus Test for symmetry with respect to both axis and the origin.?

Question

A) x/(x۸2)+1

B) x • (y۸2) + 10 = 0

Answer 1

I presume that A is f(x) = x/(x^2 + 1)? Since otherwise it's the same as 1/x + 1.

Anyway - a function is symmetric about the y-axis if f(x) = f(-x). In the case of your first problem,

f(x) = x/(x^2 + 1)

f(-x) = -x/((-x)^2 + 1) = -x/(x^2 + 1)

They are not the same, so the function is not symmetric about the y-axis.

A function is symmetric about the origin if f(x) = -f(-x), Again, for your first problem,

f(x) = x/(x^2 + 1)

-f(-x) = -(-x)/((-x)^2) + 1 = x/(x^2 + 1)

They are the same, so the function is symmetric about the origin.

A function cannot be symmetric about the x-axis, because then it wouldn't be a function. If a relation were to be symmetric about the x-axis, it would have to be the case that for every x you gave it, you got both positive and negative y out. There's no way to write that with functional notation, for reasons already noted.

Does that help?

Answer 2

a million) (-x, y). So (-x)^2 + y^2 = 9. through fact once you replace -x in for x you get the coolest comparable answer, it DOES have symmetry with the y-axis 2) (x, -y). lower back, considering which you get the comparable answer (the subject does not replace in any respect, as -y will become beneficial while squared), that's symmetrical with the x axis 3) This one is weird and wonderful. The try is -f(x) = f(-x). that's, while you're making each and every X in the subject damaging, do you get the comparable consequence as in case you have been to make each and every thing in the subject damaging? the respond is not any. (-x)^2 + y^2 = 9 and (-x)^2 (-y)^2 = -9 are no longer the comparable. So, that equation has symmetry with the x and y axes, yet no longer with the beginning place.

Answer 3

To test symmetry, put -x in for x. If the result for -x is the same function value as for x, the function is symmetric about the y axis. If the result for -x is the negative of the result for x, the function is symmetric about the origin.