- #1

ability

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1. 2x = 3y^2

(sqrt 2x/3) = 3y/3

y = sqrt (2x/3)

Symmetric to the X-axis

2. x^2 + 4y^2=16

not symmetric to any axis

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- #1

ability

- 4

- 0

1. 2x = 3y^2

(sqrt 2x/3) = 3y/3

y = sqrt (2x/3)

Symmetric to the X-axis

2. x^2 + 4y^2=16

not symmetric to any axis

- #2

OlderDan

Science Advisor

Homework Helper

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I assume you are talking about the symmetry of graphs of equations and not the symmetry of "functions". Functions have conditions that are not satisfied by all equations. Your first case allows for x to be considered a function of y, or as you have rewritten it, for y to be a function of x, but in the latter case it is only a function because you have writtenability said:

1. 2x = 3y^2

(sqrt 2x/3) = 3y/3

y = sqrt (2x/3)

Symmetric to the X-axis

2. x^2 + 4y^2=16

not symmetric to any axis

[tex] y = \sqrt {2x/3} [/tex]

rather than writing

[tex] y = \pm \sqrt {2x/3} [/tex]

The first equation satisfies the condition for y to be a function of x (only one value of y for each value of x), but without the [tex] \pm [/tex] sign the graph of the equation has no symmetry axis.The second equation has the symmetry axis you have identified, but it does not represent a function. In either case, x must be positive.

Your second case is the equation of an ellipse, which is not a function. It has two symmetry axes. If you manipulated the equation to solve for y, and then to solve for x I think you might see the symmetry.

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