Symmetry Testing: Solutions to 2 Equations

[ability]

Here's 2 more I am having difficulty with, I'll show you what I came up with or what I think the answer is

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

 

[OlderDan]

Here's 2 more I am having difficulty with, I'll show you what I came up with or what I think the answer is

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

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 written

y = \sqrt {2x/3}

rather than writing

y = \pm \sqrt {2x/3}

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 \pm 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.

 

[thepatientmental]

Your solutions for both equations seem to be correct. For the first equation, you have correctly identified that the graph is symmetric to the x-axis. This means that if you were to fold the graph along the x-axis, the two halves would perfectly overlap. This is due to the fact that the equation has a square root, which is an even function and therefore symmetric about the y-axis.

For the second equation, you have correctly identified that the graph is not symmetric to any axis. This is because the equation does not have any terms that are even functions. In order for a graph to be symmetric to an axis, it must have at least one term that is an even function. In this case, both terms are odd functions, x^2 and 4y^2, which means they are not symmetric about any axis.

Great job on solving these equations and identifying their symmetries! Keep practicing and you will become even more confident in your symmetry testing skills.