Symmetry about x-axis, y-axis or origin

[jaysquestions]

Homework Statement

determine whether the graph of the function has symmetry about the x-axis, the y-axis or the origin. Check work by graphing:
x^(2/3) + y^(2/3) = 1

Homework Equations

The Attempt at a Solution

[ x^(2/3) + y^(2/3) = 1 ] ^(3/2)

y = -x+1

Its a straight line with y-intercept at 1. I don't understand how it has any symmetry about either x or y axis?
thanks

 

[SammyS]

Homework Statement

determine whether the graph of the function has symmetry about the x-axis, the y-axis or the origin. Check work by graphing:
x^(2/3) + y^(2/3) = 1

Homework Equations

The Attempt at a Solution

[ x^(2/3) + y^(2/3) = 1 ] ^(3/2)

y = -x+1

Its a straight line with y-intercept at 1. I don't understand how it has any symmetry about either x or y axis?
thanks

Bad algebra.

$\displaystyle \ (a+b)^{p}\ne a^p+b^p \$ unless p=1 .

 

[DeldotB]

Well, you could solve for y:

$$f(-x)=f(x)?$$ or $$f(-x)=-f(x)?$$

 

[Mark44]

Well, you could solve for y:

$$f(-x)=f(x)?$$ or $$f(-x)=-f(x)?$$

This is not helpful. The equation given does not have y as a function of x.

 

[DeldotB]

This is not helpful. The equation given does not have y as a function of x.

And why can't you just solve for y? I know you will have a plus or minus...so just analyze both parts. Same thing-This will show its symmetric about the origin.

 

[yango_17]

Keep in mind that a function will have symmetry about the x-axis if you can replace all the y's with y-'s and obtain the same function, and it will have symmetry about the y-axis if you can do the same thing to the x's in the function and obtain the same function. If you change both y's with y-'s and x's with x-'s and the function remains the same, then the function will have symmetry about the origin.

 

[Mark44]

And why can't you just solve for y?

If you solve for y, you don't get a function. Some of the points on the graph of the given equation are (-1, 0), (1, 0), (0, 1), and (0, -1). The latter two points show that this graph is not that of a function, so writing f(x) and f(x) isn't applicable here.

 

[Mark44]

Keep in mind that a function will have symmetry about the x-axis if you can replace all the y's with y-'s and obtain the same function

If you can do this, then the graph you're looking at is not a function.

Corrected, the above would say "a function graph will have symmetry about the x-axis if you can replace all the y's with y-'s their opposites and obtain the same function graph"

, and it will have symmetry about the y-axis if you can do the same thing to the x's in the function and obtain the same function. If you change both y's with y-'s and x's with x-'s and the function remains the same, then the function will have symmetry about the origin.

 

[yango_17]

So, strictly speaking, one would have to label it as an equation? Or is my proposed method incorrect?

 

[Mark44]

So, strictly speaking, one would have to label it as an equation? Or is my proposed method incorrect?

See my previous post.

 

[yango_17]

If you can do this, then the graph you're looking at is not a function.

Corrected, the above would say "a function graph will have symmetry about the x-axis if you can replace all the y's with y-'s their opposites and obtain the same function graph"

Ah, okay. Thanks for clarifying.

 

[DeldotB]

If you solve for y, you don't get a function. Some of the points on the graph of the given equation are (-1, 0), (1, 0), (0, 1), and (0, -1). The latter two points show that this graph is not that of a function, so writing f(x) and f(x) isn't applicable here.

So you are saying that $$y=(1-x^{2/3})^{3/2}$$ and $$y=-(1-x^{2/3})^{3/2}$$ are not functions?

And how do (0,1) and (0,-1) make it not a function? The top and bottom half of the unit circle contain those points - this is analogous to saying that the graph of the top half of a circle is not a function because it contains these points.

 

[Mark44]

So you are saying that $$y=(1-x^{2/3})^{3/2}$$ and $$y=-(1-x^{2/3})^{3/2}$$ are not functions?

No, I'm not. Those are functions. However, for almost every value of x in the domain of the original equation, there are two values of y. For a function, each value in the domain is paired with a single value in the codomain.

For a simpler example, consider the equation $x^2 + y^2 = 1$. If you solve for y you get $y = \pm \sqrt{1 - x^2}$. Again, there are two y values for each x value in the domain (with exceptions for x = 1 and x = -1. The graph of this equation is a circle of radius 1, with center at the origin. The graph does not pass the vertical line test, another clue that the equation does not represent a function. This is the same situation that is present in the equation of this thread.

 

[DeldotB]

So you are saying that $$y=(1-x^{2/3})^{3/2}$$ and $$y=-(1-x^{2/3})^{3/2}$$ are not functions?

No, I'm not. Those are functions. However, for almost every value of x in the domain of the orgininal equation, there are two values of y. For a function, each value in the domain is paired with a single value in the codomain.

For a simpler example, consider the equation $x^2 + y^2 = 1$. If you solve for y you get $y = \pm \sqrt{1 - x^2}$. Again, there are two y values for each x value in the domain (with exceptions for x = 1 and x = -1. The graph of this equation is a circle of radius 1, with center at the origin. The graph does not pass the vertical line test, another clue that the equation does not represent a function. This is the same situation that is present in the equation of this thread.

No, for every x value in the domain there is one y if you analyze just the top or bottom half. The graph passes the verticals line test as well. I do not understand why you are saying there are two y values for one x value if you analyze just the top or bottom half. I know that a functions inverse will also be a function if it's one to one. Maybe I'm not understand what you are saying. It sounds to me right know that you are saying something like y=x^2-1 is not a function because it contains (0,1) and (0,-1)

All I was originally saying was that you can break the function into two separate functions of x and test for symmetry.

 

[Mark44]

No, for every x value in the domain there is one y if you analyze just the top or bottom half.

That's true, but the original equation includes both halves.

If we go back to the simpler example I gave, $x^2 + y^2 = 1$, if you solve for y you get $y = \pm \sqrt{1 - x^2}$. The top half (with the pos. square root) is a function, and the bottom half (neg. sq. root) is also a function, but both together do not represent a function.

The graph passes the verticals line test as well.

No it doesn't. As I said before, the graph of the equation at the start of this thread includes (0, 1) and (0, -1), as well as many other pairs of points that are on a vertical line.

The graph of $x^2 + y^2 = 1$ also includes (0, 1) and (0, -1) and a lot more pairs that are on vertical lines. The equations $x^{2/3} + y^{2/3} = 1$ and $x^2 + y^2 = 1$ do not represent functions.

I know that a functions inverse will also be a function if it's one to one. Maybe I'm not understand what you are saying. It sounds to me right know that you are saying something like y=x^2-1 is not a function because it contains (0,1) and (0,-1)

First, I am not saying this, and second, it's not true. This equation, which is different from the one in my example, is a function. If x = 0, then y = -1, so the point (0, -1) is on the graph. The point (0, 1) is not on the graph of this parabola.

 

[DeldotB]

That's true, but the original equation includes both halves.

If we go back to the simpler example I gave, $x^2 + y^2 = 1$, if you solve for y you get $y = \pm \sqrt{1 - x^2}$. The top half (with the pos. square root) is a function, and the bottom half (neg. sq. root) is also a function, but both together do not represent a function.
No it doesn't. As I said before, the graph of the equation at the start of this thread includes (0, 1) and (0, -1), as well as many other pairs of points that are on a vertical line.

The graph of $x^2 + y^2 = 1$ also includes (0, 1) and (0, -1) and a lot more pairs that are on vertical lines. The equations $x^{2/3} + y^{2/3} = 1$ and $x^2 + y^2 = 1$ do not represent functions.

First, I am not saying this, and second, it's not true. This equation, which is different from the one in my example, is a function. If x = 0, then y = -1, so the point (0, -1) is on the graph. The point (0, 1) is not on the graph of this parabola.

Ok, I feel we have just mis understood each others posts. Obviously (0,1) is not on the graph of a porabola 1-x^2. We disagreed because I was talking about the graph of ONE of the two functions while you are talking about the entire graph (+/-). I was saying it sounded like you were implying that (0,1) AND (0,-1) were on the graph of THE TOP HALF of the graph ( which is what I was talking about all along). No more needs to be said here. Simple mis communication.

 

[Mark44]

Ok, I feel we have just mis understood each others posts. Obviously (0,1) is not on the graph of a porabola 1-x^2. We disagreed because I was talking about the graph of ONE of the two functions while you are talking about the entire graph (+/-). I was saying it sounded like you were implying that (0,1) AND (0,-1) were on the graph of THE TOP HALF of the graph ( which is what I was talking about all along). No more needs to be said here. Simple mis communication.

Yes, throughout I was talking about the graph of the given equation and that of my example, not part of the graph of either equation.