Is y = | x | - 2 Symmetric About the x-axis, y-axis, or Origin?

[mathdad]

Test for symmetry about the x-axis, y-axis and origin.

y = | x | - 2

What are the rules for testing for symmetry?

 

[Evgeny.Makarov]

On this forum you are supposed to show an attempted solution or at least describe the difficulty. Surely you can plot the first graph: there is nothing that requires ingenuity about it.

In general the equation f(|g(x,y)|) = 0 for any functions f and g is equivalent to the following:
\[
\left\{\begin{aligned}f(g(x,y))&=0\\g(x,y)&\ge0\end{aligned}\right.\quad\text{or}\quad
\left\{\begin{aligned}f(-g(x,y))&=0\\g(x,y)&<0\end{aligned}\right.
\]

To make |x + y| = 2 a special case of f(|g(x,y)|) = 0 we define f(x) = x - 2 and g(x,y) = x + y; then f(|g(x,y)|) = 0 is |x + y| - 2 = 0. According to the statement above, it is equivalent to
\[
\left\{\begin{aligned}x+y-2&=0\\x+y&\ge0\end{aligned}\right.\quad\text{or}\quad
\left\{\begin{aligned}-(x+y)-2&=0\\x+y&<0\end{aligned}\right.
\]
So you need to find the set of (x, y) that lie on the line x+y-2 = 0 and also satisfy x + y >= 0, and take the union with the set of (x, y) that lie on the line x+y+2 = 0 and also satisfy x + y < 0.

 

[mathdad]

I changed the question. Forget the graph. What are the rules for testing for symmetry?

 

[skeeter]

Forget the graph.

why? ... easiest way to check symmetry is by viewing the graph.

 

[mathdad]

If you are going to reply using textbook jargon, then what's the point of asking for guidance? I changed the question. What are the rules for symmetry?

 

[mathdad]

y = | x | - 2

-y = | x | - 2

Not symmetric about the x-axis.

y = | -x | - 2

y = x - 2

Not symmetric about the y-axis.

-y = | -x | - 2

-y = x - 2

Not symmetric about the origin.

 

[MarkFL]

Let:

$$f(x)=y=|x|-2$$

We find:

$$f(-x)=|-x|-2=|x|-2=y=f(x)$$

Thus, this function is symmetric about the $y$-axis. No further tests for symmetry are needed. :D

 

[mathdad]

I thought the | -x | = x.

Does | x | = x?

Are you saying that y = | x | - 2 means y = x - 2?

 

[MarkFL]

I thought the | -x | = x.

Does | x | = x?

We can correctly state:

$$|x|=|-x|$$

But, we can only state:

$$|x|=x$$

if:

$$0\le x$$

Or we can state:

$$|x|=-x$$

if:

$$x<0$$

 

[mathdad]

How is | x | = | -x |?

- - - Updated - - -

Absolute value equations are tricky.

 

[MarkFL]

How is | x | = | -x |?

One way to show this is to think of the definition:

$$|x|\equiv\sqrt{x^2}$$

And so we have:

$$|-x|=\sqrt{(-x)^2}=\sqrt{x^2}=|x|$$

Another way to think of this is to view |x| as the distance on the number line between the origin (0) and x (the magnitude of x). Then, we observe that for all real x, -x is the same distance from the origin, even though it's on the opposite side of the origin.

 

[mathdad]

This will take more thinking on my part. Check out the other symmetry post. Right or wrong?

 

[MarkFL]

Consider the function:

$$f(x)=m|x-h|+k$$

This is much like the vertex form for a quadratic. The vertex is at:

$$(h,k)$$

The axis of symmetry is the vertical line:

$$x=h$$

If:

$$0<m$$

then the graph opens upwards, and the range is:

$$[k,\infty)$$

But if:

$$m<0$$

then graph opens downwards, and the range is:

$$(-\infty,k]$$

The magnitude of m will determine how "skinny" the graph is...the greater the magnitude, the skinnier the graph.

Here is a graph with sliders so that you can see the effect the parameters have:

[DESMOS=-10,10,-10,10]y=m\left|x-h\right|+k;k=0;h=0;m=1[/DESMOS]

 

[mathdad]

Your knowledge of math is cool.