# Thread: Absolute value and hyperbolas?

1. ## Absolute value and hyperbolas?

how do I graph these?
i have 2...

1) |x-y|+|x+y|+|x|+|y|=12
I tried graphing this but did not end up with the graph depicted in the answers....could anyone help me out with how to solve/graph this?

2) 1/|x+2|
What I did:
x can't equal -2, y can't equal 0
x > -2; y=1/x+2 ( i graphed this line)
x<-2; y=1/-x-2 (i graphed this but it fell below the x-axis)

the second graph (1/-x-2) is supposed to be flipped up
I understand why (its absolute value so everything is positive) but when i graph it it doesn't end up that way....

how to i solve/graph it properly? is there something wrong with my method?

thanks!!

2. Originally Posted by sorkii
how do I graph these?
i have 2...
1) |x-y|+|x+y|+|x|+|y|=12
I tried graphing this but did not end up with the graph depicted in the answers....could anyone help me out with how to solve/graph this?

2) 1/|x+2|
Here is a web tool.

Another one too.

3. i understand what they look like, but when im in an exam i can't just use a web tool...

i want to know how to graph them, like algebraically, but not the conventional substitution...
these graphs require me to do something with what will happen when each component is greater than or below its critical value (where it makes the term equal 0) and a few other things

i just want to know how to mathematically graph these using absolute value

thanks for the tools, but i already have microsoft maths to do the graphing for me~

4. Originally Posted by sorkii
i understand what they look like
i want to know how to graph them, like algebraically, but not the conventional substitution...
Knowing what a graph 'looks like' means that you can use the plot to find out how the algebra works.
That is the whole point of using a plotting program.
You reverse engineer the plot.

5. Divide an concur!

|x| = x for x > 0
|x| = -x for x < 0

Graph separately.

|x-y| = x-y for x > y
|x-y| = y-x for y > x

Graph separately

|x+y| = x+y for x+y > 0 ==> y > -x
|x+y| = -(x+y) for x+y < 0 ==> y < -x

Graph separately

Get a handle on the decomposition. Really, there is no magic trick. It is experience and exploration.

6. ## Re: Absolute value and hyperbolas?

Use symmetry:
Replace y with ‒y. You get the same equation. → The graph is symmetric WRT the x-axis.
Replace x with ‒x. You get the same equation. → The graph is symmetric WRT the y-axis.
So the graph is also symmetric WRT the origin.
You only need to graph this in the first quadrant.

Now look at TKHunny's post. In the first quadrant:
x≥0, so |x|=x .
y≥0, so |y|=y .
Also, y≥-x, so |x+y|=x+y .

So, we've got |x‒y|+x+y+x+y=12.

Now, for y≥x, you have |x‒y|=y‒x.

For x≥y, you have |x‒y|=x‒y.