1. Absolute Value Graphs

Hi

i) abs(x) + abs(y) = 1

ii) abs(x+y) = 1

iii) abs(x) - abs(y) = 1

iv) abs(x-y) = 1

v) 2*abs(x) + abs(y) = 1

Could someone please show me how to sketch these absolute value graphs? I've found examples on the easier versions in textbooks but they don't cover these types of graphs. (I'm just trying to learn next year's stuff so I don't fall behind.)

Thanx very much

2. You may try consider different range of y

E.g. $\displaystyle |x|+|y|=1$

Consifer $\displaystyle y\ge 0$
$\displaystyle |x|+y=1$
$\displaystyle y=1-|x|$
So you draw $\displaystyle y=1-|x|$ for $\displaystyle y\ge 0$ (above x-axis)

Consider $\displaystyle y\le 0$
$\displaystyle |x|-y=1$
$\displaystyle y=|x|-1$
So you draw $\displaystyle y=|x|-1$ for $\displaystyle y\le 0$ (below x-axis)

The graph will be a square

3. Absolute value graphs

Hello xwrathbringerx
Originally Posted by xwrathbringerx
Hi

i) abs(x) + abs(y) = 1

ii) abs(x+y) = 1

iii) abs(x) - abs(y) = 1

iv) abs(x-y) = 1

v) 2*abs(x) + abs(y) = 1
I don't think it's helpful to try to find hard and fast rules when you're sketching any graphs - there are so many variations that it's impossible to try to categorise them all. A fair bit of trial and error and a lot of practice is really essential.

BUT here are a few tips when you're sketching absolute value graphs:

Tip #1

abs(x) = x if x > 0 and abs(x) = -x if x < 0 (and of course abs(x) = 0 if x = 0).

What does this tells you about the graph if you have a function of the form y = f(abs(x))? (In other words, the formula to give the value of y is in terms of abs(x) only.)

Well, it means that if you know the value of y for any positive value of x, you'll get the same value of y for the corresponding negative value of x. What this tells you is that the graph of a function like this is symmetrical about the y-axis.

Tip #2

For equations of the form abs(y) = f(x), for some function of x (which may contain abs(x) or it may not):

• Whenever f(x) has a positive value, y will have two values: +f(x) and -f(x).
• Whenever f(x) has a negative value, y will have no value (because abs(y) can't be negative.)

So the graph of functions like this will be symmetrical about the x-axis.

Tip #3

If you really can't see where to start, try to set up a table with a few simple values of x: 3, 2, 1, 0, -1, -2, -3 (or whatever), and work out corresponding values of y. This should give you some ideas as to what the graph will look like.

Let's take one or two of your questions, and see what happens when we try these things out.

(i) abs(x) + abs(y) = 1

Re-arrange: abs(y) = 1 - abs(x)

So this is equivalent to:

abs(y) = 1 - x, if x > 0 and abs(y) = 1 + x if x < 0.

Take x > 0 first. This gives us y = 1 - x. Now, if x < 1, (1 - x) is positive so abs(y) is OK; we shall have two values of y (one positive and one negative) for each value of x.

But if x > 1, (1 - x) will be negative, so we shan't get any values of y (because, as we said in Tip #2, abs(y) can't be negative). So there's nothing on the graph to the right of x = 1.

Then look at x < 0. This gives us abs(y) = 1 + x. This gives a similar situation that we had before: if 0 > x > -1, abs(y) is positive and we get two values of y; if x< -1, there are no values of y. So there's nothing on the graph to the left of x = -1.

Can you see what the graph looks like? It will be a square joining the points (-1, 0), (0, 1), (1, 0) and (0, -1). (Notice it's symmetrical about both axes.)

(ii) abs(x + y) = 1

This is easier than (i). If abs(x + y) = 1, then (x + y) is either +1 or -1. So we simply get two parallel straight line graphs:

x + y = 1 and x + y = -1

(iii) Re-arrange to read abs(y) = abs(x) - 1. Now bear in mind that abs(y) can't be negative, and that for every positive value of x that works, there'll be a corresponding negative value of x that gives the same value of abs(y). Because the equation contains only abs(x) and abs(y) it will be symmetrical about both axes.

See if you can come up with the sketch. (If you get stuck, try Tip #3.)

(iv) Similar to (ii)

(v) Re-arrange: abs(y) = 1 - 2*abs(x).

No values of y if 1 - 2*abs(x) is negative.

If x > 0, what then? If x < 0, what then? Can you do it now?