# Graph of functions

I am trying to envision the graphs of the following equations but for some reason or the other I am not able to figure a systematic way of graphing them. I need the most basic procedure to achieve their graphs. The equations are:

1. $|y|=x$
2. $y^2=x^2$
3. $|x|+|y|=1$
4. $|x+y|=1$

If am not mistaken, non of them are graphs of functions.

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wolframalpha.com –  akkkk Dec 22 '12 at 16:56
@akkkk Interesting as it is, I am looking for more explanation and of course the graphs. Am looking to understand in depth the graphs of the functions. –  azetina Dec 22 '12 at 17:03
That's not what you asked for. –  akkkk Dec 22 '12 at 17:05
@akkkk My apologies! –  azetina Dec 22 '12 at 17:13

$y^2 = x^2$

So, for example, for any given x: $(x, x), (-x, -x), (-x, x), (x, -x)$ together represent all solutions to the equation above: each representing a line originating at the origin, and extending orthogonally, one in each quadrant.

But this can be represented by the lines:

• $y = x\quad$ $(\iff -y = -x);\;$ slope = $\;1$
• $y = -x\;$ $(\iff -y = \;\;x);\;$ slope = $-1$

$(1)$ The graph of the above is then two perpendicular lines intersecting at the origin, one with slope $1$, the other with slope $-1$:

$\quad\quad\quad$

• Note that the graph of $y^2 = x^2$ is also the graph of the equation $|y| = |x|$.
• The graph of $y = |x|$ is the portion of the graph $(1)$ at and above the x-axis;
• the graph of $|y| = x$ is the portion of the graph $(1)$ at and to the right of the y-axis.

$(2)$ The graph of $|x| + |y| = 1$ is the graph of $y = |x| - 1$ and $y = 1 - |x|$

$\quad\quad\quad$

$(3)$ The graph of $|x + y| = 1$ is two parallel lines, each of slope $-1$:

$\quad\quad\quad$

This is the graph of $y = 1 - x$, $y = -x - 1$. Note that the segments of the graph in the first and third quadrant are the same as the segments in the respective quadrants of the graph of $|x| + |y| = 1.$

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Symmetries are a good start: If $f(x)=f(-x)$, then it's even. If $f(y)=f(-y)$, then it has symmetry across the horizontal axis. If $f(x,y)=f(y,x)$ (swap $x$ and $y$), then it has symmetry across the line $y=x$, etc. For example, the first one is

$|y|=x$

Swap these around, and you get

$y=|x|$

which you should be familiar with. Since this was done by interchanging $x$ and $y$, the first one should be the graph of $|x|$ transformed that way. i.e, a v-shape opening to the right.

Another way, good for absolute values, is to break it up into cases. For $y>0, |y|=y$, so we have, as one part:

$$y>0, y=x$$

When $y<0,|y|=-y$ (i.e, $|-1|=-(-1)=1$), so the other part is

$$y<0, y=-x$$

So sticking these two graphs together in their respective domains (first one on top since it has $y>0$, and the second underneath because it has $y<0$, you get the result. You are correct in saying that none of these are functions, but you can still graph them by looking at pieces that would be functions on their own.

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I think wanting a "systematic way" that applies uniformly to all those equations is a mistake. First, such a systematic procedure would be very complex (I suspect it would end merely being a collection of rules of thumb for different special cases anyway). Second, you will learn much less from carrying out a by-the-book procedure here than you will from understanding each equation on its own terms.

1. Here $x$ is clearly a function of $y$. Draw it as that, then flip axes if you need to afterwards.

2. The important fact is to know that $x^2=y^2$ exactly when $x=y$ or $x=-y$. Draw each of those to cases separately and superpose the graphs.

3. It's easy enough to draw in the quadrant where $x$ and $y$ are both nonnegative. The other quadrants are mirror images of that (why?).

4. $|f(x,y)|=1$ exactly if $f(x,y)=1$ or $f(x,y)=-1$. As in (2), draw these cases separately.

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Interesting indeed. Serves a lot. –  azetina Dec 22 '12 at 17:21