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I'm using biconditional graphs for the lack of a better name, I don't know the name of it, if you know the real name, feel free to edit it.

Draw the set of points given by: $|x-3|\leq1$ and $|y-2|\leq5$

I have evaluated the values for which the conditions are true and I obtained $\{x:2\leq x\leq 4\}$ and $\{y:-3\leq y\leq 7 \}$, now I'm trying to plot as if it were a function $f(x):=|x-3|\leq1$, then I'm using the values of $y$ for making a correlation with $x$, for example:

$$\begin{eqnarray*} {f(2)}&:=&{|2-3|\leq1} \\ {}&:=&{1\leq 1 } \end{eqnarray*}\tag{1}$$

$$\begin{eqnarray*} {f(3)}&:=&{|3-3|\leq1} \\ {}&:=&{0\leq 1 } \end{eqnarray*}\tag{2}$$

$$\begin{eqnarray*} {f(4)}&:=&{|4-3|\leq1} \\ {}&:=&{1\leq 1 } \end{eqnarray*}\tag{3}$$

Thus $f(2)=1$, $f(3)=0$, $f(4)=1$, but I can't find a match, because there are not $0$'s nor $1$'s in the interval $2\leq x \leq 4$, is this correct?

The secondary hypothesis I have is that this kind of operation is made to plot planes, not lines, then I should just draw a plane, is that it?

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Yes, you should draw and shade in the rectangle in the Cartesian plane whose vertices are $(2,-3),(2,7),(4,-3),(4,7)$. – Adriano Jun 6 '13 at 19:29
up vote 1 down vote accepted

Note that you have two inequalities: solving one gives you the values of $x$ defined by the first inequality; solving the other gives you the values of $y$ defined by the second inequality. Neither inequality is a function, and so $x$ is certainly not a function of $y$, nor is $y$ a function of $x$.

Try graphing the lines $y = -3$ and $y = 7$: those will be the horizontal bounds of the region defined by your inequalities. Then graph the the vertical lines $x = 2, x = 4$: giving you the bounded region of interest: the rectangular region with vertices: $$(2,-3),(2,7),(4,-3),(4,7)$$

Then simply "shade" the region bounded by these lines, inclusive of the lines.

From Wolfram Alpha (though note the scale for the x, y axis are not matched. If you were to graph this to scale, you'd have a rectangle that is much "taller" than it is wide):

enter image description here

ADDED: Nicer graph of the inequalities (in that it is properly scaled):

enter image description here

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Thanks, AmWhy. $$$$$$$$ – Voyska Jun 6 '13 at 22:02
You're welcome, Gustavo! – amWhy Jun 6 '13 at 22:08
Amy, can you help me with one more thing? There's also an exercise with or instead of and. I'm thinking that it's plot is like a big plus sign. Is this correct? – Voyska Jun 11 '13 at 6:34
Think of "and" as the intersection of all those regions (territory in common), and "or" as the union of all defined regions (all territory covered by one or more inequality). – amWhy Jun 11 '13 at 17:32
Yeah. I thought about the truth tables, and would be the intersection, yes because $1\wedge 1=1$. I guess I've figured the answer, was asking just because Wolfram Alpha didn't show me the result. – Voyska Jun 11 '13 at 17:40

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