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The function I have is $$|x-1|+|y+2|\leq2$$

So I tried graphing it the same way I would $|x|+|y|\leq1$

I know that $|x-1|$ is $x-1$ when $x>0$ and $-x+1$ when $x<0$

Similarly, I know that $|y+2|$ is $y+2$ when $y>0$ and $-y-2$ when $y<0$

So, I have four cases:

I. $x>0,y>0$ (Quadrant I)

II. $x<0,y>0$ (Quadrant II)

III. $x<0,y<0$ (Quadrant III)

IV. $x>0,y<0$ (Quadrant IV)

So for Quadrant I, we need to solve $x-1+y+2\leq2$ and get $y\leq-x+1$

Quadrant II, we solve $-x+1+y+2\leq2$ and get $y\leq x-1$

Quadrant III, we solve $-x+1-y-2\leq2$ and get $y\geq -x-3$

Quadrant IV, we solve $x-1-y-2\leq2$ and get $y\geq x-5$

When I graph these lines in their respective quadrants, I do not get the correct answer. Can someone please tell me where I went wrong in my method?

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    $\begingroup$ "I know that $|x - 1| = x - 1$ when $x > 0$..." This is wrong. It should be $x - 1> 0$. You just need to remember the one inside the absolute value is positive, rather than $x$ itself. $\endgroup$
    – Azlif
    Sep 6, 2021 at 2:15
  • $\begingroup$ A better way to do this would be to plot $|x|+|y|\le2$ first and then shifting the graph along $x$ and $y$ by $1$ and $-2$ units respectively $\endgroup$
    – DatBoi
    Sep 6, 2021 at 2:49
  • $\begingroup$ A more intuitive way to see this might be to note that |𝑥 − 1| is the distance from the line 𝑥 = 1, and |𝑦 + 2| is the distance from the line 𝑦 = -2.  So you want points where the sum of those distances is ≤ 2.  (This will be a diamond, centred on the point where those lines cross.) $\endgroup$
    – gidds
    Sep 6, 2021 at 11:00

3 Answers 3

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You know that $|x|\geq 0\iff x\geq0$

Replace $x$ with $x-1$ and you get $|x-1|\geq 0\iff x-1\geq0$ which is equivalent to

$|x-1|\geq 0\iff x\geq1$

So you will need to make $4$ cases-

$x\leq1, y\geq2$

$x\leq1, y\leq2$

$x\geq1, y\leq2$

$x\geq1, y\geq2$

If you know simple graph transformation methods then from simple equation , you can reach this equation

$|x|+|y|\leq2$

Replace $x$ with $x-1$ and shift whole graph $1$ units right

Now you got graph of $|x-1|+|y|\leq2$

Now replace $y$ with $y+2$ and shift graph $2$ units down and you will get graph of $|x-1|+|y+2|\leq 2$

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  • $\begingroup$ thank you. your method of just doing $|x|+|y|\leq2$ and shifting it 1 unit to the right and 2 units down was much easier $\endgroup$ Sep 6, 2021 at 4:30
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    $\begingroup$ @user8358234 If my answer helped you , please accept it :-) $\endgroup$ Sep 6, 2021 at 4:32
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Hint: Let $L_1, L_2$ be two non parallel lines lines, then $$a|L_1|+b|L_2|=c$$ represents a parallelogram with diagonals as $L_1=0=L_2$. If $L_1,L_2$ are perpendicular then it is asquare or rhombus, otherwise rectangle/parallelogram. With sises as $$aL_1+bL_2=c, -aL_!+bL_2=c, aL_1-bL_2=c,-aL_1+bL_2=c.$$ If adjacent sides are perpendicular then it is square or rectangle. So in this case we have a square centered at $(1,-2)$.

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$|x-1|=x-1$ when $x \geqq 1$ and not 0. Same is for $y$. You may check this by putting $x=\frac{1}{2}$. You get $-0.5$ from your analysis but it should have been $+0.5$.

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