Need help graphing an absolute value function 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?
 A: 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 $|x-1|+|y+2|\leq 2$
A: 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 $(1,-2)$.
A: $|x-1|=x-1$ 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 $+0.5$.
