# 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?

• "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. Sep 6, 2021 at 2:15
• 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 Sep 6, 2021 at 2:49
• 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.) Sep 6, 2021 at 11:00

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 $$|x-1|+|y+2|\leq 2$$

• 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 Sep 6, 2021 at 4:30
• @user8358234 If my answer helped you , please accept it :-) Sep 6, 2021 at 4:32

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 $$(1,-2)$$.

$$|x-1|=x-1$$ 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 $$+0.5$$.