modulus

sketch y = | x - 1 | + |x + 2|

i got the lines y = 2x - 3, y = -2x +3 , y=3, y = -3 but the book doesn't draw y= -3 and why does the graph look like a squared U shape wouldn't the line y =3 continue?

(edited 6 years ago)

Scroll to see replies

Reply 1

RDKGames

Study Forum Helper

No, also I'm not sure where you get your

y= \pm 1

from.

That graph has two critical points;

x=-2

and

x=1

For

x<-2

we have

|x-1| = 1-x

and

|x+2|=-x-2

, so for that region we have

y=-1-2x

For

-2<x<1

we have

|x-1| = 1-x

and

|x+2| = x+2

, so for that region we have

y=3

For

x>1

we have

|x-1|=x-1

and

|x+2|=x+2

, so for that region we have

y=2x+1

Hence the 'square' U shape.

Reply 2

man111111

Original post by RDKGames

No, also I'm not sure where you get your

y= \pm 1

from.

That graph has two critical points;

x=-2

and

x=1

For

x<-2

we have

|x-1| = 1-x

and

|x+2|=-x-2

, so for that region we have

y=-1-2x

For

-2<x<1

we have

|x-1| = 1-x

and

|x+2| = x+2

, so for that region we have

y=3

For

x>1

we have

|x-1|=x-1

and

|x+2|=x+2

, so for that region we have

y=2x+1

Hence the 'square' U shape.

how did you work out the critical points?

Reply 3

RDKGames

Study Forum Helper

Original post by man111111

how did you work out the critical points?

Look at the equation - they are obvious. These are the points where x-1=0 or x+2=0. They’re important because on either side of them, the modulus of the corresponding function can be written without the mod sign.

Reply 4

man111111

Original post by RDKGames

No, also I'm not sure where you get your

y= \pm 1

from.

That graph has two critical points;

x=-2

and

x=1

For

x<-2

we have

|x-1| = 1-x

and

|x+2|=-x-2

, so for that region we have

y=-1-2x

For

-2<x<1

we have

|x-1| = 1-x

and

|x+2| = x+2

, so for that region we have

y=3

For

x>1

we have

|x-1|=x-1

and

|x+2|=x+2

, so for that region we have

y=2x+1

Hence the 'square' U shape.

Do you mind explaining this to me without any short cuts because i still don't understand?

Reply 5

RuthieG101

draw the lines on a separate graph without the modulus.
when you draw it with the modulus anything that is y less than zero is now reflected in y=0. this makes all the negative values positive.

e.g. if you had the point (3,-2) the modulus of this point is (3,2).

this is how I understand it.
You could also put the equation into a graphical calculator or software. put it with and without the modulus and it looks like a reflection in the line y=0

Reply 6

DFranklin

Original post by man111111

Do you mind explaining this to me without any short cuts because i still don't understand?

I would say RDK has already removed all the short cuts.

What is the *first* thing you don't understand in what he's written?

[META: in general, if someone posts a significant chunk of mathematics, it's really unhelpful to just say "I don't understand". Explain what's confusing you, or at least, *exactly* what part of what they posted you don't understand].

Reply 7

man111111

Original post by DFranklin

I do appreciate RDKs help but i've never been taught this method before
when RDK says For

we have

and

. i don't understand why you multiple it with -1 when x is less than -2

Reply 8

man111111

Original post by RDKGames

No, also I'm not sure where you get your

y= \pm 1

from.

That graph has two critical points;

x=-2

and

x=1

For

x<-2

we have

|x-1| = 1-x

and

|x+2|=-x-2

, so for that region we have

y=-1-2x

For

-2<x<1

we have

|x-1| = 1-x

and

|x+2| = x+2

, so for that region we have

y=3

For

x>1

we have

|x-1|=x-1

and

|x+2|=x+2

, so for that region we have

y=2x+1

Hence the 'square' U shape.

y = | x - 1 | + |x + 2|

1) y= -(x-1)+ -(x +2)
2) y = (x-1)+ -(x+2)
3) y= -(x-1)+ (x+2)
4) y= (x-1)+ (x+2)

then i simplified

1) y = -2x-1
2) y= -3
3) y= 3
4) y = 2x+1

Reply 9

DFranklin

Original post by man111111

I do appreciate RDKs help but i've never been taught this method before
when RDK says For

we have

and

. i don't understand why you multiple it with -1 when x is less than -2

If x < -2 then certainly x-1 < 0 and so |x-1| = 1-x. (Because by the definition of modulus, |y| = -y when y < 0).

Similarly, if x < -2 then x+2 < 0 and so |x-2| = -(x+2).

The basic idea is to break the entire range (

-\infty < x < \infty

) into subregions where we know the sign of each term inside modulus doesn't change, so in each subregion we can get rid of the modulus signs.

[I wasn't implying you weren't grateful to RDK or anything - simply that he's already broken this down to steps smaller than I'd use actually answering the question in an exam. So you need to specify exactly where your problem is in understanding things (as you've now done)].

Reply 10

man111111

Original post by DFranklin

-\infty < x < \infty

thanks
For

we have

and

, so for that region we have

. why do you multiply

by -1 but not

Reply 11

DFranklin

Original post by man111111

thanks
For

we have

and

, so for that region we have

. why do you multiply

by -1 but not

|x+2|=-(x+2) only when x+2 < 0.

Reply 12

man111111

Original post by DFranklin

|x+2|=-(x+2) only when x+2 < 0.

thanks for your help

Reply 13

man111111

Original post by DFranklin

|x+2|=-(x+2) only when x+2 < 0.

how would i sketch y=|x|+|x-1|+|x-2|

Reply 14

man111111

how would i sketch y=|x|+|x-1|+|x-2|

Reply 15

man111111

Original post by RDKGames

No, also I'm not sure where you get your

y= \pm 1

from.

That graph has two critical points;

x=-2

and

x=1

For

x<-2

we have

|x-1| = 1-x

and

|x+2|=-x-2

, so for that region we have

y=-1-2x

For

-2<x<1

we have

|x-1| = 1-x

and

|x+2| = x+2

, so for that region we have

y=3

For

x>1

we have

|x-1|=x-1

and

|x+2|=x+2

, so for that region we have

y=2x+1

Hence the 'square' U shape.

why is it -2<x<1 but not -2 ≤ x ≤ 1

Reply 16

RDKGames

Study Forum Helper

Original post by man111111

how would i sketch y=|x|+|x-1|+|x-2|

Play the same game as above. Consider the following regions individually:

x<0

0<x<1

1<x<2

x>2

...and state for each one whether

|x|=\pm x

|x-1|= \pm(x-1)

and

|x-2|=\pm (x-2)

before adding them together to get the line to be sketched for the corresponding region. (this is exactly the same recipe as in my prev. answer to the other question)

Original post by man111111

why is it -2<x<1 but not -2 ≤ x ≤ 1

The boundary points are trivial and not that important for the analysis of the graph.

(edited 6 years ago)

Reply 17

man111111

Original post by RDKGames

Play the same game as above. Consider the following regions individually:

x<0

0<x<1

1<x<2

x>2

...and state for each one whether

|x|=\pm x

|x-1|= \pm(x-1)

and

|x-2|=\pm (x-2)

before adding them together to get the line to be sketched for the corresponding region. (this is exactly the same recipe as in my prev. answer to the other question)

i'm stuck with the inequalities 0<x<1 and 1<x<2 do you mind helping me out with one of them