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    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?
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    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.
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    (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?
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    (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.
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    (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?
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    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
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    (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].
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    (Original post by DFranklin)
    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].
    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
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    (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
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    (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 &lt; x &lt; \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)].
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    (Original post by DFranklin)
    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 &lt; x &lt; \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)].
    thanks
    For we have and , so for that region we have . why do you multiply by -1 but not
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    (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.
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    (Original post by DFranklin)
    |x+2|=-(x+2) only when x+2 < 0.
    thanks for your help
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    (Original post by DFranklin)
    |x+2|=-(x+2) only when x+2 < 0.
    how would i sketch y=|x|+|x-1|+|x-2|
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    how would i sketch y=|x|+|x-1|+|x-2|
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    (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&lt;-2 we have |x-1| = 1-x and |x+2|=-x-2, so for that region we have y=-1-2x

    For -2&lt;x&lt;1 we have |x-1| = 1-x and |x+2| = x+2, so for that region we have y=3

    For x&gt;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
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    (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&lt;0

    0&lt;x&lt;1

    1&lt;x&lt;2

    x&gt;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.
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    [QUOTE=RDKGames;76072856]Play the same game as above. Consider the following regions individually:

    x&lt;0

    0&lt;x&lt;1

    1&lt;x&lt;2

    x&gt;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
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    (Original post by man111111)
    i'm stuck with the inequalities 0<x<1 and 1<x<2 do you mind helping me out with one of them
    For 0&lt;x&lt;1 we have:

    |x| = x
    |x-1| = 1-x
    |x-2| = 2-x
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    (Original post by RDKGames)
    For 0&lt;x&lt;1 we have:

    |x| = x
    |x-1| = 1-x
    |x-2| = 2-x
    why do you multiply x-2 with -1
 
 
 

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