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HELP c3 MODULUS GRAPH PLOTTING Watch

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    Hi, normally I can plot modulus graphs but I'm having some problems with this one. f(x) = 3 - l2x-1l

    Its the 3 in front thats confusing me and Id really appreciate if someone could give me an easy method to plod graphs such as these - it asks for where the graph crosses the coordinate axes. Any help is greatly appreciated thank you.
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    (Original post by MrToodles4)
    Hi, normally I can plot modulus graphs but I'm having some problems with this one. f(x) = 3 - l2x-1l

    Its the 3 in front thats confusing me and Id really appreciate if someone could give me an easy method to plod graphs such as these - it asks for where the graph crosses the coordinate axes. Any help is greatly appreciated thank you.
    Start with y = 2x - 1. Use this to draw y = | 2x - 1 |. Then think about the transformation required to change this to y = - | 2x - 1 |. Finally, consider the 3, but it will be easier to think of the equation you are working on as y = - | 2x - 1 | + 3, which I am sure you will agree is the same thing.
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    (Original post by MrToodles4)
    Hi, normally I can plot modulus graphs but I'm having some problems with this one. f(x) = 3 - l2x-1l

    Its the 3 in front thats confusing me and Id really appreciate if someone could give me an easy method to plod graphs such as these - it asks for where the graph crosses the coordinate axes. Any help is greatly appreciated thank you.
    Can you draw |2x-1|? Can you hence draw -|2x-1|? So drawing -|2x-1|+3 should be simple as it's just a translation up by 3 units from the previous graph.
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    The +3 means it moves up by 3 (do this translation last). Do you have a graphical calculator??? I always just use mine to help me draw graphs.
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    (Original post by U-GradeStudent)
    The +3 means it moves up by 3 (do this translation last). Do you have a graphical calculator??? I always just use mine to help me draw graphs.
    https://www.desmos.com/calculator
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    (Original post by Pangol)
    Start with y = 2x - 1. Use this to draw y = | 2x - 1 |. Then think about the transformation required to change this to y = - | 2x - 1 |. Finally, consider the 3, but it will be easier to think of the equation you are working on as y = - | 2x - 1 | + 3, which I am sure you will agree is the same thing.
    Thank you so much. So I've drawn the graph and gotten the y intercept to be (0,2) and I did 3-l2x-1l=0 to find x=2 (was this method correct??) But for the second x value I get -2 instead of the correct answer of -1... I do 3--l2x-1l - Im not sure if Im using the right method.
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    (Original post by MrToodles4)
    Thank you so much. So I've drawn the graph and gotten the y intercept to be (0,2) and I did 3-l2x-1l=0 to find x=2 (was this method correct??) But for the second x value I get -2 instead of the correct answer of -1... I do 3--l2x-1l - Im not sure if Im using the right method.
    You do need to solve 3 - l2x - 1l = 0. If you rearrange this to l2x - 1l = 3, it should be a bit easier to solve,
 
 
 
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