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    y=x(x+1)(x-1)

    Would the valueS of x be X=-1 , X=1 ???
    Or is there more needed to be done as there is an "x" in front of the first bracket.??? Confused -.-
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    (Original post by Modesty)
    y=x(x+1)(x-1)

    Would the valueS of x be X=-1 , X=1 ???
    Or is there more needed to be done as there is an "x" in front of the first bracket.??? Confused -.-
    x also equals 0
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    (Original post by kf141998)
    x also equals 0
    What do you mean?

    What do i do with the "x" in front?
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    (Original post by Modesty)
    y=x(x+1)(x-1)

    Would the valueS of x be X=-1 , X=1 ???
    Or is there more needed to be done as there is an "x" in front of the first bracket.??? Confused -.-
    What do you mean by "values of x"?

    If you mean x-intercepts, then x=0, -1, 1
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    (Original post by ubisoft)
    What do you mean by "values of x"?

    If you mean x-intercepts, then x=0, -1, 1
    Sorry i meant where the graph crosses the x axis.
    Btw how do you know x=0 ?
    I know how to get the other intercepts but got confused of the x in front of the first bracket.
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    (Original post by Modesty)
    y=x(x+1)(x-1)

    Would the valueS of x be X=-1 , X=1 ???
    Or is there more needed to be done as there is an "x" in front of the first bracket.??? Confused -.-
    It means it's a cubic graph; in this case there a three solutions. You solved (x+1)(x-1)=0, you needed to solve x(x+1)(x-1)=0
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    (Original post by Modesty)
    Sorry i meant where the graph crosses the x axis.
    Btw how do you know x=0 ?
    I know how to get the other intercepts but got confused of the x in front of the first bracket.
    x on its own is the same as (x-0)
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    (Original post by Modesty)
    Sorry i meant where the graph crosses the x axis.
    Btw how do you know x=0 ?
    I know how to get the other intercepts but got confused of the x in front of the first bracket.
    Imagine it as y=(x+0)(x+1)(x-1)
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    (Original post by Andy98)
    It means it's a cubic graph; in this case there a three solutions. You solved (x+1)(x-1)=0, you needed to solve x(x+1)(x-1)=0
    I know they are cubic graphs , but thx for clearing that up for me !
    I understand now.
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    (Original post by Modesty)
    I know they are cubic graphs , but thx for clearing that up for me !
    I understand now.
    no problemo
 
 
 
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