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    For f(x)=x^2. domain= -3<x<3.
    The range is 0 greater than or equal to f(x) <9
    Why is the range greater than or equal to 0. I get why it is less than 9 part of the range
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    (Original post by geography1294)
    For f(x)=x^2. domain= -3<x<3.
    The range is 0 greater than or equal to f(x) <9
    Why is the range greater than or equal to 0. I get why it is less than 9 part of the range
    Because y=0 is as low that parabola will go. It touches the x-axis at x=0
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    (Original post by geography1294)
    For f(x)=x^2. domain= -3<x<3.
    The range is 0 greater than or equal to f(x) <9
    Why is the range greater than or equal to 0. I get why it is less than 9 part of the range
    Any number squared, regardless of whether it is positive or negative, will be positive; therefore the lowest value of f(x) is 0 squared, which is 0, hence the function's range - as any non-zero number squared will be bigger than 0.
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    (Original post by RDKGames)
    Because y=0 is as low that parabola will go. It touches the x-axis at x=0
    Oh yeah! sorry for the silly question oops
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    (Original post by some-student)
    Any number squared, regardless of whether it is positive or negative, will be positive; therefore the lowest value of f(x) is 0 squared, which is 0, hence the function's range - as any non-zero number squared will be bigger than 0.
    Thank you!
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    (Original post by some-student)
    Any REAL number squared, regardless of whether it is positive or negative, will be positive; therefore the lowest value of f(x) is 0 squared, which is 0, hence the function's range - as any non-zero REAL number squared will be bigger than 0.
    You have to be careful saying stuff like that round here.
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    (Original post by B_9710)
    You have to be careful saying stuff like that round here.
    Sorry - I completely forgot about that
 
 
 
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