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Maths help-functions

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
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
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.
(edited 7 years ago)
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
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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B_9710
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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.
Original post by B_9710
You have to be careful saying stuff like that round here.


Sorry - I completely forgot about that :frown:

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