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OpenStudy (anonymous):
i would but not good at math lol srry bud
OpenStudy (anonymous):
Lol
OpenStudy (e.cociuba):
Just post the question.
OpenStudy (anonymous):
Which is the graph of f(x) = |x 1| 3?
OpenStudy (e.cociuba):
Is that the absolute value of 1x?
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OpenStudy (anonymous):
Sorry, it is -1 and -3
OpenStudy (e.cociuba):
Okay, so f(x) = |x -1| -3, correct?
OpenStudy (anonymous):
Correct
OpenStudy (anonymous):
I hate this section of Algrebra II.
OpenStudy (anonymous):
Me too!
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OpenStudy (anonymous):
i got an A in this part
OpenStudy (anonymous):
I just failed a test on this :/
OpenStudy (anonymous):
I hope it gets easier. lol
OpenStudy (anonymous):
lol it deffinetly does @devinthedev
OpenStudy (anonymous):
Anything involving slope makes no sense to me for some reason.
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OpenStudy (e.cociuba):
|dw:1379606831398:dw|
OpenStudy (anonymous):
May I ask another one?
OpenStudy (anonymous):
no
OpenStudy (anonymous):
lol
OpenStudy (e.cociuba):
Does that graph make sense to you? Sure.
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OpenStudy (anonymous):
Yes it does.
OpenStudy (***[isuru]***):
yo,
hence this is an absolute value u have to something like this
Absolute value means there r only positive values
so u have to make the expression I x- 1 I positive even when the value of x is negative
To do that
u'll have to substitute
x - 1 when x >= +1
-(x - 1) when x< 1
now when x is >= +1
f(x) = x-1 -3
= x -4
when x is < 1
F(x) = -(x-1) -3
= -2 -x
You can no w plot these 2 graphs in one plane and it's the answer
OpenStudy (anonymous):
|dw:1379607167416:dw|
OpenStudy (anonymous):
Yes it does.
OpenStudy (e.cociuba):
|dw:1379607420198:dw| Where did you get that from?