Mathematics
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OpenStudy (anonymous):
help again ...
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OpenStudy (anonymous):
I'm here for ya
OpenStudy (anonymous):
So you're in calculus right?
OpenStudy (anonymous):
yep
OpenStudy (anonymous):
Derive the function
OpenStudy (anonymous):
3x^2-12x
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OpenStudy (anonymous):
-3x^2 -12x
OpenStudy (anonymous):
But from here on out take out a common factor of -3x
OpenStudy (anonymous):
So it would be -3x(x+4)
OpenStudy (anonymous):
would u make it -3x^@(x+4)
OpenStudy (anonymous):
3x^2(x+4)
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OpenStudy (anonymous):
No no just 3x because you can't pull out another x from 12x
OpenStudy (anonymous):
o ok
OpenStudy (anonymous):
would you put it to zero ?
OpenStudy (anonymous):
Equate it to 0, yes you would
OpenStudy (anonymous):
So you have x+4=0
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OpenStudy (anonymous):
x=-4
OpenStudy (anonymous):
so it would be decreasing
OpenStudy (anonymous):
No, not exactly. X=-4 is the critical point where the function is not decreasing or increasing
OpenStudy (anonymous):
so how know if the graphg is increasing or decreasing
OpenStudy (anonymous):
graph
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OpenStudy (anonymous):
So if you graph it and you look at the graph of the derivative which is x+4
OpenStudy (anonymous):
On which interval is x+4 above the x axis or positive is where the function is increasing
OpenStudy (anonymous):
when i graphed it i was negative
OpenStudy (anonymous):
i think im doing it wrong
OpenStudy (anonymous):
You graphed x+4?
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OpenStudy (anonymous):
yeah i graohed it but now i think it increasing
OpenStudy (anonymous):
graphed
OpenStudy (anonymous):
|dw:1429149673747:dw|
If you look at the graph above you see that the graph is positive for every value above x=-4