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Mathematics 21 Online
OpenStudy (anonymous):

help again ...

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

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)

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

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

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?

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

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