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Mimi_x3Best ResponseYou've already chosen the best response.2
Have you differentiated it?
 one year ago

margejeanBest ResponseYou've already chosen the best response.0
No I have not. How do you do that?
 one year ago

margejeanBest ResponseYou've already chosen the best response.0
No I havent taken that class yet. This is AP Stats.
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
Statistics..? This is Calculus. Or is there an algebra method to solve this?
 one year ago

margejeanBest ResponseYou've already chosen the best response.0
This class is AP Stats but U guess it is just basic math review. (Algebra)
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
Hm, I don't know the algebra method.. maybe @lgbasallote might know
 one year ago

lgbasalloteBest ResponseYou've already chosen the best response.0
i dont know algebra sorry
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
when the first derivative is zero, there will be a minimum or maximum
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
is the algebra method got to do something with the vertex (b)/2a?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
yes Mimi_x3, that formula will give us the xcoordinate of the vertex,
 one year ago

margejeanBest ResponseYou've already chosen the best response.0
um i am not sure, but i think if I use (b)/2a the answer would be 5 or 2.5?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
the value is the f(x), so sub x=b/2a into f f(b/2a)= to get the value of f(x) that the minimum
 one year ago

Mimi_x3Best ResponseYou've already chosen the best response.2
Vertex: \[\large \left(\frac{b}{2a},a\left(\frac{b}{2a}\right)^{2}+b\left(\frac{b}{2a}\right)+C\right) \] a> 0 => its a min a<0 => its a max. however, the calculus method is more nicer :)
 one year ago

margejeanBest ResponseYou've already chosen the best response.0
I got 2.5.i hope its right. thank you for helping.
 one year ago
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