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anonymous
 5 years ago
can someone help me find the critical points for: .288x2 +3.47x +5.46?
anonymous
 5 years ago
can someone help me find the critical points for: .288x2 +3.47x +5.46?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0f' = .288(2)x +3.47 solve for f'=0 x = 3.47/(.288(2))

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0actually its even simpler than that, only one critical point exists because this graph looks like a "U". The critical point is at the bottom part of th "U"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its not multiplied by 2 thats supposed to be a power...sorry typo

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.03.47/[2(.288)] comes out the same, but you aint gotta try to derive it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0right, I assumed that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is a critical point a coordinate?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.06.024 is pretty close, it is the "x" coordinate; just plug it into the equation to get the "y" coordinate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the x coordinate is 6, so I plug this in to the original equation?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(6.024,36.81) is what I get for a coordinate if I did it right :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah.... 6... which means I plugged it in wring ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wait can you tell me how you did that?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to find a derivative?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how you find the coordinate? and is that coordinate the critical point?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0to find which value of x gets us a critical point, we can find the derivative of an equation if you do not know what the graph of the equation will look like

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y =.288x^2 +3.47x + 5.46 the derivative is gotten thru using the rules for derivatives, mainly the power rule for this equation.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the power rule states: when ever you have a variable to the power of a number like: x^4 you can multiply the power by whatever number is in front of the "x" and then reduce the power by 1 (4) x ^ 41 = 4x^3 does that make sense?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0y =.288x^2 +3.47x + 5.46 y' = (2).288x^1 + (1)3.47x^0 y' = .165888 x + 3.47 when y' = 0 we have found a critical point for "x". so we set y' = 0 0 = .165888x + 3.47 solve for "x" 3.47/.165888 = x this is the value of x that we need to use in the original equation; the y = .288x^2 +3.47x +5.46

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and I messed up that 2(.288).... redo that will ya :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = 3.47/.576 which is about 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok thanks! could the critical point and the minimum point be the same?
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