anonymous
  • anonymous
can someone help me find the critical points for: .288x2 +3.47x +5.46?
Mathematics
katieb
  • katieb
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amistre64
  • amistre64
f' = .288(2)x +3.47 solve for f'=0 x = -3.47/(.288(2))
amistre64
  • amistre64
actually 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
  • anonymous
its not multiplied by 2 thats supposed to be a power...sorry typo

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amistre64
  • amistre64
-3.47/[2(.288)] comes out the same, but you aint gotta try to derive it :)
amistre64
  • amistre64
right, I assumed that
anonymous
  • anonymous
is a critical point a coordinate?
amistre64
  • amistre64
6.024 is pretty close, it is the "x" coordinate; just plug it into the equation to get the "y" coordinate.
anonymous
  • anonymous
the x coordinate is -6, so I plug this in to the original equation?
amistre64
  • amistre64
(6.024,36.81) is what I get for a coordinate if I did it right :)
amistre64
  • amistre64
yeah.... -6... which means I plugged it in wring ...
anonymous
  • anonymous
wait can you tell me how you did that?
amistre64
  • amistre64
which part?
amistre64
  • amistre64
do you know how to find a derivative?
anonymous
  • anonymous
how you find the coordinate? and is that coordinate the critical point?
amistre64
  • amistre64
to 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
  • amistre64
y =.288x^2 +3.47x + 5.46 the derivative is gotten thru using the rules for derivatives, mainly the power rule for this equation.
amistre64
  • amistre64
the 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 ^ 4-1 = 4x^3 does that make sense?
anonymous
  • anonymous
yes...then what?
amistre64
  • amistre64
y =.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
  • amistre64
and I messed up that 2(.288).... redo that will ya :)
amistre64
  • amistre64
x = -3.47/.576 which is about -6
anonymous
  • anonymous
ok thanks! could the critical point and the minimum point be the same?

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