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FoolForMathBest ResponseYou've already chosen the best response.1
For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at \( \large \left(\frac b {2a} , \frac {b^24ac}{4a}\right) \).Equation of line of symmetry will be at \(x=\frac b {2a} \) Maximum and minimum values will be at \( \large \left(\frac b {2a} , \frac {b^24ac}{4a}\right) \) and \( \large \left(\frac b {2a} , \frac {b^24ac}{4a}\right) \) for \(a<0\) and \(a>0\) respectively.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
@FoolForMath but that does not lead to the correct answer
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
i think there is no minimum value
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
it's unbounded
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
For this problem there will be no minimum values.
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
can we do this like : 45x4x^2 44(x^2+5x/4) 44(x^2+5x4 +25/64) + 25/16 ?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
is this the correct method ?
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
dw:1340428823754:dw
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
just maximum value
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
44(x^2+5x4 +25/64) + 25/16 89/164((x+5/8)^2) hence it as min. value of 89/16 at x =5/8 is it right ?
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
THIS FUNCTION HAS NO MINIMUM VALUE
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
http://www.wolframalpha.com/input/?i=45x4x%5E2&t=crmtb01 i am getting right according to this but it may be wrong please have a look
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Lets try to plot this function graphically,
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
ok try x = 10000
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
when do you think the minimum value will be achieved?
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
can u tell m not soo good
 one year ago

RaphaelFilgueirasBest ResponseYou've already chosen the best response.1
every x you get i will pick one,that that my function will be less than yours
 one year ago

FoolForMathBest ResponseYou've already chosen the best response.1
Yes, for every minimum you find I can find another x which gives a lesser value ...
 one year ago

mathsloverBest ResponseYou've already chosen the best response.0
thanks a lot every one
 one year ago
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