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find the minimum value of \( 4-5x-4x^2\).

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For any parabola of the form \(ax^2+bx+c=0 \) the vertex will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \).Equation of line of symmetry will be at \(x=-\frac b {2a} \) Maximum and minimum values will be at \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) and \( \large \left(-\frac b {2a} , -\frac {b^2-4ac}{4a}\right) \) for \(a<0\) and \(a>0\) respectively.
@FoolForMath but that does not lead to the correct answer
i think there is no minimum value

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Other answers:

it's unbounded
For this problem there will be no minimum values.
can we do this like : 4-5x-4x^2 4-4(x^2+5x/4) 4-4(x^2+5x4 +25/64) + 25/16 ?
is this the correct method ?
just maximum value
4-4(x^2+5x4 +25/64) + 25/16 89/16-4((x+5/8)^2) hence it as min. value of 89/16 at x =-5/8 is it right ?
why ?
THIS FUNCTION HAS NO MINIMUM VALUE i am getting right according to this but it may be wrong please have a look
Lets try to plot this function graphically,
1 Attachment
ok try x = 10000
when do you think the minimum value will be achieved?
can u tell m not soo good
every x you get i will pick one,that that my function will be less than yours
Yes, for every minimum you find I can find another x which gives a lesser value ...
thanks a lot every one

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