Question

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

Answer 1

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.

Answer 2

@FoolForMath but that does not lead to the correct answer

Answer 3

i think there is no minimum value

Answer 4

it's unbounded

Answer 5

For this problem there will be no minimum values.

Answer 6

can we do this like :
4-5x-4x^2
4-4(x^2+5x/4)
4-4(x^2+5x4 +25/64) + 25/16 ?

Answer 7

is this the correct method ?

Answer 8

just maximum value

Answer 9

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 ?

Answer 10

nope

Answer 11

why ?

Answer 12

THIS FUNCTION HAS NO MINIMUM VALUE

Answer 13

http://www.wolframalpha.com/input/?i=4-5x-4x%5E2&t=crmtb01
i am getting right according to this but it may be wrong please have a look

Answer 14

Lets try to plot this function graphically,

Answer 15

k

Answer 16

ok try x = 10000

Answer 17

when do you think the minimum value will be achieved?

Answer 18

why?

Answer 19

can u tell m not soo good

Answer 20

every x you get i will pick one,that that my function will be less than yours

Answer 21

Yes, for every minimum you find I can find another x which gives a lesser value ...

Answer 22

thanks a lot every one