## mathslover 4 years ago find the minimum value of $$4-5x-4x^2$$.

1. anonymous

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.

2. mathslover

3. anonymous

i think there is no minimum value

4. anonymous

it's unbounded

5. anonymous

For this problem there will be no minimum values.

6. mathslover

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

7. mathslover

is this the correct method ?

8. anonymous

|dw:1340428823754:dw|

9. anonymous

just maximum value

10. mathslover

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 ?

11. anonymous

nope

12. mathslover

why ?

13. anonymous

THIS FUNCTION HAS NO MINIMUM VALUE

14. mathslover

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

15. anonymous

Lets try to plot this function graphically,

16. mathslover

k

17. anonymous

ok try x = 10000

18. anonymous

when do you think the minimum value will be achieved?

19. anonymous

why?

20. mathslover

can u tell m not soo good

21. anonymous

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

22. anonymous

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

23. mathslover

thanks a lot every one