## anonymous one year ago um i dont understand this part...when finding the max/min.vertex etc. P (x) = 16x - x2 P (x) = -x2 + 16x = -(x2 - 16x + 64 - 64) P (x) = -(x - 8)2 + 64

1. anonymous

okay

2. anonymous

How did this = -(x2 - 16x + 64 - 64) turn to this P (x) = -(x - 8)2 + 64

3. anonymous

thats the only part i dont get

4. anonymous

you distribute the 2....

5. anonymous

after the perenthesis

6. anonymous

hm?

7. anonymous

P (x) = -(x - 8)2 + 64 ^------this two, you multiply the two *s what ever is inside the ()

8. anonymous

thats x^2 - 16... but what about the 64?

9. anonymous

64 minus 64 is 0

10. anonymous

:(

11. anonymous

P (x) = -(x - 8)2 + 64, i believe this step is suppose to come before the first equation in the sentence...

12. anonymous

it would still be -(x - 8)^2 +0

13. anonymous

help

14. perl

Do you agree that 16x - x^2 = -(x-8)^2 + 64?

15. anonymous

no

16. perl

How about this step, does this make sense $$\Large { -x^2 + 16x = -(x^2 - 16x + 64 - 64) }$$

17. anonymous

uhum i guess so

18. anonymous

well no actually

19. perl

Let me add a few steps $\Large { -x^2 + 16x \\ = -x^2 + 16x + 0 \\ = -(x^2 - 16x + 0) \\ = -(x^2 - 16x + 64 - 64) }$

20. anonymous

hm

21. anonymous

why did u change the + to a minus

22. perl

we factor out the negative

23. anonymous

ok?

24. perl

$\Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) }$

25. anonymous

oh ok i see

26. anonymous

but how does it become a (x +8) ^2 + 64

27. perl

ok so far you agree with the steps above ?

28. anonymous

yes

29. perl

$\Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) }$

30. perl

I distributed the negative on the last term -64

31. anonymous

ok i see but how does it become a (x +8) ^2 + 64

32. perl

$\Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - (x^2 -16x +64) + 64 \\ = - (x-8)^2 +64 }$

33. perl

the reason why we added 64 in the first place was to complete the square

34. anonymous

sorry i couldnt help :(

35. perl

To complete the square we use the identity $\Large x^2 + bx + \left( b/2 \right)^2 = (x+ b/2)^2$Here we have $$\large b=-16$$ $\Large x^2 + (\text{-}16)x + \left( \text{-}16/2 \right)^2 = (x+ (\text{-}16)/2)^2$We can simplify $\Large x^2 -16x + \left( \text{-}8 \right)^2 = \left( x+ (\text{-}8) \right)^2$$\Large x^2 -16x + 64 = \left( x - 8 \right)^2$

36. perl

Can you verify that this is always true. $$\Large x^2 + bx + \left( b/2 \right)^2 = (x+ b/2)^2$$ Start from the right side and expand

37. anonymous

@perl oh ok

38. anonymous

i mean i still dont understand this =−(x2−16x+64−64)=−(x2−16x+64)−(−64)=−(x2−16x+64)+64=−(x−8)2+64

39. perl

x^2 -16x + 64 = (x-8)^2 so we substituted

40. anonymous

why do u only multiply the - by 64 nd not all the numbers?

41. perl

The expressions in red are exactly equal $$\Large −\color{red}{(x^2−16x+64)}+64=−\color{red}{(x−8)^2}+64$$

42. perl

$\Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - \color{red}{(x^2 -16x +64)} + 64 \\ = - \color{red}{(x-8)^2} +64 }$

43. anonymous

?, but x squared = x^2 and 8 squared equals 64, so it would be x^2 - 64... and why do u only subtact the -64

44. anonymous

$-(x^2 - 16x + 64) (+64)..(why single out)( -64)$

45. anonymous

and... is there a method to turn this$-(x^2 - 16x + 64)$ into the one u had said, if so ive prob forgot it

46. anonymous

@perl

47. perl

The goal here is to find the 'vertex form' of the quadratic function $$\Large y = a(x-h)^2 + k$$

48. perl

The vertex form tells us the max/min is at (h,k)

49. anonymous

this is what they did... The sum of two numbers is 16. Let x and y be the two numbers. x + y = 16 -----------> (1) y = 16 - x P (x) is a product function. P (x) = xy is maximum. P (x) = x(16 - x) P (x) = 16x - x2 P (x) = -x2 + 16x = -(x2 - 16x + 64 - 64) P (x) = -(x - 8)2 + 64 The values of a is negative. So, the function is maximum. The vertex of the function is (8, 64) and the maximum value is 64.

50. anonymous

and i dont understand how this = -(x2 - 16x + 64 - 64) turned ito this P (x) = -(x - 8)2 + 64

51. perl

did you follow the steps above

52. anonymous

why do u multiply the negative by the -64 but not the other numbers, why are the other units in perenthesis and -64 isnt?

53. anonymous

hello?

54. anonymous

@perl

55. perl

I will label the steps $\Large { \\ 1) ~~-x^2 + 16x \\ 2) ~~-(x^2 - 16x ) \\ 3) ~~~-(x^2 - 16 x + 0) \\ 4)~~ -(x^2 - 16x + 64 - 64) \\ 5)~~ - (x^2 -16x +64) - (-64) \\ 6)~~ - \color{red}{(x^2 -16x +64)} + 64 \\ 7)~~ - \color{red}{(x-8)^2} +64 }$

56. perl

which step is unclear , let me know :)

57. anonymous

ive told u 3 times what step is unclear

58. anonymous

why do u multiply the negative by the -64 but not the other numbers, why are the other units in perenthesis and -64 isnt?

59. perl

in order to complete the square

60. perl

We can wait to multiply by -64 if you like.

61. anonymous

o,o...what part dont u understand? why arent u answering my question....? i will figure it out by myself

62. perl

Here is alternative steps. $\Large { -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = -( (x^2 - 16x + 64) - 64) \\ = - ((x-8)^2 -64) \\ = -(x-8)^2-(-64) \\ = -(x-8)^2+ 64 }$