Question

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

Answer 1

okay

Answer 2

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

Answer 3

thats the only part i dont get

Answer 4

you distribute the 2....

Answer 5

after the perenthesis

Answer 6

hm?

Answer 7

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

Answer 8

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

Answer 9

64 minus 64 is 0

Answer 10

:(

Answer 11

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

Answer 12

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

Answer 13

help

Answer 14

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

Answer 15

no

Answer 16

How about this step, does this make sense
$$ \Large { -x^2 + 16x

= -(x^2 - 16x + 64 - 64)


}$$

Answer 17

uhum i guess so

Answer 18

well no actually

Answer 19

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


}\]

Answer 20

hm

Answer 21

why did u change the + to a minus

Answer 22

we factor out the negative

Answer 23

ok?

Answer 24

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


}\]

Answer 25

oh ok i see

Answer 26

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

Answer 27

ok so far you agree with the steps above ?

Answer 28

yes

Answer 29

\[ \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)

}\]

Answer 30

I distributed the negative on the last term -64

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

sorry i couldnt help :(

Answer 35

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
\[ \Large x^2 -16x + \left( \text{-}8 \right)^2 = \left( x+ (\text{-}8) \right)^2 \]\[ \Large x^2 -16x + 64 = \left( x - 8 \right)^2 \]

Answer 36

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

Answer 37

@perl oh ok

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

?, 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

Answer 44

\[-(x^2 - 16x + 64) (+64)..(why single out)( -64)\]

Answer 45

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

Answer 46

@perl

Answer 47

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

Answer 48

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

Answer 49

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.

Answer 50

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

Answer 51

did you follow the steps above

Answer 52

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

Answer 53

hello?

Answer 54

@perl

Answer 55

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

Answer 56

which step is unclear , let me know :)

Answer 57

ive told u 3 times what step is unclear

Answer 58

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

Answer 59

in order to complete the square

Answer 60

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

Answer 61

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

Answer 62

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