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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
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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How did this = (x2  16x + 64  64) turn to this P (x) = (x  8)2 + 64

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats the only part i dont get

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you distribute the 2....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0after the perenthesis

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0P (x) = (x  8)2 + 64 ^this two, you multiply the two *s what ever is inside the ()

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thats x^2  16... but what about the 64?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0P (x) = (x  8)2 + 64, i believe this step is suppose to come before the first equation in the sentence...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it would still be (x  8)^2 +0

perl
 one year ago
Best ResponseYou've already chosen the best response.0Do you agree that 16x  x^2 = (x8)^2 + 64?

perl
 one year ago
Best ResponseYou've already chosen the best response.0How about this step, does this make sense $$ \Large { x^2 + 16x = (x^2  16x + 64  64) }$$

perl
 one year ago
Best ResponseYou've already chosen the best response.0Let me add a few steps \[ \Large { x^2 + 16x \\ = x^2 + 16x + 0 \\ = (x^2  16x + 0) \\ = (x^2  16x + 64  64) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why did u change the + to a minus

perl
 one year ago
Best ResponseYou've already chosen the best response.0we factor out the negative

perl
 one year ago
Best ResponseYou've already chosen the best response.0\[ \Large { x^2 + 16x \\ = x^2 + 16x \\ = (x^2  16x ) \\ = (x^2  16 x + 0) \\ = (x^2  16x + 64  64) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but how does it become a (x +8) ^2 + 64

perl
 one year ago
Best ResponseYou've already chosen the best response.0ok so far you agree with the steps above ?

perl
 one year ago
Best ResponseYou've already chosen the best response.0\[ \Large { x^2 + 16x \\ = x^2 + 16x \\ = (x^2  16x ) \\ = (x^2  16 x + 0) \\ = (x^2  16x + 64  64) \\ =  (x^2 16x +64)  (64) }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.0I distributed the negative on the last term 64

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i see but how does it become a (x +8) ^2 + 64

perl
 one year ago
Best ResponseYou've already chosen the best response.0\[ \Large { x^2 + 16x \\ = x^2 + 16x \\ = (x^2  16x ) \\ = (x^2  16 x + 0) \\ = (x^2  16x + 64  64) \\ =  (x^2 16x +64)  (64) \\ =  (x^2 16x +64) + 64 \\ =  (x8)^2 +64 }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.0the reason why we added 64 in the first place was to complete the square

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry i couldnt help :(

perl
 one year ago
Best ResponseYou've already chosen the best response.0To 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 \[ \Large x^2 16x + \left( \text{}8 \right)^2 = \left( x+ (\text{}8) \right)^2 \]\[ \Large x^2 16x + 64 = \left( x  8 \right)^2 \]

perl
 one year ago
Best ResponseYou've already chosen the best response.0Can 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i mean i still dont understand this =−(x2−16x+64−64)=−(x2−16x+64)−(−64)=−(x2−16x+64)+64=−(x−8)2+64

perl
 one year ago
Best ResponseYou've already chosen the best response.0x^2 16x + 64 = (x8)^2 so we substituted

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do u only multiply the  by 64 nd not all the numbers?

perl
 one year ago
Best ResponseYou've already chosen the best response.0The expressions in red are exactly equal $$\Large −\color{red}{(x^2−16x+64)}+64=−\color{red}{(x−8)^2}+64$$

perl
 one year ago
Best ResponseYou've already chosen the best response.0\[ \Large { x^2 + 16x \\ = x^2 + 16x \\ = (x^2  16x ) \\ = (x^2  16 x + 0) \\ = (x^2  16x + 64  64) \\ =  (x^2 16x +64)  (64) \\ =  \color{red}{(x^2 16x +64)} + 64 \\ =  \color{red}{(x8)^2} +64 }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0?, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(x^2  16x + 64) (+64)..(why single out)( 64)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and... is there a method to turn this\[(x^2  16x + 64)\] into the one u had said, if so ive prob forgot it

perl
 one year ago
Best ResponseYou've already chosen the best response.0The goal here is to find the 'vertex form' of the quadratic function $$ \Large y = a(xh)^2 + k $$

perl
 one year ago
Best ResponseYou've already chosen the best response.0The vertex form tells us the max/min is at (h,k)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and i dont understand how this = (x2  16x + 64  64) turned ito this P (x) = (x  8)2 + 64

perl
 one year ago
Best ResponseYou've already chosen the best response.0did you follow the steps above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do u multiply the negative by the 64 but not the other numbers, why are the other units in perenthesis and 64 isnt?

perl
 one year ago
Best ResponseYou've already chosen the best response.0I will label the steps \[ \Large { \\ 1) ~~x^2 + 16x \\ 2) ~~(x^2  16x ) \\ 3) ~~~(x^2  16 x + 0) \\ 4)~~ (x^2  16x + 64  64) \\ 5)~~  (x^2 16x +64)  (64) \\ 6)~~  \color{red}{(x^2 16x +64)} + 64 \\ 7)~~  \color{red}{(x8)^2} +64 }\]

perl
 one year ago
Best ResponseYou've already chosen the best response.0which step is unclear , let me know :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ive told u 3 times what step is unclear

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why do u multiply the negative by the 64 but not the other numbers, why are the other units in perenthesis and 64 isnt?

perl
 one year ago
Best ResponseYou've already chosen the best response.0in order to complete the square

perl
 one year ago
Best ResponseYou've already chosen the best response.0We can wait to multiply by 64 if you like.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0o,o...what part dont u understand? why arent u answering my question....? i will figure it out by myself

perl
 one year ago
Best ResponseYou've already chosen the best response.0Here is alternative steps. \[ \Large { x^2 + 16x \\ = (x^2  16x ) \\ = (x^2  16 x + 0) \\ = (x^2  16x + 64  64) \\ = ( (x^2  16x + 64)  64) \\ =  ((x8)^2 64) \\ = (x8)^2(64) \\ = (x8)^2+ 64 }\]
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