anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
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anonymous
  • anonymous
okay
anonymous
  • anonymous
How did this = -(x2 - 16x + 64 - 64) turn to this P (x) = -(x - 8)2 + 64
anonymous
  • anonymous
thats the only part i dont get

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anonymous
  • anonymous
you distribute the 2....
anonymous
  • anonymous
after the perenthesis
anonymous
  • anonymous
hm?
anonymous
  • anonymous
P (x) = -(x - 8)2 + 64 ^------this two, you multiply the two *s what ever is inside the ()
anonymous
  • anonymous
thats x^2 - 16... but what about the 64?
anonymous
  • anonymous
64 minus 64 is 0
anonymous
  • anonymous
:(
anonymous
  • anonymous
P (x) = -(x - 8)2 + 64, i believe this step is suppose to come before the first equation in the sentence...
anonymous
  • anonymous
it would still be -(x - 8)^2 +0
anonymous
  • anonymous
help
perl
  • perl
Do you agree that 16x - x^2 = -(x-8)^2 + 64?
anonymous
  • anonymous
no
perl
  • perl
How about this step, does this make sense $$ \Large { -x^2 + 16x = -(x^2 - 16x + 64 - 64) }$$
anonymous
  • anonymous
uhum i guess so
anonymous
  • anonymous
well no actually
perl
  • perl
Let me add a few steps \[ \Large { -x^2 + 16x \\ = -x^2 + 16x + 0 \\ = -(x^2 - 16x + 0) \\ = -(x^2 - 16x + 64 - 64) }\]
anonymous
  • anonymous
hm
anonymous
  • anonymous
why did u change the + to a minus
perl
  • perl
we factor out the negative
anonymous
  • anonymous
ok?
perl
  • perl
\[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) }\]
anonymous
  • anonymous
oh ok i see
anonymous
  • anonymous
but how does it become a (x +8) ^2 + 64
perl
  • perl
ok so far you agree with the steps above ?
anonymous
  • anonymous
yes
perl
  • perl
\[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) }\]
perl
  • perl
I distributed the negative on the last term -64
anonymous
  • anonymous
ok i see but how does it become a (x +8) ^2 + 64
perl
  • perl
\[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - (x^2 -16x +64) + 64 \\ = - (x-8)^2 +64 }\]
perl
  • perl
the reason why we added 64 in the first place was to complete the square
anonymous
  • anonymous
sorry i couldnt help :(
perl
  • 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 \[ \Large x^2 -16x + \left( \text{-}8 \right)^2 = \left( x+ (\text{-}8) \right)^2 \]\[ \Large x^2 -16x + 64 = \left( x - 8 \right)^2 \]
perl
  • 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
anonymous
  • anonymous
@perl oh ok
anonymous
  • anonymous
i mean i still dont understand this =−(x2−16x+64−64)=−(x2−16x+64)−(−64)=−(x2−16x+64)+64=−(x−8)2+64
perl
  • perl
x^2 -16x + 64 = (x-8)^2 so we substituted
anonymous
  • anonymous
why do u only multiply the - by 64 nd not all the numbers?
perl
  • perl
The expressions in red are exactly equal $$\Large −\color{red}{(x^2−16x+64)}+64=−\color{red}{(x−8)^2}+64$$
perl
  • perl
\[ \Large { -x^2 + 16x \\ = -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = - (x^2 -16x +64) - (-64) \\ = - \color{red}{(x^2 -16x +64)} + 64 \\ = - \color{red}{(x-8)^2} +64 }\]
anonymous
  • 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
anonymous
  • anonymous
\[-(x^2 - 16x + 64) (+64)..(why single out)( -64)\]
anonymous
  • 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
anonymous
  • anonymous
@perl
perl
  • perl
The goal here is to find the 'vertex form' of the quadratic function $$ \Large y = a(x-h)^2 + k $$
perl
  • perl
The vertex form tells us the max/min is at (h,k)
anonymous
  • 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.
anonymous
  • anonymous
and i dont understand how this = -(x2 - 16x + 64 - 64) turned ito this P (x) = -(x - 8)2 + 64
perl
  • perl
did you follow the steps above
anonymous
  • 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?
anonymous
  • anonymous
hello?
anonymous
  • anonymous
@perl
perl
  • perl
I will label the steps \[ \Large { \\ 1) ~~-x^2 + 16x \\ 2) ~~-(x^2 - 16x ) \\ 3) ~~~-(x^2 - 16 x + 0) \\ 4)~~ -(x^2 - 16x + 64 - 64) \\ 5)~~ - (x^2 -16x +64) - (-64) \\ 6)~~ - \color{red}{(x^2 -16x +64)} + 64 \\ 7)~~ - \color{red}{(x-8)^2} +64 }\]
perl
  • perl
which step is unclear , let me know :)
anonymous
  • anonymous
ive told u 3 times what step is unclear
anonymous
  • 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?
perl
  • perl
in order to complete the square
perl
  • perl
We can wait to multiply by -64 if you like.
anonymous
  • anonymous
o,o...what part dont u understand? why arent u answering my question....? i will figure it out by myself
perl
  • perl
Here is alternative steps. \[ \Large { -x^2 + 16x \\ = -(x^2 - 16x ) \\ = -(x^2 - 16 x + 0) \\ = -(x^2 - 16x + 64 - 64) \\ = -( (x^2 - 16x + 64) - 64) \\ = - ((x-8)^2 -64) \\ = -(x-8)^2-(-64) \\ = -(x-8)^2+ 64 }\]

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