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i really need help with this by today im just really confused bc of the complex number

 

so this is what i have so far:

 

f(x) = 1/4(x-3)^2 - 11

      = 1/4(x^2 -6x + 9) - 11

      = 1/4(x^2 -6x - 2)

 

I get stuck on this next step

i know its supposed to be f(x) = a (x +/- p)(x +/- q) 

but i dont know the last step

Some help would be awesome

 Dec 5, 2019
edited by Guest  Dec 5, 2019
edited by Guest  Dec 5, 2019
 #1
avatar+106539 
+2

f(x)   =  (1/4) ( x - 3)^2  - 11       

 

y  =  ( 1/4) ( x - 3)^2  -  11

 

y  =  (1/4)(x - 6x + 9)  -  11

 

y  =  (1/4)x^2 - (3/2)x  + 9/4  -  44/4

 

y  = (1/4)x^2 - (3/2)x  -  35 / 4

 

y =  (1/4)( x^2 - 6x  - 35)

 

a  =1/4

 

And we can solve this

 

x^2  - 6x  -  35  = 0       add   35 to both sides

 

x^2  - 6x   = 35          complete the square on x

 

x^2 - 6x + 9  =  35  + 9

 

(x - 3)^2   =   44   take both roots

 

x - 3  =   ±sqrt (44)

 

x - 3  =   ±2 sqrt (11)   add  3  to both sides

 

x =  3+ 2sqrt (11)    or    x =  3 - 2sqrt (11)

 

So....in factored form we have

 

y = (1/4)  [ x - ( 3 + 2sqrt (11) ) ]  [ x - (3 - 2sqrt (11) )   ]

 

See the graph here to verify  this  :  https://www.desmos.com/calculator/wx3tlpbvuk

 

 

cool cool cool

 Dec 5, 2019
 #2
avatar+19820 
0

Another way is to use the QUadratic Formula....

you got pretty far

 

1/4 (x^2-6x+9) - 11

1/4 (x^2-6x-35)         use quadratic formula

a=1 b = -6 c = -35

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

x=3±2sqrt(11)        Then continue as Cphil did.......don't forget the 1/4 in front....

 Dec 5, 2019

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