+14 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
+129894 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
+37150 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....
