i have two more that i need to be in factored form
f(x) = -1/10(x-3)^2 +10
f(x) = -1/5(x-3)^2 + 7
I dont really understand the completing the square part even tho i was taught quad equaton. Some help would be greatly appreciated.
-1/10 (x-3)^2 +10
-1/10 [ (x-3)^2-100]
-1/10 [ x^2 - 6x+9 -100 ]
-1/10 (x^2-6x-91)
-1/10 (X-13)(X+7)
Second one
-1/5 (x-3)^2 +7 (are you sure about the red part....did you enter it corrrectly?)
-1/5 [ (x-3)^2 -35]
-1/5 (x^2-6x+9-35)
-1/5 (x^2-6x-26)
-1/5 (x-8.92)(x+2.92) using Quadratic Formula (this is the reason why I question the red part)
and yes im sure because x - 3 squared is the 3 to the right of zero on the x axis
OK.....that just seems like an odd answer....is the +7 portion correct too?
Second one
f(x) = - (1/5) ( x - 3)^2 + 7
a = (-1/5)
And we can solve this :
(-1/5) ( x^2 + 6x + 9) + 7 = 0 ...... multiply through by -5
(x^2 + 6x + 9) - 35 = 0
x^2 + 6x - 26 = 0 add 26 to both sides
x^2 + 6x = 26
Complete the square on x......take (1/2) of 6 = 3, square it = 9 ......add this to both sides
x^2 + 6x + 9 = 26 + 9 factor the left side and simplify
(x- 3)^2 = 35 take both roots
x - 3 = ±sqrt(35) add 3 to both sides and we have that
x = 3 + sqrt (35) and x= 3 - sqrt (35)
So......the factored form is
f(x) = (-1/5) ( x - ( 3 + sqrt(35) ) ( x - (3 - sqrt (35) )
See the graph here to confirm this : https://www.desmos.com/calculator/pppmfkal6k