+0

# i need help again... The Same Vertex to Factored

+1
122
6
+14

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.

Dec 6, 2019

#1
+22096
+1

-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)

Dec 6, 2019
#3
+14
0

thank you so much you really helped me

erydaybeats  Dec 6, 2019
#2
+22096
0

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)

Dec 6, 2019
#4
+14
0

and yes im sure because x - 3 squared is the 3 to the right of zero on the x axis

erydaybeats  Dec 6, 2019
#5
+22096
0

OK.....that just seems like an odd answer....is the +7 portion correct too?

ElectricPavlov  Dec 6, 2019
#6
+109563
+1

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

Dec 6, 2019