Question

Using this f(x) = x2 + 6x – 16
This is algebra 2 I get how to find the vertex once its put in the right form but i dont know how to put it in this form f(x) = a(x - h)2 + k

Answer 1

complete the square. Have you learned how?

Answer 2

yes i have i have to get my notes real quick

Answer 3

okay i got my notes but im not sure what to do with it i think what you would do is get rid of the coefficient

Answer 4

if you have time, this goes into details
https://www.khanacademy.org/math/algebra/quadratics/completing_the_square/v/solving-quadratic-equations-by-completing-the-square

meanwhile, if you have
f(x) = x^2 + bx + c
(and you do here)
find the b (number in front of the x) , and divide it by 2
then square it. (b/2)^2
then put in
f(x) = x^2 + bx + (b/2)^2 - (b/2)^2 + c

Answer 5

the first three terms in
f(x) = x^2 + bx + (b/2)^2 - (b/2)^2 + c

i.e. the x^2 + bx + (b/2)^2
can be re-written as (x+b/2)^2

Answer 6

f(x) = x2 + 6x – 16
so 6 is the b and c is the -16 ?

Answer 7

that was supposed to come out a -16 but it didn't show up right on my screen

Answer 8

OS is a bit screwed up. It shows up if you re-fresh the browser (which seems to take forever)

meanwhile, yes b is 6
you now need b/2

Answer 9

so 3

Answer 10

now square it

Answer 11

9

Answer 12

f(x) = x^2 + 6x – 16

now add in +9 - 9 (which is zero, so it's ok) into the equation
f(x) = x^2 + 6x +9 - 9 – 16

and look at only the first 3 terms
f(x) = (x^2 + 6x +9) - 9 – 16

Answer 13

refresh your browser (reload this page) if you can't read that

Answer 14

why would i need do what i just did though ?
like with the 6/2 and squaring that

Answer 15

and why wouldn't it just cancel out ?

Answer 16

I posted a link to a Khan video that explains why this works. meanwhile, we are just doing it.

Answer 17

wouldn't you add 9 to both sides thats what it showed in my example

Answer 18

we could cancel it out. but we don't have to. Instead , we will group the first 3 terms
and re-write them

f(x) = (x^2 + 6x + 9) - 9 -16

the -9 - 16 can be simplified to -25
f(x) = (x^2 + 6x + 9) -25

and because of adding in +9, the first part is a perfect square, and can be written
as (x+ 3)^2

notice: (x+3)(x+3)= x^2 +3x +3x + 9 = x^2 +6x + 9

Answer 19

im not getting this at all ill add a picture of what my example says

Answer 20

yes we could add +9 to both sides
which makes more sense if we had
x^2 + 6x -16 = 0
and we did

x^2 + 6x+9 -16 = 0+9

but notice if we now add - 9 to both sides it would look like this
x^2 + 6x+9 -9 -16 = 0+9-9
or
x^2 + 6x+9 -9 -16 = 0

which is ok.

Answer 21

thats what i got it from

Answer 22

yes, that is fine.

if you like we can do that:

f(x) = x^2 + 6x -16

add +9 to both sides (9 is from (6/2)^2 i.e. (b/2)^2 )
f(x)+9 = x^2 + 6x +9 -16

if we want to "get rid of the -16" we can add +16 to both sides
f(x)+9+16= x^2 + 6x +9 -16 + 16
and simplify
on the left 9+16 is 25
on the right 16-16 is 0
we get
f(x) +25 = x^2 + 6x +9

Answer 23

ok ?

Answer 24

and the 9 was from the 6/2 squared ?

Answer 25

yes

Answer 26

okay i got that

Answer 27

the important idea is if you have

y = x^2 + bx + other junk
if we add in (b/2)^2 (and subtract it somewhere's else to keep things balanced)
like this
y + (b/2)^2 = x^2 + bx + (b/2)^2 + other junk
this part: x^2 + bx + (b/2)^2 can be written as (x+b/2)^2
y + (b/2)^2 = (x+b/2)^2 + other junk

Answer 28

if that is not clear, when you have time, watch the video.

meanwhile, you should have for your problem
f(x) +25 = x^2 + 6x +9
f(x) + 25 = (x+3)^2

and add -25 on both sides to "clean up"
f(x) +25 - 25 = (x+3)^2 - 25

f(x) = (x+3)^2 - 25

Answer 29

so that was it or is there more ?

Answer 30

f(x) = a(x - h)2 + k
in this form it would be f(x)=(x-3)^2 -25
but i dont see where a is at if thats what that equation is for

Answer 31

if you need the vertex, you can match up that equation with

f(x) = a(x-h)^2 + k

vertex is (h,k)

Answer 32

did i do it right up there

Answer 33

h would be -3 and k would be -25 ?

Answer 34

I get what you did but i still dont see where the letters match up at

Answer 35

(x) = (x+3)^2 - 25
to make it "match up" with
f(x) = a(x - h)2 + k

we write f(x) = (x+3)^2 - 25
as
f(x) = 1*(x- (-3) )^2 + -25

match to a*(x- h)^2 + k

Answer 36

a matches 1
h matches (-3) or just -3
k matches -25

Answer 37

okay so the vertex would be (-3,-25) right ?

Answer 38

yes

Answer 39

and since a didnt show up that just got replaced with a 1?

Answer 40

I would say we don't bother to write down the 1*stuff
in other words if we have 1*x
we would just say x
rather than 1x

so if you see (x- -3)^2
it's the same as 1*(x- -3)^2

Answer 41

would a always be like that or could it have a higher number than one

Answer 42

"a" will make the parabola fatter (or skinnier) than normal if it's not 1

Answer 43

we would get a not 1 if we started with for example
2x^2 +6 x -16

but it makes "completing the square" harder. (so you don't want to see a not be 1!)

Answer 44

so a in that would be the 2

Answer 45

yes, in that example, a would end up being 2


btw, if you have
f(x) = a x^2 + bx +c
the x value of the vertex is -b/(2a)
for your problem we get -6/2= -3 (much faster than completing the square)
to find the y value of the vertex, we would set x= -3 in the equation and solve for f(-3)

Answer 46

so to summarize how to figure it out i would say what i did was take 6x and divided it by 2 then squared the answer to get 9 and with to get rid of -16 i added +16 on both sides and got f(x) 25 = x^2 (x+3)^2 the to clean it up i subtracted 25 from both sides and ended up with f(x)=(x+3)^2-25

Answer 47

take the 6 from 6x (just to be picky)
divide by 2 to get 3
square to get 9
next we add 9 to the right-side of the equation.
to keep things balanced, we *either* add 9 to the left-side OR add -9 also to the right-side

Answer 48

Sorry it kicked me out so i had to go on my other account