Question

how to get vertex form of f(x)=4x^2+8x+2

Answer 1

\[
f(x)=4x^2+8x+2\\
f(x) = 4x^2 + 8x +4-4+2\\
f(x) = (2 x+2)^2-2\\
\]
Can you finish it now?

Answer 2

eh I dont think so

Answer 3

what x makes 2 x+2 =0?

Answer 4

x=-1

Answer 5

f(-1)=-2

Answer 6

vertex=(-1,-2)

Answer 7

I need it in vertex form like
f(x)=4(x+2)^2-6

Answer 8

\[

f(x)=4x^2+8x+2\\
f(x) = 4x^2 + 8x +4-4+2\\
f(x) = (2 x+2)^2-2\\
f(x) = 4( x+1)^2-2\\
\]

Answer 9

Did you get it now?

Answer 10

ok why did you turn the positive 2 too a negative in the 3rd step

Answer 11

-4+2=-2

Answer 12

give me a sec

Answer 13

the second step is whats getting me, why did you put +4-4+2

Answer 14

@eliassaab

Answer 15

0=4-4
I added zero

Answer 16

Have you heard about completing the square?

Answer 17

thats what I was trying to find out previously, I have no clue on how to do it

Answer 18

\[
4 x^2 +8 x + 4=(2x+2)^2\\
(2x)^2 + 2(2x)2+ 2^2\\
a^2+ 2 a b + b^2=(a+b)^2
\]
I have to add 4 to make it a perfect square and to not do any damage I take away -4

Answer 19

Forget it i'm wasting time here not mine only but yours as well nevermind. Thanks for helping me. @eliassaab

Answer 20

YW