Question

Can someone show me how to complete the square of: \[f(t) = -16t^2 - 32t + 128\]

Answer 1

first thing u should do is take out common any coefficients if possible

Answer 2

I think you have to set f(t) to 0 right?

Answer 3

hey do you got this one Al Capone?

Answer 4

So um,
\[0=16t^2+32t+c?\]

Answer 5

Then don't you have to find a c value that completes the square?

Answer 6

Hey Cammi still need help? ^-^

Answer 7

I do

Answer 8

sorry froze ok so..... lets see...

Answer 9

\[0 = 16t _{2} + 32t + c\]

Answer 10

You mean t^2 right?

Answer 11

I think thats right ._. and no Idk but im getting the answer with 16t_(2)

Answer 12

Thats not the soution though right? cx
Im completing the square, not factoring

Answer 13

oh xD sorry wrong steps

Answer 14

I'll just do it laterz
Thanx for trying though cx

Answer 15

hey ill start solving it if you want c: so when "laterz" comes i can have it ready

Answer 16

why did u replace \(c\) with \(128\)

Answer 17

f(t)=−16t2−32t+128

Answer 18

Thats how I remember it from my textbook
I might've interperted it wrongly

Answer 19

@camerondoherty you do not change what is given you set the expression equal to 0

Answer 20

so -16t^2 - 32 t + 128 = 0

Answer 21

So
\[0=16t^2+32t+128 \]

Answer 22

as suggested by someone already since you see a common factor you can factor that out

Answer 23

@mathmath333 will continue

Answer 24

Yes, but the question asks to complete the square.
I've already factored it cx

Answer 25

see if this makes sense

\(\large\tt \begin{align} \color{black}{f(t) = -16(t^2 + 2t -8)\hspace{.33em}\\~\\}\end{align}\)

Answer 26

this \(\Huge \uparrow \) makes it easy

Answer 27

But I have already factored it out, I just wanted to find the vertex by completing the square

Answer 28

yes m telling the method of completing the square
we know that

\(\large\tt \begin{align} \color{black}{t^2 + 2t+1=(t+1)^2\hspace{.33em}\\~\\}\end{align}\)

so we just adjust that in
\(\large\tt \begin{align} \color{black}{f(t) = -16(t^2 + 2t -8)\hspace{.33em}\\~\\}\end{align}\)

Answer 29

this can be done by writting it as

\(\large\tt \begin{align} \color{black}{f(t) = -16(t^2 + 2t \color{red}{+1-1}-8)\hspace{.33em}\\~\\}\end{align}\)

see if it makes sense

Answer 30

Not really >.<
Sorry...

Answer 31

I have to go too... sorry cx

Answer 32

we just want to put this
\(\large\tt \begin{align} \color{black}{t^2 + 2t+1=(t+1)^2\hspace{.33em}\\~\\}\end{align}\)