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For part b I completed the square and got (t+1)^2 = 25 but how do I find the vertex?

I said, post your next question separately. Anyone can help with these.

Yeah I did post it separately and you said you would help!

I need help on finding the vertex and I am confused on how to

For Part A, you are correct (almost)

What did I leave out? @TheSmartOne could you please explain

so the t in t-4 is supposed to be x-4? @TheSmartOne

\(\sf\Large y=a(x-h)^2+k\)
The vertex is \(\sf\Large (h,k)\)

so how would I incorporate (t+1)^2 = 25 in to that?

y = a(t-1)^2 + 25 ???? @TheSmartOne

one sec

You didn't properly complete the square.

?? what do you mean?

\(\sf\Large -16t^2 - 32t + 384 \color{red}{\neq} (t-1)^2+25\)

so the completed square form is (t - 1)^2 + 25

it isn't. That's wrong.

oh ok i got it. How would I find it then?

could you show me step by step

-16

correct; (-16)^2

whats next ??? @TheSmartOne

you have to add it to both sides:
\(\sf\Large -16t^2 - 32t+(-16^2) =-384+(-16)^2\)

can you factor the left hand side?
Hint:
\(\sf\Large a^2+2ab+b^2=(a+b)^2\)

I am so sorry i lost wifi! @TheSmartOne

ok so you get -16t^2 - 32t + (-256) = -384 + (-256) @TheSmartOne

-16t^2 - 32t + (-256) = -640 @TheSmartOne

hmmm

One sec, we'll need to backtrack over here

ok no problem

do you know what went wrong?

so the factored form is -16(t+6)(t-4)

*disappears*

https://mathway.com/examples/Algebra/Quadratic-Equations/Solve-by-Completing-the-Square?id=29
mhmmm

i just dont understand completing the square?

the link was an example of how to do it

oh ok so what would I do ??

yes, the symmetry of axis is x=-1

ok I go that part and I know that the maximum is 400 i just need to show my work of how I got that

@zepdrix could you help us to complete the square for
\(\sf\Large f(t)=-16t^2-32t+384\)

I can tell you how to do the first step

divide all the numbers by -16

the calculator gives you the final answer, but no steps: http://prntscr.com/7metrk

@mathmate could you help us complete the square?

t^2 + 2t +24

OMG EVERYONE I THINK I MAY HAVE FIGURED IT OUT! GIVE ME ONE SEC!

we could have just as easily found the vertex by -b/2a ¯\_(ツ)_/¯

a(x+d)2+e
d=−32 / 2⋅(−16)
d=1
e=384− (−32)^2 / 4⋅(−16)
e=400
−16(t+1)2+400

that's it, I've been working it out over here as well. nicely done. :)

well we got the completing square out of the way

haha yep! so to find the vertex I wold just do y = -16(x-1)^2 +400??

or in other words the vertex is (-1,400)

y = -16(x+1)^2 +400

correct

and that would be a maximum or a minimum?

maximum!

correct :)

I had to go to lunch. I would have helped otherwise.

yeah yeah whatever

I probably wouldn't have helped if it wasn't for Hero: http://prntscr.com/7mezsj

at least hero considered it...

@kaite_mcgowan
Is the question resolved?