A sandbag was thrown downward from a building. The function f(t) = -16t2 - 32t + 384 shows the height f(t), in feet, of the sandbag after t seconds:
Part A: Factor the function f(t) and use the factors to interpret the meaning of the x-intercept of the function.
Part B: Complete the square of the expression for f(x) to determine the vertex of the graph of f(x). Would this be a maximum or minimum on the graph?
Part C: Use your answer in part B to determine the axis of symmetry for f(x)?

- anonymous

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- katieb

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- anonymous

for part A i got -16(t+6)(x-4) when factored and said that the x-intercepts of the functions is (4,0) which represents the time in seconds that the sandbag was thrown to where it lands.

- anonymous

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

- anonymous

@Hero @Hero you said you would help me!

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## More answers

- Hero

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

- anonymous

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

- anonymous

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

- TheSmartOne

For Part A, you are correct (almost)

- anonymous

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

- TheSmartOne

It's a simply mistake, but:
You were given \(\sf\Large f(t) = -16t^2 - 32t + 384\)
however you factored it to \(\sf\large -16(t+6)(\color{red}{x}-4)\)

- anonymous

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

- TheSmartOne

The variable you were given @kaite_mcgowan was \(\bf t\) so I don't understand where you brought in the \(\bf x\) from.
So you just need to change the x to a t to make it correct :)

- anonymous

oh ok thanks so much! could you please help me on b and finding the vertex with the completed square of (t+1)^2 = 25 @TheSmartOne

- TheSmartOne

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

- anonymous

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

- anonymous

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

- TheSmartOne

one sec

- TheSmartOne

You didn't properly complete the square.

- anonymous

?? what do you mean?

- TheSmartOne

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

- anonymous

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

- TheSmartOne

it isn't. That's wrong.

- anonymous

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

- anonymous

could you show me step by step

- TheSmartOne

\(\sf\Large -16t^2 - 32t =-384\)
What is \(\sf\Large\left( \frac{b}{2}\right)^2=\left( \frac{-32}{2}\right)^2=?\)

- anonymous

-16

- TheSmartOne

correct; (-16)^2

- anonymous

whats next ??? @TheSmartOne

- TheSmartOne

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

- TheSmartOne

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

- anonymous

I am so sorry i lost wifi! @TheSmartOne

- anonymous

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

- anonymous

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

- TheSmartOne

hmmm

- anonymous

@TheSmartOne

- TheSmartOne

One sec, we'll need to backtrack over here

- anonymous

ok no problem

- anonymous

do you know what went wrong?

- TheSmartOne

we have to first factor the 16 out of the equation, so:
\(\sf\Large -16t^2 - 32t + 384 =-16(t^2+2t-24) \)

- TheSmartOne

@Mehek14 @paki I'm not sure how to complete the square :/

- anonymous

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

- TheSmartOne

there are so many ways to solve math, I don't know why they limit us with this completing the square lol

- Mehek14

*disappears*

- TheSmartOne

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

- anonymous

wait thats the answer the link above? I know that the highest point or the maximum is 400 and the symmetry of axis is -1???

- anonymous

i just dont understand completing the square?

- TheSmartOne

the link was an example of how to do it

- anonymous

oh ok so what would I do ??

- TheSmartOne

yes, the symmetry of axis is x=-1

- anonymous

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

- TheSmartOne

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

- anonymous

@zepdrix please I really need help!

- Mehek14

I can tell you how to do the first step

- Mehek14

divide all the numbers by -16

- TheSmartOne

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

- TheSmartOne

@mathmate could you help us complete the square?

- anonymous

t^2 + 2t +24

- anonymous

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

- TheSmartOne

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

- anonymous

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

- anonymous

@TheSmartOne

- jtvatsim

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

- anonymous

thanks so much!!!! @jtvatsim

- TheSmartOne

well we got the completing square out of the way

- TheSmartOne

\(\color{blue}{\text{Originally Posted by}}\) @TheSmartOne
\(\sf\Large y=a(x-h)^2+k\)
The vertex is \(\sf\Large (h,k)\)
\(\color{blue}{\text{End of Quote}}\)

- anonymous

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

- anonymous

@TheSmartOne

- anonymous

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

- anonymous

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

- TheSmartOne

correct

- TheSmartOne

and that would be a maximum or a minimum?

- anonymous

maximum!

- TheSmartOne

correct :)

- anonymous

Thanks for all of your help @TheSmartOne and sticking with me through the whole equation unlike other people who promised before **cough cough @Hero cough cough** thanks again @TheSmartOne

- Hero

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

- anonymous

yeah yeah whatever

- TheSmartOne

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

- anonymous

at least hero considered it...

- mathmate

@kaite_mcgowan
Is the question resolved?

- anonymous

yes it is thanks for asking! @mathmate

- mathmate

Sorry, I came a little late.
Here's what I would have done anyway:
Given f(t)=\(-16t^2-32t+384\)
(a) factorization
f(t)=\(-16(t^2+2t-24)\)
f(t)=-16(t+6)(t-4)
(b) Complete square
f(t)=-16(t+1 +5)(t+1 -5)
f(t)=-16[(t+1)^2-25]
maximum is at x=-1 (solve for t in t+1=0), f(-1)=400.
Maximum because leading coefficient (of t^2) is negative.
(c) Axis of symmetry
Axis of symmetry of a quadratic is location of maximum/minimum, i.e.
x=-1

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