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)?

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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)?

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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.
For part b I completed the square and got (t+1)^2 = 25 but how do I find the vertex?
@Hero @Hero you said you would help me!

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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
It's a simply mistake, but: You were given \(\sf\Large f(t) = -16t^2 - 32t + 384\) however you factored it to \(\sf\large -16(t+6)(\color{red}{x}-4)\)
so the t in t-4 is supposed to be x-4? @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 :)
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
\(\sf\Large y=a(x-h)^2+k\) 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
\(\sf\Large -16t^2 - 32t =-384\) What is \(\sf\Large\left( \frac{b}{2}\right)^2=\left( \frac{-32}{2}\right)^2=?\)
-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?
we have to first factor the 16 out of the equation, so: \(\sf\Large -16t^2 - 32t + 384 =-16(t^2+2t-24) \)
@Mehek14 @paki I'm not sure how to complete the square :/
so the factored form is -16(t+6)(t-4)
there are so many ways to solve math, I don't know why they limit us with this completing the square lol
*disappears*
https://mathway.com/examples/Algebra/Quadratic-Equations/Solve-by-Completing-the-Square?id=29 mhmmm
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???
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 \(\sf\Large f(t)=-16t^2-32t+384\)
@zepdrix please I really need help!
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. :)
thanks so much!!!! @jtvatsim
well we got the completing square out of the way
\(\color{blue}{\text{Originally Posted by}}\) @TheSmartOne \(\sf\Large y=a(x-h)^2+k\) The vertex is \(\sf\Large (h,k)\) \(\color{blue}{\text{End of Quote}}\)
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 :)
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
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?
yes it is thanks for asking! @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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