anonymous
  • anonymous
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)?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • 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
  • anonymous
For part b I completed the square and got (t+1)^2 = 25 but how do I find the vertex?
anonymous
  • anonymous
@Hero @Hero you said you would help me!

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Hero
  • Hero
I said, post your next question separately. Anyone can help with these.
anonymous
  • anonymous
Yeah I did post it separately and you said you would help!
anonymous
  • anonymous
I need help on finding the vertex and I am confused on how to
TheSmartOne
  • TheSmartOne
For Part A, you are correct (almost)
anonymous
  • anonymous
What did I leave out? @TheSmartOne could you please explain
TheSmartOne
  • 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
  • anonymous
so the t in t-4 is supposed to be x-4? @TheSmartOne
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
  • 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
  • TheSmartOne
\(\sf\Large y=a(x-h)^2+k\) The vertex is \(\sf\Large (h,k)\)
anonymous
  • anonymous
so how would I incorporate (t+1)^2 = 25 in to that?
anonymous
  • anonymous
y = a(t-1)^2 + 25 ???? @TheSmartOne
TheSmartOne
  • TheSmartOne
one sec
TheSmartOne
  • TheSmartOne
You didn't properly complete the square.
anonymous
  • anonymous
?? what do you mean?
TheSmartOne
  • TheSmartOne
\(\sf\Large -16t^2 - 32t + 384 \color{red}{\neq} (t-1)^2+25\)
anonymous
  • anonymous
so the completed square form is (t - 1)^2 + 25
TheSmartOne
  • TheSmartOne
it isn't. That's wrong.
anonymous
  • anonymous
oh ok i got it. How would I find it then?
anonymous
  • anonymous
could you show me step by step
TheSmartOne
  • TheSmartOne
\(\sf\Large -16t^2 - 32t =-384\) What is \(\sf\Large\left( \frac{b}{2}\right)^2=\left( \frac{-32}{2}\right)^2=?\)
anonymous
  • anonymous
-16
TheSmartOne
  • TheSmartOne
correct; (-16)^2
anonymous
  • anonymous
whats next ??? @TheSmartOne
TheSmartOne
  • TheSmartOne
you have to add it to both sides: \(\sf\Large -16t^2 - 32t+(-16^2) =-384+(-16)^2\)
TheSmartOne
  • TheSmartOne
can you factor the left hand side? Hint: \(\sf\Large a^2+2ab+b^2=(a+b)^2\)
anonymous
  • anonymous
I am so sorry i lost wifi! @TheSmartOne
anonymous
  • anonymous
ok so you get -16t^2 - 32t + (-256) = -384 + (-256) @TheSmartOne
anonymous
  • anonymous
-16t^2 - 32t + (-256) = -640 @TheSmartOne
TheSmartOne
  • TheSmartOne
hmmm
anonymous
  • anonymous
@TheSmartOne
TheSmartOne
  • TheSmartOne
One sec, we'll need to backtrack over here
anonymous
  • anonymous
ok no problem
anonymous
  • anonymous
do you know what went wrong?
TheSmartOne
  • 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
  • TheSmartOne
@Mehek14 @paki I'm not sure how to complete the square :/
anonymous
  • anonymous
so the factored form is -16(t+6)(t-4)
TheSmartOne
  • TheSmartOne
there are so many ways to solve math, I don't know why they limit us with this completing the square lol
Mehek14
  • Mehek14
*disappears*
TheSmartOne
  • TheSmartOne
https://mathway.com/examples/Algebra/Quadratic-Equations/Solve-by-Completing-the-Square?id=29 mhmmm
anonymous
  • 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
  • anonymous
i just dont understand completing the square?
TheSmartOne
  • TheSmartOne
the link was an example of how to do it
anonymous
  • anonymous
oh ok so what would I do ??
TheSmartOne
  • TheSmartOne
yes, the symmetry of axis is x=-1
anonymous
  • 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
  • TheSmartOne
@zepdrix could you help us to complete the square for \(\sf\Large f(t)=-16t^2-32t+384\)
anonymous
  • anonymous
@zepdrix please I really need help!
Mehek14
  • Mehek14
I can tell you how to do the first step
Mehek14
  • Mehek14
divide all the numbers by -16
TheSmartOne
  • TheSmartOne
the calculator gives you the final answer, but no steps: http://prntscr.com/7metrk
TheSmartOne
  • TheSmartOne
@mathmate could you help us complete the square?
anonymous
  • anonymous
t^2 + 2t +24
anonymous
  • anonymous
OMG EVERYONE I THINK I MAY HAVE FIGURED IT OUT! GIVE ME ONE SEC!
TheSmartOne
  • TheSmartOne
we could have just as easily found the vertex by -b/2a ¯\_(ツ)_/¯
anonymous
  • 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
  • anonymous
@TheSmartOne
jtvatsim
  • jtvatsim
that's it, I've been working it out over here as well. nicely done. :)
anonymous
  • anonymous
thanks so much!!!! @jtvatsim
TheSmartOne
  • TheSmartOne
well we got the completing square out of the way
TheSmartOne
  • 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
  • anonymous
haha yep! so to find the vertex I wold just do y = -16(x-1)^2 +400??
anonymous
  • anonymous
@TheSmartOne
anonymous
  • anonymous
or in other words the vertex is (-1,400)
anonymous
  • anonymous
y = -16(x+1)^2 +400
TheSmartOne
  • TheSmartOne
correct
TheSmartOne
  • TheSmartOne
and that would be a maximum or a minimum?
anonymous
  • anonymous
maximum!
TheSmartOne
  • TheSmartOne
correct :)
anonymous
  • 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
  • Hero
I had to go to lunch. I would have helped otherwise.
anonymous
  • anonymous
yeah yeah whatever
TheSmartOne
  • TheSmartOne
I probably wouldn't have helped if it wasn't for Hero: http://prntscr.com/7mezsj
anonymous
  • anonymous
at least hero considered it...
mathmate
  • mathmate
@kaite_mcgowan Is the question resolved?
anonymous
  • anonymous
yes it is thanks for asking! @mathmate
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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