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anonymous

  • one year ago

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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  1. anonymous
    • one year ago
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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.

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

  3. anonymous
    • one year ago
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    @Hero @Hero you said you would help me!

  4. Hero
    • one year ago
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    I said, post your next question separately. Anyone can help with these.

  5. anonymous
    • one year ago
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    Yeah I did post it separately and you said you would help!

  6. anonymous
    • one year ago
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    I need help on finding the vertex and I am confused on how to

  7. TheSmartOne
    • one year ago
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    For Part A, you are correct (almost)

  8. anonymous
    • one year ago
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    What did I leave out? @TheSmartOne could you please explain

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

  10. anonymous
    • one year ago
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    so the t in t-4 is supposed to be x-4? @TheSmartOne

  11. TheSmartOne
    • one year ago
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    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 :)

  12. anonymous
    • one year ago
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    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

  13. TheSmartOne
    • one year ago
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    \(\sf\Large y=a(x-h)^2+k\) The vertex is \(\sf\Large (h,k)\)

  14. anonymous
    • one year ago
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    so how would I incorporate (t+1)^2 = 25 in to that?

  15. anonymous
    • one year ago
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    y = a(t-1)^2 + 25 ???? @TheSmartOne

  16. TheSmartOne
    • one year ago
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    one sec

  17. TheSmartOne
    • one year ago
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    You didn't properly complete the square.

  18. anonymous
    • one year ago
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    ?? what do you mean?

  19. TheSmartOne
    • one year ago
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    \(\sf\Large -16t^2 - 32t + 384 \color{red}{\neq} (t-1)^2+25\)

  20. anonymous
    • one year ago
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    so the completed square form is (t - 1)^2 + 25

  21. TheSmartOne
    • one year ago
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    it isn't. That's wrong.

  22. anonymous
    • one year ago
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    oh ok i got it. How would I find it then?

  23. anonymous
    • one year ago
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    could you show me step by step

  24. TheSmartOne
    • one year ago
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    \(\sf\Large -16t^2 - 32t =-384\) What is \(\sf\Large\left( \frac{b}{2}\right)^2=\left( \frac{-32}{2}\right)^2=?\)

  25. anonymous
    • one year ago
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    -16

  26. TheSmartOne
    • one year ago
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    correct; (-16)^2

  27. anonymous
    • one year ago
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    whats next ??? @TheSmartOne

  28. TheSmartOne
    • one year ago
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    you have to add it to both sides: \(\sf\Large -16t^2 - 32t+(-16^2) =-384+(-16)^2\)

  29. TheSmartOne
    • one year ago
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    can you factor the left hand side? Hint: \(\sf\Large a^2+2ab+b^2=(a+b)^2\)

  30. anonymous
    • one year ago
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    I am so sorry i lost wifi! @TheSmartOne

  31. anonymous
    • one year ago
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    ok so you get -16t^2 - 32t + (-256) = -384 + (-256) @TheSmartOne

  32. anonymous
    • one year ago
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    -16t^2 - 32t + (-256) = -640 @TheSmartOne

  33. TheSmartOne
    • one year ago
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    hmmm

  34. anonymous
    • one year ago
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    @TheSmartOne

  35. TheSmartOne
    • one year ago
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    One sec, we'll need to backtrack over here

  36. anonymous
    • one year ago
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    ok no problem

  37. anonymous
    • one year ago
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    do you know what went wrong?

  38. TheSmartOne
    • one year ago
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    we have to first factor the 16 out of the equation, so: \(\sf\Large -16t^2 - 32t + 384 =-16(t^2+2t-24) \)

  39. TheSmartOne
    • one year ago
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    @Mehek14 @paki I'm not sure how to complete the square :/

  40. anonymous
    • one year ago
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    so the factored form is -16(t+6)(t-4)

  41. TheSmartOne
    • one year ago
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    there are so many ways to solve math, I don't know why they limit us with this completing the square lol

  42. Mehek14
    • one year ago
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    *disappears*

  43. TheSmartOne
    • one year ago
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    https://mathway.com/examples/Algebra/Quadratic-Equations/Solve-by-Completing-the-Square?id=29 mhmmm

  44. anonymous
    • one year ago
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    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???

  45. anonymous
    • one year ago
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    i just dont understand completing the square?

  46. TheSmartOne
    • one year ago
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    the link was an example of how to do it

  47. anonymous
    • one year ago
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    oh ok so what would I do ??

  48. TheSmartOne
    • one year ago
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    yes, the symmetry of axis is x=-1

  49. anonymous
    • one year ago
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    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

  50. TheSmartOne
    • one year ago
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    @zepdrix could you help us to complete the square for \(\sf\Large f(t)=-16t^2-32t+384\)

  51. anonymous
    • one year ago
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    @zepdrix please I really need help!

  52. Mehek14
    • one year ago
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    I can tell you how to do the first step

  53. Mehek14
    • one year ago
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    divide all the numbers by -16

  54. TheSmartOne
    • one year ago
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    the calculator gives you the final answer, but no steps: http://prntscr.com/7metrk

  55. TheSmartOne
    • one year ago
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    @mathmate could you help us complete the square?

  56. anonymous
    • one year ago
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    t^2 + 2t +24

  57. anonymous
    • one year ago
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    OMG EVERYONE I THINK I MAY HAVE FIGURED IT OUT! GIVE ME ONE SEC!

  58. TheSmartOne
    • one year ago
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    we could have just as easily found the vertex by -b/2a ¯\_(ツ)_/¯

  59. anonymous
    • one year ago
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    a(x+d)2+e d=−32 / 2⋅(−16) d=1 e=384− (−32)^2 / 4⋅(−16) e=400 −16(t+1)2+400

  60. anonymous
    • one year ago
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    @TheSmartOne

  61. jtvatsim
    • one year ago
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    that's it, I've been working it out over here as well. nicely done. :)

  62. anonymous
    • one year ago
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    thanks so much!!!! @jtvatsim

  63. TheSmartOne
    • one year ago
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    well we got the completing square out of the way

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

  65. anonymous
    • one year ago
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    haha yep! so to find the vertex I wold just do y = -16(x-1)^2 +400??

  66. anonymous
    • one year ago
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    @TheSmartOne

  67. anonymous
    • one year ago
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    or in other words the vertex is (-1,400)

  68. anonymous
    • one year ago
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    y = -16(x+1)^2 +400

  69. TheSmartOne
    • one year ago
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    correct

  70. TheSmartOne
    • one year ago
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    and that would be a maximum or a minimum?

  71. anonymous
    • one year ago
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    maximum!

  72. TheSmartOne
    • one year ago
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    correct :)

  73. anonymous
    • one year ago
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    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

  74. Hero
    • one year ago
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    I had to go to lunch. I would have helped otherwise.

  75. anonymous
    • one year ago
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    yeah yeah whatever

  76. TheSmartOne
    • one year ago
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    I probably wouldn't have helped if it wasn't for Hero: http://prntscr.com/7mezsj

  77. anonymous
    • one year ago
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    at least hero considered it...

  78. mathmate
    • one year ago
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    @kaite_mcgowan Is the question resolved?

  79. anonymous
    • one year ago
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    yes it is thanks for asking! @mathmate

  80. mathmate
    • one year ago
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    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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