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anonymous

  • one year ago

f(x) = -16t^2 + 96t what is the x and y intercept

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  1. freckles
    • one year ago
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    where is the function of x? is t suppose to be x?

  2. freckles
    • one year ago
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    well I mean that could be a function of x it would just be a constant function of x

  3. anonymous
    • one year ago
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    http://assets.openstudy.com/updates/attachments/53ce7ea6e4b081ba963957d9-coltjoco-1406041829668-paulkicks.jpg this is the graph

  4. Vocaloid
    • one year ago
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    |dw:1436976311744:dw|

  5. Vocaloid
    • one year ago
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    |dw:1436976350942:dw|

  6. Vocaloid
    • one year ago
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    the x-intercepts are the points where the function is equal to 0 the y-intercept is the point where x = 0

  7. anonymous
    • one year ago
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    is the x intercept 1,6?

  8. Vocaloid
    • one year ago
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    yes, or more specifically (1,0) and (6,0)

  9. anonymous
    • one year ago
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    also could you help me with a few more questions

  10. anonymous
    • one year ago
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    Find the vertex algebraically. Write as an ordered pair.

  11. anonymous
    • one year ago
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    What is the maximum height of the ball? Write your answer as an integer.

  12. freckles
    • one year ago
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    (1,0) isn't on the graph though :p

  13. anonymous
    • one year ago
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    so it would 6,0

  14. freckles
    • one year ago
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    (0,0) and (6,0) if we are talking about the function in the graph

  15. anonymous
    • one year ago
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    okay

  16. freckles
    • one year ago
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    \[f(x) = -16t^2 + 96t \] so I guess that that one x was suppose to be a t :p

  17. anonymous
    • one year ago
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    oh

  18. anonymous
    • one year ago
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    Find the vertex algebraically. Write as an ordered pair. would the vertex 6,0 ?

  19. freckles
    • one year ago
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    wait and it looks like they called the function h(t) instead of f(t)

  20. freckles
    • one year ago
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    but whatever

  21. freckles
    • one year ago
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    the vertex would be that peak point

  22. freckles
    • one year ago
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    if you wanted to find it using that graph you could observe that the vertex occurs midway between the x-intercepts

  23. anonymous
    • one year ago
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    which is 3?

  24. freckles
    • one year ago
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    I will also show you how to find it algebraically since you asked

  25. freckles
    • one year ago
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    but first using the graph the vertex appears to be at (3,a little above 140) I will let you approximate that y value

  26. freckles
    • one year ago
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    but anyways algebraically ...

  27. freckles
    • one year ago
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    \[y=at^2+bt+c \\ \text{ we want to put this in vertex form } \\\] \[y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c\] the vertex is given by: \[(-\frac{b}{2a},-a(\frac{b}{2a})^2+c)\]

  28. anonymous
    • one year ago
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    3o would the vertex be 3,14 or 3,140?

  29. freckles
    • one year ago
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    neither

  30. freckles
    • one year ago
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    look at the graph you should see where that mountain's peak is just a hair above 140 the y value isn't exactly 140 it is just above 140

  31. anonymous
    • one year ago
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    160?

  32. freckles
    • one year ago
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    160 is too high

  33. freckles
    • one year ago
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    it is closer to 140 than 160

  34. freckles
    • one year ago
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    |dw:1436977378401:dw|

  35. freckles
    • one year ago
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    |dw:1436977410220:dw|

  36. anonymous
    • one year ago
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    120

  37. freckles
    • one year ago
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    how is 120 just a little above 140 ?

  38. freckles
    • one year ago
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    you can find the y-coordinate of the vertex exactly by using the steps I gave you above

  39. anonymous
    • one year ago
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    sorry I was confused

  40. freckles
    • one year ago
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    but you can approximate it using the visual it looks like it could be maybe less than 150 but greater than 140 and I say less than 150 because the peak doesn't reach halfway between 140 and 160 it is less than halfway I would approximate saying 142 but that could be off a little because I did find this visually and not algebraically as the instructions|dw:1436977561263:dw| suggested

  41. freckles
    • one year ago
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    |dw:1436977596987:dw| so you know the vertex is gonna be somewhere close to (3,142)

  42. freckles
    • one year ago
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    you can find the answer algebraically using the method above I suggested

  43. freckles
    • one year ago
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    this one: \[y-c =at^2+bt \text{ I subtract } c \text{ on both sides } \\ \frac{y-c}{a} =t^2+\frac{b}{a} t \text{ I divided both sides by } a \\ \frac{y-c}{a} +(\frac{b}{2a})^2=t^2+\frac{b}{a}t +(\frac{b}{2a})^2 \\ \text{ I threw in a number on both sides so I could } \\ \text{ complete the square on one side } \\ \frac{y-c}{a}+(\frac{b}{2a})^2= (t+\frac{b}{2a})^2 \\ \text{ now \let's re-isolate } y \\ \text{ subtract} (\frac{b}{2a})^2 \text{ on both sides } \\ \frac{y-c}{a}=(t+\frac{b}{2a})^2-(\frac{b}{2a})^2 \\ \text{ multiply } a \text{ on both sides } \\ y-c=a(t+\frac{b}{2a})^2-a (\frac{b}{2a})^2 \\ \text{ add } c \text{ on both sides } \\ y=a(t+\frac{b}{2a})^2-a(\frac{b}{2a})^2+c\]

  44. anonymous
    • one year ago
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    I got 141 ?

  45. anonymous
    • one year ago
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    though I think it may be -3 I'm not sure

  46. freckles
    • one year ago
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    how did you get 141?

  47. freckles
    • one year ago
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    are you trying to put your quadratic in vertex from yet?

  48. anonymous
    • one year ago
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    I'm still using the graph for it

  49. freckles
    • one year ago
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    oh okay well don't forget it does say to find the vertex algebraically

  50. freckles
    • one year ago
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    using a graph to obtain the vertex is not finding it algebraically

  51. anonymous
    • one year ago
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    oh

  52. anonymous
    • one year ago
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    so it's not -3

  53. freckles
    • one year ago
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    are you talking about the x-coordinate of the vertex?

  54. anonymous
    • one year ago
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    vertex

  55. freckles
    • one year ago
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    no -3 is not the x-coordinate

  56. freckles
    • one year ago
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    |dw:1436978591061:dw|

  57. anonymous
    • one year ago
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    141 ?

  58. freckles
    • one year ago
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    the vertex is pretty close to (3,142) or I guess you could say (3,141) but we don't know the vertex exactly until you find it algebraically

  59. freckles
    • one year ago
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    it could be (3,144)

  60. freckles
    • one year ago
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    or (3,143)

  61. freckles
    • one year ago
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    but to find out we need to put your quadratic in vertex form

  62. freckles
    • one year ago
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    I gave a general way to do that above

  63. freckles
    • one year ago
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    |dw:1436979098053:dw| please look at this graph... |dw:1436979115849:dw|

  64. freckles
    • one year ago
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    again this is not finding the vertex algebraically this is approximating the vertex with the help of visual aid you can find the exact vertex by putting the quadratic into vertex form you will need to complete the square

  65. freckles
    • one year ago
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    \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+?)+16(?)\] what do you need to add inside the ( ) to complete the square

  66. anonymous
    • one year ago
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    1 ?

  67. freckles
    • one year ago
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    no recall that: \[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2 \]

  68. freckles
    • one year ago
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    the number in front of t is -6 let's call k, -6 so if k=-6 then k/2=?

  69. anonymous
    • one year ago
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    it would -6/2

  70. freckles
    • one year ago
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    \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2\] right can you finish putting into vertex form

  71. freckles
    • one year ago
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    by the way I didn't put the - in because (6/2)^2 is the same as (-6/2)^2 but you could have

  72. anonymous
    • one year ago
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    so the answer is -6

  73. freckles
    • one year ago
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    no

  74. freckles
    • one year ago
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    we already know we are suppose to get something close to (3,142) because this what we found visually

  75. freckles
    • one year ago
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    so I don't know what -6 is suppose to be

  76. anonymous
    • one year ago
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    would it 3,2

  77. freckles
    • one year ago
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    but anyways I was asking if you could finish putting into the vertex form

  78. freckles
    • one year ago
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    i have to ask are you playing with me?

  79. freckles
    • one year ago
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    like you keep giving me these y values that aren't closer to 140

  80. anonymous
    • one year ago
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    no I'm not i'm really stuck on this equation

  81. freckles
    • one year ago
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    I have shown you visually the y-value is a little above 140 we know the x-value to be exactly 3 the vertex is definitely somewhere close to (3, 142)

  82. anonymous
    • one year ago
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    which I had gotten 141

  83. freckles
    • one year ago
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    \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \] there I completed the square for you

  84. freckles
    • one year ago
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    as we see the x-coordinate of the vertex is 3 just evaluate 16(6/2)^2 to find the y coordinate of the vertex

  85. freckles
    • one year ago
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    @Aliypop yes I see you said 141 as approximate but then you asked me if it was 2,14,... and so on...

  86. anonymous
    • one year ago
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    I confused on what the x value is

  87. freckles
    • one year ago
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    the x-coordinate of the vertex is 3 we have found this both visually and algebraically now

  88. freckles
    • one year ago
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    the y-coordinate of the vertex is given by 16(6/2)^2 which we have approximated to be somewhere just above 140

  89. anonymous
    • one year ago
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    so is it (3,3)

  90. freckles
    • one year ago
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    no...

  91. freckles
    • one year ago
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    again 3 is nowhere near 140

  92. freckles
    • one year ago
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    the x-coordinate of the vertex is 3 the y-coordinate of the vertex is something just above 140 we have found this information visually we have also found algebraically the x-coordinate of the vertex is 3 the y-coordinate of the vertex is 16(6/2)^2

  93. freckles
    • one year ago
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    \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2)\]

  94. freckles
    • one year ago
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    all you have to do is finishing simplifying

  95. freckles
    • one year ago
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    do you need help doing 16(6/2)^2?

  96. freckles
    • one year ago
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    6 divided by 2 is 3 3 squared is 9 16*9 is...

  97. anonymous
    • one year ago
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    144

  98. freckles
    • one year ago
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    yes \[y=-16t^2+96t \\ \text{ factor out -16 } \\ y=-16(t^2-6t) \\ y=-16(t^2-6t+(\frac{6}{2})^2)+16(\frac{6}{2})^2 \\ y=-16(t-3)^2+16(\frac{6}{2})^2 \\ \text{ this says the vertex is } (3,16(\frac{6}{2})^2) =(3,144)\]

  99. freckles
    • one year ago
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    the vertex is exactly (3,144)

  100. freckles
    • one year ago
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    as we see this makes sense from our visual answer because we did say the vertex had x-coordinate 3 and y-coordinate of something just above 140

  101. anonymous
    • one year ago
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    okay

  102. anonymous
    • one year ago
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    What is the maximum height of the ball? Write your answer as an integer. this was the other question i need help on

  103. freckles
    • one year ago
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    you actually already found this when you found the vertex the vertex is the highest point and you know the y-value represents the height

  104. anonymous
    • one year ago
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    so it would 144

  105. freckles
    • one year ago
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    |dw:1436980560407:dw| yes the height is 144 and the time in which the height is 144 is t=3 sec if the time unit is second

  106. freckles
    • one year ago
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    though usually you would say 144 units where units is some kind of distance measurement but that isn't really given here

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