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anonymous

  • one year ago

A ball is thrown into the air with an upward velocity of 32 ft/s. Its height h in feet after t seconds is given by the function h = -16t^2 + 32t + 6. In how many seconds does the ball reach its maximum height? Round to the nearest hundredth if necessary. What is the ball’s maximum height? A) 1 s; 22 ft B) 2 s; 22 ft C) 2 s; 6 ft D) b1 s; 54 ft

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  1. freckles
    • one year ago
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    find the vertex of the parabola

  2. freckles
    • one year ago
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    oh wait your function isn't a parabola

  3. freckles
    • one year ago
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    and it should be interesting

  4. freckles
    • one year ago
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    \[\text{ maybe you meant } h=-16t^2+32t+6 ?\]

  5. anonymous
    • one year ago
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    wait let me recheck

  6. anonymous
    • one year ago
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    yeah sorry forgot the ^2

  7. freckles
    • one year ago
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    \[h=at^2+bt+c \\ h=a(t^2+\frac{b}{a}t)+c \\ h=a(t^2+\frac{b}{a}t+(\frac{b}{2a})^2)+c-a(\frac{b}{2a})^2 \\ h=a(t+\frac{b}{2a})^2+c-a (\frac{b}{2a})^2 \text{ is \in vertex form } \\ \text{ so the vertex is } (\frac{-b}{2a},c-a (\frac{b}{2a})^2)\]

  8. freckles
    • one year ago
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    so if a<0 then you have a max and if a>0 then you have a min we have the first case so \[c-a(\frac{b}{2a})^2 \text{ the max } \\ \text{ and } \\ \frac{-b}{2a} \text{ will be where the max occurs }\]

  9. anonymous
    • one year ago
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    I'm kinda confused.. I don't really get it

  10. freckles
    • one year ago
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    do you need an example of writing in vertex form?

  11. anonymous
    • one year ago
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    there's too much letters, I really got confused

  12. freckles
    • one year ago
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    \[h=2t^2+5t+5\] we are going to follow the same steps above: \[h=(2t^2+5t)+5 \\ h=2(t^2+\frac{5}{2}t)+5 \\ h=2(t^2+\frac{5}{2}t+(\frac{5}{2 \cdot 2})^2)+5 -2 \cdot (\frac{5}{2 \cdot 2})^2 \\ h=2(t+\frac{5}{2 \cdot 2 })^2+5-2 (\frac{5 }{ 2 \cdot 2})^2 \\ \text{ so the vertex here is } (\frac{-5}{2 \cdot 2 },5-2 (\frac{5}{2 \cdot 2})^2) \\ \text{ you can simplify } (\frac{-5}{4},5-2 (\frac{25}{16})) \\ \text{ still simplifying } (\frac{-5}{4},5-\frac{25}{8}) \\ \text{ sill simplifying } (\frac{-5}{4}, \frac{40-25}{8}) \\ \text{ and finally } (\frac{-5}{4},\frac{15}{8})\]

  13. freckles
    • one year ago
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    i used completing the square to do this if it wasn't obvious \[x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2\]

  14. freckles
    • one year ago
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    \[h=-16t^2+32t+6 \\ h=(-16t^2+32t)+6\] first step look at the terms in ( ) and factor out what is in front of your first term inside

  15. freckles
    • one year ago
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    notice that is -16

  16. freckles
    • one year ago
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    \[h=-16(t^2-2t)+6 \\ h=-16(t^2-2t+?)+6+16(?) \text{ this step notice I added in a zero } \\ \text{ this zero it to keep the equation equivalent and to complete the square } \\ \text{ the zero I added in was of the form } -16(?)+16(?) \]

  17. freckles
    • one year ago
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    compare \[t^2-2t+? \text{ to the left hand side of } \\ t^2+kt+(\frac{k}{2})^2=(t+\frac{k}{2})^2\] what do we need to make k and therefore the ? mark is ?

  18. freckles
    • one year ago
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    oops I should have ended that question with punctuation because that might be confusing :p

  19. freckles
    • one year ago
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    you know since I'm already using a ? for a constant

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