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A volleyball is served from a height of 6 feet with an initial velocity of 33 feet per second. The situation is modeled using the function h=−6^t2+33t+6

, where h is the height in feet and t is time in seconds.

Which of the following uses the function to correctly identify the corresponding y-values to the given x-values of 0, 1, 2, 3, and 4?

(1 point)
Responses

(0, 6), (1, 33), (2, 48), (3, 51) and (4, 42)
(0, 6), (1, 33), (2, 48), (3, 51) and (4, 42)

(6, 0), (33, 1), (48, 2), (51, 3), and (42, 4)
(6, 0), (33, 1), (48, 2), (51, 3), and (42, 4)

(0, 0), (1, 27), (2, 42), (3, 45), and (4, 36)
(0, 0), (1, 27), (2, 42), (3, 45), and (4, 36)

(0, 6), (1, 21), (2, 48), (3, 69), and (4, 90)

1 answer

To evaluate the given function \( h(t) = -6t^2 + 33t + 6 \) for the specified values of \( t \), we need to compute the height \( h \) at \( t = 0, 1, 2, 3, \) and \( 4 \) seconds.

  1. For \( t = 0 \): \[ h(0) = -6(0)^2 + 33(0) + 6 = 6 \]

  2. For \( t = 1 \): \[ h(1) = -6(1)^2 + 33(1) + 6 = -6 + 33 + 6 = 33 \]

  3. For \( t = 2 \): \[ h(2) = -6(2)^2 + 33(2) + 6 = -6(4) + 66 + 6 = -24 + 66 + 6 = 48 \]

  4. For \( t = 3 \): \[ h(3) = -6(3)^2 + 33(3) + 6 = -6(9) + 99 + 6 = -54 + 99 + 6 = 51 \]

  5. For \( t = 4 \): \[ h(4) = -6(4)^2 + 33(4) + 6 = -6(16) + 132 + 6 = -96 + 132 + 6 = 42 \]

Now we can summarize the results:

  • \( (0, 6) \)
  • \( (1, 33) \)
  • \( (2, 48) \)
  • \( (3, 51) \)
  • \( (4, 42) \)

Looking at the options provided:

Correct Answers:

  • (0, 6), (1, 33), (2, 48), (3, 51), and (4, 42)

So the correct response is: (0, 6), (1, 33), (2, 48), (3, 51) and (4, 42).