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14. 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=−6t2+33t+6
, where h is the height in feet and t is the time in seconds.

Use the function to calculate the h(x) values in the table below.

(3 points)
x h(x)
0
2
4


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1 answer

To find the height \( h(x) \) at different time values \( t \), we can plug in the values of \( t \) into the function \( h(t) = -6t^2 + 33t + 6 \).

Let's compute \( h(t) \) for the given \( t \) values:

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

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

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

Now we can fill in the values in the table:

| x | h(x) | |---|---| | 0 | 6 | | 2 | 48 | | 4 | 42 |

Thus, the final values of the table are:

  • \( h(0) = 6 \)
  • \( h(2) = 48 \)
  • \( h(4) = 42 \)