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 time in seconds. Complete the input-output table: x h(𝑥) 0 1 2 3 4

Answer 1

To complete the input-output table for the function \( h(t) = -6t^2 + 33t + 6 \), we need to calculate the height \( h(t) \) at each given time \( t \) (where \( t \) is 0, 1, 2, 3, and 4 seconds).

Let's calculate \( h(t) \) for each value of \( t \):

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 fill in the input-output table:

\[ \begin{array}{|c|c|} \hline t & h(t) \ \hline 0 & 6 \ 1 & 33 \ 2 & 48 \ 3 & 51 \ 4 & 42 \ \hline \end{array} \]

Here’s the completed input-output table:

\[ \begin{array}{|c|c|} \hline t & h(t) \ \hline 0 & 6 \ 1 & 33 \ 2 & 48 \ 3 & 51 \ 4 & 42 \ \hline \end{array} \]