A volleyball is served from a height of 6 feet with an initial velocity of 31 feet per second. The situation is modeled using the equation h=−16t2+31t+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, and 2?(1 point) Responses (0,6), (1,21), (2,4) (0,6), (1,21), (2,4) (0,0), (1,27), (2,2) (0,0), (1,27), (2,2) (0,6), (1,31), (2,4) (0,6), (1,31), (2,4) (6,0), (31,1), (4,2) (6,0), (31,1), (4,2)

1 answer

To determine the correct y-values (heights) for the given x-values (time in seconds) using the function \( h = -16t^2 + 31t + 6 \), we will evaluate the function at \( t = 0 \), \( t = 1 \), and \( t = 2 \).

  1. When \( t = 0 \): \[ h = -16(0)^2 + 31(0) + 6 = 6 \] So the point is \( (0, 6) \).

  2. When \( t = 1 \): \[ h = -16(1)^2 + 31(1) + 6 = -16 + 31 + 6 = 21 \] So the point is \( (1, 21) \).

  3. When \( t = 2 \): \[ h = -16(2)^2 + 31(2) + 6 = -16(4) + 62 + 6 = -64 + 62 + 6 = 4 \] So the point is \( (2, 4) \).

Combining these results, we have the points \( (0, 6) \), \( (1, 21) \), and \( (2, 4) \).

Thus, the correct response is: (0, 6), (1, 21), (2, 4).