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A basketball player shoots a ball towards the hoop.

The trajectory of the ball can be modeled by the quadratic functionf(x) = −(x−1.4)2+6
, where x represents the horizontal distance in feet from the player, and f(x) represents the height of the ball in feet.



(1 point)
Part A: Identify the vertex (write your answer as an ordered pair):

Part B: Complete the table

x f(x)
0
1
2
3


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2 answers

To analyze the quadratic function \( f(x) = -(x - 1.4)^2 + 6 \), let's start with Part A and find the vertex of the function.

Part A: Identify the Vertex

The vertex form of a quadratic function is given by \( f(x) = a(x - h)^2 + k \), where \( (h, k) \) is the vertex.

In the given function:

  • \( h = 1.4 \)
  • \( k = 6 \)

Thus, the vertex is: Vertex: \( (1.4, 6) \)

Part B: Complete the Table

Next, we will calculate the values of \( f(x) \) for given values of \( x \) (0, 1, 2, and 3).

  1. For \( x = 0 \): \[ f(0) = -((0 - 1.4)^2) + 6 = -(1.96) + 6 = 4.04 \]

  2. For \( x = 1 \): \[ f(1) = -((1 - 1.4)^2) + 6 = -(-0.4^2) + 6 = -0.16 + 6 = 5.84 \]

  3. For \( x = 2 \): \[ f(2) = -((2 - 1.4)^2) + 6 = -(0.6^2) + 6 = -0.36 + 6 = 5.64 \]

  4. For \( x = 3 \): \[ f(3) = -((3 - 1.4)^2) + 6 = -1.6^2 + 6 = -2.56 + 6 = 3.44 \]

Now we can fill in the table:

\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 0 & 4.04 \ 1 & 5.84 \ 2 & 5.64 \ 3 & 3.44 \ \hline \end{array} \]

Final answers:

Part A: Vertex: \( (1.4, 6) \)

Part B: Table: \[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 0 & 4.04 \ 1 & 5.84 \ 2 & 5.64 \ 3 & 3.44 \ \hline \end{array} \]

the table says misplaced /hline