To determine the equation of the quadratic function that models the height of the ball \( h(t) \) at time \( t \), we need to use the given points and the standard form of the vertex equation of a parabola.
Given points:
- After 1 second, the ball is 15 feet high: (1, 15)
- After 4 seconds, the ball reaches its maximum height of 42 feet: (4, 42)
- After 7 seconds, it returns to a height of 15 feet: (7, 15)
We are given three key points on the parabolic flight path: the vertex \( (4, 42) \) and two other points \( (1, 15) \) and \( (7, 15) \).
The vertex form of a quadratic equation is:
[tex]\[
h(t) = a(t - h_{\text{shift}})^2 + k
\][/tex]
where \( (h_{\text{shift}}, k) \) is the vertex of the parabola. From the problem, we know the vertex \((h_{\text{shift}}, k)\) is \( (4, 42) \).
Thus, we substitute \( h_{\text{shift}} = 4 \) and \( k = 42 \):
[tex]\[
h(t) = a(t - 4)^2 + 42
\][/tex]
Next, we need to find the coefficient \( a \) by plugging in one of the other points, say \( (1, 15) \).
Plugging in \( t = 1 \) and \( h(t) = 15 \):
[tex]\[
15 = a(1 - 4)^2 + 42
\][/tex]
Simplify the equation:
[tex]\[
15 = a(-3)^2 + 42
\][/tex]
[tex]\[
15 = 9a + 42
\][/tex]
Subtract 42 from both sides to solve for \( a \):
[tex]\[
15 - 42 = 9a
\][/tex]
[tex]\[
-27 = 9a
\][/tex]
[tex]\[
a = -3
\][/tex]
Thus, the quadratic function is:
[tex]\[
h(t) = -3(t - 4)^2 + 42
\][/tex]
To confirm, we verify with another point, \( (7, 15) \):
[tex]\[
h(7) = -3(7 - 4)^2 + 42
\][/tex]
[tex]\[
h(7) = -3(3)^2 + 42
\][/tex]
[tex]\[
h(7) = -3(9) + 42
\][/tex]
[tex]\[
h(7) = -27 + 42
\][/tex]
[tex]\[
h(7) = 15
\][/tex]
Both points satisfy the equation, confirming our solution is correct. Therefore, the equation modeling the height of the ball is:
[tex]\[
h(t) = -3(t - 4)^2 + 42
\][/tex]
Hence, the correct option is:
[tex]\[
h(t) = -3(t - 4)^2 + 42
\][/tex]
