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Dominique throws a softball from the outfield. After 1 second, the ball is 30 feet high. After 3 seconds, the ball reaches its maximum height of 38 feet. After 5 seconds, it returns to a height of 30 feet.

What is the equation of the quadratic function that models the height of the ball h(t) at time t?

1 Answer

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Answer: h(t) = -2(t - 3)^2 + 38

Explanation:


To find the equation of the quadratic function that models the height of the ball h(t) at time t, we need to use the vertex form of a quadratic equation.


The vertex form of a quadratic equation is given by:


h(t) = a(t - h)^2 + k where (h, k) is the vertex of the parabola.


Given the information:


After 1 second, the ball is 30 feet high. This gives us the point (1, 30). After 3 seconds, the ball reaches its maximum height of 38 feet. This gives us the vertex (3, 38). After 5 seconds, it returns to a height of 30 feet. This gives us the point (5, 30).

We can use these three points to find the equation of the quadratic function.

1. Find the value of a: Using the vertex form, we substitute the vertex point (3, 38) into the equation: 38 = a(3 - 3)^2 + k 38 = a(0)^2 + k 38 = k


2. Substitute the vertex point and another point into the equation:


Using the point (1, 30), we substitute it into the equation: 30 = a(1 - 3)^2 + 38 30 = a(-2)^2 + 38 30 = 4a + 38 4a = -8 a = -2 3.


Substitute the value of a and the vertex point into the equation: The equation becomes: h(t) = -2(t - 3)^2 + 38

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