Question

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? h(t) = −2(t − 3)2 + 38 h(t) = 2(t + 3)2 + 38 h(t) = −(t − 3)2 + 38 h(t) = (t + 3)2 + 38

Answer 1

(a) h(t) = -2(t -3)² +38

Explanation:

You want to know the equation that models the height of a softball that is 30 ft high 1 second after being thrown, and a maximum of 38 ft high 3 seconds after being thrown.

Vertex form

The height of the ball is modeled by a quadratic equation. Since we are given the maximum height and time, we can use the vertex form of the equation:

h(t) = c(t -a)² +b . . . . . . . . where (a, b) is the vertex of the path

h(t) = c(t -3)² +38 . . . . . . . using (3, 38) as the vertex

To find the constant c, we can substitute the other point values given.

(t, h) = (1, 30)

30 = c(1 -3)²) +38 = 4c +38

-8 = 4c . . . . . . subtract 38

c = -2 . . . . . . . divide by 4

The equation that models the height is h(t) = -2(t -3)² +38, choice A.

<95141404393>