Question

A rock is thrown upward from a bridge that is 22 feet above a road. The rock reaches its maximum height above the road 0.69 seconds after it is thrown and contacts the road 2.53 seconds after it was thrown.
Write a function ff that determines the rock's height above the road (in feet) in terms of the number of seconds tt since the rock was thrown.

f(t)=_________.

Hint: the function ff can be written in the form f(t)=c⋅(t−t1)(t−t2)f(t)=c⋅(t-t1)(t-t2) for fixed numbers cc, t1t1, and t2t2.

Answer 1

Answer:

• f(t) = -7.561(t + 1.15)(t - 2.53)

Explanation:

The hint is that the function f can be written in the form:

• f(t) = c (t - t₁) (t - t₂)

It is a quadratic function, i.e. a parabola.

The maximum point, maximum height, of a parabola is its vertex.

And the x-coordinate of the vertex is at the midpoint between the two x-intercepts.

The time when the rock reaches the road is one x-intercept: t = 2.53 seconds: (2.53, 0).

The other x-intercept is sucht that (x + 2.53) / 2 = 0.69

• (x + 2.53) / 2 = 0.69
• x + 2.53 = 1.38
• x = 1.38 - 2.53
• x = -1.15

Thus, the equation is:

• f(t) = c(t + 1.15) (t - 2.53).

You can find c by doing t = 0, because at t = 0 the height, f(t), is 22 feet:

• f(0) = 22 = c (0 +1.15)(0 - 2.53)
• 22 = c(1.15)(-2.53)
• 22 = c (-2.9095)
• c = - 22/(2.9095)
• c = - 7.561

Then, the function can be written as:

• f(t) = -7.561(t + 1.15) (t - 2.53) ← answer