**Answers: a) 6.3m b) 1.0s c) 1.4m**

Given the quadratic, `h(t) =-4.9text(m)/text(s^2)(t-1text(s))^2+6.3text(m)` , you are to determine the following:

1. Maximum height.

2. Time to maximum height.

3. Height of trampoline.

The quadratic graphs a parabola. Knowing a bit about its characteristics will help understand questions like this.

This equation is already in vertex-format (h,k):

`y(t) = a(t-h)^2+k`

Where h is *time* to maximum height, and k is the maximum. (This is the case because a is negative - otherwise k is the minimum).

The time to the maximum is 1.0 seconds, and the heighest point is 6.3 m

So far so good. The height of the trampoline is another story. Let's take a look at the graph and see what we find:

The graph verifies our first answers! Notice that the graph intersects the y-axis. Remember that y represents height and x represents time. The y-intercept happens when x is zero. In terms of the problem, this is where the trampoline artist is at time=0, probably at the trampoline.

So, evaluate h(0) and you will have your last answer!

`h(0)=-4.9text(m)/text(s^2)((0)-1)^2+6.3text(m) `

`h(0)=-4.9text(m)/text(s^2)(1text(s^2))+6.3text(m) `

`h(0) = 1.4text(m)`

The trampoline must be 1.4 m high off the ground. Now as the graph and equation suggests, at 2s, either the trampoline artist misses the trampoline or the graph starts over again - you decide :) Good job!

**Answers: a) 6.3 m b) 1.0s c) 1.4m**

**Further Reading**

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