## height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

### height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3

A trampoline artist bounces on a trampoline.Her height above ground is modelled by the function h=-4.9(t-1)^2+6.3.

how long after she leaves the trampoline does she reach her maximum height.
frankie10

Posts: 24
Joined: Wed Jun 17, 2009 3:49 pm

### Re: height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3

This type of function is called a quadratic whose graph is a parabola. The highest(lowest) point of a parabola is called the vertex. Do you know how to find the vertex of a parabola?
Honeysuckle588

Posts: 23
Joined: Wed Jul 15, 2009 4:07 pm

### Re: height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3

no,i'm not sure how
frankie10

Posts: 24
Joined: Wed Jun 17, 2009 3:49 pm

### Re: height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3

Ok...This particular quadratic is written in the form
$y = a(x-b)^2+c$

which is just a transformation of the basic $y = x^2$ quadratic which has a vertex at (0, 0). The two most important transformations for your question are the value of b and c.

The b value tells you how far to the right the vertex has moved, and the c value tells you how far up the vertex has moved.

In your case, the b value is 1 and the c value is 6.3.

So your vertex is at (1, 6.3). Since this is a coordinate it has the form (t, h) for your equation.

With this information, can you tell me when the highest point (i.e. the vertex) is reached?
Honeysuckle588

Posts: 23
Joined: Wed Jul 15, 2009 4:07 pm