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

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

Postby frankie10 on Wed Jul 22, 2009 7:15 pm

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

Postby Honeysuckle588 on Thu Jul 23, 2009 4:46 am

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

Postby frankie10 on Thu Jul 23, 2009 5:11 pm

no,i'm not sure how
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Re: height above trampoline modelled by h = -4.9(t - 1)^2 + 6.3  TOPIC_SOLVED

Postby Honeysuckle588 on Thu Jul 23, 2009 8:15 pm

Ok...This particular quadratic is written in the form


which is just a transformation of the basic 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?
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