A ball is thrown straight up from a height of 3 ft with a speed of 32 ft/s. Its height above the ground after x seconds is given by the quadratic function y = - 16x^2 + 32x + 3. Explain the steps you would use to determine the path of the ball in terms of a transformation of the graph of y = x2.

To answer this question you need to transform the given equation of the parabola to the vertex form.

You do that in this way:

y = - 16x^2 + 32x + 3

Extract common factor - 16 for the first two terms

y = - 16 (x^2 - 2x) + 3

Complete a perfect square

y = - 16 (x - 1) ^2 + 16 + 3 = - 16 (x - 1) ^2 + 19

Which using f (x) = x^2 is - 16[f (x-1) ] + 19

That means that you use these steps:

1) shift f (x) one unit to the right,

2) reflect it over the x-axys, due to the negative sign in front of 16 (this changes the parabola from open up ward to open down ward)

3) multiply by the factor 16 (this stretches the graph)

4) shift the graph19 units up.