Aball 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 = x^2.

The height above ground (y) as function of time (x) is given as
y = -16x² + 32x + 3

Write the equation in the standard form for a parabola.
y = -16[x² - 2x] + 3
   = -16[(x-1)² - 1] + 3
   = -16(x-1)² + 16 + 3
y = -16(x-1)² + 19
This equation shows that the parabola is downward, has its vertex at (1,19), and has its axis of symmetry at x = 1.
A plot of the graph is shown below.

Transformations of y = x².
The equation for the parabola can be obtained from y = x² in the following steps:
1. Shift right by 1 to obtain y = (x-1)².
2. Dilate by -16 to obtain y = -16(x-1)².
3. Shift up by 19 to obtain y = -16(x-1)² + 19.

Two because two of one thing is two.