Please help!! 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 = x^2.
Accepted Solution
A:
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.