A ball is thrown straight up from 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+3Explain 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 given equation has a y-intercept at (0, 3).
y = -16x^2 + 32x + 3 = -16(x^2 - 2x) + 3 = -16(x - 1)^2 + 19. This means the vertex is at (1, 19).
To transform the y = x^2 graph:
First we invert the graph with respect to the x-axis, maxing it a downward parabola y = -x^2.
Next, we move its vertex from the origin (0, 0) to (1, 19), making the equation y = -(x - 1)^2 + 19.
Third, we "expand" the opening of the parabola such that it passes through the y-intercept of (0, 3). The right-side of the parabola should also be expanded similarly, since it is symmetric.