A ball is thrown straight up from the height of 3 ft with a speed of 32 ft/s. It’s 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.

Answer:

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.

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Will mark brainliest if correct

GUYS THIS IS DUE SOON I NEED HELP

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} and the range of the relation is {
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for m(x)=-2x+7, find the value of x for which m(x)=13

Please help will give brainliest 19-26

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