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 = -16x2 + 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.

Question

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 = -16x2 + 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.

anonymous · Answer

Please help! I'm having a hard time understanding how to do this and i have a test tomorrow.

anonymous · Answer

I'm sorry, those x2's are suppost to be x^2

anonymous · Answer

Ok, well this is going to be a parabola opening downwards. Let's draw a picture so you can see the path of the ball. Now in your equation: -16x^2 + 32x + 3, this is the equation of the formula that says: $$-16t ^{2}  + V _{0}t + h _{0}$$ Where t = time, V naught (the little 0) is your initial velocity, and h naught is your height.

anonymous · Answer

If your initial height is 3ft then h naught is = 3
And if your initial velocity is 32 ft/sec then V naught = 32

This is how we get the equation -16x^2 + 32x + 3, but in this case it is "x" instead of "t".

Are you with me so far?

anonymous · Answer

Yep, I understand so far.

anonymous · Answer

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