Aball is hit straight up in the air from a height of 4 feet with an initial speed of 64 feet per second. the height of the ball as a function of time can be modeled by the function h(t) = -16t•-16t + 64t + 4. what is the maximum height of the ball?

Step-by-step explanation:

The max height of the ball can be found by putting the position equation into vertex form.  This is done by completing the square.  We begin by setting the quadratic equal to 0 and moving over the constant to get:

-16t² + 64t = -4

The rule for completing the square is that the leading coefficient has to be a positive 1.  Ours is a -16, so we factor it out:

-16(t² - 4t) = -4

Next rule is to take half of the linear term, square it, and add it to both sides.  Our linear term is 4.  Half of 4 is 2, and 2 squared is 4.  BUT we cannot forget about the -16 siitting out front.  It is a multiplier.  That means we added in -16(4) which is -64.  That gives us:

-16(t² -4t + 4) = -4 - 64

The reason for doing this is to create a perfect square binomial on the left which will serve as the h value of our vertex (h, k).  Writing the left side in terms of the perfect square binomial and at the same time simplifying the right:

-16(t - 2)² = -68

Last we move the -68 back over by addition and set the quadratic back equal to y:

y = -16(t - 2)² + 68

From this we can ascertain that the vertex is located at (2, 68).  That means that at a time of 2 seconds, the height of the ball will be at its max value which is 68 feet.

after simplifying it i got: -36.25 as my answer. hope that !