The height of a ball thrown into the air is a quadratic function of time, the ball is thrown from a height of 6 ft above the ground. After 1 second, the ball reaches its maximum height of 22 ft above the ground, write the equation of the function in vertex form.

The correct answer is: Hence, the equation of the function in vertex form is -16(t-1)2 + 22.

Let’s say that the quadratic equation representing the height of a ball thrown into the air as a function of time is h = at² + bt + c

         At t = 0, h = 6 ft

             6 = 0 + 0 + c

c = 6

                            At t = 1 second, h = 22 ft

                          22 = a(1)² + b(1) + 6
a + b = 16 …..(1)
Now, after 2 seconds, the ball will reach the height from which it is thrown i.e. 6 ft

                   At t = 2 seconds, h = 6 ft

                 6 = a(2)² + b(2) + 6

  4a + 2b = 0

        b = -2a   (2)
From equations 1 and 2
a + (-2a) = 16
a = -16
and b = 32
So, the quadratic equation is h = -16t² + 32t + 6 = -16(t-1)² + 22
Final Answer:
Hence, the equation of the function in vertex form is -16(t-1)² + 22.