An object is launched directly in the air at a speed of 48 feet per second from a platform located 12 feet above the ground. The position of the object can be modeled using the function f(t)=−16t2+48t+12, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

9514 1404 393

  48 ft

Step-by-step explanation:

For the quadratic ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For the given quadratic, which defines a parabola opening downward, the axis of symmetry defines the time at which the maximum height is reached.

  t = -48/(2(-16)) = 1.5

Then the maximum height is ...

  f(1.5) = (-16·1.5 +48)1.5 +12 = (24·1.5) +12

  f(1.5) = 48

The maximum height the object will reach is 48 feet.

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )

the 2 figures are congruent bc if you reflect abcd over the y axis then over the d axis the figures overlap