Aball is thrown in the air that is 48 feet above the ground with an initial vertical velocity of 32 feet per second. the height of the ball, in feet, can be represented by the function h(t)= -16t^2 + 32t + 48, where t is the time, in seconds, since the ball was thrown. what function rewritten in the form that would be best to use to identify the maximum height of the ball?

h(t) = -16(t - 1)² + 64

Step-by-step explanation:

The function that would best to use to identify the maximum height of the ball is the vertex form of the parabola.

The standard form of a quadratic function is  

y = ax² + bx + c

The vertex form is

y = a(x - h)² + k

where (h, k) is the vertex of the parabola.

h = -b/(2a) and k = f(h)

In your equation, h(t) = -16t^2 + 32t + 48

a = -16; b = 32; c = 48

Calculate h

h = -32/[2(-16)]

h = (-32)/(-32)

h = 1

Calculate k

k = -16(1)² + 32×1 + 48

k = -16      + 32    + 48

k =  64

So, h = 1, k = 64, a = -16

The vertex form of the equation is  h(t) = -16(t - 1)² + 64.

The graph below shows h(t) with the ball at its maximum height of 64 ft one second after being thrown.

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!