a quarterback throws a ball in the air from an initial height of 5 feet. The football has an initial vertical velocity of 50 feet per second. Another player catches the ball when it is 2 feet above the ground. How long is the ball in the air?

The range of the function is from 2 feet to 12.25 feet

Step-by-step explanation:

In a function f(x) = y

x is the domain of the function

y is the range of the function

The height (h) of the ball at time (t) seconds can be represented by the equation h(t) = - 16 t² + 20 t + 6

∵ h(t) = - 16 t² + 20 t + 6

∴ The domain is t

∴ The range is h(t)

- To find the range of the quadratic function find the maximum or

minimum value of it

∵ The leading coefficient of the function is -16

∴ The function has a maximum value

To find the maximum value differentiate h(t) with respect to t and equate it by 0 to find the value of t for the maximum height

∵ h'(t) = -16(2) t + 20(1)

∴ h'(t) = -32 t + 20

- Equate it by 0

∵ h'(t) = 0

∴ -32 t + 20 = 0

- Subtract 20 from both sides

∴ -32 t = - 20

- Divide both sides by -32

∴ t = 0.625 seconds ⇒ time for the maximum height

Substitute the value of t in h(t) to find the maximum height

∵ h(0.625) = -16(0.625)² + 20(0.625) + 6

∴ h(t) = 12.25 feet

∴ The maximum height of the ball is 12.25 feet

∵ The ball is caught at 2 feet

∴ The range of the function is 2 ≤ h(t) ≤ 12.25

The range of the function is from 2 feet to 12.25 feet

option a.

step-by-step explanation:

if you look closely at the graph, we can notice that the function is not defined at two different points.

if you look at functions b and c, you will notice that both functions are not defined at just one point. the function b is not defined at x= -5/2 and function c is not defined at x=-5. both options are discarded, given that our function should not be defined at two fiferent points.

now, if you look closely at the graph you will notice that the function sketched is not defined at x=5 and x=-2.