Question

a quarterback throws a football from a height of 6 feet with an initial velocity of 20 feet per seconds. the height (h) of the ball at time (t) seconds can be represented by the equation h(t)=-16(t)^2 + 20t + 6. if the ball is caught at 2 feet, find the range of the function

Answer 1

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

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You can learn more about the quadratic equations in brainly.com/question/9390381

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Answer 2

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

The given problem is based on the fundamentals of function. Consider a function, f(x) = y. Then in the given function,  

• 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,

$h(t) = -16t^{2}+20t+6$

Here, the domain is t and the range is h(t).

 

Now, calculate the range of the quadratic function find the maximum or

minimum value.  

• The leading coefficient of the function is -16.

• 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.

Solving as,

$h'(t) = -16(2) t + 20(1) \\\\h'(t) = -32 t + 20$

Now,

$h'(t) = 0 \\\\-32 t + 20 = 0\\\\-32 t = - 20\\\\t = 0.625 \;\rm seconds$

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

$h(0.625) = -16(0.625)^{2} + 20(0.625) + 6 \\\\h(t) = 12.25 \;\rm feet$

The maximum height of the ball is 12.25 feet

The ball is caught at 2 feet

The range of the function is $2 \leq h(t) \leq 12.25$.

Thus, we can conclude the range of the function is from 2 feet to 12.25 feet.

Learn more about the range of function here:

https://brainly.com/question/20899336