The function shown models the height (in feet) of a softball 1 seconds after it
is pitched in an underhand motion. Describe the domain and range. Find the
maximum height of the softball. (Section 8.3)
H(t)= - 16t^2 - 32t- 2

Part 1) The maximum height of the softball is 18 feet

Part 2) The range for H(t) is the interval [0,18]

Part 3) The domain is the interval [0,2.06]

Step-by-step explanation:

step 1

Find the maximum height of the softball

we have

This is a vertical parabola open downward

The vertex represent a maximum

step 1

Find the maximum height of the ball

we know that

The maximum height of the ball correspond to the y-coordinate of the vertex

Convert the quadratic equation into vertex form

Factor -16

Complete the square

Rewrite as perfect squares

The vertex is the point (1,18)

therefore

The maximum height of the softball is 18 feet

step 2

Find the range of the function

The range for H(t) is the interval [0,18]

step 3

Find the domain

To find out the domain we need to determine the x-intercepts

For H(t)=0

The solutions are

---> is not solution because the time cannot be negative

so

The domain is the interval [0,2.06]

option 1 is correct.

step-by-step explanation:

y varies inversely with x and constant of variation is 4

the equation will be:

y = k/x

where k =4 so,

y = 4/x

if we put values of x of option 1 in the above equation and find values of y

y = 4/-2 => y =-2

y = 4/-1 => y = -4

y = 4/1 => y = 4

y = 4/2 => y =2

so, all the values of x and y are correct.

if we put values of x given in other options and find value of y we cannot find whole table satisfying the equation.

hence option 1 is correct.

72 is the answer to this problem