Algebra 2: the height y(in feet) of a basketball t seconds after it is thrown can be modeled by the function: y=-16t^2+32t+2
a) find the maximum height of the basketball
b) how long is the basketball in the air

Part A) The maximum height of the basketball is 18 feet

Part B) The basketball is in the air 2.06 seconds

Step-by-step explanation:

Part A) Find the maximum height of the basketball

Let

y is the height of a basketball in feet

t is the time in seconds

we have

This is a vertical parabola open downward (the leading coefficient is negative)

The vertex represent a maximum    

Find 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)

That means

The maximum height is 18 ft when t=1 sec

therefore

The maximum height of the basketball is 18 feet

Part B) How long is the basketball in the air

Find the x-intercepts of the quadratic equation

The x-intercepts are the values of x when the value of y is equal to zero

In this context, the x-intercept is the time in second when the height of the basketball is zero (on the floor)

so

For y=0

simplify

square root both sides

---> is not a solution (time cannot be negative)

therefore

The basketball is in the air 2.06 seconds

24

step-by-step explanation:

so, converting firstly the mixed fractions to improper fractions and then "adding".