Water is shot straight up out of a water soaker toy. The quadratic function y = –16x2 + 32x models the
height in feet of a water droplet after x seconds. How
long is the water droplet in the air?

Help!

The water droplet is in the air for two seconds.

Step-by-step explanation:

You can solve this by finding the times at which its height is zero.  The lesser of the two values will be upon launch, and the greater upon landing:

y = -16x² + 32x

Let y = 0

-16x² + 32x = 0

factor -16 out:

x² - 2x = 0

complete the square:

x² - 2x + 1 = 1

and solve for x

(x - 1)² = 1

x - 1 = ±1

x = 1 ± 1

x = 0, 2

So the water is at a height of zero at zero seconds and two seconds.  It's thus in the air for two seconds.

3x^3-x^2+6x-9

step-by-step explanation:

first, eliminate the parenthesis:

3

x

+

2

−

4

x

second, group like terms:

3

x

−

4

x

+

2

third, combine like terms:

(

3

−

4

)

x

+

2

→

−

1

x

+

2

→

−

x

+

2