Question

h(x)=−5(x−4) 2 +180 what is the height of the object at time of launch

Answer 1

Answer:

x=10

Step-by-step explanation:

h(x)=-5(x-4)^2+180

Set y = 0 to find when it hits the ground

0 = -5(x-4)^2+180

Subtract 180 from each side

-180 =-5(x-4)^2+180-180

-180 =-5(x-4)^2

Divide by -5

-180/-5 =-5/-5(x-4)^2

36 = (x-4)^2

Take the square root of each side

±sqrt(36) = sqrt( (x-4)^2)

±6 =  (x-4)

Add 4  to each side

4±6 =  (x-4)+4

4±6=x

x = 10 or -2

But time cannot be negative so x=10

Answer 2

Answer:

In conclusion, the height of the object at the time of launch is 100 meters.

Step-by-step explanation:

The height of the object at the time of the launch is given by h(0).

h(0) = -5 (0 - 4)^2 + 180

       = -5 (16) + 180

       = 100