Question

Question 317 ptsA model rocket is launched straight up. Its height in feet (y) above theground x seconds after launch is modeled by the quadratic function: y =1-16x2 + 208x + 12.How many feet above the ground was the rocket 3 seconds after launch?nse471 feet 492 feet510 feet537 feet

Answer 1

492 feet

1) Since the curve of the rocket is modeled by the function

y= -16x²+208x+12

y = height

x= seconds

2) Let's find out how many feet

f(x) =-16x²+208x+12

f(3) = -16(3)²+208(3)+12

f(3) = 492 feet