Question 1201821: A hovercraft takes off from a platform.Its height (in meters), x seconds after takeoff, is modeled by:h(x)=−3(x−3)^2 +108 What is the height of the hovercraft at the time of takeoff?
Found 4 solutions by carrollmary87, ikleyn, Alan3354, greenestamps:
Answer by carrollmary87(4)   (Show Source):
You can put this solution on YOUR website! -3(x - 3)^2 + 108 First FOIL (x - 3)(x - 3) = x^2 - 6x + 9
Then -3(x^2 - 6x + 9) + 108 Distribute the -3
-3x^2 + 18x - 27 + 108 Now combine the like terms.
-27 + 108 = 81 This is the y-intercept, so this is where the plane takes off. It is the initial height.
What is the height of the hovercraft at the time of takeoff? 81 meters. You could have also graphed it on Desmos.com and solved it as well.
Answer by ikleyn(52886)  (Show Source):
You can put this solution on YOUR website! .
To answer this question, simply substitute x= 0 into the height function,
because x= 0 seconds is the starting moment. You will get
h(0) = -3*(0-3)^2 + 108 = -3*3^2 + 108 = -3*9 + 108 = -27 + 108 = 81 meters.
ANSWER. The height of the hovercraft at the time of takeoff was 81 meters.
Solved.
Answer by Alan3354(69443)  (Show Source):
Answer by greenestamps(13209)  (Show Source):
You can put this solution on YOUR website!
The inverted parabola equation for the height of the hovercraft is absurd. With the given equation, the hovercraft...
.. is at its maximum height of 108m at x=3 seconds after takeoff;
.. is at a height of 81m at takeoff and again at x=6 seconds after takeoff;
.. is at height 0 at x=9 seconds after takeoff; and
.. is at a negative height (i.e., below the surface of the water) at all times after x=9 seconds after takeoff
We can of course get an answer to the question that is asked by blindly plugging x=0 in the equation -- but that answer is absurd....
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