A missile is launched from the ground. Its height, h(x), can be represented by a quadratic function in terms of time, x, in seconds.
After 1 second, the missile is 58 feet in the air; after 2 seconds, it is 112 feet in the air.
Complete the height function, h(x), for this situation.
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Since h(x) is a quadratic, we have
Where a,b and c are constants to be determined by the information in the question, x is the time in seconds and h(x) is the height of the missile at time x.
We are given that at time x=1 that h(1)=58, that is, the missile after 1 second is 58 feet above the surface.
Using our 1st equation above and plugging in the given values for x and h(x):
We see that we have 3 constants to solve for but we only have one equation so far.
We also can assume that at time x=0 that h(0)=0. This is saying that the missile starts out on the ground. This implies that c=0 because using the 1st equation above:
0=a\times 0^2+b\times 0 + c\\
We are also given that at time x=2, h(2)=112. So now we have
We now have two equations and two unknowns: Solvable:
Now subtract them:
Using any of our new equations, we can now get \(b\):
Does this make sense?