Question

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.

Answer 1

Since h(x) is a quadratic, we have
$$
h(x)=ax^2+bx+c
$$
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):
$$
58=a1^2+1b+c=a+b+c
$$
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:
$$
0=a\times 0^2+b\times 0 + c\\
\implies c=0
$$
We are also given that at time x=2, h(2)=112. So now we have
$$
112=a2^2+b2=4a+2b\\
\implies 56=2a+b
$$
We now have two equations and two unknowns: Solvable:
$$
58=a+b\\
56=2a+b
$$
Now subtract them:
$$
58-56=a-2a+b-b=-a\\
a=-2
$$
Using any of our new equations, we can now get \(b\):
$$
58=-2+b\\
56=-4+b
$$

Does this make sense?