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anonymous
 one year ago
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.
anonymous
 one year ago
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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ybarrap
 one year ago
Best ResponseYou've already chosen the best response.1Since 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: $$ 5856=a2a+bb=a\\ a=2 $$ Using any of our new equations, we can now get \(b\): $$ 58=2+b\\ 56=4+b $$ Does this make sense?
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