• anonymous
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.
  • Stacey Warren - Expert
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
  • katieb
I got my questions answered at in under 10 minutes. Go to now for free help!
  • ybarrap
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?

Looking for something else?

Not the answer you are looking for? Search for more explanations.