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A missile is launched from the ground. It’s height, h(x), can be represented by a quadratic function in terms of time, x, in seconds

After 1 second, the missile is 110 feet in the air; after 2 seconds, it is 200 feet in the air

2 Answers

1 vote

Answer:

the anwser is 320

Explanation:

User Pyg
by
9.0k points
4 votes

Answer:


f(x)=-10x^2 + 120x

Explanation:

Let

x ----> the time in seconds

f(x) ----> the height in feet

we know that

The standard form for the quadratic, which is the function that models parabolic motion, is


f(x)=ax^2 + bx + c

Remember that

The problem tells us that the missile was launched from the ground, so c (the initial height of the missile) is 0.

so


f(x)=ax^2 + bx

We have two given points

(1,110) and (2,200)

substitute


110=a(1)^2 + b(1)


110=a + b ----> equation A


200=a(2)^2 + b(2)


200=4a + 2b ----> equation B

Solve the system of equation s by graphing

Remember that the solution is the intersection point both lines

using a graphing tool

The solution is (-10,120)

see the attached figure

therefore

a=, b=

The quadratic equation is


f(x)=-10x^2 + 120x

A missile is launched from the ground. It’s height, h(x), can be represented by a-example-1
User NavaneethaKrishnan
by
8.6k points

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