Question

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

Answer 1

Answer:

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

Step-by-step 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

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