Answer:
[tex]f(x)=-10x^2 + 120x[/tex]
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
[tex]f(x)=ax^2 + bx + c[/tex]
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
[tex]f(x)=ax^2 + bx[/tex]
We have two given points
(1,110) and (2,200)
substitute
[tex]110=a(1)^2 + b(1)[/tex]
[tex]110=a + b[/tex] ----> equation A
[tex]200=a(2)^2 + b(2)[/tex]
[tex]200=4a + 2b[/tex] ----> 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
[tex]f(x)=-10x^2 + 120x[/tex]
