Abasketball is launched from a sling shot. its height, h(x), can be represented by a quadratic function in terms of time x, in seconds.

after 1 second the ball is 121 feet in the air; after 2 seconds, it is 224 feet in the air.

find the height in feet of the ball after 3 seconds in the air

309 feet

Step-by-step explanation:

given that the height can be represented by a quadratic equation, we can say that the general form of the equation will look something like this:

h(x) = Ax² + Bx + C

we know that at the starting point of the launch that time = 0 (i.e x = 0) and hence h (x) = 0, if we substitute this into our equation, we can find the value for C

h(0) = A(0)² + B(0) + C = 0

C = 0

Hence the equation becomes

h(x) = Ax² + Bx

Given when x = 1, h(1) = 121,

121 = A(1)² + B(1)

A + B = 121 > eq 1

Given when x = 2, h(2) = 224,

224 = A(2)² + B(2)

4A + 2B = 224 (divide both sides by 2)

2A + B = 112> eq 2

Solving the system of equations which comprise eq 1 and eq 2 using your favorite method, we end up with A = -9 and B = 130

our equation becomes:

h(x) = -9x² + 130x

when x = 3

h(3) = -9(3)² + 130(3) = 309 feet

answer:

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step-by-step explanation: