David throws a ball vertically upward from a height of 3 meters. after 1 second, the ball is at a height of 12.5 meters. after 2 seconds, it’s at a height of 11.8 meters. write an equation to model the ball's height, f(x), at different times, x.

f(x) = -5.1x² +14.6x +3

Step-by-step explanation:

The three given points are obviously not on a line, but can be modeled exactly by a 2nd degree polynomial. (The degree is 1 less than the number of points.) I find it convenient to let a graphing calculator's quadratic regression function tell me the coefficients.

Solving by hand

Since the "y-intercept" is 3, the equation can be written

... f(x) = ax² +bx + 3

Filling in the given numbers, we have 2 equations in the 2 unknowns, a and b.

... f(1) = 12.5 = a·1² +b·1 +3

... f(2) = 11.8 = a·2² +b·2 +3

Simplifying these gives ...

Subtracting the first equation from the second gives ...

... (2a +b) -(a +b) = (4.4) -(9.5)

... a = -5.1

The first equation tells us

... b = 9.5 -a = 9.5 +5.1

... b = 14.6

The equation is f(x) = -5.1x² +14.6x +3.