Question

The graph below shows the height of a kicked soccer ball f(x), in feet, depending on the distance from the kicker x, in feet:

Answer 1

Because the vertex of the parabola is at (10,15), its equation is of the form
y = a(x-10)² + 15

The graph goes through (0,0), therefore
a(0 - 10)² + 15 = 0
100a = -15
a = -0.15

The equation is 
y = f(x) = -0.15(x - 10)² + 15

The graph is shown below.

Part A
Note that y = f(x).

The x-intercepts identify values where the function or  y=0. The x-intercepts occur at x=0 and x=20, or at (0,0) and (20,0).

The maximum value of y occurs at the vertex (10, 15) because the curve is down due to the negative leading coefficient of -0.15.

The curve increases in the interval x = (-∞, 10) and it decreases in the interval x = (10, ∞).

Part B
When x=12, y = -0.15(12 - 10)² + 15 = 14.4
When x=15, y = -0.15(15 - 10)² + 15 = 11.25

The average rate of change between x =12 to x = 15 is
(11.25 - 14.4)/(15 - 12) = -1.05

This rate of change represents the slope of the secant line from A to B. It approximates the rate at which f(x) decreases in the interval x =[12, 15].

Answer 2

The x-intercepts are clearly (0, 0) and (20, 0), not (0, 0), (0, 20). Because if we have (0, 0) and (0, 20) then this is no longer a function since we have two similar values of x for different y’s. It should be similar y for different x’s.

Moreover, we can also determine the quadratic function as follows.
The quadratic function with roots a and b, is given by the equation:
f(x) = c (x-a) (x-b).

Since f(a)=0, f(b)=0, f(x) must have one factor (x-a) and one (x-b)

The x coordinates of the x intercepts are the roots of the function so with the roots given:
f(x)=c(x-0)(x-20)=cx(x-20), the vertex is at (10, 15) so

15=f(10)=c*10*(10-20)=c*10*(-10)=-100c
c=-15/100=-3/20

The quadratic function (expressed in factorized form) is
f(x)=(-3/20) x (x-20)

Part A. the x intercept = 0 tells us the point wherein the ball was hit, while the x intercept = 20 tells us the point wherein the ball fell back to the ground, that is 20 feet away from the kicker

The function is increasing in the interval x∈(0, 10) and decreasing in x∈(10, 20) . This only tells us that the height  increases in the interval from  (0, 10) and then decreases as x goes through the interval (10,20). While the distance from the kicker is constantly increasing during the whole interval (0, 20)

Part B:

at x1=15,

f(x)=f(15) = y1=(-3/20)*15(15-20)= 45/4

at x2=12

f(x)=f(12)= y2=-3/20*12(12-20)= 72/5

The average rate of change is then calculated as:

average rate of change = (y2-y1)/(x2 –x1) = (72/5 – 45/4) / (12-15) = -1.05

 

Therefore this means that the height is decreasing at 1.05 ft per feet in horizontal distance from x=12 to x=15.