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: Graph of quadratic function f of x having x intercepts at ordered pairs 0, 0 and 16, 0. The vertex is at 8, 10. Part A: What do the x-intercepts and maximum value of the graph represent? What are the intervals where the function is increasing and decreasing, and what do they represent about the distance and height? (6 points) Part B: What is an approximate average rate of change of the graph from x = 8 to x = 13, and what does this rate represent? (4 points)

Answer 1

Part A: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  

X Intercept=(0,0) it represents the time the ball was not kicked at all  

X intercept=(16,0)  

the feet away from the starting point is 16ft the location or distance the ball is away from the starting point once landed.  

Maximum: is 10  

intervals:it increased(0 to 8) decreased (8 to 16)  

when the ball was 8 ft away from the kicker horizontally, it was 10 feet in the air then it landed an additional 8 feet away from the kicker that was 16 feet away one it landed after going a maximum of 10 feet.  

Part B: m=(y2 - y1) / (x2 - x1)  

(x1, y1) (x2, y2)  

(8,10) (13, 7)  

m = (7-10)/(13-8) = -3/5

Answer 2

Because the vertex of the parabola is at (16,0), its equation is of the formy = a(x-10)² + 15
The graph goes through (0,0), thereforea(0 - 10)² + 15 = 0100a = -15a = -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 = 14.4When 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].