Respuesta :
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].
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].
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