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In order to find the average you have to add up all the data values and then find the sum of the data values.
So on average he ran about 82.5 miles.
The function of the path of the punted ball is a quadratic function which
follows the path of a parabola.
The correct responses are;
Part A: The coordinates of the vertex is
Part B: The maximum height of the punt is <u>36.81 ft.</u>
Part C: The defensive player must reach up to <u>7.65 feet</u> to block the punt.
Part D: The distance down the field the ball will go without being blocked is approximately <u>119.67 ft.</u>
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Reasons:
The function for the height of the punted ball is; h = -0.01·x² + 1.18·x + 2
Assumption; The distances are feet.
Part A: By completing the square, we have;
f(x) = -0.01·x² + 1.18·x + 2
100·f(x) = -x² + 118·x + 200
-100·f(x) = x² - 118·x - 200
x² - 118·x + (118/2)²= 200 + (118/2)²
(x - 59)² = 200 + (59)² = 3681
(x - 59)² - 3681
At the vertex, -3281 = -100·f(x)
∴ f(x) at the vertex = -3681/-100 = 36.81
Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.
Part C: When <em>x</em> = 5, we have;
h = -0.01·x² + 1.18·x + 2
h = -0.01 × 5² + 1.18 × 5 + 2 = 7.65
The defensive player must reach up to 7.65 feet to block the punt
Part D: The distance the ball will go before it hits the ground is given by
the function, for the height as follows;
h = -0.01·x² + 1.18·x + 2 = 0
From the completing the square method, above, we get;
-0.01·x² + 1.18·x + 2 = 0
x² - 118·x - 200 = 0
x² - 118·x + (118/2)²= 200 + (118/2)²
x² - 118·x + (59)²= 200 + (59)² = 3681
(x - 59)² = 3681
x - 59 = ±√3681
x = 59 ± √3681
x = 59 + √3681 ≈ 119.67
The distance down the field the ball will go without being blocked, x ≈ <u>119.67 ft.</u>
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