Question

The height of a punted football can be modeled with the quadratic function 0.01x^2 + 1.18x+2.
Part A: Find the vertex of the graph of the function by completing the square.

Part B: What is the maximum height of the punt?

Part C: The nearest defensive player is 5 feet horizontally from the point of impact. How high must the player reach to block the punt?

Part D: Suppose the ball was not blocked but continued on its path. How far downfield would the ball go before hitting the ground?

Answer 1

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 $\underline {(59, \, 36.81)}$

Part B: The maximum height of the punt is 36.81 ft.

Part C: The defensive player must reach up to 7.65 feet to block the punt.

Part D: The distance down the field the ball will go without being blocked is approximately 119.67 ft.

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

$\mathrm{\underline{Coordinate \ of \ the \ vertex = (59, \, 36.81)}}$

Part B: The maximum height is given by the y-value at the vertex = 36.81 ft.

Part C: When x = 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 ≈ 119.67 ft.

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