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The path of a ball is given by y = -1/20x^2 + 3x + 5, where y is the height (in feet) of the ball and x is the horizontal distance (in feet) from where the ball was thrown.

a) Find the maximum height of the ball.
b) Which number defines the height at which the ball was thrown? Does changing this value change the coordinates of the maximum height of the ball? Explain.

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User Max Li
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1 Answer

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Explanation:

a) To find the maximum height of the ball, we need to find the vertex of the parabola. The vertex is located at the point (-b/2a, f(-b/2a)), where a, b, and c are the coefficients of the quadratic equation. In this case, the equation is y = -1/20x^2 + 3x + 5, so a = -1/20 and b = 3.

The x-coordinate of the vertex is given by -b/2a = -3/(2*(-1/20)) = 30, and the y-coordinate is f(30) = -1/20(30)^2 + 3(30) + 5 = 80.5.

Therefore, the maximum height of the ball is 80.5 feet.

b) The height at which the ball was thrown is given by the y-intercept of the parabola, which occurs when x = 0. Plugging in x = 0 into the equation y = -1/20x^2 + 3x + 5, we get y = 5. Therefore, the ball was thrown at a height of 5 feet.

Changing the height at which the ball was thrown does not affect the coordinates of the maximum height of the ball. The vertex of the parabola depends only on the coefficients of the quadratic equation and not on the y-intercept. The y-intercept only shifts the entire parabola up or down

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User Tianle
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