Question

an athlete whose event is the shot put releases a shot. when the shot whose path is shown by the graph to the right is released at an angle of 40°, its height, f(x), in feet, can be modeled by f(x)= - 0.01x² + 0.8x + 5.6, where x is the shots horizontal distance, in feet, from its point of release. use this model to solve parts (a) through (c) and verify your answers using the graph. a. what is the maximum height of the shot and how far from its point of release does this occur? the maximum height is 21.6000, which occurs 40 feet from the point of release. (type an integer or decimal rounded to four decimal places as needed.) b. what is the shots maximum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? feet (type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Find the roots of the quadratic function

The function is \(f(x)=- 0.01x^{2}+0.8x + 5.6\). For a quadratic function \(y = ax^{2}+bx + c\) (\(a=-0.01\), \(b = 0.8\), \(c = 5.6\)), the roots are given by the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\).

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$$x_{1}=\frac{-0.8+\sqrt{0.864}}{-0.02}\approx - 6.5\quad x_{2}=\frac{-0.8-\sqrt{0.864}}{-0.02}$$

$$x_{2}=\frac{-0.8 - 0.93}{-0.02}=\frac{-1.73}{-0.02}=86.5$$

Answer:

\(86.5\)