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 30°, its height, f(x), in feet, can be modeled by f(x)= - 0.01x² + 0.6x + 5.3, where x is the shots horizontal distance, in feet, from its point of release. use the 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 14.3, which occurs 30 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? (type an integer or decimal rounded to the nearest tenth as needed.)

Explanation:

Step1: Find the x - coordinate of the vertex for part (a)

For a quadratic function \(y = ax^{2}+bx + c\), the x - coordinate of the vertex is given by \(x=-\frac{b}{2a}\). In the function \(f(x)=-0.01x^{2}+0.6x + 5.3\), \(a=-0.01\) and \(b = 0.6\).

$$x=-\frac{0.6}{2\times(-0.01)}=\frac{- 0.6}{-0.02}=30$$

Step2: Find the x - intercept for part (b)

Set \(y = f(x)=0\), so we have the quadratic equation \(-0.01x^{2}+0.6x + 5.3=0\). Multiply through by \(-100\) to get \(x^{2}-60x - 530=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=-60\), \(c=-530\)

$$x=\frac{60\pm\sqrt{(-60)^{2}-4\times1\times(-530)}}{2\times1}=\frac{60\pm\sqrt{3600 + 2120}}{2}=\frac{60\pm\sqrt{5720}}{2}=\frac{60\pm75.63}{2}$$

We take the positive root \(x=\frac{60 + 75.63}{2}=\frac{135.63}{2}=67.815\approx67.8\)

Answer:

a. \(30\) feet
b. \(67.8\) feet