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 65°, its height, f(x), in feet, can be modeled by f(x)=-0.03x² + 2.1x + 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 , which occurs 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.)
c. from what height was the shot released?

Explanation:

Step1: Recall vertex - formula for parabola

The function of the shot - put's height is a quadratic function \(y = ax^{2}+bx + c\), where \(a=-0.03\), \(b = 2.1\), \(c = 5.6\). The \(x\) - coordinate of the vertex of a parabola \(y=ax^{2}+bx + c\) is given by \(x=-\frac{b}{2a}\).

Step2: Calculate the horizontal distance \(x\) at maximum height

Substitute \(a=-0.03\) and \(b = 2.1\) into \(x =-\frac{b}{2a}\).
\[x=-\frac{2.1}{2\times(-0.03)}=\frac{2.1}{0.06}=35\]

Step3: Calculate the maximum height

Substitute \(x = 35\) into the function \(y=-0.03x^{2}+2.1x + 5.6\).
\[y=-0.03\times35^{2}+2.1\times35 + 5.6=-0.03\times1225+73.5 + 5.6=-36.75+73.5 + 5.6=42.35\approx42\]

Step4: Find the height at release (\(x = 0\))

Substitute \(x = 0\) into the function \(y=-0.03x^{2}+2.1x + 5.6\). When \(x = 0\), \(y=5.6\approx5.6\)

Answer:

a. 42 feet, 35 feet
b. 35.0000 feet
c. 5.6 feet