Question

an athlete whose event is the shot - put releases the shot. when the shot is released at an angle of 35°, its path can be modeled by the formula y = - 0.01x²+0.7x + 6.1. in which x is the shots horizontal distance, in feet, and y is its height, in feet. this formula is shown by one of the graphs, (a) or (b), in the figure. use the formula to answer the questions below. use the formula to determine the shots maximum distance. the maximum distance is approximately □ feet. (round to the nearest tenth as needed.)

Explanation:

Step1: Identify the quadratic - function form

The given function is $y = - 0.01x^{2}+0.7x + 6.1$, which is in the form $y = ax^{2}+bx + c$ where $a=-0.01$, $b = 0.7$, and $c = 6.1$. The shot's maximum distance occurs when $y = 0$.

Step2: Use the quadratic formula

The quadratic formula for $ax^{2}+bx + c=0$ is $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Substitute $a=-0.01$, $b = 0.7$, and $c = 6.1$ into the formula:
\[

$$\begin{align*} x&=\frac{-0.7\pm\sqrt{(0.7)^{2}-4\times(-0.01)\times6.1}}{2\times(-0.01)}\\ &=\frac{-0.7\pm\sqrt{0.49 + 0.244}}{-0.02}\\ &=\frac{-0.7\pm\sqrt{0.734}}{-0.02} \end{align*}$$

\]

Step3: Calculate the two solutions for x

First, find $\sqrt{0.734}\approx0.857$. Then:
\[

$$\begin{align*} x_1&=\frac{-0.7 + 0.857}{-0.02}=\frac{0.157}{-0.02}=-7.85\\ x_2&=\frac{-0.7 - 0.857}{-0.02}=\frac{-1.557}{-0.02}=77.85 \end{align*}$$

\]
We discard the negative value since distance $x\geq0$.

Answer:

$77.9$