Questions: JAL EXAM REVIEW Question 8, 3.1.57 HW Score: 11.25%, 4.95 of 44 points AL Part 3 of 3 Points: 0 of 1 Save 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 degrees, its height, f(x), in feet, can be modeled by f(x)=-0.01 x^2+0.8 x+5.5, where x is the shot's 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 21.5, 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 shot's maximum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? 86.4 feet (Type an integer or decimal rounded to the nearest tenth as needed.) c. From what height was the shot released? square feet (Type an integer or decimal rounded to the nearest tenth as needed.)

Solution Steps

The function representing the height of the shot put is given by $f(x) = -0.01x^2 + 0.8x + 5.5$. This is a quadratic function in the form $ax^2 + bx + c$, where $a = -0.01$, $b = 0.8$, and $c = 5.5$. Since $a$ is negative, the parabola opens downwards, and the vertex represents the maximum point.

The x-coordinate of the vertex is given by $x = -\frac{b}{2a}$. Substituting the values, we get $x = -\frac{0.8}{2(-0.01)} = \frac{0.8}{0.02} = 40$. This means the maximum height occurs at a horizontal distance of 40 feet.

To find the maximum height, substitute $x = 40$ into the function: $f(40) = -0.01(40^2) + 0.8(40) + 5.5 = -0.01(1600) + 32 + 5.5 = -16 + 32 + 5.5 = 21.5$. So, the maximum height is 21.5 feet.

The maximum horizontal distance occurs when the shot hits the ground, meaning $f(x) = 0$. We need to solve the quadratic equation $-0.01x^2 + 0.8x + 5.5 = 0$. We can use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substituting the values, we get $x = \frac{-0.8 \pm \sqrt{0.8^2 - 4(-0.01)(5.5)}}{2(-0.01)} = \frac{-0.8 \pm \sqrt{0.64 + 0.22}}{-0.02} = \frac{-0.8 \pm \sqrt{0.86}}{-0.02}$.

We are looking for the positive value of $x$, so we take the negative square root: $x = \frac{-0.8 - \sqrt{0.86}}{-0.02} \approx \frac{-0.8 - 0.927}{-0.02} \approx \frac{-1.727}{-0.02} \approx 86.35$. Rounded to the nearest tenth, the maximum horizontal distance is 86.4 feet.

The release height is the initial height when the horizontal distance $x = 0$. Substituting $x = 0$ into the equation, we get $f(0) = -0.01(0^2) + 0.8(0) + 5.5 = 5.5$.

Final Answer