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 35°, its height, f(x), in feet, can be modeled by f(x)=-0.01x² + 0.7x + 5.4, 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.)

Explanation:

Step1: Identify the function type

The height - function $f(x)=-0.01x^{2}+0.7x + 5.4$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-0.01$, $b = 0.7$, and $c = 5.4$.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is given by $x=-\frac{b}{2a}$. Substitute $a=-0.01$ and $b = 0.7$ into the formula:
$x=-\frac{0.7}{2\times(-0.01)}=\frac{0.7}{0.02}=35$.

Step3: Find the y - coordinate of the vertex

Substitute $x = 35$ into the function $f(x)=-0.01x^{2}+0.7x + 5.4$:
$f(35)=-0.01\times(35)^{2}+0.7\times35 + 5.4$.
First, calculate $-0.01\times(35)^{2}=-0.01\times1225=-12.25$.
Then, calculate $0.7\times35 = 24.5$.
So, $f(35)=-12.25+24.5 + 5.4=17.65$.

Answer:

The maximum height is $17.6500$ feet, which occurs $35.0000$ feet from the point of release.