Question

3.1 Q15
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 70°, it's height, f(x), in feet, can be modeled by:
$f(x)=-0.06x^2 + 2.7x + 5.9$
where x is the shot's horizontal distance, in feet, from its point of release.

What is the maximum height of the shot and how far from it's point of release does this occur?

Answer 1

The maximum height of the shot is represented by the y-coordinate of the vertex, and the horizontal distance when the maximum height occurs is represented by the x-coordinate of the vertex.

Find x-coordinate of the vertex (horizontal distance)
The function above is in standard form of ax² + bx + c = 0
In standard form of ax² + bx + c = 0, we could find the x-coordinate of the vertex by the formula of
$x= -\cfrac{b}{2a}$

Extract the information of a and b from the function f(x) = -0.06x² + 2.7x + 5.9
a = -0.06
b = 2.7

Input the value of a and b to the formula of x-coordinate of the vertex
x = $-\dfrac{b}{2a}$
x = $- \dfrac{2.7}{2 \times (-0.06) \\}$
x = $- \dfrac{2.7}{-0.12}$
x = $\dfrac{2.7}{0.12}$
x = 22.5
The x-coordinate represents the horizontal distance. So the maximum height occurs when the distance is 22.5 feet far from its point of release.

Find y-coordinate of the vertex (the maximum height)
To find the maximum height, input the value of x-coordinate of the vertex to the function.
f(x) = -0.06x² + 2.7x + 5.9
f(22.5) = -0.06(22.5)² + 2.7(22.5) + 5.9
f(22.5) = -0.06(506.25) + 60.75 + 5.9
f(22.5) = -30.375 + 60.75 + 5.9
f(22.5) = 36.275
The maximum height is 36.275 feet