Question

a ball is thrown upward and outward from a height of 8 feet. the height of the ball, f(x), in feet, can be modeled by f(x) = -0.2x² + 1.4x + 8 where x is the balls horizontal distance, in feet, from where it was thrown. use this model to solve parts (a) through (c). a. what is the maximum height of the ball and how far from where it was thrown does this occur? the maximum height is \\(\square\\) feet, which occurs \\(\square\\) feet from the point of release. (round to the nearest tenth as needed.)

Explanation:

Step1: Identify the vertex of the parabola

The function \( f(x) = -0.2x^2 + 1.4x + 8 \) is a quadratic function in the form \( f(x) = ax^2 + bx + c \), where \( a = -0.2 \), \( b = 1.4 \), and \( c = 8 \). The x - coordinate of the vertex of a parabola given by \( f(x)=ax^{2}+bx + c \) is \( x=-\frac{b}{2a} \).
Substitute \( a=-0.2 \) and \( b = 1.4 \) into the formula for the x - coordinate of the vertex:
\( x=-\frac{1.4}{2\times(-0.2)}=\frac{- 1.4}{-0.4} = 3.5 \)

Step2: Find the maximum height

To find the maximum height, substitute \( x = 3.5 \) into the function \( f(x)=-0.2x^{2}+1.4x + 8 \)
\( f(3.5)=-0.2\times(3.5)^{2}+1.4\times3.5 + 8 \)
First, calculate \( (3.5)^{2}=12.25 \)
Then, \( -0.2\times12.25=-2.45 \)
\( 1.4\times3.5 = 4.9 \)
Now, \( f(3.5)=-2.45 + 4.9+8=10.45\approx10.5 \) (rounded to the nearest tenth)

Answer:

The maximum height is \( 10.5 \) feet, which occurs \( 3.5 \) feet from the point of release.