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) = -0.1x² +0.6x+8
where x is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c).
WEB
a. What is the maximum height of the ball and how far from where it was thrown does this occur?
The maximum height is feet, which occurs feet from the point of release.
(Round to the nearest tenth as needed.)

Answer 1

Check the picture below.

let's change the decimal values to fractions, say 0.1 to 1/10 and 0.6 to 6/10 = 3/5, so then

$\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-\frac{1}{10}}x^2\stackrel{\stackrel{b}{\downarrow }}{+\frac{3}{5}}x\stackrel{\stackrel{c}{\downarrow }}{+8} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\left(-\cfrac{~~ \frac{3}{5}~~}{2\frac{-1}{10}}~~,~~8-\cfrac{\left( \frac{3}{5} \right)^2}{4\left(-\frac{1}{10} \right)} \right)\implies \left(\cfrac{~~ \frac{3}{5}~~}{\frac{1}{5}}~~,~~8+\cfrac{\frac{9}{25}}{\frac{2}{5}} \right)$

$\left( \cfrac{3}{5}\cdot \cfrac{5}{1}~~,~~8+\left[ \cfrac{9}{25}\cdot \cfrac{5}{2} \right] \right)\implies \left(3~~,~~8+\cfrac{9}{10} \right)\implies \stackrel{\stackrel{\textit{maximum height}}{\qquad \downarrow }}{\underset{\underset{\textit{horizontal distance}}{\uparrow \qquad \quad }}{\left( 3~~,~~8\frac{9}{10} \right)}}$