Question

the path of a ball is modeled by the function, (f(x)=-\frac{1}{20}x^{2}+3x + 5), where (f(x)) is the height (in feet) of the ball and (x) is the horizontal distance (in feet) from where the ball was thrown. the maximum height of the ball is enter answer.

Explanation:

Step1: Identify the coefficients

The function is $f(x)=-\frac{1}{20}x^{2}+3x + 5$, where $a =-\frac{1}{20}$, $b = 3$, $c = 5$.

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function $y = ax^{2}+bx + c$ is $x=-\frac{b}{2a}$. Substitute $a =-\frac{1}{20}$ and $b = 3$ into the formula: $x=-\frac{3}{2\times(-\frac{1}{20})}=30$.

Step3: Find the maximum height

Substitute $x = 30$ into the function $f(x)=-\frac{1}{20}x^{2}+3x + 5$.
$f(30)=-\frac{1}{20}\times30^{2}+3\times30 + 5=-\frac{1}{20}\times900+90 + 5=-45+90 + 5=50$.

Answer:

50