Answer: OPTION A.
Step-by-step explanation:
Given the following function:
[tex]h(x)=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{2}[/tex]
You know that it represents the the height of the ball (in meters) when it is a distance "x" meters away from Rowan.
Since it is a Quadratic function its graph is parabola.
So, the maximum point of the graph modeling the height of the ball is the Vertex of the parabola.
You can find the x-coordinate of the Vertex with this formula:
[tex]x=\frac{-b}{2a}[/tex]
In this case:
[tex]a=-\frac{1}{4}\\\\b=\frac{1}{2}[/tex]
Then, substituting values, you get:
[tex]x=\frac{-\frac{1}{2}}{(2)(-\frac{1}{4}))}\\\\x=1[/tex]
Finally, substitute the value of "x" into the function in order to get the y-coordinate of the Vertex:
[tex]h(1)=y=-\frac{1}{4}(1)^2+\frac{1}{2}(1)+\frac{1}{2}\\\\y=0.75[/tex]
Therefore, you can conclude that:
The maximum height of the ball is 0.75 of a meter, which occurs when it is approximately 1 meter away from Rowan.