Question

Hi boys can you help me pls!!!!!!!

Answer 1

Answer: OPTION A.

Step-by-step explanation:

Given the following function:

$h(x)=-\frac{1}{4}x^2+\frac{1}{2}x+\frac{1}{2}$

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:

$x=\frac{-b}{2a}$

In this case:

$a=-\frac{1}{4}\\\\b=\frac{1}{2}$

Then, substituting values, you get:

$x=\frac{-\frac{1}{2}}{(2)(-\frac{1}{4}))}\\\\x=1$

Finally, substitute the value of "x" into the function in order to get the y-coordinate of the Vertex:

$h(1)=y=-\frac{1}{4}(1)^2+\frac{1}{2}(1)+\frac{1}{2}\\\\y=0.75$

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.