The vertex is the maximum or minimum point of the equation's parabola
The general form of a quadratic equation is :
![\text{ y = ax}^2\text{ + bx + c}](https://img.qammunity.org/2023/formulas/mathematics/college/og4idotdnmlzy0prs4rad9v46t8dkv1ffy.png)
If we select the first two points i.e
![(0,\text{ 5) and (0.25, 8)}](https://img.qammunity.org/2023/formulas/mathematics/college/fglxkp2xsgxcrcoj0ngjxk9u2zz5c3uez2.png)
We substitute each point to get:
![5\text{ = c}](https://img.qammunity.org/2023/formulas/mathematics/college/i2h4rdmkc9nqgavhr05iy962bnz84n7jr9.png)
![\begin{gathered} 8\text{ = 0.0625a + 0.25b + 5} \\ 0.0625a\text{ + 0.25b = 3 eqn (1)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/99vxd2bzu11qkps3210hm7v7cn64fjjg34.png)
Select point (0.50, 9) and substitute into our general expression:
![\begin{gathered} 9\text{ = 0.25a + 0.5b + 5} \\ 0.25a\text{ + 0.5b = 4 eqn (2)} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i5caf8jp6ahy0hlhn2dmfmz54p6bk9335w.png)
Solving eqn (1) and (2) simultaneously, we have
![\begin{gathered} a\text{ = -16} \\ b\text{ =16} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hz8h70l5ymglktwrvwuuwape1t5q91bifx.png)
The quadratic equation is given as :
![H=-16t^2\text{ + 16t + 5}](https://img.qammunity.org/2023/formulas/mathematics/college/8jp4r0es43oc3scpwuesiuppznf94id737.png)
The x-coordinate of the vertex can be obtained as:
![\begin{gathered} =\text{ }(-b)/(2a) \\ =\text{ }\frac{-16}{2\text{ }*\text{ -16}} \\ =\text{ 0.5} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ekuie8h6hvimiwgq08vh8fd0hils8zb9z1.png)
y-coordinate is obtained as :
![\begin{gathered} H=-16(0.5)^2\text{ + 16(0.5) + 5} \\ =\text{ 9} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/skoddngmr3wkcml4vqj15f23dlfsloincx.png)
The correct option should be the time in seconds when the ball reaches the maximum height